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Totals ==>246251013129005016
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Scope
YearAnswerIDPromptExpectedOriginRationale
3
12020x1 is eventually true if you run many many trials.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
many x outside showing many different trials.
4
12020After state 4, x1 will eventually be true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
The eventually clause does not apply to the first 4 states.
5
12020There is eventually a true state in x1 which occurs after four transitionsX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
The four transitions comes from the four X's. Then, eventually (F), x1 is true
6
12020x1 takes on the value true four values after x1 eventually becomes trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
F(x1) evaluates to true whenever there is an instance where x1 evaluates the true in some state in the future. X(X(X(X(F(x1)))) states that four values after x1, x1 must evaluate to true.
7
12020x1 is eventually true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
This more directly translates as "at the fourth subtrace, x1 is eventually true," however this is the same as just "x1 is eventually true."
8
12020x1 will hold eventually after the fourth time step.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
Using four X operations outputs the subset of the trace after the first four time steps.
9
12020Four states later, x1 will eventually hold.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
The X(X(X(X))) specifies the 'four states later', and F(x1) says that at that point x1 will eventually be true
10
12020Four states from now, x1 will eventually become trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
I think applying X multiple times moves the constraint further into the future.
11
12020Starting four states ahead, x1 must eventually become true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
The x's move it 4 states ahead and the f means eventually.
12
12020In the 4th next state, x1 is eventually true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
4 X means the 4th next state, which wraps F(x1), meaning x1 is eventually true.
13
12020x1 is eventually true in the next 4 timestepX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
each X means next timestep F means eventually
14
12020x1 is true at some point after 4 iterations.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
F(x1): x1 is at some point true in the future. XXXX: after 4 iterations from the current state.
15
12020There exists a case where eventually x1 becomes true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
I thought that the X's since they all mean the same thing would basically cancel each other out to just say exist. F(x1) means eventually x1 becomes true.
16
12020After the fourth state x1 will eventually hold.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
The X's just mean next state, so we use them to progress through possible states, until we reach the F, which states that x1 will finally/eventually hold.
17
12020After the 4th state, x1 will eventually hold.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
Four applications of X gets us to the fifth state, and when we apply the finally operator we state the there will exist some state after and including the 5th state where x1 is true. This allows x1 to be true in states 1-4 but still have the expression be false if it is never true again.
18
12020After the next forth state, x1 will be trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
Its logic is clear and it can be followed by the formula to imply it's meaning
19
12020After the fourth time step, it eventually holds that x1 is trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
F(x1) means that x1 is eventually true and each X means that this occurs at the next time step
20
12020It eventually holds that x1 is true starting at four states later.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
21
12020x1 is true at or after the fifth stateX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
I saw this one in the other section, but anyways... "X(X(X(X(" basically says to ignore the first four states, and focus on the fifth one. F(x1) says x1 is eventually true, so x1 is true at or after the fifth state.
22
12020x1 is false on the fourth stateX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
four X's
23
12020x1 will eventually happen in some states after the first four states,X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
Straightforward.
24
12020after 4 transitions, eventually x1 will be trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
X means one transition, and after 4 X, F means x1's eventual feature
25
12020x1 must be eventually true after the 4th state.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
There are 4 X's which means that F(x1) = 'x1 is eventually true' should hold after 4th state.
26
12020For every 4th state, it is eventually true that x1 holds.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
F means eventually and X means the next state. When I substitute these symbols to words, I get the above answer.
27
12020In four steps it holds that x1 is eventually true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
I think the four "X"s will look 4 steps ahead and then the F(x1) will check that x1 is eventually true
28
12020Starting from the fifth entry of the trace on, it eventually holds that x1 is trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
29
12020Traces with a state occurring after the 4th entry that satisfies x1X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
We start looking at the F(x1) after chopping off the first four entries
30
12020it eventually holds that x1 is true in the next next next next trace (four traces ahead)X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
the four X's denote that F(x1) applies to the fourth next trace
31
12020in 5 states, eventually x1 will be trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
idk how the X operators stack
32
12020There are four states, until, at some point, x1 is true.X(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
I think the nested "X"s (next state) indicate a sequence of four states, with the innermost state being the last state.
33
12020x1 next next next next finally is trueX(X(X(X(F(x1)))))
This is a question that Tim thought would be interesting.
34
12020Always if x1 is true, x2 finally is trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
35
12020x1 eventually implies x2G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
the formula means it always holds that x1 implies that x2 is eventually true. If this always hold, x1 would imply x2 at some state
36
12020If x1 is true, x2 will always eventually become true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
37
12020It always holds that, if x1 is true x2 will eventually be true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G gives the always, F gives the eventually and -> gives the contingency on the value of x1.
38
12020It always holds that, if x1 is true, x2 is eventually trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G = always, -> = if ... then ..., F(x2) = eventually x2
39
12020it always holds that x1 eventually implies x2G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G is always true, -> means implies, and F is eventually true
40
12020It is always true that if x1 is initially true then it will eventually lead to x2 being true..G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G() means always true. F(x2) means that x2 is eventually true. -> means leads to. Thus if x1 is initially true then it will eventually lead to x2 being true.
41
12020it always holds that when x1 is true, x2 will eventually be trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
42
12020it always holds that if x1 is true, x2 must eventually be trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
x1 -> F(x2) means that if x1 is true in the current state then x2 is eventually true. This must always be true which means that if x1 is true at some point, x2 must eventually be true
43
12020It is always the case that x1 implies that x2 is eventually true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G means always, -> means implies, F means eventually. As a result, the above can be translated in English as shown above.
44
12020it is always true that if x1 is true, then x2 will eventually be trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
45
12020It is always true that x1 implies x2 is eventually true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
This is my answer because "G( )" means always, "->" means implies, and "F( )" means eventually true.
46
12020it always holds that if x1 is true then eventaully x2 must be true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
x1 -> F(x2) means that if the first state in the sequence is true then eventually x2 must be true. We put on the global constraint stating that if x1 is ever true, then there must exist a point where x2 is true.
47
12020Every state that satisfies x1 is eventually followed by one that satisfies x2G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
Globally, a state either doesn't satisfy x1 or has some state after it that satisfies x2
48
12020It is always true that if x1 is true then x2 will eventually be trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
The G signifies always and the F signifies eventually, so if x1 then x2 is eventually true all the time
49
12020Whenever x1 holds, x2 will eventually hold then or after.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G(x1 -> ...) says that the conclusion should hold in any state where x1 holds. The conclusion is just eventually x2 holds.
50
12020it's always true that if x1 is true, x2 will be true at some point after thatG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G is always, and F is at some point
51
12020It is always holds that if x1 is true x2 is eventually true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G = always, so for all states x1 should imply that x2 is (F = eventually) true sometime in the future.
52
12020Always, x1 being true eventually implies x2 is eventually true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
x1 could be true at any number of states, and x2 has to be true sometime after the last time x1 is true (and, if x1 is true in an infinite number of states, x2 must also be true in a countable number of states)
53
12020it always holds that when x1 is true, eventually x2 is true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
the nested expression shows an implication that the status of x1 implies the future status of x2. This relationship between x1 and x2 always holds.
54
12020It always holds that if x1 is true, x2 is eventually true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
Once again, like with a previous one, I'm not 100% clear on "when" x1 would be.
55
12020It always holds that if x1 starts out true then x2 will eventually become true in the trace, otherwise if x1 starts out false then x2 can have any value (true or false) at any point in the trace.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
56
12020It always holds that if x1 is true, then it eventually holds x2 is true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
57
12020It is always true that if x1 holds, then eventually x2 holds.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
F means eventually, -> means implies, and G means always. Therefore, when I substitute these symbols to words, I get the above answer.
58
12020it always holds that x1 being true at the first time step means that x2 is eventually trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G means always / globally, x1 means x1 is true at the first time step, F(x2) means that x2 is eventually true
59
12020It always holds that x1 implies eventually x2 is true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
It has to be always true that if x1, then at some points x2 has to be true.
60
12020always, if x1 then eventually x2G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
61
12020It always holds that if x1 is true, then x2 will eventually be true.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
x1 -> F(x2) means that x1 being true implies that x2 will eventually be true. Surrounding with the G tag makes it always.
62
12020It is globally true that if x1 is true, then x2 is eventually true thereafter.G(x1 -> F(x2))
This is a question that Tim thought would be interesting.
G: it is globally true that ... x1 ->: if x1 is true, then... F(x2): x2 is true eventually true in the future
63
12020It always holds that x1 being true implies that x2 is eventually trueG(x1 -> F(x2))
This is a question that Tim thought would be interesting.
The formula is wrapped in a "globally" operator, that says x1 implies F(x2). This inner part means that x1 true somewhere in the path implies that x2 will be true in the subsequent path
64
112020x1 implies not x1 in the next state which is always true, and x1 is true for the next two states
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
I am not sure what the X is so I assumed it means states/transitions, and its essentially two statements joined together with an AND so I translated them separately
65
12020It always holds that whenever x1 is true, it is false in the next state, and x1 is true in the state after the next.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
x1 -> X(!x1) means that when x is true, x is not true in the next state, so G(x1 -> X(!x1)) means that it always holds. X(x1) means that x1 is true in the next state, so X(X(x1)) means that it is true that in the next state, x1 is true in that state's next state.
66
12020The empty set.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
It says that if x1 is true, the next state must be false, and two states ahead of any state must be true. But this is impossible, as X(X(x1)) will soon require every state to be true and x1 -> X(!x1) will require true states to be followed by false states.
67
12020It always holds that x1 is true 2 entries from now and if x1 is true in this entry then it must be false in the next entry, otherwise if x1 is false in this entry then it can be any value (true or false) in the next entry
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
My Internet was cutting in and out so I might have accidentally submitted no answers for some questions :(
68
12020It is always holds that, if x1 is true, x1 is false in the next state and true again in the following state
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
69
12020Always, x1 being true implies the next state is false and the one after that is true again
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
G means this implication always holds. X refers to the next state, and the double X's refers to the state after that. Finally, !x1 means the next state is false, and the no exclamation point means the one after that is true.
70
12020The sequence alternates between states that satisfy x1 and those that do not.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
x1 -> X(!x1) and X(X(x1)) means that if a state satisfies x1, the next one will not and the state two after it will.
71
12020It always holds that if x1 is true at some state, then x1 is false at next state and x1 is true at the state after next state.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
This formula describes that x1 is true at some state followed by x1 is false at the next state, and so on.
72
12020It always holds that if x1 is true in the current state, then x1 is false in the next state and true again in the state after that.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
Because x1 implies that x1 is false in the next state and true again in the state after
73
12020I always holds that if x1 is true then x1 is false in the next state, and that in every other state x1 is true.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
This one is cool -- it simulates a light switch flipping on and off, but by only constraining x true -> x false, and forcing that every other state x is true with X(X(x1))
74
12020It's always true that if x1 is true then in the next state x1 is false, and the following state x1 is true again
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
I'm not entirely sure what the -> means, but assume that it is approximately implies. The G() means that the inner proposition is always true. So if x is true at sequence i, then (X(!x1) and X(X(x1))) is true. So X(!x1) means that x1 is false in i+1, and X(X(x1)) means that x1 is true in i+2
75
12020It is always such that when x1 is true, the next step is that x1 is false and that the step after that, x1 is true again
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
Globally (x1 being true implies that the next x1 is false and the next next x1 is true)
76
12020It always holds that if x1 is true, then the next state should include x1 to be false and in the state after that, x1 should be true again
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
See answer
77
12020It always holds that if x1 is true, the next x1 is false and the next next x1 is true.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
I'm assuming here that the order of operations means that everything after the implication is grouped together, meaning this is saying that x1 is going to alternate in truth value from state to state.
78
12020It always holds that x1 implies x1 is false in the next subtrace and true in the subtrace after that
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
This is my answer because I translated "G( )" to mean always, "->" to mean implies, "X( )" refers to the next subtrace, and "X(X( ))" applies to the subtrace after that.
79
12020it always holds that the sequence is alternating between true and false.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
x1 -> X(!x1) evaluates true for when either x1 evaluates to false or when x1 & X(!x1) evaluates to true and we know that the third element must be true. Since we require this to be true for globally we know that there cannot be two sequential states that evaluate to true or false.
80
12020Whenever x1 is true, the next x1 is false and two states ahead it is true. It alternates
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
The x functions look one state ahead, making this one self explanatory. The g function makes it so it is always the case that this holds.
81
12020For every state, if x1 is true, then x1's next state would be false and the next of its next state would be true
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
it actually creates a interleaving sequence with “true false true false true false ...” or “false true false true false true ..."
82
12020it is always true that if x1 at time t x1 must be false at time t+1 and true at time t+2 (x1 oscillate between true and false)
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
83
12020It always holds that when x1 is true then in the next state x1 will be false and in the next next sate x1 will be true.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
G means that the expression inside of it is always true. Then the x1 -> parts means that when x1 is true the stuff to the right of the arrow occurs, and I think that the X represents the next state so when it says X(!x1) that's saying that x is not true in the next state and then two Xs means its the next next state.
84
12020It is always true that "if x1 is true x1 is false in the next state", and "x1 is true in the next next state".
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
G: it is always true that... x1 -> X(!x1): if x1, then NOT x1 in the next state. X(Xx1) x1 is true in the next next state.
85
12020it always holds that when x1 is true, x1 is false in the next state and then true in the following state
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
X(!x1) is the next state and X(X(x1)) is the one after that
86
12020If x1 is true, it will always be false next and true in the state after that.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
We have that for all states, if x1 is true, the next state will have x1 false (by application of the next operator X), and the state 2 states ahead will have x1 true (by double application of X). This would set up a consistent oscillation between x1 being true and false, with no breaks.
87
12020It always holds that if x1 is true, the next x1 will be false, and the next next state of x1 will be true
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
Its logic is clear and it can be followed by the formula to imply it's meaning
88
12020It always holds that x1 will alternate between true and false, unless x1 is always false.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
When x1 is true, that implies the next state will have x1 as false, and the state after that will have x1 true again. That holds always. Therefore, if there is a single true, it will alternate true / false.
89
12020Again, same as last two questions, with a different formula: x1 switches off being true and not true, in a cycle forever.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
It is always true that, after x1 being true, this happens: in the next state, x1 will be not true. And in the state after that, x1 will be true again. And since x1 always implies that this sequence of states will happen, the cycle will repeat globally (forever).
90
12020Upon x1 being true for the first time (if ever), it alternates true, false, true, false, ... from then on.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
G(x1 -> ...) says for the conclusion to hold in any state in which x1 is true. The conclusion says for the next state to have x1 false, and the following state for it to be true again. So, in any state where x1 is true, the next state should have it false, and the following state it should be true again. However, in that following state, the condition then holds again, and so this process repeats forever.
91
12020If it holds at any point that x1 is true, then from that point on x1 will vacillate between being true and not true in each subsequent state
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
Assuming that the given formula is meant to be G(x1 -> (... and ...)), instead of G((x1 -> ... ) and ...), this formula holds that for any state where x1 is true, it must be true that in the next state, !x1 holds, and in the next, next state, x1 holds. Thus, given some first state where x1 is true, the following states must follow a x1, !x1, x1, !x1 pattern.
92
12020it is always true that if x1 is true then the next time x1 is false and the time after that x1 is true.
G(x1 -> X(!x1) and X(X(x1)))
This is a question that Tim thought would be interesting.
Based on the implies we know that x1 alternates between its next, and then the time after it's next it goes back to what it was.
93
12020Traces with a state that satisfies x1 and with all states not satisfying x2F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
We can think of the two constraints separately
94
12020x1 is eventually true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F = eventually, so x1 is going to be eventually true in the future. G = always, so x2 is false in all states.
95
12020Eventually, x1 will be true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F(x1) means that x1 is eventually true and G(!x2) means that x2 is always false. The "and" combining them just means that both of these must be true
96
12020Eventually x1 is true and x2 is always not true.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
As Professor Nelson talked about in lecture, F means eventually and G means globally or always, so since it is F(x1) that means eventually x1 is true and !x2 means x2 is false, so G(!x2) means that x2 is always false.
97
12020x1 eventually becomes true, and x2 is always false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F(x1) implies x1 eventually becomes true, G(!x2) implies x2 is always false globally
98
12020x1 is eventually True and x2 is always FalseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
This is a direct translation of the above predicate since F means eventually and G means always
99
12020Eventually x1 is true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
x2 is a state of all falses (as G means always), and x1 has an infinite number of true states since you can always find a future true state (F)
100
12020It eventually holds that x1 is true. And x2 is always false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
G means always and F means eventually.
101
12020x1 is eventually true and x2 is always false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F means eventually and G means always. Therefore, when I substitute these symbols to words, I get the above answer.
102
12020eventually x1 is true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F( ) means eventually, and "G( )" means always. So F(x1) means that x1 is eventually true, and G(!x2) means that x2 is always false.
103
12020There is a case where eventually x1 is true and x2 is always false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F(x1) means eventually x1 is true. G(!x2) means x2 is always false.
104
12020The set of traces where x1 is eventually true, and x2 is never trueF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F is the operator for eventually, so x1 is eventually true; and G is the operator for always, so G(!x2) means that it always holds that x2 is false
105
12020x1 is eventually true and x2 is always false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F(x1): x1 is eventually true G(!x2): x2 is false globally
106
12020Eventually x1 is true and always x2 is false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F is eventually and G is always. The "and" is outside of both, so both statements independently hold.
107
12020x1 is true eventually, but x2 is never true.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
This is direct translation.
108
12020x1 eventually becomes true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
109
12020Eventually x1 will be true, while x2 will never be true.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
Pretty simple.
110
12020x1 will eventually happen, and x2 will never happen.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
Straightforward.
111
12020x1 is finally true and x2 is always is trueF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
112
12020x1 eventually holds at some point and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
x1 has to be true at some point, and in the mean time x2 has to be always false.
113
12020x1 is true eventually and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F is eventually and G is global
114
12020It eventually holds that x1 is true AND x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
x1 is in a "finally" formula, and is in conjunction with "not x2" in a global formula
115
12020x1 is eventually true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F(x1) -> eventually x1, G(!x2) -> always not x2
116
12020It eventually holds that x1 is true, and it always holds that x2 is false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
117
12020it eventually holds that x1 is true and it always holds that x2 is falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
F(x1) says that it eventually holds that x1, and G(!x2) says that !x2 is always true, which means that x2 is always false
118
12020x1 will eventually be true, and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
eventually x1 and always not x2.
119
12020x1 is eventually true and x2 is always not (never) true.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
x1 is true in the future implied by F and x2 is always not true, I am not positive but I think the conjunction of G and ! = never
120
12020x1 is eventually true and x2 is always falseF(x1) and G(!x2)
This is a question that Tim thought would be interesting.
121
12020It eventually holds that x1 is true and always holds that x2 is false.F(x1) and G(!x2)
This is a question that Tim thought would be interesting.
122
12020It always holds that when x1 is true then in the next state x1 is false until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G means that the expression inside always holds and U means that the expression holds until the the expression on the right of it becomes true. So since the whole the formula is inside of the G, it means that the expression always holds and the x1 -> part means that when x1 is true then the expression on the right occurs. Then the X refers to the next state, so it is saying that x1 is false in the next state until x2 is true.
123
2020It always holds that x1 is true if in the first entry (i=1) x2 is true and x1 is false later on.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G means always. a U b means that b must happen before a. X() tells us to look at the i=1 entry in the sequence.
124
12020Whenever x1 is true, x1 will be false thereafter until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G = (always), so for all states x1 implies that in the next state x1 is false until x2 is true.
125
12020It is always true that if x1 starts true then eventually x1 is false until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G() means that this always happens. x1 -> means that x1 starting true implies something. What it implies is that X(!x1 U x2), meaning that eventually x1 is false until x2 is true.
126
12020It is always true that if x1 is true then the next state x1 is false until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
X signifies the next state, so until x2 is true, x1 must be false
127
12020This formula accepts traces where, once x1 is true, it can never be true again until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
For every state, x1 implies that in the next state, x1 is never true until x2 is true.
128
12020It always holds the x1 implies x1 will be false in the future until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
It has to always be the case that if x1 is true, then in the next state x1 is false unless x2 is true.
129
12020It is always true that if x1 holds, then in the next state, x1 does not hold until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G means always, '->' means implies, and U means until. When I substitute these symbols to words, I get the above answer.
130
12020It is always true that if x1 is true then at the next timestep x1 is false until a certain point where x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
If x1 is true then at the next step then (!x1 U x2) is true, or x1 is false up until a state where x2 is true.
131
12020it always holds x1 is true implies x1 is false until x2 is true in the next state
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
g means always, and X denotes the next state and U denotes "until"
132
12020it is always true that when x1 is true, it implies that in the next state, x1 is false until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G() means always true for the entire statement and within G() is an implied statement. I don't know what X means so I assumed it means in the next state
133
12020It always holds that if x1 is true at some state, then x1 is false or x2 is true at the next state
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
134
2020it always holds that while x1 is true, x2 is true (or x1 is true until x2 is false)
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
Ohhh I did the other quiz first and did not know how to do this really but this seems cleaner than whatever I did--x1 being true implies that on the next step either x2 is true or x1 switches to false.
135
12020It is always true that x1 implies after the first entry, not x1 is true before x1 becomes true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
X operator talks about a formula being true for all entries after the first.
136
12020Whenever x1 holds, it will not hold again until x2 holds.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
This is just direct translation. "G(x1 -> ...)" says that the conclusion holds in any state where x1 is true, and the conclusion says for x1 to not hold until x2 holds. The X only tells us to ignore the state in which x1 is true, as we only care about x2 not holding in any _following_ state.
137
12020It is globally true that if x1 is true, then from the next iteration x1 is false until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G: it is always true that...\nx1 ->: if x1 is true then,...\nX(!x1 U x2): from the next iteration, x1 is false until x2 is true
138
12020It always holds that x1 being true implies that for the next step x1 is false up until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G indicates always so what goes inside of it always holds. The rest is based on X referring to the next step and U signifying until.
139
112020the following always happens: if x1 is true, then either x1is initially false or x2 initially true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
140
12020Whenever x1 is true, in the next state, x1 is false until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G() ~ whenever; X() ~ in the next state, U ~ until
141
12020it always holds that if x1 is true, x1 is false until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
U is until
142
2020It always holds that x1 being true implies that the next x1 states are false until x2 is true.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G means that this is always true, x1 implies (->), X represents the next traces. !x1 means that x1 is false, the U means this is the case until x2 becomes true.
143
12020After a state satisfying x1, no state can satisfy x1 until one appears that satisfies x2
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
The X(!x1 U x2) means that the trace with the first element chopped off has x1 false until x2 is true, and this requirement is brought about whenever a state satisfies x1
144
12020x1 being true always implies that in the next step x1 becomes false until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
Globally (x1 implies the next step is (x1 is false until x2))
145
2020This formula accepts traces in which x1 is false
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
The G function would then always contain essentially a null interior, being satisfied
146
12020if x1 is true it will never be true afterwards until x2 is true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
147
12020It is always true that if x1 is true, then in the next state x1 should be false or x2 should be true
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
148
12020It always holds that x1 implies that after that entry x1 will be false and then x2 will be true after that.
G(x1 -> X(!x1 U x2))
This is a question that Tim thought would be interesting.
G is always. X refers to the sequence not including the "current" entry. !x1 U x2 is that not x1 will occur before x2.
149
12020If x1 holds at some point, x2 will hold from that point onward.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
The eventually shows that the following expression will hold at some point, and the implication can only evaluate to false if x1 is true, so the constraint on x2 being true for all subsequent states (G) only becomes relevant when x1 is true for some state.
150
12020it eventually holds that x1 implies x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
example traces: once you understand basic addition, you can add any numbers
151
12020Eventually, x1 being true will imply that x2 is always true.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
At some point, x1 will become true, which will mean that from then on, x2 is true. I think this might be similar to !x1 U x2. x1 -> G(x2) means that x1 being true implies x2 is always true, and surrounding it with F() means that this is an "eventually" case.
152
12020Eventually x1 doesn't hold, or else eventually x2 always holds.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
With implication, x1 -> G(x2) is equivalent to ~x1 v G(x2). Since F distributes over v, we can say that F(~x1 v G(x2)) is equivalent to F(~x1) v F(G(x2)). This is translated to what I have in my answer.
153
12020Traces that either have a point at which x1 holds and x2 holds from that point on, or have no states that satisfy x1F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
Because we are working with implication, we also have to think that there could be no states satisfying x1
154
2020It eventually holds that if x1 is true (in this entry) then x2 is always true, otherwise if x1 is false then x2 can have any value at any point in the traceF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
155
12020It eventually holds that x1 implies that x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
G(x2) means x2 is always true, this is implied by x1 eventually because the whole statement is wrapped in an "eventually" formula
156
12020Eventually if x1 holds then x2 always holds.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
Describes set of traces where eventually an implication holds. The implication describes that if x1 holds then x2 will always hold.
157
toplevel G2020It is eventually true that x1 is true implies x2 would always be true.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
It means there is a state, where after such state where x1 is true, then for all states after that state, x2 would retain true.
158
12020it eventually holds that if x1 is true it implies that x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
The F specifies that it eventually holds. And the G specifies "always". So combining them with the implies and order of operations produced my answer.
159
12020it is eventually true that x1 implies x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
-> means implies and the entire statement is eventually true
160
12020It eventually holds that x1 implies that it always holds that x2 is true.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
I based this off of F meaning eventually and G meaning always.
161
12020At some point when x1 is true, x2 will be always true thereafter.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
F = (eventually), x1 implies that G = (always) x2 will be always true.
162
12020It eventually holds that, if x1 is true, x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
F = eventually, -> = if and only if, G(x2) = always x2
163
12020Eventually if x1 is true, x2 will always be trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
It's straightforward
164
12020it eventually holds that if x1 is true then x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
the F operator states that we will eventually reach a condition, and x1 -> G(x2) states that if x1 is true in initial states then x2 is true in all states.
165
12020If x1 is eventually true, then x2 is true in all states.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
One thing I'm unsure about is if the G for x2 applies retroactively (i.e. states before the state that x1 is true). My intuition would be that it does.
166
12020Eventually there will be a point where if x1 is true, x2 is always true thereafterF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
F corresponds to eventually, and the -> says that if x1 is true, x2 has to be true thereafter
167
12020at some time in the future, if x1, x2 will always holdF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
168
12020It is eventually true that if x1 holds then x2 always holds.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
F means eventually, -> means implies, and G means always. Therefore, when I substitute these symbols to words, I get the above answer.
169
12020Finally, if x1 is true, x2 always is trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
170
12020Eventually, x1 is true, implying x2 is always trueF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
F means that eventually x1 is true, and G means that x2 is always true, which is implied by x1 (the arrow means implies)
171
12020it is eventually true that if x1 then x2 is always true"F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
I translate what is inside the F: x1 implies globally x2 is true.\nThen, I added it is eventually true because of "F"
172
12020It is eventually true that x1 implies that x2 is always true.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
F means finally, so the fact that in this trace, eventually some property will hold. This property is that x1 being true implies something about x2 -- namely, that x2 will be globally (always) true.
173
12020If x1 is always true, then eventually x2 will always be true.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
If x1 is ever false, then the statement is satisfied (x1 being false satisfies the implies). G requires the "always" true.
174
12020it always eventually holds that when x1 is true, x2 is true for all future statesF(x1 -> G(x2))
This is a question that Tim thought would be interesting.
175
12020It eventually holds that if x1 is true then it always holds that x2 is true.F(x1 -> G(x2))
This is a question that Tim thought would be interesting.
176
12020eventually x1 is true and x2 is always falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
G(!x2) means that x2 is always false, and F means eventually
177
12020Eventually, we will reach a state in which x1 is true and x2 is always false since then.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
Pretty straightforward.
178
12020Eventually, x1 is true and x2 will always be falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
I first translated the statements inside the F() which means eventually, and how g() means always.
179
12020It eventually holds that x1 is true and x2 is always false.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
180
12020It eventually holds that x1 is true and x2 is always false.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
181
12020There is eventually a state in which x1 holds, and in that state and all after, x2 does not hold.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
This is just direct translation.
182
12020Traces that have a state where x1 holds and x2 does not from that state onF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
Everything inside of F only applies to the subtrace it looks for, so G(!x2) only applies after the state that satisfies x1
183
12020Eventually x1 will be true and x2 will always be falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
F indicates that what goes in it will eventually hold. x1 is inside this so it will eventually be true. G indicates always and ! indicates not so x2 will eventually always be false.
184
12020It is eventually true that x1 is initially true and that x2 is always falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
G(~x2) means for all time i, x2 is always false\nx1 means x1 is initially true \nso F(x1 and G(!x2)) means "It is eventually true that x1 is initially true and that x2 is always false"
185
12020It eventually holds that x1 is true, and x2 will always be falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
x1 and G(!x2) describes a trace in which x1 is true, and !x2 will always be true, meaning x2 will always be false. Wrapping this in F gives a trace in which this condition eventually holds
186
2020at some time in the future x1 will be true while x2 will be falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
187
12020It is eventually true that x1 is true and x2 is false thereafterF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
F: eventually true \nG(!x2): x2 is false globally
188
12020It eventually holds that x1 is true and x2 is false and is false thereafterF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
At some point it has to be the case that x1 is true and x2 is always false, which means x2 is false and is false thereafter
189
12020eventually x1 will be true and, from that state onwards, x2 will always be falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
190
12020it always eventually holds that x1 is true and from there on, x2 is always falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
F is eventually holds and G is always from there
191
12020At some state x1 will be true, and from that point onward (inclusive) x2 will be falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
The eventually shows that the following expression will hold at some point, and we only have x1 holding at the first state at that point (because there are no qualifiers), while !x2 will hold for all subsequent states (including the state where x1 was true because there is no X).
192
12020in at least one state, x2 is always false and x1 is always trueF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
f function makes it so that it is for at least one state, everything else is self explanatory.
193
12020It will eventually hold that x1 is true while x2 is always falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
F(___) means that eventually, the inner variable will eventually be true, at some w^n. Given that n, and the inner G(!x2), we know that x2 must be false for all w^i where i >= n
194
12020A state will eventually be reached where x1 is true and all states afterwards (including the state where x1 is true) hold x2 to be false.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
Since the "F" formula is being applied to "x1 and G(!x2)", this formula is describing a trace where, eventually, "x1 and G(!x2)" is satisfied. The subformula "x1 and G(!x2)" is satisfied in a subtrace w where w(0) sets x1 to true, and all states in w hold x2 to be false.
195
12020It will eventually hold that x1 is true and x2 always holds false.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
F means eventually, G means globally/always, and ! means not.
196
12020It eventually holds that x1 is true and it always holds that x2 is false.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
The whole formula is wrapped in an F, meaning "eventually". Inside we have a bare x1, meaning that x1 is true. We also have G(!x2), which means that it always holds that x2 is false. Therefore it eventually holds that 1: x1 is true, and 2: it always holds that x2 is false.
197
12020it is eventually true that x1 is true and x2 is always falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
x1 and G(!x2) : means that in the current state, x1 is true and !x2 is always false in all the rest of the states
198
12020It eventually holds that x1 is true and that x2 is false always.F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
I'm unsure if the G will refer to the subtrace that the F is continuing off of, not just the trace as a whole.
199
12020eventually, x1 is true and x2 is always falseF(x1 and G(!x2))
This is a question that Tim thought would be interesting.
Eventually comes from F. So, eventually, x1 must be true, and then x1 must always (G) be false (!)
200
12020Eventually, x1 will be true and x2 will be always false (from that point forward)F(x1 and G(!x2))
This is a question that Tim thought would be interesting.
F = eventually, G(!x2) = always not x2 (from the point of eventuality)
201
12020x1 eventually holds. When x1 holds, it implies that x2 always holds.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F means eventually, -> means implies, and G means always. When I substitute them, I get the above answer.
202
12020If x1 is eventually true, then x2 is globally trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F(x1) is x1 is eventually true.\nWe have IMPLIES.\nG(x2) is x2 is globally true.
203
12020It eventually holds that x1 is true which implies that x2 is true alwaysF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
204
12020if x1 is true at some point then x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F is eventually and G is always
205
12020If eventually x1 becomes true, then x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
If the first property is satisfied (F(x1)), then G(x2) is satisfied.
206
2020If x1 will always eventually be true, then x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F(x1) will hold if x1 will always eventually be true, implying that x2 is true globally.
207
2020This always eventually holds: when x1 is true, x2 must be true in all i cases.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
They are separate variables, thus their dependence on each other is outlined in the function.
208
12020x1 eventually being true implies that x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
If x1 is true at some point, then x2 has to be always true.
209
12020If it eventually holds that x1 is true then x2 is always true in the trace, otherwise if x1 is never true then x2 can have any value (true or false) at any point in the trace.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
An implication statement (i.e. p->q) is true when p is true and q is true, and also true when p is false and q is any value.
210
12020Eventually x1 is true implies x2 is always true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F(x1) means eventually x1. G(x2) means x2 is always true. The -> between them means that x1 eventually being true implies that x2 is always true.
211
12020Eventually, x1 being true implies x2 is always true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
If x1 is a trace with infinite true states, this implies that x2 is always true (just the trace of true states)
212
12020Traces that contain no entries that satisfy x1 or have all entries satisfy x2F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
Desugared the implication: ~F(x1) v G(x2)
213
12020if x1 will happen at some point in the future, x2 will always holdsF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
214
12020Either x1 is always false or x2 is always true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
Using definition of implies, this is equivalent to ~F(x1) v G(x2), and using the property that ~F~ = G, we get that this is equivalent to G(~x1) v G(x2), which directly translates to what I have above.
215
12020if x1 is eventually true, then x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F(x1) means that x1 is eventually true and G(x2) means that x2 is always true
216
12020If it eventually holds that x1 is true, then it always holds that x2 is trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F(x1) says that it eventually holds that x1 is true, -> is implies, and G(x2) says that it always holds that x2 is true
217
12020if x1 is eventually true than x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
This is my answer because "F(x1)" means that x1 is eventually true, "->" means implies, and "G(x2)" means that x2 is always true.
218
12020If x1 eventually evaluates to true, then x2 must always be true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F(x1) is true if x1 ever becomes true. G(x2) states that x2 must be true for all states.
219
12020If x1 is eventually true, then x2 must always be, and have always been true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
This is using the F operator which indicates eventual truth and saying that this implies something about the G operator which indicates global truth.
220
12020If there is at least one state in which x1 is true then x2 is true in all states.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F = (eventually), so if x1 is eventually true then G = (always) x2 is always true.
221
12020If it eventually holds that x1 is true, then it always holds that x2 is true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
222
12020It holds for all traces except ones where x1 is eventually true, but x2 is not always true.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
I read this as F(x1) implies G(x2). Therefore it holds for all traces except ones where F(x1) is true, but G(x2) is not true. F(x1) is true for all traces where x1 is eventually and G(x2) holds for traces where x2 is always true in all states. Thus the state described above is one where F(x1) is satisfied but G(x2) is not.
223
112020Eventually x1 is true and x2 is always true thereafter.F(x1) -> G(x2)
This is a question that Tim thought would be interesting.
F means eventually, so this F(x1) means that x1 will eventually be true and so the -> means basically thereafter, meaning once this happens then x2 is always true since G means always.
224
12020if x1 is eventually true then x2 is always trueF(x1) -> G(x2)
This is a question that Tim thought would be interesting.
implies can be thought of as if-then\nF is eventually\nG is always
225
2020It always holds that x1 is false at some state and eventually x1 is true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
226
2020No traces satisfy this formulaG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
To satisfy the requirement that eventually x1 will be satisfied for all entries in the trace, it must be true that there is an entry that satisfies x1. This is ruled out by the fact that x1 must not be satisfied by every trace
227
2020it always holds that x1 is set to false and eventually x1 is trueG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
!x1 implies x1 is set to false \n F(x1) eventually x1
228
2020It always holds that x1 is false in one state and x1 is eventually true in the subsequent statesG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
!x1 and F(x1) mean that x1 does not hold now but it eventually holds, and this is always true because it is wrapped in a "globally" operator
229
2020It always holds that x1 is false and eventually x1 is true (note: this is a contradiction)G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
G = always, !x1 = not x1, F(x1) = eventually x1
230
2020It always holds that x1 is false and eventually x1 is true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
This directly translates as the above because "G( )" means always holds, and "F( )" means eventually holds. However I don\'t think this LTL formula describes any traces because if there is a trace where x1 is always false, x1 cannot eventually be true.
231
2020It is always true that x1 does not hold and x1 is eventually true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
F means eventually, '!' means does not hold, and G means always. When I substitute these symbols to words, I get the above answer.
232
2020X1 is always false and it will eventually be trueG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
nothing complicated
233
2020It always holds that x1 is not true now but will be in the future.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
The G implies the always of the relationship between x1 now and the future of x1. The not x1 works in conjunction with x1 being true in the future
234
2020It is always the case that x1 is false and x1 is eventually true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
G means always, and F means eventually. As a result, the above expression can be translated to it is always the case that x1 is false and x1 is eventually true.
235
2020Always, !x1 is false and eventually (F) x1 is trueG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
This wouldn't describe any traces... this says both that every single x1 is false, and there always exists some future x1 is true, which is a contradiction.
236
2020x1 is always false and always eventually true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
G(!x1) = x1 is always false\nG(F(x1) = x1 is always eventually true\nSeems like it should be unsat
237
2020always, x1 is false and eventually x1 is trueG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
this is a contradiction
238
2020x1 is always false and x1 always becomes true eventuallyG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
in alway G, we see that x1's value and the eventual state of x1
239
2020it always holds that x1 is false, and eventually will be true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
is this possible? if x1 is always false, then it can never eventually be true.
240
2020The empty set.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
It says x1 is always false, yet x1 is always eventually true, which is a contradiction.
241
2020It always holds that x1 is false and will eventually become trueG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
242
2020No traces.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
This formula says that x1 never holds, but that, at all states, it eventually holds. G(F(x1)) implies F(x1), and F(x1) implies ~G(~x1), so the original formula is equivalent to G(~x1) AND ~G(~x1), an absurdity.
243
2020It's always true that x1 is false, but x1 will eventually be true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
This doesn't seem possible. Any potential time for x1 to be true, x1 will be false.
244
2020It always holds that x1 is false and that x1 will eventually be true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
This is the "procrastination" predicate... eventually your work will get done, but not right now. Unfortunately, it also means your work can definitionally never get done.\n\nThis predicate says that, for all states, x1 is false, but that x1 will eventually be true in some state following the current one. The addition of the global operator makes this an impossibility, because the clause enforcing x1 is false prevents the eventually clause from ever being true.\n\nI think this is my favorite LTL expression of all time.
245
2020This formula is unsatisfiableG(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
It says that it is always the case that x1 is false, but also that x1 will eventually be true
246
2020It always holds that if x1 evaluates to False and some time in the future x1 must evaluate to True.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
F(x1) states that sometime in the future x1 evaluates to true. The G implies that this must hold for all subsequences.
247
2020it always holds that x1 is false and eventually holds in the future?G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
248
2020It is always that x1 is false, and it is always that x1 will eventually become true, which is impossible.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
We know that G is distributive under conjunction, so the clause could be simplified to G(!x1)&GF(x1). However, if x1 is always false, then x1 will never become true.
249
2020When x1 eventually holds true, it always holds that x1 is not true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
Following order of operations, the innermost parentheses were expressed first.
250
2020It's always true that x1 is false and eventually x1 is true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
I simply read the LTL and translated each part by part by definition. \n\nG -> always\n! x1-> not x1 ->x1 is false\nF(x1) -> eventually x1
251
2020x1 is always false. Also, at any point, there is always a later point at which x1 is true.G(!x1 and F(x1))
I don't think there are any sequences that satisfy this formula, which is interesting.
We can distribute the always around the "and". The first part is just an always, the second is just an always eventually.
252
12020Its always true that if x1 is true then x1 is false in the next state and its always true that if x1 is false then x1 is true in the next state
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
The and means that both statements must be true. G(x1 -> X(!x1)) means that if x1 is true the next x1 is false. While G(!x1 -> X(x1)) means that if x1 is false that x1 is true in next
253
12020x1 and !x1 alternate
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
it always holds true that x1 implies that the next must be !x1 and it always holds true that x1! implies that the next must be x1
254
12020It is always true that x1 being true implies x1 eventually becomes false. It is also always true that x1 being false implies eventually x1 becomes true.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
This means that if x1 starts true then it must become false and if it starts false then it must become true.
255
12020It is globally true that if x1 then x1 is false next, and it is always true that if x1 is false, then x1 is true next.\nThe value of x1 changes every time.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
G(x1 -> X(NOT x1)): It is always true that x1 implies x1 is false next.\nThe second part is pretty much that same expect we have not first.
256
12020if x1 is true, it's next value is always false; if x1 is false, it's next value is always true
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
in the always, we see that how x1's value implies its next value
257
12020x1 always flips values over each transition
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
The more logically complete/literal answer would probably be it is always true that if x1 is true, x1 is false in the next state, and if x1 is false, it is true in the next state. But I like the plain english version more.
258
12020It always holds that if x1 is true then x1 is not true at the next state, and it always holds that if x1 is not true then x1 is true at the next state. Therefore x1 alternates being true and false for each state.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
The formula starts with a G, meaning always true. Inside we start with x1 -> X(!x1), which means that x1 implies X(!x1), or that x1 being true implies that x1 is not true at the next state. We also have G(!x1 -> X(x1)), which means that it always holds that if x1 is not true then x1 must be true at the next state. Thus x1 -> X(!x1) and !x1 -> X(x1), meaning that x1 has to alternate between being true and false for each successive state.
259
12020It is always true that x1's truth value switches between each state.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
The first part of the formula says that if x1 is true, then in the next state it false always and the second part says that if x1 is false it is true in the next state, therefore the truth value switches each state.
260
2020Always satisfied
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
If x1 is true, X(!x1) cannot be true, then it is false in all cases. The same is true for the opposite direction. But the G functions will always be True as they contain the same inputs in all traces (and is commutative so though this statement is not entirely true it accurately sums up my logic here)
261
12020This describes a trace where x1 being true and x1 being not true switch off sequentially, forever.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
It is always true that x1 being true implies that, in the next state, x1 is not true, and that x1 not being true implies that x1 is true in the next state.
262
12020It always holds that if x1 is true then x1 is false in the first state and it always holds that if x1 is false then x1 is true in the first state.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
X means in the first State.
263
12020Sequence is alternating true, false, globally.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
G(x1->X(!x1)) implies either that for every state x1 can either be true and X(!x1)be True (alternating). Or x1 is false and X(!x1) is anything. The second condition constrains it to the alternating condition because G(!x1->X(x1)) implies that if x1 is False, X(x1) must be True globally.
264
12020For every state, x1 is not equivalent to its next state.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
265
12020x1 alternates true, false, true, false, ... (or false, true, false, true, ...) forever.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
G(x1 -> ...) says for the conclusion to hold in any state where x1 is true. X(!x1) says for x1 to not hold in the next state. So, whenever x1 is true, it's false in the following state. However, the second half says the exact inverse: whenever x1 is false, it is true in the following state. This leads to an alternating pattern of true/false forever.
266
12020This describes the trace where x1 alternates between true and false
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
The first clause says that whenever x1 is true, it will be false in the next state. The second clause says that whenever x1 is false it will be true in the next state
267
12020It always holds that if x1 is true at some state then x1 is false at the next state, vice versa.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
268
12020It always holds that if x1 is true, x1 will be false at the next state, and it is always true that if x1 is false, x1 will be true at the next state
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
x1 -> X(!x1) describes a trace in which x1 being true implies !x1 will be true at the next state, meaning x1 will be false at the next state. Wrapping this in G means that this property always holds. The next clause is just the opposite condition.
269
12020It always holds that x1 being true implies that in the next step x1 will be false. It always holds that x1 being false implies that x1 will be true at the next step.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
I think this is saying that x1 will always alternate.
270
12020IT always holds that x1 is true implies that the next state of x1 is false, and it always holds that x1 is false implies that the next state of x1 is true
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
This statement seems to be reciprocal, as it basically describes true to false, and false to true always holds
271
2020It is always true that x1 being true means that x1 will become false and that x1 being false means that x1 will become true
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
This just says that x1 will never settle to a single value, it will always oscillate.
272
2020It always holds that x1 implies that the next subtrace contains x1 being false. And it always holds that if x1 is false then in the next subtrace x1 is true.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
This is my answer because I translated "G( )" to mean always, "->" to mean implies, and "X( )" to mean something is true about the next subtrace.
273
12020It is always true that when x1 holds, in the next state, x1 does not hold. It is also always true that when x1 does not hold, in the next state x1 holds.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
F means eventually, -> means implies, G means always, X means next state, and ! means does not hold. When I substitute these symbols to words, I get the above answer.
274
12020It always holds that, if x1 is true x1 is false in the next state; and it always holds that, if x1 is false x1 is true in the next state
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
G() = always, -> = if ... then ..., !x1 = not x1, X(x1) = x1 in next state
275
12020This is the same as the previous, but without the eventually. Traces in which x1 alternates should satisfy.
G(x1 -> X(!x1)) and G(!x1 -> X(x1))
This describes only traces alternating true and false statements, which is an interesting and very specific example.
The last explanation holds here.
276
2020It always holds that in that when x1 is true in the next state that when x2 is true then it is not true in the next state. All of these expressions are bidirectional, meaning that when x2 is false in the next state then it is true at the current state and that this then means that x1 is true in the next state.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G means that the formula inside always holds. X refers to the next state and <=> means that the relation is bidirectional, meaning that it holds for both ways, which is how I arrived at my answer above.
277
2020It always holds that the next state of x1 if and only if x2 if and only if the next state of x2 is false
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
The double arrows in this case represent the if and only if clause, which describes the relationship between x2 and x1.
278
2020it is always true that x1 will be true in the next timestep if and only if x2 has the opposite value to x2 in the next step
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
279
2020It is always the case that: x1 is true in the next state if and only if (x2 is true if and only if x2 is false in the next state).
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
I essentially unpacked the formula piece by piece.
280
2020It is always true that x1 will be true in the next states if and only if x2 is true if and only if x2 is false in the next state.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
X(x1) means x1 is true in the next state. X(!x2) means !x2 is true in the next state, so x2 is false in the next state. So, X(x1) <=> (x2 <=> X(!x2) means that x1 is true in the next state iff x2 is true iff x2 is false in the next state. Wrapping all of this in G says that it's always true, which means it is always true that x1 will be true in the next states if and only if x2 is true if and only if x2 is false in the next state.
281
2020x2 is always true if and only if the next time x2 is not true, always if and only if the next time x1 is true.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
Beginning with the nested expression, x2 is related to its next iteration, and this is related to x1's following occurrence. This relationship between x1 and x2 always holds.
282
2020Always, if and only if the next x1 is true, then it is also true that x2 being true implies and is implied when the next x2 is false
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
Globally (the Next x1 is true if and only if (x2 is true if and only if the next x2 is false))
283
2020always ,x1 next is true if and only if x2 is true if and only if x2 next is not true
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
284
2020It is always the case that x1 being true in the next step implies that x2 implies x2 is false in the next step and vice-versa, and vice-versa.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G means always, <=> is a two ways implies, meaning that a <=> b means that a implies b and b implies a (or "vice-versa"). X means in the next step.
285
2020For every state, either x2 is true in the current state and false in the next and x1 is true in the next state, or x1 is false in the next state and x2 is false in the current state and true in the next.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
It's always true that x1 is true in the next state if and only if (x2 is true if and only if x2 is not true in the next state.
286
2020(iff x1 holds true for the next then iff( x2 then x2 holds false for the next.))
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
x1 must be followed by another x1 which also means that there must be an x2 then x2! followed by another x2!
287
2020it is always true that x1 is true iff x2 alternates being true and false
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
X(x1) : x1 is true in the next state\n(x2 <=> X(!x2)) : x2 is true iff it is false in the next state,\nwhich means that x2 alternates being true and false\n
288
2020x2 is true if and only if initially x2 being false. This will happen if and only if x1 initially being true.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
double arrow gives if and only if.
289
2020Always x2 iff in the previous entry x2 iff in this entry not x2.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G is always and the iff is interpreted directly.
290
2020It is always true that the next state after x1 is true if and only if this is also true: that x2 holds if and only if, in the subsequent state, x2 does not hold.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G means "always," and X means the subsequent state. The rest of the formula is just nested statements of bidirectional implies (iff). So the formula describes a trace where certain properties (x1 and x2) holding is contingent on what their values would be in the subsequent states.
291
2020It always holds that x1 is true at the next state iff the following: x2 is true iff x2 is not true at the next state.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
The formula starts with a G, meaning the contents are always true. We have a bi-implication that starts with X(x1), meaning x1 is true at the next state. The second half of the bi-implication is another bi-implication: x2 <=> X(!x2), or x2 is true iff x2 is not true at the next state. Therefore x1 is true at the next state iff x2 implies that x2 is not true at the next state, and vice versa.
292
2020x2 is true iff the next time x2 happens is not true, iff it is always true that the next time x1 happens is true.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
Following the order of operations, the innermost parentheses is expressed first, which is then connected to the rest of the expression using iff.
293
2020It is always true that x1 eventually becomes true if and only if x2 becomes true if and only if x2 is eventually false.
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
x2 <=> X(!x2) means that x2 can only start initially true if and only if it eventually becomes false. X(x1) <=> means that x1 can only become true if and only if the other thing is true. Thus, this means that x2 must start true then eventually become false and x1 must eventually become true. This always holds because of G.
294
2020it always holds that x2 becomes false at some point and that x1 is always true after the first state
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
The condition x2 <=> X(!x2) has two cases associated with it. Either x2 is True, meaning that the first entry in the sequence evaluates to true, implying that X(!x2) must evaluate to true (there exists one state where x2 must evaluate to false). Or x2 is False, and X(!x2) must evaluate to false meaning that everything past the first entry must be false. Therefore this condition ensures that x2 takes on a False value at some point. X(x1) ensures that x1 must be true for the subsequence after the first state.
295
2020It always holds that x1 is true in the next entry if and only if x2 is true in this entry and false in the next entry
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
296
2020it always holds that x1 is true in the first entry, if and only if x2 is true, if and only if x2 is not true in the first entry
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G means always, and X means to look at the 1st index (as in after the 0th index)
297
2020It always holds that, x1 is true in the next step if and only if it holds that, x2 is true if and only if x2 is false in the next step (yeah, that's a mouthful)
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G = always, X(x1) = x1 in next step, !x2 = not x2, <=> = if and only if
298
2020x1 will always be true in the next state if and only if x2 is false proceeding x2 being true
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
It's always true that x1 is true in the next state if and only if x2 is true if and only if x2 is false in the next state.
299
2020it always holds true that x1 being true at the next time step means that x2 is true now and x2 is false at the next time step (and that x2 being true now and x2 being false at the next time step means that x1 is true at the next time step)
G(X(x1) <=> (x2 <=> X(!x2))
When I came up with this formula, I was imagining a light switch turning a light on and off. My idea was that, whenever x1 is true, it flips the state of x2 like a person pressing a light button. I had to go through a few different ideas before coming upon this one. One of the interesting ideas here is that, intuitively, we want to talk about the "previous" state to say that, if the button is pressed, the light should switch from its previous state, and otherwise should stay the same. To get around this, I made the current state the "previous" state, and considered what happens if the button is pressed in the next state, instead.
G means that this will always be true, <=> means iff, X(x1) means x1 is true at the next time step, x2 means x2 is currently true and X(!x2) means that x2 is false at the next time step
300
112020if x1 is true then x1 is also true in the next time step\nif in the next two timestep x1 is true, then x1 is false in the next three timesteps
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
X(x1) means the next timestep of x1, \n-> means if-then\nso we can translate word by word and obtain the above result
301
12020If x1 is true, then x1 is also true next. Moreover, x1 is true in the next next sate and x1 is false in the next next next next state.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
x1 -> X(x1): x1 implies x1 in the next state\nX(X(x1) and X(X(X(!x1))) is just a bunch of next...
302
12020if x1 is true in the first state, then it is true in the second and third and false in the fourth
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
The first statement after the implication means that x1 is true in the successor state; the second statement means that x1 is true in the successor to the successor state; the third means that it is false in the successor to the successor to the successor state.
303
112020it holds when if the second state of x1 is true if the first state of x1 is true and if the 3rd and 5th states of x1 evaluate to true
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
in order for x1 -> X(x1) to evaluate true, either x1 is False, or both X(x1) and and x1 is true. X(X(x1) and X(X(X(!x1))) means that we are looking at making sure the third element evaluate true due to the X(X(x1)) and the fifth element evaluate false due to X(X(X(X(!x1)))) (assuming the distributive property holds).
304
12020If x1 is true now, then it will be true in the next two states, but not in three states.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
This is direct translation.
305
112020x1 is true implies x1 is true in the next state, and x1 is true for the next two states, and the next 3 states x1 is false
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
I assume X() means next state, -> means implies and I think nested X()s means the next state
306
12020If x1 in the first state, it is also true in the second, third, and fourth
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
There is no wrapping G or F, so this only applies at the start
307
12020if x1 is true in the current state, then it is true in the next, and true in the state after, and false in the state after that
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
X(x1) : means x1 is true in the next\nX(X(x1)) : means x1 is true in the next next state\nX(X(X(!x1))) : means x1 is true in the next next next state
308
12020if x1 is true at some state, then x1 is true at the next state, and x1 is true at the state after next state and x1 is false at the fourth state.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
309
12020If x1 is true (in the very first state), it will be true in true in the second state, and in the third, and false in the fourth
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
-> = if ... then ..., X(x1) = x1 in next state, X(X(x1)) = x1 in next next state, X(X(X(!x1))) = not x1 in next next next state
310
12020x1 implies its next, and the time after x1's next must be true and the 3rd time after x1 must be false
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
I got this from mainly following the order of operations and counting all of the nested Xs to figure out which "next\'s next" the statement was referring to.
311
12020If x1 is true, then x1 is true in the next and after next states, but is false in the state after that.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
312
2020this statement evaluates to false
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
X(X(x1)) equals X(x1). So this statement is equivalent to x1 -> X(x1) and X(X(x1) and X(!x1)). X(x1) and X(!x1) is always false.
313
12020if x1 is true, then x1 is true in the next two states, but false in the 3rd state afterwards
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
314
12020x1 is true then in the next state x1 is true and in the next next state x1 is true and in the next next next state x1 is false.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
I know that the x1 -> means that x1 is true then the part of the formula to the right of the arrow happens after this. I think that X represents like the next state, but I posted on Piazza about was X meant and I never received an answer so that is why I went with that. So I took it to mean that X(x1) means that x1 in the next state will be true and so on and so forth for the remaining part of the formula.
315
12020If x1 is true in the first state, then x1 is true in the next state, and x1 is true in the second state and x1 is false in the third state.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
The lack of wrappers like G or F make me a bit unsure, but I believe without these it defaults to rule 1 which points to state 0. The wrapped X's seem to look ahead one extra state for each additional X.
316
112020If x1 is initially true, then at the next step x1 is true. Also, at the third step (step 2), x1 is true. At the fourth step (step 3), x2 is true.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
The "ands" can each be separated and then the implies can be separted. The remainder is just saying that x1 is true or false at some step.
317
12020if x1 is true, the next time and time after that x1 is true, but the following time x1 is false.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
I reasoned through this sequentially going through the x1 and the turns of it to see when it changes
318
12020if x1 is true in entry 0, then it is also true in entry 1 and 2, and false in entry 3
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
I may have misunderstood how X(a) works. I thought it meant that a must hold for every entry after the first one in the sequence. If this is the case, then x1 in the first entry can only be false, because X(x1) contradicts X(X(X(!x1))). If X(a) means that a must hold in the next entry after the first one (and we do not know/care about the rest), then we may get the sequence TTTF... ('...' signifies that we do not know/care about the rest of the sequence).
319
12020if x1 is true in state 0, then x1 is true in state 1. x1 is true in state 2 and x1 is false in state 3.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
x1 -> X(x1) : x1 is true in state 0 implies x1 is true in state 1\nX(X(x1): x1 is true in state 2\nX(X(X(!x1))): x1 is false in state 3
320
12020If x1 starts true then it is eventually true. If eventually x1 eventually becomes true, also eventually eventually eventually x1 becomes false.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
x1 -> X(x1) is saying that if x1 starts it must become true again later. X(X(x1) and X(X(X(!x1))) is saying that there is an even later point where once again x1 is true and a further later point where x1 is false. This is saying that if x1 starts true then there are later points where it is true and even later points where it is false.
321
12020if x1 is true, then in the next state it will be true, in the state after that it will also be true, but in the state after that it will be false.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
I think this just dictates a particular sequence, though I think it is impossible, because XX(x1) would imply that the next state has x1 being true, but this stipulates that x1 is false.
322
12020When x1 is true, the next two states are also true, and the third is false.
x1 -> X(x1) and X(X(x1) and X(X(X(!x1)))
in a row is interesting
Each X refers to the next state. So, x1 implies that the two next states are true, and the next after that is false (!)
323
12020It always holds that x1 being true implies that on the next step, x1 will be falseG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
Any time x1 is true, the very next state is that x1 becomes false
324
12020if x1 ever becomes True, there must always be a state after it that evaluates to false (i.e, there are never two states that evaluate to true in a row)G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
I explained this a few questions ago, x1 either takes the value true and X(!x1) takes on the value true. If x1 is False, then X(!x1) can take on any value. Therefore if we ever encounter x1 as true, the next state must evaluate to false.
325
12020If x1 is true in the current state, x1 is false in the next state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
Globally (for all states), x1 implies not x1 in the next state.
326
12020Whenever x1 is true, in the next state, x1 will be falseG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
G() ~ Whenever, X() ~ in the next state
327
12020In this trace, it is always true that x1 being true implies that in the subsequent state, x1 is false.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
G means globally (always), the arrow means implies, X means "next", so the next state.
328
12020it is always true that if x1 is true, it must be false in the next stateG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
x1 -> X(!x1) : means that if x1 is currently true, \nthen in the next state x1 is false\n\nG(x1 -> X(!x1)) : means that this must always be true\n
329
12020It is globally true that if x1 is true, then x is false in the next iteration.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
G: it is globally true that...\nx1 -> X(NOT x1): x1 implies x is false next
330
12020It is always true that x1 starting true implies that eventually x1 becomes false.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
This means that if x1 is initially true it must eventually become false. x1 means initially true, -> means implies, and X(!x1) means eventually false.
331
12020It always holds that if x1 is true at some state, then x1 is false following at the next state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
It is possible that x1 is true just once, then followed by x1 is false starting at next state, instead of alternating between satisfying and not satisfying x1.
332
12020In any state where x1 is true, x1 is false in the following state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
This is really just a direct translation. "G(x1 -> ...)" says for the conclusion to hold in all states where x1 is true, and the conclusion says for x1 to not hold in the next state.
333
12020It always holds that if x1 is true, then x1 is false in the next subtrace.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
This is my answer because "G( )" means always, "->" means implies, and "X( )" refers to the next subtrace.
334
12020It is always that if a state of x1 is true, then the next state of x1 must be false.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
It means the following statement is always true: if the current state of x1 is true, then the next state of !x1 must be true, which means such state of x1 must be false.
335
12020It always holds that x1 is true implies x1 is false in the next state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
x1->X(!x1) represents x1 flipping from true to false in the next state. The G wrapper indicates that this always holds true. This type of formula might represent a trace where you continuously flip a light switch on and off.
336
12020Every time x1 is true, the next state is falseG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
These traces have at least one false state after each true state.
337
12020It is always true that when x1 is true, x1 is false in the next state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
x1 -> X(!x1) means x1 will be false in the next state when x1 is true. That holds always.
338
12020x1 cannot be true twice in a rowG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
for all states if x1 then in the next state not x1
339
12020It always holds that, if x1 is true, in the next state x1 is falseG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
G = always, -> = if ... then ..., X(!x1) = not x1 in the next state
340
12020It always holds that if x1 is true in one state then x1 is false in the next state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
341
12020it's always true that if x1 is true, it is false in the next stateG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
G is always, X is next
342
12020it always holds that if x1 is true, the next state of x1 is falseG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
Its logic is clear and it can be followed by the formula to imply it's meaning
343
12020It's always true that following x1 being true, x1 will be false in the next state.G(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
x1 implies !x1 in the next state.
344
12020it's always true that if x1 holds x1 will not hold in the next stepG(x1 -> X(!x1))
I think it would be easy to look at the formula and think it means that states alternate between satisfying and not satisfying x1, but if you look closer it only requires that satisfying states are followed by non-satisfying states
345
2020No traces satisfy this formula
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
For every state, x1 is true until it is not true and at least once, x1 is not true until it is.
346
2020\nThere is at least one state where x1 is false and one state where x1 is true.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
x1 U !x1 is always true anyways because x1 is true until x1 is false. Similarly x1! U x1 is also always true, but the fact there is F = (eventually) means that there should be a state with !x1 and another state with x1.
347
2020It is always true that x1 will be true until x1 is false (which is a tautology), and there is a state where after such state x1 will be false until x1 becomes true again.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
It means there is a global property that we can always make it true. It shows that we can always say that x1 is true until it turns to false, but this is always correct. Also, for each state, we can always find a state S, and x1 becomes false until it turns to x1 again.
348
2020for all values of x1 it's true
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
Since x1 U !x1 is always true, then both internal formulas are always true, and taking the AND of the two will always be true. That means the trace is independent of x1 and will be true always.
349
2020It holds for all possible traces
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
I am not sure about this one. It seems like it should hold for all possible cases because obviously x1 U !x1.
350
2020it is always true that: (either x1 is true/false) and eventually the opposite is true
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
351
2020x1 will switch between true and false infinitely often
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
It always holds that x1 is true until x1 is false and eventually x1 is false until x1 is true; the second part ensures that whenever x is false, there will be a state after that such that x becomes true, and the first part ensures that x will become false after starting true. Combined with the G, this means that x will be true and false infinitely often
352
2020It always holds that x1 is either true or false and that eventually x1 is either true or faluse.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
G means always/globally, F means eventually, and U means union while AND means set intersection.
353
2020it always eventually holds that x2 is true and always eventually holds that x2 is false
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
I think that the above expression has a lot of redundancy. Saying x1 is true until !x1 is almost vacuously true. The only thing it requires is that eventually !x1 is true.
354
2020it is always the case that x1 is true until it is never true again, and it is always true that eventually x1 is not true until it is always true
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
this formula is definitely confusing
355
2020It's always true that x1 or not x1 and eventually x1 or not x1
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
G means always true, x1 U !x1 means x1 or not x1 the and means that both are true and F(!x1 U x1) means eventually x1 or not x1
356
2020it is always true that x1 is true until x1 is false, and it is eventually true that x1 is false until x1 is true.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
U means something holds until something else happens, and G() means always true and F() means eventually true.
357
2020It is always true that x1 is true until it is not true and that eventually x1 is not true until it is true.\nx1 will start true then become not true then become true again.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
x1 starts true. Then because of (x1 U !x1) it will become not true. Then because of F(!x1 U x1) it will become true again.
358
2020x1 is globally true.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
The LHS of the and is a tautology, as x1 will always hold true until it doesn't. The right-hand side says that it finally not x1 will hold until it does; this also seems like a tautology. Because of these tautologies, it seems to be that it is just saying that x1 is globally just itself.
359
2020Always, x1 will be true until x1 is false and x1 will be false until x1 is eventually be true
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
I think this LTL is false
360
2020It always holds that x1 is true until x1 is false, and it is eventually true that x1 will be true again after x1 is false. More succinctly, this means that x1 is true until x1 is false, and eventually x1 will be true again.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
This is my answer because "G( )" means always, "a U b" means that a is true until b is true, and "F( )" means that something is eventually true.
361
2020It always holds that x1 is true until it is not true and that eventually x1 is false until x1 is true.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
Following the example provided in the question above, since the whole expressions is included in G, this means that the expressions inside always hold and so U means until and !x1 I took to understand that x1 is false, so x1 U !x1 would be that it is true until it is false and then since the other expression is included in F, this means that this expression will eventually hold, so eventually x1 is false until x1 is true.
362
2020This expression is unsatisfiable
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
The F function will always be satisfiable. We know from the U that the ith (for some i) x1 will be true, and for all jth (j < i) x1 will be false. Then, the opposite U function makes this unsatisfiable in combination with the G which requires all to evaluate to true.
363
2020It always holds true that x1 or not x1 is true when not x1 or x1 is eventually true.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
The two portions of the expression are connected using the AND operator.
364
2020it is always true that x1 always holds until it's not and there is some time in the future that x1 doesn't holds until it does
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
365
2020Always eventually x1 holds and always eventually x1 doesn't hold.
G((x1 U !x1) AND (F(!x1 U x1)) )
I think it is an interesting combination of G, U, and F on two boolean variables.
(x1 U !x1) is equivalent to F(!x1) since anything before the first occurrence of !x1 by definition of first must have x1 hold. Similarly, (!x1 U x1) is equivalent to F(x1), so F(!x1 U x1) is equivalent to F(F(x1)) is equivalent to F(x1). Thus, this reduces to G(F(!x1) AND F(x1)). We can further distribute the global over the AND since G(a AND b) iff G(a) AND G(b), so this is just G(F(!x)) AND G(F(x1)). So, this says "always eventually x1 holds and always eventually x1 doesn\'t hold", i.e. there is never an infinite chain of x1 or an infinite chain of ~x1.
366
2020Eventually x1 being true implies that starting 2 steps from now x1 will be false, or eventually x1 will always be false.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
Eventually either x1 will go from true to false in two steps or x1 will always be false.
367
2020once, when x1 is true it implies that x1 is always false (which i don't see how it could happen) or x1 is always false two states ahead (which could happen). So, at least once, when x1 is true, x1 is always false two states ahead.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
I think my explanation is enough in the answer box.
368
2020At some point, either x1 will be always false or when x1 is true x1 will be always false after 2 states.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F = (eventually), so at some point in the future one of 2 things should happen: (1) 2 states after x1 is true x1 should be always false; (2) x1 is always false.
369
2020It eventually holds that x1 is true implies that the second time x1 happens, it always holds that x1 is not true or x1 is always not true.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
Following the order of operations, the nexts needed to be broken down and then preceded by the x1 implies expression.
370
2020it eventually holds that x1 is true implies that either it always holds that x1 is true in the next next trace (two traces ahead) or it always holds that x1 is false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
the F shows that this entire clause "eventually" holds true, and the XX G shows that this sub-clause always holds true for the next next trace, and G ! shows it always holds false
371
2020It eventually holds that x1 implies either x1 is always false or the state after the next state is always false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
X(X(G(!x1))) indicates that the state after the next state is always false. Then we see that x1 implies either that, or that x1 is always false (G(!x1)). And that is all wrapped in the F, which means it will eventually be true.
372
2020Either it eventually holds that, if x1 is true, then after two states x1 will always be false, or x1 is always false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F = eventually, G = always, => = if ... then ..., !x1 = not x1, X(X(G(!x1))) = always not x1 after two states
373
2020If x1 is true at some point, x1 will either be false for all following states, or be false for all states following the next two.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
Note: I believe that the G(!x1) side of the OR will always be false, because the implication means that x1 is true in the current state, so x1 cannot be false in all states including this one (this depends on how the natural numbers are defined in the handout, but I believe the intention is that G means all states including the current).\n\nOtherwise, the formula states that there must be some state where this implication holds (note that it's fine if there is no state where x1 is true). Given that the implication holds, we must either have that !x1 holds in all states including and after the state where x1 was true, or that it holds in all states including and after two states after the current state (by the double application of X).
374
2020It eventually holds that either x1 is always false or once x1 becomes true then starting from 2 entries after that point x1 is always false.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
375
2020it eventually holds true that if x1 is true at the first time step then after two time steps x1 is always false or x1 is always false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F means eventually, x1 means x1 is true at the first time step, X(X(G(!x1))) means that x1 is always false after two time steps, G(!x1) means that x1 is always false
376
2020at sometime in the future x1 implies that x1 will always be false or at the next two step x1 will always be false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
X(X(G(!x1))) at the next two step x1 will always be false\nG(!x1)) 1 will always be false
377
2020It eventually holds that x1 implies that the time after the next is never true or that x1is never true.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
The fact that x1 implies X(X(... this refers to the time after the next. And G(!x1) means that x1 is never true. By combining all of these statements with their corresponding operators and wrapping them under an "it eventually holds..",that is how I arrived at my final answer.
378
2020it eventually always holds that x1 is false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
Either case ends up with G(!x1), so whether that occurs after x1 has been true and then two steps pass or it just happens, the resting state of the traces is the same
379
2020It eventually holds that x1 implies x1 is always false (to simplify, it means x1 is false)
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
X(X(G(!x1))) equals G(!x1), so it means F(x1 => G(!x1) or G(!x1))
380
2020eventually the case that if x1 is true, then either 1) starting from the state 2 steps after, x1 is always false, or 2) it is immediately the case that x1 is always false
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F means eventually, and x1 => means that if x1 is true the following is true. Thus, it is eventually the case that if x1 is true then starting from the next next state, x1 will be false, or immediately x1 is false. I'm not really sure that this makes sense because stating that an implication eventually holds is counterintuitive
381
2020eventually either x1 becomes always False.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F(G(!x1)) states that x1 must become false, x1=>X(X(G(!x1))) states that if x1 becomes true, then 2 states after that x1 must always be false (implying that x1) will become always False at some point.
382
2020It is eventually true that when x1 holds, in the 2nd next state, x1 is always false or x1 is always false in the next state.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F means eventually, -> means implies, G means always, X means next state and ! means does not hold. When I substitute these symbols to words, I get the above answer.
383
2020it will eventually hold that x1 implies not x1 is always true in 2 stages or that not x1 is always true
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
F is eventually true, X is true in the next stage, and G is always true
384
2020x1 is eventually false.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
I am assuming the expression is: F(x1=>[(X(X(G(!x1)))) or G(!x1)]). The entire right side of the implies is grouped I think.\nIf x1 is ever false, then the expression is satisfied. If x1 is always true, then the right part might be satisfied. The X(X(G(!x1)) is that G(!x1) holds 2 entries after x1 is eventually true. However, in this case this is equivalent to G(!x1), so we can simplify to x1 being eventually false. However, this is only implies if x1 is always true. Thus, this can't occur and x1 must eventually false.
385
2020Eventually if x1 is true at the current time step then for every following time step x1 is false or x1 is always false two timesteps in the future.
F(x1=>(X(X(G(!x1)))) or G(!x1))
The interesting part about this formula is the alternating between satisfying or not satisfying x1 under different circumstances.
The formula describes traces where an implication eventually happens. That implication seems to describe that if x1 holds, then x1 will be false after forever or will be false after two more timesteps.
386
2020x1 and x2 will be true until x3 is
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
It will always holds true that x1 and x2 will be true until x3 is true
387
2020It is always true that x1 and x2 are true, until x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
In this trace, it is always (G means globally, always) true that first x1 and x2 are true (they are both true at the same time), until, at a later state, x3 is true.
388
2020x1 and x2 will be true until a certain point when x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
It will always be true that x1 and x2 are true up until a certain point which satisfies the condition that x3 is true.
389
2020For every state, x1 and x2 are true until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
(x1 AND x2) Until x3 = x1 and x2 are both true in every state before x3 is true.\nNormally we ignore the states after x3 is true, but when we add the global predicate this applies it to all states.\nIt seems like a long way of saying G( (x1 and x2) or x3) since one of the two must be true in every state but I didn't put that as my answer to play it safe.
390
2020it is always true that x1 and x2 are true until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
391
2020x1 and x2 are always true until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
By order of operations, the expression left of the until must be expressed first, and thus depends on the value of the expression to the right of the until.
392
2020it always holds that x1 and x2 are both true, until x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
fairly self-explanatory, i think? this describes traces where x1 and x2 both stay true until the next time x3 is true.
393
2020it always holds x1 and x2 are true until x3 becomes true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
G means always, and A Until B means from some point in time, B becomes true, and A has to be true since any time before that time.
394
2020It is always the case that x1 and x2 are true until x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
G means always, and until, or U in LTL, is read as in the English definition.
395
2020It always holds that x1 and x2 are true until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
The G symbol encodes that for all states, x1 and x2 are only both true in states before x3 is true.
396
2020It always holds that x1 and x2 are true until x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
This is based off G meaning always and U meaning until.
397
2020It is always true that x1 and x2 are true until x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
The outer G implies everything inside is always true. Then, x1 and x2 must be true together signified by the parentheses until x3 is true
398
2020for every state, x1 and x2 will remain true, util x3 is true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
399
2020It always holds that x1 and x2 are true at least until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
The entire statement is nested in an ALWAYS formula, and the UNTIL operator says the left side is true at least until the right side is true, which it will be in some current or future position
400
2020it always holds that x1 and x2 are true until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
an example of traces can be [x1x2, x1x2 AND !x3, x3, ...]
401
2020It will always occur that x1 and x2 will hold to be true until x3 becomes true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
G indicates global always, and the statement inside is that some clause (x1 & x2) are true until the case where x3 becomes true.
402
2020x1 and x2 is equivalent to x3
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
I think this is what it means? Starting from every state, x1 and x2 must be true unless x3 is true, they can't be false after an x3, as they'll still have to be true until the next x3.
403
2020Globally, x1 AND x2 is true until x3 is true
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
Globally, x1 AND x2 is true until x3 is true
404
2020It always holds that x1 and x2 are true until x3 is true, and x3 will be true.
G((x1 AND x2) UNTIL x3)
This formula is interesting because it describes the fact that, if you specify something at the beginning, that thing will always be true until something else (that must also be pre-specified) also happens. That is, we can see the very strict linear causality that is demonstrated in the above case.
Using the order of operations with the nested parenthesis I address x1 and x2 status first and how they are no longer true when x3 is true.
405
12020at some point, x1 is eventually trueF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
406
12020It eventually holds that it eventually holds that x1 is true.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
F means that the expression inside eventually holds. Since x1 is inside two Fs, this basically means that the expression eventually holds that it eventually holds that x1 is true.
407
2020If you run the trial many many times, you will get x1 true at least once.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
F(x1) is eventually true. If you run n times, there should be once that F(x1) is satisfied, giving F(F(x1))
408
12020x1 will eventually be true sometimes in the futureF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
F(F()) ~ F()
409
12020It eventually holds that x1 is eventually trueF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
We have a nested "eventually" formula
410
12020In the future, so eventually, it will hold that x1 is true.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
F means "eventually" or "finally." I don\'t think that the doubled "F" adds any meaning to this formula!
411
12020It is eventually eventually true that x1 is true.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
I see the double eventually and I'm wondering if it should just be one? But I think this may change the meaning
412
12020eventually, eventually x1 will be trueF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
pretty sure the second F doesn't do anything
413
12020it eventually holds that x1 is eventually trueF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
The reason for the two eventually's is that there are two Fs that appear in the formula
414
12020x1 will be true at two points in the futureF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
The first eventually ensures that x1 is true at some point in the future. The second eventually makes it so that from the state where x1 is first true, x1 will be true at some point in the future.
415
12020x1 will eventually happen.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
The two F's are unnecessary since F(x) <=> F(F(x))
416
12020eventually x1 is trueF(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
F() stands for eventually, so I think that a nested F() would still mean eventually.
417
12020x1 is eventually true.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
less succinctly, this translates to "it is eventually true that x1 is eventually true." But if it is eventually true that something is eventually true, that is the same as just saying it is eventually true.
418
12020Eventually x1 is true.F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
419
12020for some time in the future, there is some time after that that x1 is true (this should be equivalent to Fx1)F(F(x1))
This formula ensures that there must be at least two instances where x1 becomes True.
420
2020It eventually holds that always on the entire subsequent path, x1 being true implies that x1 is false in the next step, and it always holds that x1 being false implies that x1 is true in the next step
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
The left-hand side of the conjunction says that eventually, something is always true in the subsequent path. That something is that x1 being true at one step implies that x1 is false in the next step (as denoted by X(!x1)). The right-hand side says that !x1 (x1 being false) implies that x1 is true in the next step
421
2020Eventually, x1 always switches signs between states
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
This is similar to an earlier one, but instead of saying that this is true for all states, we eventually get to a series of states where x flips signs.
422
2020It is eventually true that when x1 is true, the true/false values of x1 alternate.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
Eventually, it will globally hold that x1 being true implies that in the next state it will be false and that it globally holds that if x1 is false, then in the next state it will be true.
423
2020It eventually holds that x1 always flips value in the next state
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
I believe the F wrapper is arbitrary here, and that removing it would provide the same result. Saying something "eventually is always true" when the always of G starts at the first state seems silly. The nested part-- that x1 always flips value, I got since x1 is true implies x1 is false in the next state, and x1 is false implies x1 is true in the next state.
424
2020It is eventually true that its always true that if x1 is true, then x1 is false in the next state and if x1 is false, then x1 is true in the next state
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
425
2020It eventually always holds that if x1 in state0 then !x1 in state 1 and always holds that !x in state 0 then x1 in state1.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
F(G(x1->X!x1) eventually always that x1 is true in state0 implies x1 is false in state1\n G(!x1->Xx1)) it always holds that x1 is false in state0 implies that x1 is true in state 1
426
2020Eventually, it will always hold true that if x1 is true, then in the next state, x1 will be false, and that it was always hold true that if x1 is false, then in the next state, x1 will be true.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
This statement states an oscillation of the values of x1 and x2, mainly that it will switch from one to the other, whch I got from the x1 -> !X1 and the other way around.
427
2020It will eventually hold that "x1\'s truth value always implies that the i=1 x1 in the sequence is the opposite of that truth value"
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
F is eventually, G is globally/always. In the parentheses of G, x1 being true implies that the first (i=1) entry of the sequence has a false x1, and that the opposite is true if x1 is false (that is, x1 when i=1 is true in this case)
428
2020it eventually holds that x1 always implies not x1 in 2 states and that not x1 always implies x1 in 2 states
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
F is eventually true, G is always true, and X is true in the next state; X being after the implication means it holds in the state afterwards
429
2020It will eventually hold that all x1 become not x1 in the next step and that all not x1 become x1 in the next step
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
430
2020At some point, x1 will switch its truth value at every step of the trace.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
It eventually always holds that x1 in state i implies not x1 in state i+1 and not x1 in state j implies x1 in state j+1.
431
2020it is eventually true that: it is always true that x1 being true implies that in the next timestep x1 will be false AND it is always true that if x1 is false then that implies at the next timestep x1 will be true
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
we see this eventually wrapping two globals - one saying that x1 implies that the next timestep x1 will be false and vice versa for the second global
432
2020Eventually, we will reach a state such that x1 alternates between true and false afterward.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
Good problem.
433
2020Eventually, it will always be true that x1 implies x1 is false in the next state. Also, x1 being false always implies x1 is true in the next state.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
This alternates x1 and !x1.
434
2020x1 alternates true/false
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
G(x1->X!x1) means that if x1 is true, it is always implied that x1 is false in the next state. G(!x1->Xx1) means that if x1 is false, it is always implied that x1 is true in the next state. F(...) means that this will eventually hold, which means that it will always hold (since G(...) is global).
435
2020It eventually always holds that x1 implies that in the next state x1 is not true, and it always holds that x1 is false implies that the next state of x1 is true
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
The F comes before the G, and the states of x1 are described above.
436
2020It is eventually true that x1 alternates every state step.
F(G(x1->X!x1) and G(!x1->Xx1))
I think it's interesting to define an "end" to a trace sort of like how decimals can be infinitely repeating, which is somehow both infinite and quantifiable, so having it be eventually true that two connected and totally defined situations occur effectively creates a cyclic resting state (this example defines a state for x1 which extends as TFTFTFTFT... forever)
F gives eventually, x1->X(!x1) and !x1->X(x1) gives alternating and G gives globally.
437
12020At some point in time it will be true that x2 is true and two instances after x2 will be true again. Leading up to that time, x1 will always be true
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
The U means that the right side will be true at some point in time and the left side is always true until that time. So at some point x1 and X(X(x1)) will be true meaning that x1 is true and next of next of the instance is true. Up until that point, x1 is true
438
2020x1 true until x2, x2's next next will be x2 and true
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
at current time, x1 is true until x2 will be true and its next next will also be true
439
2020For x1 is true, x2 is true and the next of next state of x2 is also true.
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
440
12020x1 until x2 and the state-after-next is also valid for x2
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
The initial U indicates an until, and the two X's mean the two next states.
441
112020x1 is true until (x2 becomes true for one entry, then remains true from two entries after that and on)
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
X(X(x2)): starting entry 2 (counting from 0) x2 is true. (x2 and X(X(x2))): x2 is true for entry 0 and remains true starting entry 2. x1 U (x2 and X(X(x2))): x1 is true until eventually the previous property is satisfied.
442
flip U2020at some point, x1 is true, but until then and in the state immediately after, x2 is true
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
there exists a state at which x1 is true such that before that state it is always the case that x2 is true two states in the future from that one
443
12020x1 always happen until we reach a state in which x2 happen and x2 happen in two states after.
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
Straightforward
444
12020x1 is true until x2 in the current state and x2 two states away are true
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
The u function indicates the the x2 piece ends it, and the X(X(x2)) function i think is looking at x2 two steps in the future.
445
12020x1 is true in every state until a state where x2 is true in both that state and two states later.
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
Since the formula "U" is being used here with x1 and x2, traces where all states before the first occurrence of (x2 and X(X(x2))) (i.e., a state where x2 is true and then x2 is true two states later as well) must set x1 to be true.
446
12020x1 is true until x2 is true and x2 next next is true
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
447
12020Trace where up to a subtrace index, x1 is always true and eventually an entry and two following that entry have x2 as true.
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
Again, using the subtrace index idea, but now the second "eventually" clause has to eventually have a set of entries where x2 is true in two sandwiching one where x2 may or may not be true.
448
uneg2020x1 is always true for a while at the start, and once x1 becomes false, x2 will immediately be true and will be true two states later.
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
x1 U (...) ensures that x1 is true for all states up to the start of the state sequence satisfying (...). In the latter sequence, the formula states that x2 will hold for the first state, and the double application of X on x2 states that x2 will be true two states later (note that this makes no assertions about the second state of the three).
449
12020x1 is true must hold until x2 is true for every 3rd state
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
This is where "every *x* state" comes into play, but i\'m still unsure of how that works modulo wise
450
12020x1 is true until x2 is true and will be true again in two states
x1 U (x2 and X(X(x2)))
I think it's interesting that you can use more complex conditions with weak until
451
12020It always holds that either x1 is always true or x1 is eventually false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
The first G wrapper makes one of the clauses always true, since they are combined by an OR.
452
2020x1 is always true or x1 is not sometimes in the futureG(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
G(G()) ~ G()
453
2020Either x1 is always true or x1 is always eventually false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
The formula distributes to G(x1) or G(F(!x1))
454
2020Either it always holds that x1 is true or it eventually holds that x1 is false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
x1 is either true all the time or it is false at some point
455
2020Always, either x1 is always true or eventually x1 holds as falseG(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
Globally (globally x1 is true or eventually x1 holds false)
456
2020it's always true that either (or both) x1 is always true or eventually x1 becomes falseG(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
global statement wrapping a global statement of x2 and an eventuality statement of !x1
457
2020It always holds that x1 is always true or it is eventually true that x1 is false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
The outer G signifies that the inner is always true that x1 is always true or the latter that x1 is eventually false
458
2020Any arbitrary trace.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
This is a valid formula since it is always the case that x1 is always true or there is some state that makes x1 false.
459
2020It always holds true that x1 always holds true, or that x1 will eventually be false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
The nested outside G, indicates always. The conditions for this always is when G(x1) (x1 is true) or eventuality (not x1 is true eventually, aka x1 is false).
460
2020All traces.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
It is always true that x1 is always true or x1 is eventually false; then a trace that is always true satisfies the former, while a trace that is false at any point satisfies the latter.
461
2020It always holds that it always holds that x1 union it eventually holds that not x1G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
This statement seems to describe how the union of x1 with not x1 always will hold
462
2020It is always that x1 would be true, or eventually x1 would be false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
Decompose the whole clause, we see that it is always that G(x1) or F(!x1). It means it is always x1 is true, or it is always eventually x1 becomes false. This could be rewritten to: G(x1) or GF(!x1).
463
2020it is always true that either x1 is always true or x1 is eventually falseG(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
F means eventually, and x1 can't be always true and eventually false at the same time (hence the OR)
464
2020for all states, x1 must be true and stay true or eventually be falseG(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
I think this formula is true for any trail?
465
2020It always is true that x1 is always true or eventually x1 is false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
G(x1) says that x1 is always true. F(!x1) says that eventually !x1 is true, so eventually x1 is false. Combining these, G(x1) or F(!x1) says that x1 is always true or x1 is eventually false. So, G(G(x1) or F(!x1)) says that it is always true that x1 is always true or eventually x1 is false.
466
2020It is always true that x1 is either always true or eventually false.G(G(x1) or F(!x1))
I think this is interesting because I don't think any traces can exist where x1 is always true and x1 is eventually false.
I like this one a lot! G(x1) or F(!x1) is the meat and potatoes (the "x1 is either always true or eventually false"), and a question I have is how the outer G() changes meaning (Of course adding the "It is always true" part, but is there much of a logical/semantic difference between the two sentences?)
467
2020It always holds that x1 or x2 is true, and x1 is true before x2 is true, and x2 is true before x1 is true.
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
The formula is a little bit misunderstanding because of its unknown priority inference. (Maybe need some brackets)
468
2020Its always true that x1 or x2 is true
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
The ands mean that all three statements are true. The first one G(x1 or x2) means that x1 or x2 is always true, while both (x1 U x2) and (x2 U x1) mean that x1 or x2 are true in the current state, which is contained in the first statement.
469
2020x1 and x2 are both always true
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
G(x1 or x2) means that at least one of x1 or x2 must always be true. x1 U x2 means that x2 is true in some state, and x1 is true in all previous states. x2 U x1 means that x1 is true in some state, and x2 is true in all previous states. So, both x1 and x2 must always be true.
470
2020It always holds x1 or x2 is true when x1 or x2 is true and x2 or x1 is true.
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
There are no surrounding parentheses, so you have to start with the leftmost expression.
471
2020x1 and x2 are always true.
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
It says that x1 or x2 must always be true, that x1 must be true until a point where x2 must be true, and x2 must be true until a point where x1 must be true. Thus, x1 and x2 must be true all the time.
472
2020it is always true that x1 or x2 is true, and that x1 will be true until x2 is true and x2 will be true until x1 is true
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
It's three separate statements joined together with and's, and I translated left to right
473
2020This describes a trace with two variables/features -- x1 and x2. One of them is always true, but not both at the same time. I believe that this describes that x1 and x2 switch off being true in an infinite cycle -- this is what is described by the "U" (until).
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
The literal translation of the formula would be something like: It is always true (G) that either x1 or x2 holds, and, in any time-step, x1 is true until x2 is true, and x2 is true until x1 is true.
474
2020This formula accepts traces where it always holds that x1 or x2 is true, and that it also holds that eventually, x1 will be true, and eventually, x2 will be true.
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
Since this formula enforces that each state of a given trace either has x1 or x2 true, as long as, at every point in the trace, x1 will eventually be true and x2 will eventually be true, x1 U x2 and x2 U x1 are satisfied naturally. For example, given some trace w, if w(1) = (x2), then x1 U x2 is satisfied, and as long as x1 is eventually true, x2 U x1 will be satisfied as well, since each state between w(1) and the next state w(i) (where x1 is true) must satisfy "x1 or x2", and thus have x2 set as true.
475
2020it always holds that x1 or x2 is true and x2 is true until x1 is true and x1 is true until x2 is true
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
G indicates always and "x1 or x2" is inside it so x1 or x2 will always be true. For the other parts, I based my answer on U signifying "until".
476
2020it always holds true that x1 or x2 alternate being true at current time
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
G says one of them must be true, and the U statements show that they alternate
477
2020X1 and X2 both hold
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
The first part says that it is global that x1 or x2 hold; however, the second part says that x1 holds until x2 doesn't and vice versa, which locks x1 and x2 to be true forever. That overrides the first part.
478
2020It is always true that x1 or x2 is true. Also, x1 is true until x2 is true, and x2 is true until x1 is true.
G(x1 or x2) and x1 U x2 and x2 U x1
This formula is interesting, because it can never be x1 and x2 at the same time, but one must always be true.
G(x1 or x2) says that "x1 or x2" is always true. x1 U x2 says that x1 is true until x2 becomes true, and x2 U x1 says that x2 is true until x1 is true. Combining them all together, we get that it is always true that x1 or x2 is true, and x1 is true until x2 is true and x2 is true until x1.
479
2020it is always the case that if x1 is true x2 and x3 won't be, but x2 or x3 will be true in the next state
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
The first clause says that it is always the case that the current state being x1 implies that in the next state x2 or x3 will be true. The second clause in the and says that it is always the case that if x1 is true, both x2 and x3 are false
480
2020It is always true that x1 implies x2 or x3 in the next state and that x1 implies both not x2 and and not x3
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
G corresponds to always true, X means that it is true in the next state, -> means implies, and ! means not
481
2020it always holds that if x1 is true, then x2 or x3 will be true in the next state, and that x2 and x3 will both be false in the next state.
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
this seems to be saying that x1 can never be true, or else it would imply the impossible.
482
2020It always holds that x2 and x3 are false when x1 is true and x1 is proceeded by x2 or x3.
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
If x1 is true, then x2 and x3 are false. The next state after x1 is true is one where x2 or x3 is true.
483
2020It will always hold true that x1 implies that x2 or x3 will be true in the next state. Also, that it will always hold true that x1 will imply that x2 and x3 are false in the same state.
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
This statement means that given the nature of x1, the values of x2 and x3 will always depend on that, and that they will always be equal. Something like TFF, TTT would be an example trace.
484
2020It is always true that for each state where x1 retains true, we have in such states both x2 and x3 retain false, also in the next state of such states, x2 or x3 retains true.
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
It is actually a conjunction of two properties. for each state of assigning x1 to be true: both x2 and x3 would be false in such states; also, in every state where x1 to be true, x2 or x3 would be true in the next state.
485
2020It is always the case that 1) x1 true implies that x2 or x3 are true in the next state, and 2) x1 true implies that x2 and x3 are both false in the next state
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
I think this is unsatisfiable because it says that it is always the case that x1 implies that both x2 and x3 are true and both x2 and x3 are false in the next state
486
2020It always holds that if x1 is set to true in state 0, then x2 and x3 are set to true in state 1 and x2 and x3 are set to false in state 0\n
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
G(x1 -> X(x2 or X3)) it always holds that if x1 is set to true in state 0 then x2 and x3 are set to true in state 1\nG(x1-> (!x2 and !x3)) it always holds that if x1 is set to true in state 0 then x2 and x3 are set to false in state 0
487
2020it always holds that when x1 is true, x2 and x3 are false but in the next state at least one of x2 and x3 is true
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
the first says x1 being true implies next state, while second is about this state
488
2020it is always true that: if x1 is true that implies that at the next temporal step, either or both of x2 or x3 will be true OR if x1 is true then at the next temporal step, x2 and x3 will both be false (aka x1 being true results in either/both x2 or x3 being true at the next time step or it results in both x2 and x3 being false)
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
You can see that there are these two global conditions and that x1 implies these two cases
489
2020x1 implies x2 or x3 initially true.. Also, x1 implies x2 and x3 both being false
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
490
2020eventually, if x1 is true, one of x2 and x3's next values should be, and x2 and x3's current value should both be false
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
X(x2 or x3) implies the value of x2 and x3's next value, and (!x2 and !x3)) implies the value of x2 and x3's current value
491
2020it always holds that if x1 is true then x2 or x3 needs to be true and then the next state both x2 and x3 have to be false
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
492
2020it is always true that if x1 is true, then x2 or x3 are true in the next state and it is also always true that if x1 is true, then x2 and x3 are false in the current state
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
this one would have been a contradiction if not for the next operator
493
2020x1 is true for at most one entry, and in that entry x2 and x3 are both false, but for all following entries at least one of x2, x3 is true
G(x1 -> X(x2 or X3)) and G(x1-> (!x2 and !x3))
I am trying to say that if x1 is true it means at the next step x2 or x3 will be true. But also when x1 is true, x2 and x3 will be false. So, when x1 is true it will signal at least one of these to become true.
G(x1 -> X(x2 or x3)): always if x1 is true then from the next state at least one of x2, x3 is true (property 1). G(x1->(!x2 and !x3)): always if x1 is true then x2 and x3 are false (property 2). Putting these properties together, we get that in one entry x2 and x3 cannot be true if x1 is true, but one of them (x2, x3) will be true starting next entry. x1 cannot be true for more than one entry: the second time we find x1=true, according to property 1, x2 or x3 will be true, and this contradicts property 2.
494
2020If it eventually holds that x2 is true, then x2 is false in this state or x1 and x3 are true.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
This seems to be a straightforward translation of a truth functional expression.
495
2020If x2 is eventually true, then x2 does not hold until both x1 and x3 hold.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
F means eventually, => means implies, and ! means does not hold. When I substitute these symbols to words, I get the above answer.
496
2020if eventually x2 is true then that implies that x2 must remain false until both x1 and x3 are true
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
we can see that f(x2) means that eventually x2 becomes true, and then that implies the case where x2 remains false until x1 and x3 become true
497
2020If eventually x2 is true, then there exists an i such that for all entries before i x2 is false and x1 and x3 is true at i.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
The first part is just an eventually. !x2 must hold for all values less than some i and then (x1 and x3) hold for the sequence starting at i. This is just x1 and x3 at i.
498
2020if x2 is eventually true then x2 is false until x1 and x3 are true
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
The F signifies eventually and the arrow means an implication, so if the left is true then the right must be true.
499
2020if x2 will happen in the future, x2 is false now or x1 and x3 is true now
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
500
2020If x2 is eventually true, then x2 is not true initially and will not be true until x1 and x3 are true.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
If x2 is eventually true, then x2 is not true at the beginning, and x2 cannot be true until x1 and x3 are true
501
2020if x2 is eventually true, then x2 is not true in the first time step or x1 and x3 are both true in the first time step
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
F(x2) means x2 is eventually true, !x2 means x1 is not true in the first time step, x1 and x3 means x1 and x3 are both true in the first time step
502
2020If it eventually holds that x2 is true, then x2 must be false until x1 and x3 are both true in some state.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
F(x2) means that x2 must be eventually true, so the overall formula states that if this first condition holds, then it must be true that "!x2 U (x1 and x3)". Thus, if x2 is eventually true for the given trace, then the trace must hold x2 to be false until a state where x1 and x3 are simultaneously true is reached. Traces where x2 is never true also pass this formula!
503
2020If x2 is eventually true, then x2 is false until x1 and x3 are true.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
!x2 U (x1 and x3) means x2 is false until x1 and x3 are true. That is implied by the x2 eventually being true.
504
2020x2 is eventually true implies that there is a subtrace such that x2 is not true for all sequences preceding and x1 and x3 are true eventually for the sequences following.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
The implication is showing that if the first half is true, then the second half must be true, basically all setups excluding F => T. The second part of the clause is a similar breakdown to the first two questions, delineating a subtrace where the variables hold true always and eventually on either side.
505
2020X2 is eventually true if and only if x2 is false before one state and after that state x1 and x3 are both true.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
If α is of the form β U γ, then w \U0010fc00 α iff there exists an i ∈ N such that wi \U0010fc00 γ and for all 0≤j<i,wj \U0010fc00β.
506
2020if x2 is eventually true, that implies that either x2 is false or x1 and x3 are true
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
507
2020If x2 holds at some point, then x1 and x3 hold when x2 first holds.
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
The LHS of the implication evaluates to true if there is some state where x2 is true. The RHS states that there must be some non-empty set of states at the start where x2 is false, and in the state following that set (i.e. the first state where x2 is true), x1 and x3 will hold for one state. Note that this does not necessarily mean that the state where x2 eventually holds corresponds to the _first_ state where x2 holds, but this is not important for the expression. This form simply allows for x2 to never be true.
508
2020if x2 is eventually true, this implies that not x2 holds until x1 and x3 are true
F(x2) => !x2 U (x1 and x3)
It is interesting because it almost uses all the different LTL symbols that we learned in class.
F means eventually, so x2 evntually is true. For the right side, we are saying that this (!x2) will happen until [something] (x1 and x3).
509
2020Eventually x1 is false is false.!F(!x1)
This formula is equivalent to G(x1).
F means eventually, and ! means false. As a result, I think that the above is the same as saying x1 is true forever, as it isn't true that eventually x1 is false.
510
2020It will not eventually hold that x1 is false (or, it will eventually hold that x1 is true)!F(!x1)
This formula is equivalent to G(x1).
F means eventually, and ! means false.
511
2020It is not true that x1 is eventually not true (that is, x1 is always true)!F(!x1)
This formula is equivalent to G(x1).
never eventually never is synonymous with "always", according to the rules of LTL
512
2020It does not eventually hold that x1 is not true.!F(!x1)
This formula is equivalent to G(x1).
The formula starts with !F, which is the opposite of F, or "eventually". Inside the formula we have !x1, which means that x1 is not true. Therefore it does not eventually hold that x1 is not true.
513
2020x1 is not eventually not true!F(!x1)
This formula is equivalent to G(x1).
The !F means "not eventually" and since !x1 that means there is a double not, which is why it\'s "not true" at the end
514
2020it is not true that eventually x1 is false!F(!x1)
This formula is equivalent to G(x1).
F() means eventually and !F() means not that, and !x1 implies x1 is false
515
2020It does not eventually hold that x1 evaluates to false.!F(!x1)
This formula is equivalent to G(x1).
The "not eventually" would mean that there is no sequence where that !x1 exists in the sequence.
516
2020x1 is always true.!F(!x1)
This formula is equivalent to G(x1).
The formula is saying x1 will never be false.
517
2020eventually, x1 will not be false!F(!x1)
This formula is equivalent to G(x1).
F means the final state of x1, and ! before the F make it opposite
518
2020x1 will always be true.!F(!x1)
This formula is equivalent to G(x1).
The negation of an eventually implies that there is the parameter will never be true. Thus, x1 will never be false.
519
2020x1 will never be false!F(!x1)
This formula is equivalent to G(x1).
520
2020This formula accepts traces where x1 is always true.!F(!x1)
This formula is equivalent to G(x1).
It is not the case that sometimes, x1 is false.
521
2020x1 is always true!F(!x1)
This formula is equivalent to G(x1).
If x1 never eventually becomes false, it must always be true.
522
2020It is always true that x1 is true and that x2 is always true until x1 is true and that x2 is eventually true.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
G(x1 and (G(x2) U x1) and F(x2)) tells us that "x1 and (G(x2) U x1) and F(x2)" is always true. F(x2) says that x2 is eventually true. G(x2) U x1 says that G(x2) is true until x1 is true, and G(x2) says that x2 is always true. Wrapping it all together, we get that it is always true the x1 is true, x2 is always true until x1 is true, and x2 is eventually true.
523
2020x1 always holds, and x2 always eventually holds.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
Since x1 always holds, the middle term is always vacuously true.
524
2020x1 would be always true, and x2 is eventually true, and x2 is true until x1 is false.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
By distribution law, we can decompose it into three clauses: G(x1), GF(x2), and G(x2)Ux1. The first two are trivial, so the point is the third part, which means that G(x2) holds until x1 becomes true in some states. However, G(x2) means x2 remains true until forever, which is !F!x2. Then, G(x2)Ux1 means (!F!x2)Ux1. which means we could not find a state where x2 is false until x1 means to be true.
525
2020It always holds that x1 is true and that is always holds x2 is true until x1 is true and that eventually x2 is true.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
G means that the expression inside must always hold. U means until and F means that the expression eventually holds. Since this whole formula occurs inside of the G, it means that the whole expression always holds. Then since x2 is inside F it means that x2 must eventually be false and the G(x2) U x1 part means that x2 is always true until x1 is true.
526
2020It always holds that x1 is true, and x2 is true, and x2 is eventually true.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
because there is a beginning G(x1), this equation can be simplified and the following x1 in G(x2) U x1 can be deducted.
527
2020\nx2 , x2 , x1 (current time)
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
It always holds true that x1 must be true at current time, t, x2 must eventually hold true, and x2 will always hold true until x1 comes after it.
528
2020It is always true that x1 is initially true and x2 is always true until x1 is true and x2 is eventually true
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
F(x2) = x2 is eventually true\n(G(x2) U x1) = x2 is always true until x1 is true\nx1 = x1 is initially true\nthe front G = It is always true
529
2020It always holds that x1 is true and x2 is always true and x2 is eventually true.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
This is weird. x1 must always be true and x2 is globally true. I dont get the point of x2 eventually being true if it is globally true.
530
2020it always holds that x1 is true when it always hold that x2 is true or x1 when x2 eventually holds true.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
Following the order of operations, the innermost parentheses were expressed first, preceded by the initial x1 expression.
531
2020Always x1 is true and x2 always is true until x1 is true, x2 finally is true
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
532
2020it's always true that x2 will happen at some point in the future and x1 is true noe and x2 is always true or x1 is true
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
533
2020x1 is always true and x2 is always eventually true
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
G(x1) = x1 is always true\nG(F(x2) = x2 is always eventually true\n(G(x2) U x1) = (???) not sure but we know x1 is always true so it doesn't seem logically contribute anything\nSo we pretty much just have G(x1 and F(x2)
534
2020This will definitely hold for traces where x1 and x2 are globally true. I suspect this is only a subset of the correct answer. See below for more detail.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
This either says that x1 and x2 must globally be true (in which case F(x2) comes for free) or it says that x1 must be true and x2 must be true unless x1 is (and x1 is so x2 is unbound). I am not sure how to parse G(x2) U x1, which is what the distinction hinges on.
535
2020It is always the case that x1 is true, x2 is eventually true, and it is always the case that x2 is always true until x2 is true.
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
G means always, a U b is read as a until b, and F means eventually.
536
2020It always holds that x1 intersections it always holds that x2 until x1 and it eventually holds that x2 is true
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
This statement seems to describe how the relationship between x1 and x2 influences the state of x2
537
2020x1 must always be true and x2 must always be true up until some point
G(x1 and (G(x2) U x1) and F(x2))
I'm not exactly sure whether this formula is contradictory, so I thought it would be a good example. I think it's saying that x1 is true at one point, and also x2 is always true until x1 is true. So x1 and x2 can't be true at the same time. But also, x2 will eventually be true. I think this formula illustrates the difference between global/always states, and individual states, well.
x1 : means that x1 is true currently\nG(x2) U x1 : means x2 is always true until x1 is true\nF(x2) : means that x2 is eventually true\n\nthese must all be always true (because G()), which means \nthat x1 must always be true and x2 must be true at some point\n\n
538
2020It always holds that x1 is true when x2 is not true, and x3 eventually holds true.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
G denotes something always being true, while F means it is eventually true.
539
2020it always holds that x1 is true and x2 is false, and eventually x3 will be true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
I think this one is fairly straightforward...
540
2020It is always true that x1 will be true and x2 will be false. Also, x3 will eventually be true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
541
2020x2! and x1 will always be true, x3 is sometimes true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
it always holds true that x1 is true and x2 is false, and x3 will eventually be true
542
2020it always holds that x1 is true and x2 is false and it eventually holds that x3 is true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
G(x1 and !x2) means that x1 is always true and x2 is always false, F(x3) means that x3 is eventually true
543
2020Eventually x3 becomes true, x1 is always true, and x2 is always false.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
There is a conjunction of two statements: the first one states that in all states x1 is true and x2 is false; the second is eventually x2 becomes true.
544
2020x1 is always true, x2 is always false, and x3 is eventually true.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
Straightforward.
545
2020x1 is always true, x2 is always false, and x3 will eventually be true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
Nothing complicated
546
2020It always holds that x1 is true, x2 is false, and that x3 will eventually be true.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
Globally, x1 must be true and x2 must be false and x3 will eventually be true because it is in F()
547
2020it always holds that x1 is true and x2 is false but for x3 it will eventually be true.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
G means always and F means eventually.
548
2020It always holds that x1 is true or x2 is false. And it eventually holds that x3 is true.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
I based this off of G meaning always and F meaning eventually.
549
2020It always holds x1 is true and x2 is false, and will eventually holds x3 is true.
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
G means always and F means eventually
550
2020it always holds that x1 is true and x2 is false, and that eventually x3 is true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
The G signified "it always holds" and inside was x1 and !x2 meaning that x1 has to true and x2 has to be false. Then the next part means it eventually holds that x3 must true.
551
2020It always holds that x1 is true and x2 is false and eventually x3 is true
G(x1and !x2) and F(x3)
I think that this question is interesting because it makes sure that people understand what G and F mean, which is critical to be able to successfully write LTL formulas.
552
2020For every state x1 is eventually true implies eventually for every state x1 is true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
I don't actually have an idea what's the purpose of the phrase.
553
2020It always eventually holds that x1 is true, which implies that x1 is eventually always true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
The two expressions are semantically reversed but one implies the other.
554
2020if it is always true that x1 is eventually true, then it is eventually true that x1 is always true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
555
2020it is always true that if x1 is eventually true, then eventually x1 is always true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
This is an interesting formula because it seems as though before and after the implies say the same thing. That x1 is always true.
556
2020It always holds that x1 is eventually true implies that eventually, x1 is always true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
The above answer unwraps the G(F) and F(G) operators
557
2020If it always eventually holds that x1 is true, then eventually, x1 will always be true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
Globally (eventually x1 will be true) meaning that eventually (globally x1 is true)
558
2020if it always eventually holds that x1 is true, then it eventually always holds that x1 is true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
it\'s an implication between "always eventually" to "eventually always"
559
2020if it is always the case that x1 is eventually true, then eventually x1 is always true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
translating => as "if then"
560
2020x1 always eventually being true implies that x1 will eventually always become true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
Very similar to the example description
561
2020The fact that is it always the case that x1 is true implies that eventually it is always the case that x1 is true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
G means always, F means eventually, and => means implies. As a result, the above can be translated to English as done above. It is not a two way implies because of the order of always and the eventually.
562
2020It will always hold true that if x1 is eventually true, this implies that it will be eventually true that x1 will always be true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
This means that the eventuality of something being true, will show that it will always be true. This is because of the nature in which the G nests the entire phrase.
563
2020If x1 will eventually be true in any state, then after some point, x1 will always be true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
Always eventually x1 implies eventually always x1.
564
2020If it always eventually holds that x1 is true, then it eventually always holds that x1 is true.
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
I'm a bit confused by this one -- I think F(G(x1) is the same as simply G(x1) since G is more constraining than F. Then, I think it basically saying that if x1 is eventually true, then x1 is always true.
565
2020it always eventually holds that x1 is true implies that it eventually always holds that x1 is true
G(F(x1)) => F(G(x1))
I think this is interesting because I think the combinations of F and G are a little trickly in LTL and so having both ways of combining the two makes this formula kinda confusing conceptually
the G and F are switched in the clauses before and after "implies", so it is reflected in the swap of always and eventually in the English translation
566
2020If x1 and not x2 initially holds, then eventually x1 will be false and afterward x2 will be true.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
The first part of the statement refers to w(0) (the initial state). F refers to eventually and U requires that !x1 occurs before x2.
567
2020if x1 is true and x2 is false, then it is eventually true that either x1 is false or x2 is true
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
568
2020x1 is true and x2 is false implies that eventually x1 will become false until x2 becomes true
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
This is an if-then situation, and I translated it left to right just as the LTL statement says.
569
2020if x1 is true and x2 is false, then eventually x1 will be false until x2 is true
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
the order of operations with AND and => is a bit ambiguous here, I interpreted AND as being first
570
2020If the first state of x1 to be true and the first state of x2 to be false, then finally there is a state where after it x1 becomes false until x2 to be true.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
I feel this is quite confusing: I have to add parenthesis to make things more clear. I assume it means (x1 & !x2) -> F(!x1 U x2). It means if the first state shows that x1 is true and x2 is false, then there is a state that after such a state, x1 remains false until x2 to be true.
571
2020x1 is true and when x2 is false, x1 will be false until x2 eventually be true
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
572
2020x1 is true and x2 is false in the initial state implies that eventually, x1 will be false until x2 is true.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
The U operator means until, so x1 will become false at some point, and x2 will become true at the same point.
573
2020If x1 is true and x2 is false initially, then eventually, x1 will be false until x2 is true.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
x1 and not x2 implies eventually not x1 until x2.
574
2020when x1 is true, x2 is not true and eventually either x1 will not be true or x2 will be true.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
The first half of the expression relates x1 to only be true when x2 is not true, this implies in the future the relationship will change and either x1 will not be true or x2 will be true.
575
2020for the state that x1 is true and x2 is false, it eventually holds that for x1 to be false in all previous states, x2 would be true.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
576
2020If x1 holds and x2 does not hold initially, then eventually there will exist a sequence of states where x1 does not hold, followed by a state where x2 holds.
x1 AND !x2 => F( !x1 U x2 )
I think my formula uses a good combinations of the operators like the implies and U (until), and this can mirror real life situations where eventually variables that had true or false can flip up until a certain condition
Assuming that the expression is (x1 AND !x2) => ..., we have an eventually predicated on a first state where x1 holds and x2 doesn't (only in the first state). The eventually contains an until operator U, which states that !x1 must (eventually) hold for some non-zero number of states, before there is a single state where x2 holds (there are no constraints on further states).
577
2020x1 and x2 eventually being true implies that x1 always implies x2 and x2 always implies x1
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
I believe this scenario entails that once x1 or x2 becomes true, they are both always true, since they have a circular implication between them. They also both must become true in the same state, otherwise the implication would not hold true in that state, and G means it must hold true in every state. The F aspect means that at some point in the trace, both become true.
578
2020if x1 will be true at some point in the future and x2 will be true at some point in the future (not necessary at the same time). it is always true that x1 = x2
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
579
2020If x1 and x2 are eventually true, then it holds that x1 is equivalent to x2.
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
x1=>x2 and x2=>x1 means x1 is equivalent to x2, and it's an implication
580
2020if x1 and x2 are eventually true, then it always holds that if x1 is true at the first time step, then x2 is also true and if x2 is true at the first time step, then x1 is also true
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
F(x1) and F(x2) means that x1 and x2 are eventually true, G means always / globally, x1=>x2 means that if x1 is true at the first time step, then x2 is also true and x2 =>x1 means if x2 is true at the first time step, then x1 is also true
581
2020x1 and x2 both eventually being true implies that x1 always implies x2 and x2 always implies x1
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
582
2020If x1 is eventually true and x2 is eventually true, then x1 and x2 always imply each other.
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
x1 => x2 and x2 => x1 is an if and only if. The first clause before implies is that x1 is eventually true and x2 is eventually true. That implies x1 iff x2 always.
583
2020If x1 is eventually true and x2 is eventually true then x1 always implies x2 and x2 always implies x1.
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
x1 and x2 eventually being true implies that it is always true that x1 implies x2 and x2 implies x1.
584
2020if eventually x1 will be true and (separately) eventually x2 will be true, x1 and x2 always have the same value
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
The global bi-implication just kind of states that they are always the same--if one is true then the other is true, so the only way for one to be false is if they are both false
585
2020If both x1 and x2 hold true, x1 and x2 become true at the same time
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
If x1 and x2 will eventually be true, then it always holds that x1 is true implies x2 is true and x2 is true implies x1 is true. This means that one cannot be true without the other being true
586
2020Eventually it holds that x1 is true and eventually it holds that x2 is true. This implies that it always holds that x1 implies x2 and x2 implies x1.
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
I think this is saying that if it is the case that x1 and x2 are eventually true then it is also the case that x1 and x2 are always both true or both false, because they imply each other.
587
2020x1 and x2 will always be true
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
F says that at least once x1 and x2 will be true. If that is the case, then it is always the case that they imply each other. Therefore, they will always be true
588
2020Eventually, x1 will be true, and eventually, x2 will be true, which implies that it will always be true that x1 implies x2 and x2 will imply x1.
(F(x1) and F(x2)) => G(x1 => x2 and x2 => x1)
Meaning: if x1 is eventually true and x2 is eventually true, they are only ever true when both are true. I think this is interesting because it shows LTLs power in applying constraints on the state space that are _dependent_ on the characteristics of the state space. So in this example, a property can be made to hold (i.e. only instances which satisfy that property will be provided) given an existing property of the instance. It seems impressive because we usually can't observe the future and use that observation to change it. This extends to any use of binary logical operation at the top level of an LTL expression where a global property of a trace is checked independently of another clause in the expression.
This states that once we hit the case that x1 and x2 are true (which will eventually occur), we can observe that T->T and T->T, thus x\\x1 => x2 and x2 => x1
589
2020x1 is never true until x1 is always false! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
!(G(x1)) means x1 is never true. G(!x1) means x1 is always false.
590
2020x1 is not always true or x1 is always not true! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
While the right hand side and left hand side of the or sound the same, they are actually different. One says that it is not the case that x1 is always true, while the other says that x1 is always false
591
2020x1 cannot always be true and false at the same time! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
Interpretation: !(G(x1) U G(!x1)). G(x1) U G(!x1): always x1 is true, until it is always false. !(G(x1) U G(!x1)): It cannot hold that x1 is always true until it is always also false.
592
2020x1 is not always true or x1 is always false -- all traces where x1 is false once or always! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
!(G(x1)) just means F(!x1), and F(!x1) or G(!x1) just states that x1 is false once or false always
593
2020It does not hold that x1 is always true before x1 is always false! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
I think the ! has lower precedence than U, and G(x1) states that x1 is always true
594
2020It does not always hold that x1 is true and then it always holds that x1 is false! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
! (G(x1)) it always does not hold that x1 is true\nU \nG(!x1) it always holds that x1 is false
595
2020x1 can neither be always true nor always false! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
G(x1)) => x1 is always true\nG(!x1) => x1 is always false\n! (G(x1)) U G(!x1)) => x1 can neither be always true nor always false
596
2020x1 is never true until not x1 is always true! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
G is always true, x U y is x is true until y is true
597
2020x1 is never true! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
I think x1 is not always true or always x1 is not true and since union is symmetric I believe this simplifies to x1 is never true because there is no difference between G(! and !G
598
2020x1 will eventually be always false.! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
The not always true at the beginning is true if x1 is always false.
599
2020x1 always is true until x1 is not true! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
600
2020it is not the case that x1 is always true forever until x1 is false forever! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
I believe this is the same as F(G(!x1)), though I might be misinterpreting
601
2020x1 isn't always true until it is always false.! (G(x1)) U G(!x1)
I thought it was interesting to write ! (G(x1)), which means that x1 is not always true.
This seems tautological. Once x1 is never true it will not always be true.
602
2020Whenever x1 is true, x1 will always be true until x1 is false
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
I think this LTL is always false
603
2020If it eventually holds that x1 is true, then it must always hold that !x1 will eventually be true.
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
This formula states that if x1 is eventually true, then it always holds that x1 is true until x1 is false. Since x1 can only be true or false, "x1 U !x1" is always true as long as a future state exists where !x1 holds. Thus, this formula is satisfied by any trace where, given that x1 is eventually true, for all states !x1 is eventually true as well.
604
2020if x1 is at some point true, it must be false eventually (and it is true until it is false)
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
F(x1) : means that x1 is eventually true\nG(x1 U !x1) : means that it is always true that x1 is true until x1 is false (this would always be true as long as there is a point where it is false)\n\nthis means that if x1 is at some point true, then x1 is true until until it is false (but it must be false at some point)
605
2020This trace holds for all traces.
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
This formula has an implies so it holds for all traces where except for traces where F(x1) is true and G(x1 U !x1) is false. However, G(x1 U !x1) holds when x1 is always true until it's not which would be true for all traces.
606
2020This accepts all traces except the trace where x1 is always true
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
If x1 is ever true, then for every state, x1 is true until x1 is not true.
607
2020if eventually holds x1 is true, then x1 can always be true or false.
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
the arrow means imply. The x1 union !x1 will always be True.
608
2020it eventually holds that x1 is true implies that it always holds that x1 is true until x1 is false
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
it makes sense that x1 is true until x1 is no longer true (false)
609
2020It eventually holds that x1 is true
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
Again, I am not really sure if G(x1 U !x1) has any effect because it should be true for all traces
610
2020eventually x1 implies that there is always ~x1 than precedes x1.
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
Eventually x1 (is true) means that there are always states where x1 is false before there is a state where x1 is true.
611
2020if x1 is eventually true, x1 will always not be true until x1 is false
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
if x1 satisfies the F, then in G, it has an until formula to constraint x1
612
2020Invalid statement...?
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
x1 being eventually true implies that x1 is always true until it is not. However, if it is eventually true, then when it is not true, it will no longer be true. Contradiction.
613
2020If x1 is eventually true, then it is always true that x1 is true until x1 is false
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
614
2020If x1 is eventually true, x1 is always true until it is false.
F(x1) => G(x1 U !x1)
relation between F, G, U, and true and false of the same variable
F(x1) means x1 is eventually true. We see that case implies G(x1 U !x1), which means it always is the case that x1 is true until x2 is false.
615
2020This formula accepts traces where x1 alternates between being true and false.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
For every state, x1 being true implies that x1 is not true in the next state, and x1 being false implies that x1 is true in the next state.
616
2020it always holds that if x1 is true, the following time it is not true and vice versa (if x1 is not true the following time it is true)
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
The relationship that holds true is that x1 is not the same status between this state and the next one
617
2020it always holds that x1 implies not x1 in 2 states and that not x1 implies x1 in 2 states
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
G is always true, X is true in the next state; X after the implication means it is true in the state afterward
618
2020It always holds that if x1 is true, the next x1 is false, and if x1 is false, the next sate of x1 is true.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
Its logic is clear and it can be followed by the formula to imply it's meaning
619
2020it always holds that if x1 and x2 is true, x1 and x2 will be false in the next state
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
example trace: it is always true that if you clean a dirty dish with a clean towel, the dish will be clean and the towel is dirty in the next state
620
2020I think this is the same idea as in the previous question: x1 always turns "on" or "off" sequentially, in a cycle, forever.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
It is always true that x1 being true implies that, in the next state, x1 is not true, and that x1 being not true implies x1 being true in the subsequent state.
621
2020It is always the case that x1 is followed by not x1, and not x1 is followed by x1.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
x1 implies next (!x1) means if x1 is true, the next state x1 is false, etc.
622
2020If x1 is true, it will always be false next, and if x1 is false, it will always be true next.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
We have that for all states, if x1 is true, it is false in the next state (after application of X), and if x1 is false, it is always true in the next state. This would set up a consistent oscillation between x1 being true and false.
623
2020It always holds that x1 being true implies that x1 will be false at the next step. And it always holds that x1 being false implies that x1 is true at the next step.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
I think this is saying that x1 will alternate between true/false.
624
2020For every state, x1 has a different truth value than in the next state.
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
(x1 => X(!x1)) = x1 is true in current, then it is false in the next\n(!x1 => X(x1)) = x1 is false in current, then it is true in the next\nGlobal predicate makes it apply to all states
625
2020it is initially true that , not x1 implies that x1 will be true in the future. It is always true that x1 implies that not x1 will be true
G((x1 => X(!x1)) and (!x1 => X(x1)))
I like that repeating patterns can be expressed quite easily with LTL
bare variables only apply to the first state
626
2020It always eventually holds that the x1 will hold in the next state.G(F(X(x1)))
The G and X are redundant
Really similar to the example (given) G(F(x2), except the X refers to the next state, rather than the current state.
627
2020It always eventually holds that the next sequence in the current subtrace x1 is true.G(F(X(x1)))
The G and X are redundant
Again, this is how I evaluate the X as, referring to the "frame" or current trace.
628
2020It always eventually holds that x1's next is trueG(F(X(x1)))
The G and X are redundant
To be completely honest this is almost the exact one in the example description but has an additional nested X, making the statement change to x1's next being what must be true.
629
2020Initially x1 is true would happen once in many trials.G(F(X(x1)))
The G and X are redundant
630
2020It always eventually holds that x1 is true.G(F(X(x1)))
The G and X are redundant
This seems fairly similar to example. I do not know what impact the X(x1) has on this.
631
2020It always eventually holds that in the next state x1 is true.G(F(X(x1)))
The G and X are redundant
G means that the expression always holds, F means that the expression inside eventually holds and I think X refers to the next state, so X(x1) means that x1 is true in the next state. So when all of this is combined, I get the expression I provided in the answer above.
632
2020It's always the case that eventually x1 will hold in the next time step.G(F(X(x1)))
The G and X are redundant
The operators apply in the order they are nested.
633
2020x1 will always eventually be trueG(F(X(x1)))
The G and X are redundant
the X() only means you ignore the first state, which doesn't matter as the property must continue being true
634
2020it always eventually holds that x1 is true at the next time stepG(F(X(x1)))
The G and X are redundant
G means always / globally, F means eventually, and X(x1) means x1 is true at the next time step
635
2020It always eventually holds that x1 is true in all states.G(F(X(x1)))
The G and X are redundant
G = (always), F = (Eventually), x1 is true for all next states eventually so starting from some point x1 will be always true.
636
2020It always eventually holds that x1 is true in the next entryG(F(X(x1)))
The G and X are redundant
637
2020No traces satisfy this formulaG(x1) and G(!x1)
Some LTL formula represent contradictions.
This formula requires x1 to always be true, and for !x1 to always be true, which by logical contradiction is not possible! Thus, no traces will satisfy this formula.
638
2020x1 is always true and always false, aka unsatisfiableG(x1) and G(!x1)
Some LTL formula represent contradictions.
639
2020nullG(x1) and G(!x1)
Some LTL formula represent contradictions.
this reads "it always holds that x1 is true and that x1 is false" which doesn\'t make sense
640
2020It always holds that x1 is true and it always holds that x1 is false.G(x1) and G(!x1)
Some LTL formula represent contradictions.
I don't believe this is satisfiable, as the two parts contradict each other. Unless there is some weird edge case where x1 is neither true or false? But to my understanding, a boolean must always be true or false, so it cannot be both true and false at the same time.
641
2020it is always true that x1 is true and it is always true that x1 is falseG(x1) and G(!x1)
Some LTL formula represent contradictions.
two global statements --> this must be a contradiction?
642
2020It always holds that x1 is true and it always holds that x1 is falseG(x1) and G(!x1)
Some LTL formula represent contradictions.
This is a contradiction!
643
2020x1 will always hold and x1 will always be false.G(x1) and G(!x1)
Some LTL formula represent contradictions.
G is used to define that something is always true so x1 will always hold and x1 will always be false. This will accept no traces.
644
2020always true that x1 holds and always true that "not" x1 holds (is false)G(x1) and G(!x1)
Some LTL formula represent contradictions.
G(x1) and G(!x1) contradict each other.
645
2020This is unsatisfiableG(x1) and G(!x1)
Some LTL formula represent contradictions.
it requires x1 to always be true and always be false
646
2020Its always true that x1 is true and its always true that x1 is falseG(x1) and G(!x1)
Some LTL formula represent contradictions.
The and means that both are true, and the G(x1) means that x1 is always true, and G(!x1) means x1 is always false
647
2020it is always the case that x1 and not x1 are trueG(x1) and G(!x1)
Some LTL formula represent contradictions.
first we see it is always the case that x1 is true and it is always the case that not x1 is true, and from there go to the above
648
2020there are no traces that satisfy this (it is a contradiction)G(x1) and G(!x1)
Some LTL formula represent contradictions.
this means that x1 is always true and x1 is always false : which can't happen
649
2020it always holds that x1 is true and it always holds that x1 is falseG(x1) and G(!x1)
Some LTL formula represent contradictions.
this is a contradiction, as x1 cannot be always true and false
650
2020two states from the beginning x1 is true, which implies that on the next step x1 is still trueXXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
651
2020If the i=2 entry of sequence holds that x1 is true, that implies that the i=1 entry of the sequence holds x1 to be trueXXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
X tells us to look at the i=1 index.
652
2020x1 is always true (if at least one x1 is true) or always false.XXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
If x1 is true two states ahead, it is true in the next state. If it is false, then it must be false in the next state. Therefore, x1 will always be true (other than potentially the first state) if it starts as true two steps ahead. If is is false two steps ahead, this same process repeats. If an x1 true is ever hit, x1 stays true.
653
2020If x1 is true in the state after the next state, then it is true in the next state.XXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
x1 = x1 is true in current state\nX(x1) = x1 is true in next state\nX(X(x1) = x1 is true in the state after the next state.
654
2020x1 being true in the next next step implies that x1 is true in the next next stepXXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
X is the next step, and -> is implies, therefore. the above means that if x1 is true in the next next step, then x1 is true in the next next step.
655
2020if x1 is true two step in the future, it will be true in the next step tooXXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
656
2020if x1 is true two states from now, then it must also be true next state.XXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
simply requires that, for any trace, the first two states can be TT, TF, FF, but not FT.
657
2020x1 in the next state will be the same as x1 in the next 2 statesXXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
658
2020This holds for all traces except ones where x1 is true in the third state but false in the second state.XXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
This read XXx1 implies Xx1, so it holds in all traces except if XXx1 holds but Xx1 does not hold. XXx1 holds if the trace obtained from removing the first two states satisfies x1 i.e x1 is true is its first state. So XXx1 holds if x1 is true in the third state. Similarly, Xx1 satisfies x1 if x1 is true in the remaining trace after removing the first state, i.e if x1 is true in the second state. So the description above describes when XXx1 holds but Xx1 doesn\'t based on the definitions of "holds" given above.
659
2020Two states from now, x1 being true implies that x1 will be true 1 state from now.XXx1->Xx1
if we write this as a theorem, it would implies time density which is pretty cool.
More X mean more steps in the future.
660
2020eventually, x1 is true but x1 is always falseF(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
F set the eventual state, and G(!x1) means x1 will always be false
661
2020eventually x1 will be true and always x1 will be falseF(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
i think this is a contradiction
662
2020The empty set.F(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
The formula says eventually x1 is true and not x1 is always true, which is a contradiction.
663
2020It eventually holds that x1 intersection it always holds that not x1F(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
this statement seems to describe that x1 will eventually be logically equivalent with not x1
664
2020Eventually x1 will hold and x1 will always be false.F(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
Using F establishes that "something" should happen eventually in the trace. The conjunction establishes that x1 is true and that x1 will always be false.
665
2020It will eventually be the case that x1 holds and x1 is always false (impossible)F(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
x1 means that x1 is true, and G(!x1) means that x1 is always false. F(...) means that these will eventually hold in tandem, which is not possible.
666
2020it is eventually true that both x1 is true and is false in all subsequent statesF(x1 and G(!x1))
This is actually impossible. e.g. unsatisfiable. I think this is an interesting case because it deals with the edge case where x1 and G(!x1) requires x1 and !x1 (which is impossible).
667
2020After the first time x2 is true, x2 will be false, x1 will be true in all states thereafter, and x2 will never be true againG(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
it always holds that there will eventually be an x2 until there is a state where it is true that x1 is true in all successor states; the until means that F(x2) can never be true after that
668
2020This holds for traces where in all states x2 is eventually true until the second state has x1 as true. So this holds for traces where x2 is true in the first state and x1 is true in the second.G(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
The global variable means that the formula inside it must always hold. The formula F(x2) U(X(x1)) means F(x2) holds until X(x1) holds. So that means F(x2) holds if in any of the state x2 eventually becomes true, and X(x1) holds for traces where the second state is true. So F(x2) has to hold until the second state, so x2 has to be true in the first state and x1 has to be true in the second state for the trace to satisfy the given formula.
669
2020It is always true that x2 is eventually true until x1 is true in the timestep immediately afterG(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
X is the timestep immediately after\nF is eventually\nG is always\nU can be think of as \'until\'\nthen we can translate the formula and obtain "It is always true that x2 is eventually true until x1 is true in the timestep immediately after"
670
2020It always holds that: x2 will eventually be true, until the next state satisfies x1. (Alternatively, it always holds that: either x2 will eventually be true, or the next state satisfies x1)G(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
This formula requires each state of the given trace to satisfy "F(x2) U X(x1)", which means that for each state of the trace, if X(x1) is not true, x2 must be eventually true.
671
2020It always holds that eventually x2 is true or x1 is true in the next timestep.G(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
Use of G implies that something is always true. The disjunction, then must always hold.
672
2020it always holds that x2 eventually holds before the state before x1 holdsG(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
Sorry for the word vomit here, but essentially I'm saying that always, x2 must be satisfiable at some point 2 states before x1 becomes satisfiable.
673
2020It always holds that: x2 is eventually true until x1 is true in the next stepG(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
The entire formula is inside a global operator, and says that eventually x2 being true only holds until x1 is true in the next step
674
2020it always holds that before x1's next state is true, x2 is eventually trueG(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
Its\u200b logic is clear and it can be followed by the formula to imply it's meaning
675
2020it always holds that x1 is true since t1, after x2 is true.G(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
F(x2) U means whatever after x2 is true, and X(x1) means x1 is true since t1. (the first entry)
676
2020x2 is always eventually true up until a point, after which x1 is true one state after.G(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
G(F(x2)) is true up to one point at which X(x1) is true so x2 is always eventually true up until a point. One step after that point x1 is true.
677
2020it always holds that x2 will eventually be true until x1 is true in the next state. At some point, x1 will be true in the next state.G(F(x2) U (X(x1))
It uses all four of G, F, X, and U. The meaning is "it is always the case that we will a state in which x1 is going to happen in the next state, and we know in every previous state that x2 is eventually going to happen".
The outer G specifies always, F(x2) says x2 will be true eventually, the U means until, and X(x1)) means x1 is true in the next state.
678
2020If starting from the next state x1 is always true, then x1 will eventually be true.XG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
G(x1) says that x1 is always true. X(G(x1)) says that G(x1) will be true in the next state. So, starting from the next state, x1 is always true. F(x1) means x1 is eventually true. Adding the implication, we get that if x1 is always true starting from the next state, then x1 will eventually be true.
679
2020All traces where x1 is always true starting at the next state.XG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
If starting from the next state, x1 is always true, then this implies that x1 is eventually true, so traces that satisfy this are traces where x1 is always true starting at the next state.
680
2020if x1 is always true (starting in the next state) then this implies that x1 is eventually trueXG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
XG(x1) : means that in the next state, x1 must always be true. \nthis implies F(x1) : which means that x1 is eventually true
681
2020Next time it happens, it always holds that x1 is true implies that x1 eventually holds true.XG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
Since implies connects the two parts of the expression, the left side is expressed first.
682
2020if next value of x1is always be true, it means x1 is eventually trueXG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
XG(x1) means x1is always be true, then F(x1) means x1 eventually be true
683
2020in the next state it always holds that x1 is true implies that x1 is eventually trueXG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
if x1 is true in the next state, then it makes sense that x1 will eventually be true
684
2020x1 always being true implies it will finally be true.XG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
I don't think the next operator impacts anything. Also, this is a tautology?
685
2020For every state if its next state is true, then eventually x1 is true.XG(x1) => F(x1)
I think that this is a cool LTL formula to translate to English because it requires an understanding of four operators: X (in the next step), G (always), => (implies), and F (eventually). The above formula says that in the next step x1 is true implies that eventually x1 is true.
686
2020X3 is true until some state and after that state it holds that x1 is true or x3 is always true and x2 will eventually be true.
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
G means always and F means always. If α is of the form β U γ, then w \U0010fc00 α iff there exists an i ∈ N such that wi \U0010fc00 γ and for all 0≤j<i,wj \U0010fc00β.
687
2020it holds that x3 is true UNION it holds that x1 is true or it always holds that x3 is true and it eventually holds that x2 is true
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
translation is straightforward using logical operator U (Union) and G/F (always/eventually)
688
2020x3 is true or, x1 is true or x3 is always true and it eventually holds that x2 is true
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
I used the comma to signify that the statement after the or is parenthetically combined.
689
2020x3 has to remain true until x1 is true and x3 is always true and x2 is eventually true.
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
A U B means A has to remain true until B becomes true
690
2020x3 is always true and x2 is eventually true, or x3 must be true until x1 is true, and then at some point after x2 is true.
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
x3 is true until x2 is eventually true and either x1 is true or x3 is always true.
691
2020x3 is true until x1 is true or x3 is always true and x2 is finally true.
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
I'm not sure if this until statement can really work. I think F(x2) should always be true or always be false? I don't think it will change value.
692
2020x3 is true until x1 is true or x3 is always true and x2 is eventually true
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
693
2020x3 is true until: x1 is true or it always holds that x3 is true, and it eventually holds that x2 is true.
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
This formula uses the U operator, which means "until": x3 must be true until the second half of the formula is satisfied. On the right side of the U operator, we have ((x1 or G(x3)) and F(x2)). This means that x1 is true or x3 is globally true, AND x2 is eventually true. Thus x3 is true until both x1 is true or x3 is globally true, AND x2 is eventually true.
694
2020There is eventually a split in the trace such that x3 is always true for one subtrace and that eventually either x1 is true or x3 is always true and eventually x2 is true in the other subtrace.
x3 U ((x1 or G(x3)) and F(x2))
until concept is hard to understand and confusing
I was hoping to write this explanation devoid of a "trace" mention, but it was the way I can conceptualize it. I understand the U as G for a subtrace and F for the everything after it, so I tried to apply the "eventually" and "always" to the particulars of each side.
695
12020if x1 is true, x1 must be true in the next statex1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
Not sure if -> is the same thing as =>, but if it is, this means that x1 true implies that x1 must be true in the next state
696
12020x1 now implies x1 holds 2 states from nowx1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
X means the state after the current one, so X(x1) after the implication means x1 holds in the state after that
697
12020if x1 is true in state 0, x1 is true in state 1x1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
x1 in the state 0 is set to true. \nX(x1) suggests that x1 in state 1 is set to true
698
12020If the trace's first state sets x1 to be true, then the second state will set x1 to be true as well.x1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
x1 simply means that the first state of the trace must set x1 to be true, and "X(x1)" means that the second state of the trace must set x1 to be true as well. Traces w where w(0) ⊭ x1 would also pass this formula.
699
12020If x1 is true in state 0, it will be true in state 1x1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
x1 on L0, L1, ... implies x1 on L1, L2, ...
700
12020x1 implies x1 in the next statex1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
this is the infinite loop where whatever occurs, the next state is the same
701
12020x1 is always truex1 -> X(x1)
I didn't understand how X works until I do this quiz so I want to submit a quiz which is the basic usage of X (resubmit because my browser freezes)
If x1 implies that x1 will also be true in the next state, then it will always be true.
702
2020Eventually it will always hold that x1 switches in value every two steps
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
the cases are: x1 -> x1 -> !x1 (in the stepping order) and !x1 -> !x1 -> x1, either of which would at that point bounce over to the other case, so the trace is infinitely defined with this cycle as its "resting state(s)"
703
2020eventually, it will always hold, that if x1 is true, then it will be true in the next state and then false, and, if x1 is false, then it will be false in the next state, and then true.
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
i assumed some parentheses, but it seems like this one is trying to detail the following trace for x1:\n\nTTFF TTFF TTFF...\n\nbut it looks unsatisfiable? because if x1 is true true false, the second true would require that the false actually also be true (due to x1 -> X(x1)).
704
2020Eventually, x1 always implies that x1 is true in the next step and x1 is false in the next next step and it is always true that x1 being false implies that x1 is false in the next step and x1 is true in the next next step.
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
F means eventually, G means always, and X means in the next step. As a result, the above statement can be described in English by saying eventually, x1 always implies that x1 is true in the next step and x1 is false in the next next step and it is always true that x1 being false implies that x1 is false in the next step and x1 is true in the next next step.
705
2020it eventually always holds that x1 implies (x1 is true in the next trace and x1 is false in the next next trace) and it always holds that (x1 is false implies x1 is false in the next trace and x1 is true in the next next trace)
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
using the F/G (eventually/always) and X (next trace) and -> (implies), this formula basically means that the true/false condition is the same in the next trace and flips in the next next trace
706
2020It will eventually hold that x1 is true always implies that x1 is true in the first entry in the sequence and false in the following entry and that x1 is false implies that the first entry of x1 is false and the following is true.
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
F means eventually, G means always, X tells us to look at i=1 and XX tells us to look at i=2
707
2020\nThey alternate in pairs !x1 !x1 x1 x1
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
eventually it will hold true that( x1 next will always be x1 and its next next will be !x1) and ( !x1's next will be x1 and its next next will be x1)
708
2020It will eventually hold that x1 always implies that x1 is true in the next stage and not x1 is true in 2 stages; and it will also eventually hold that not x1 always implies not x1 in the next stage and x1 in 2 stages
F(G(x1 -> Xx1 and XX!x1) and G(!x1->X!x1 and XXx1))
I think the above formula is interesting because it utilizes the next operator to make a pattern. It is always eventually true that the pattern of ...x1, x1, !x1, !x1... will continue for the remainder of the states. However, it does not necessarily have to be true at the beginning.
F is eventually true, G is always true, and X is true in the next stage; XX means true in 2 stages
709
2020If x1 is true in the first entry, then: x2 is always false and x1 is false in some entries before reverting back to true.
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
(a U b): a must remain true until b becomes true. So (!x1 U x1) means that x1 will be false for a while, but will eventually be true. The F is added because this happens if x1 is true in the first entry (so G(!x1 U x1) would be a contradiction). From G(!x2 and ...) we also get that x2 will always be false.
710
2020if x1 is true then it is always true that x2 is false and x1 is false until it is eventually true
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
we do literal translation on the above formula
711
2020X1 is true if and only if x2 is always false.
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
!x1 U x1 should be always true so we don't need to take care of it.
712
2020If x1 starts true, then it always holds that x2 is false and eventually, x is false until it is true.
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
x1 means that x1 is true holds for w(0), =>(implies) can be written as if... then ___, !x2 is just x2 being false, and !x1 U x1 is x1 being false and then becoming true.
713
2020If x1 is true then globally x2 is false and eventually it will hold that x1 is false until x1 is true
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
If x1 is true then globally x2 is false and eventually it will hold that x1 is false until x1 is true
714
2020x1 implies x2 is always false and eventually x1 is false until x1 is true.
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
F(!x1 U x1) means eventually x1 is false until it's true. So it's saying x1 implies x2 is always false as well as the previous statement.
715
2020x1 is true, x2 is false and eventually x1 will be false then true
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
x1 is true at current time which implies that it always holds true that x2 will be false and eventually x1 will be false followed by true
716
2020if x1 is true then it always holds that x2 is false and eventually x1 is true or false
(x1 => G(!x2 and F(!x1 U x1))
I think my formula is interesting because it nests the global symbol within an implies and determines the status of x2 as well as the future status of x1 given the current status of x1.
x1 => ... means "if x1 is true then ...", and G(!x2 and ...) means it always holds that x2 is false and ..., and F(!x1 or x1) means eventually x1 is true or x1 is false
717
2020It always holds that x1 U not x1 is equivalent to not x1 U x1
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
This statement seems to describe the logical equivalence is a transitive form.
718
2020x1 is always sometimes true.
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
Im a bit confused on this one, but I think it simplifies down to sometimes in the middle, based on the fact that there are no parenthesis on the terms inside G, and then the G implies always.
719
2020It is always true that x1 is true until it is no longer true and x1 is False until it is true
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
a U b means 'a is true until b is true'\nSo literal translation gives the above meaning, which is a contradiction. \n
720
2020It is said that x1 is always true until x1 is not true, and x1 is always false until x1 is not false, which is always true.
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
By using distributivity, we can see it is equivalent to G(x1U!x1)&G(!x1Ux1). However, both parts would be always true: If the first state of x1 is true, then the second statement would automatically true, and vice versa. Also, each of the two statements are actually tautology, then it would be always true.
721
2020it is always true that x1 is true until it is false, and x1 is false until it is true
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
this seems kind of redundant?
722
2020it always holds that x1 stays true until it's not, and it stays false until it's not
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
this seems like a tautology.
723
2020It is always true that x1 is true until x1 is false and that x1 is false until it is true
G(x1 U !x1 and !x1 U x1)
This is the solution to the question I wrote for the other part. I think that it would lead to x1 always flipping between true and false which I found to be an interesting thing.
Globally (x1 is true until not x1 is true and not x1 is true until x1 is true)
724
2020x2 initially is False if and only if x1 is true eventually. Otherwise x2 is always true.
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
double arrow means iff and X means initially
725
2020It eventually holds that: x1 is true iff x2 is false at the next state, or it always holds that x2 is true.
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
The whole formula is wrapped in an F, meaning that it eventually holds. Inside we have (x1 <=> X(!x2)), meaning bi-implication: x1 implies X(!x2), and X(!x2) implies x1 (i.e. x1 is true iff X(!x2)). X(!x2) means that x2 is false at the next state. Additionally, we have a G(x2), which means that it always holds that x2 is true.
726
2020Finally, x1 is true if and only if x2 next is true or x2 always is true
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
727
2020it eventually holds that x1 is true if and only if x2 is false in the next state, or it eventually holds that x2 is always true
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
Example traces: [x2, x1x2, !x2, ...]
728
2020This formula accepts traces where, for some state, either x2 is always true after that state, or x1 is true in that state and x2 is false in the state after.
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
At least once, x1 is true followed by x2 being false in the next state or x2 is always true
729
2020either x2 is always true, or whenever x2 is not true, x1 was true before it
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
730
2020Either x1 is true if and only if x2 is false in the next state, or x2 is always true.
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
This statement appears to be consistent. Either x2 is always true or it can only be true if in the previous state x1 was false.
731
2020Eventually, either x1 is true only when the next state of x2 is false, or x2 is always true.
F((x1 <=> X(!x2)) or G(x2))
This formula demonstrates an interesting condition for how and when x2 would be true
This one is interesting, because uses a logical or but feels like the left side of the or negates the right side of the or. (If x2 is false in the next state, it can't always be true)
732
2020always, x2 being true will imply that x1 is not necessarily eventually true and x3 being true implies x2 is not necessarily eventually true
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
i think i answered one question blank
733
2020it is always the case that if x2, then x1 will never be true again and if x3 is true, then x2 will never be true again
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
this is what I got by translating each operator to English
734
2020It always holds that if x2 is true, x1 is not eventually true and if x3 is true, x2 is not eventually true.
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
The whole formula is wrapped in a "globally", which means that the contents always hold true. Inside, there are two subsections (like x2 => !F(x1)) where F means "eventually", and ! negates the following property. Thus, x2 being true implies that x1 is not eventually true. The same logic can be applied to the second part of the formula, x3 => !F(x2).
735
2020This formula accepts traces where, once x3 is true, x2 can never be true afterwards, and once x2 is true, x1 can never be true afterwards.
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
For every state, if x2 is true then x1 is never true, and if x3 is true then x2 is never true.
736
2020it always holds that if x2 is true, then x1 can never be true after that. and, if x3 is true, then x2 can never be true after that.
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
this describes traces where all states with x1 true occur before the first x2 true. and, all x2 trues occur before the first x3 true.
737
2020It's always true that x2 being true implies that x1 won't finally be true. Also, x3 being true implies that x2 will not finally be true.
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
Any presence of x2 means x1 won't happen after that. Any presence of x3 means that x2 won't happen after that.
738
2020It always holds that x2 is true if and only if x1 will not eventually be true. It also always holds that x3 is true if and only if x2 will not eventually be true.
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
F means eventually and G means always.
739
2020it is always true that: if x2 if true, x1 is always false and if x3 is true x2 is never true. This is true therefore, in traces where x2 is always true and x1 is always false and x3 is always false OR if x2 is always false (so x1 is unbound) and x3 is true.
G(x2 => !F(x1) and x3 => !F(x2) )
I think it is interesting that you can give an ordering to the events without fully defining how many times they will each happen
The above explains the idea fairly well. Essentially we need to unwind the related implications.
740
2020Finally, there will be always that before x2 is true, x1 is trueF(G(x1 U x2))
This formula is equivalent to F(G(x1)) and G(F(x2)).
Its logic is clear and it can be followed by the formula to imply it's meaning
741
2020Eventually, it will hold that x1 will always be true until x2 is true at which point x1 is unbound.F(G(x1 U x2))
This formula is equivalent to F(G(x1)) and G(F(x2)).
The outer most operator is an F which means were are looking for something to be eventually true. That thing, specifically, is that x1 U x2 is always true.
742
2020It eventually holds that it always holds that x1 until x2F(G(x1 U x2))
This formula is equivalent to F(G(x1)) and G(F(x2)).
this statement seems to describe how eventually it will hold when x1 U x2 always holds
743
2020It eventually always holds that x1 is true until x2 is true.F(G(x1 U x2))
This formula is equivalent to F(G(x1)) and G(F(x2)).
744
2020if x1 and x2 are both true in the current state, then x3 is true in the next state and false in the state after that.
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
(x1 and x2) = x1 and x2 are true in the current state\nX(x3) = x3 is true in the next state\nX(X(!x3)) = x3 is false in the state after the next state
745
2020if x1 and x2 are true then in the next state x3 will be true and the state after that x3 is false.
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
x1 and x2 implies that x3 is true in the next state and the double X represents that in the next next state x3 is false.
746
2020if x1 and x2 are both true, then x3 will be true in the next state and false in the one after that
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
(x1 and x2) simply means that both x1 and x2 are true. This implies that X(x3) (x3 is true in the next state) and XX(!x3) (x3 is false in the next next state)
747
2020if x1 and x2 is true, then x3 is true in the next timestep and x3 is False in the next two timestep
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
X is the next time step, so X(X(a)) is the next two timestep of a \n=> implies can be think of as 'if-then'
748
2020x1 and x2 now imply that x3 holds in the next stage and that not x3 holds in 2 stages
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
X is true in the next stage; XX means true in 2 stages
749
2020If x1 and x2 holds then x3 will hold for the next time step and x3 will be false after the next time step.
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
This implication establishes that if x1 and x2 holds then the conjunction using X to reference the following time steps will be true.
750
2020in the current state, if x1 and x2 are true this implies that x3 is true in the next state and x3 is false in the state after that
(x1 and x2) => (X(x3) and X(X(!x3)))
If x1 and x2 are true in the next state x3 must be true and in the next next state x3 must be false
(x1 and x2) : mean that x1 and x2 are true in the current state\n(X(x3) and X(X(!x3)) : mean that in the next state 3 is true and in the state after that x3 is not true
751
2020If x1 is true then it stays true for the next 2 states as well.
x1->X(x1) ^ (X(x1))->(X(X(x1))
I didn't know how X works until I read through this quiz so I think a quiz with a focus on X would be useful
If x1 is true then the next state also has x1 being true and if the next state has x1 as true then the state after that also has x1 as true (I'm not sure I exactly know what the arrows mean though so it might not be this)
752
2020(if x1 is true in entry 0, then x1 is always true) and (if x1 is true starting entry 1, it is also true starting entry 2)
x1->X(x1) ^ (X(x1))->(X(X(x1))
I didn't know how X works until I read through this quiz so I think a quiz with a focus on X would be useful
I am interpreting the formula as (x1->X(x1)) and (X(x1) -> X(X(x1))), and also X(x1) means that x1 holds for every entry after the first one.
753
2020If x1 holds at the current time step, then x1 will hold for the next time step. If x1 holds for the next time step, then x1 will hold for the time step after the next one.
x1->X(x1) ^ (X(x1))->(X(X(x1))
I didn't know how X works until I read through this quiz so I think a quiz with a focus on X would be useful
I read this as a conjunction. The first implication establishes that if x1 holds then the next time step must have x1 hold too. The second implication establishes that if in the next time step x1 holds then x1 will hold in the time step after that, x1 will hold.
754
2020If x1 is true then x1 is also true at the next state, and x1 being true at the next state implies that x1 is also true at the next next state. Therefore x1 being true implies that x1 is also true for the next two states.
x1->X(x1) ^ (X(x1))->(X(X(x1))
I didn't know how X works until I read through this quiz so I think a quiz with a focus on X would be useful
We start with an implication: x1 implies the second half of the formula. The second half starts with X(x1), meaning that x1 is true at the next state. Additionally, X(x1) implies X(X(x1)), meaning that if x1 is true at the next state it must also be true at the state after that. Therefore if x1 is true, then we can deduce that X(X(x1)) is also true, meaning that x1 being true implies it is also true for the next two states.
755
2020If x1 is true in state 0, it will be true in states 1 and 2, and if x1 is true in state 1, it will be true in state 2.
x1->X(x1) ^ (X(x1))->(X(X(x1))
I didn't know how X works until I read through this quiz so I think a quiz with a focus on X would be useful
x1 is true in state 0 implies that x1 is true in state 1, and x1 is true in state 1 implies that x1 is true in state 2.
756
2020if x1 is true, then eventually x2 is true until x1 is falsex1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
757
2020if x1 is true then eventually for there will be a x1 to be false where all x2 in previous state would be true.x1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
758
2020x1 implies that, eventually, x2 is true or x1 is falsex1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
I mainly followed the order of operations and used the commas to signify how the eventually statement refers to both x2 being true OR x1 being false.
759
2020if x1 is true, x2 will be true until !x1 isx1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
x1 implies that eventually x2 will be true until !x1 is true
760
2020x1 in this state implies that eventually x2 will hold until x1 is falsex1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
761
2020If x1 is true in the first entry, then eventually x2 will be true until x1 becomes falsex1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
x2 U !x1: x2 is true until x1 is false. F(x2 U !x1): eventually this property will hold. x1 -> F(x2 U !x1): if x1 is true, then eventually this property will hold.
762
2020If x1 is true, eventually, before x1 is false, x2 is true.x1 -> F(x2 U !x1)
This means that if x1 is true now, then eventually, x2 is true until x1 is false. We can think of this as x1 starting something, and then eventually x2 joins x1 and they are "on" together until x1 turns "off" so x2 also turns off in the same state.
Its logic is clear and it can be followed by the formula to imply it's meaning
763
2020If the next, next, next state's x1 is true, then eventually x2 will be false, and x3 in this state is true.
X(X(X(x1))) => (F(!x2) AND x3)
The third next state of x1 implies that eventually x2 is not true when x3 is true. I think that this is interesting because the expression deals with x1 and x2's future states (but expressed in different ways), while dealing with x3's current truth value.
764
2020If the third next time step from now, x1 holds then eventually x2 will be false and x3 holds.
X(X(X(x1))) => (F(!x2) AND x3)
The third next state of x1 implies that eventually x2 is not true when x3 is true. I think that this is interesting because the expression deals with x1 and x2's future states (but expressed in different ways), while dealing with x3's current truth value.
This establishes an implication that if the third next time step x1 is true, then the conjunction must hold.
765
2020If the 3rd next state has x1 as true, then x3 is true and x2 is eventually false.
X(X(X(x1))) => (F(!x2) AND x3)
The third next state of x1 implies that eventually x2 is not true when x3 is true. I think that this is interesting because the expression deals with x1 and x2's future states (but expressed in different ways), while dealing with x3's current truth value.
X(X(X(x1))) means the third next state has x1 as true. F(!x2) means x2 is eventually false.
766
2020If the state three from the current one satisfies x1 then in the current state x3 is true and eventually it will hold that x2 is not true
X(X(X(x1))) => (F(!x2) AND x3)
The third next state of x1 implies that eventually x2 is not true when x3 is true. I think that this is interesting because the expression deals with x1 and x2's future states (but expressed in different ways), while dealing with x3's current truth value.
Three applications of X gets the state three later, and the clause after the implies says that eventually not x2 and currently x3
767
2020it always holds that x1 will be true and x2 will be false until x2 is true and x1 is false
G(x1 U x2) and G(!(x1 and x2))
I think this formula is interesting because I think it is interesting to think about what the set of satisfying traces for this trace means! Because "x1 U x2" is defined as "x1 being true until x2", and this formula requires "x1 U x2" to be true globally, but also that no state in the trace may have both "x1" and "x2" true simultaneously, I believe the above formula is equivalent to "G((x1 or x2) and !(x1 and x2)) and G(F(x2))", or "G(x1 xor x2) and G(F(x2))", if an "xor" formula was given for LTL. In other words, the satisfying traces for this formula would look like states where only x1 or x2 were true, with the caveat that at any state, a future state must choose x2 to be true.
The first clause says that there will always be some number of states satisfying x1 followed by a state satisfying x2, and the second clause says x1 and x2 can never be true at the same time
768
2020It always holds that exactly one of x1 and x1 is true.
G(x1 U x2) and G(!(x1 and x2))
I think this formula is interesting because I think it is interesting to think about what the set of satisfying traces for this trace means! Because "x1 U x2" is defined as "x1 being true until x2", and this formula requires "x1 U x2" to be true globally, but also that no state in the trace may have both "x1" and "x2" true simultaneously, I believe the above formula is equivalent to "G((x1 or x2) and !(x1 and x2)) and G(F(x2))", or "G(x1 xor x2) and G(F(x2))", if an "xor" formula was given for LTL. In other words, the satisfying traces for this formula would look like states where only x1 or x2 were true, with the caveat that at any state, a future state must choose x2 to be true.
This is essentially Xor!
769
2020No traces satisfy this
G(x1 U x2) and G(!(x1 and x2))
I think this formula is interesting because I think it is interesting to think about what the set of satisfying traces for this trace means! Because "x1 U x2" is defined as "x1 being true until x2", and this formula requires "x1 U x2" to be true globally, but also that no state in the trace may have both "x1" and "x2" true simultaneously, I believe the above formula is equivalent to "G((x1 or x2) and !(x1 and x2)) and G(F(x2))", or "G(x1 xor x2) and G(F(x2))", if an "xor" formula was given for LTL. In other words, the satisfying traces for this formula would look like states where only x1 or x2 were true, with the caveat that at any state, a future state must choose x2 to be true.
Since it cant always be the case that a formula is true and false at the same time, no traces should be able to satisfy this.
770
2020x1 is always true and eventually x2 will also be true, but x2 is not always true
G(x1 U x2) and G(!(x1 and x2))
I think this formula is interesting because I think it is interesting to think about what the set of satisfying traces for this trace means! Because "x1 U x2" is defined as "x1 being true until x2", and this formula requires "x1 U x2" to be true globally, but also that no state in the trace may have both "x1" and "x2" true simultaneously, I believe the above formula is equivalent to "G((x1 or x2) and !(x1 and x2)) and G(F(x2))", or "G(x1 xor x2) and G(F(x2))", if an "xor" formula was given for LTL. In other words, the satisfying traces for this formula would look like states where only x1 or x2 were true, with the caveat that at any state, a future state must choose x2 to be true.
G(x1 U x2): there is a subtrace where x2 is true (aka eventually x2 is true) and for every sequence that includes it x1 is true (aka x1 is always true). G(!(x1 and x2)): x1 and x2 cannot always both be true.
771
2020It always true that x1 is true until x2 is true. Also, It's always true that x1 and x2 are never true in the same state.
G(x1 U x2) and G(!(x1 and x2))
I think this formula is interesting because I think it is interesting to think about what the set of satisfying traces for this trace means! Because "x1 U x2" is defined as "x1 being true until x2", and this formula requires "x1 U x2" to be true globally, but also that no state in the trace may have both "x1" and "x2" true simultaneously, I believe the above formula is equivalent to "G((x1 or x2) and !(x1 and x2)) and G(F(x2))", or "G(x1 xor x2) and G(F(x2))", if an "xor" formula was given for LTL. In other words, the satisfying traces for this formula would look like states where only x1 or x2 were true, with the caveat that at any state, a future state must choose x2 to be true.
I'm pretty sure this means that x1 and x2 can be true at any point that the other is not true.
772
2020From entry 3 (starting from 0) x1 will eventually be true. We do not care about entries 0, 1 and 2G(F(XXX(x1)))
For me the interesting part about the formula is that there is the "always eventually" part of the formula and that I use three "next" operations to say that x1 must hold three time steps in the future. The combination of these two parts of the formula makes it an interesting way to define traces.
XXX(x1): We 'take three steps in' and x1 is true. F(XXX(x1)): eventually the previous property will hold. G(F(XXX(x1))): We do not care about the first three entries, but eventually x1 will be true. (I do not believe G is needed at this point)
773
2020It always eventually holds that x1 is true in the next next next state.G(F(XXX(x1)))
For me the interesting part about the formula is that there is the "always eventually" part of the formula and that I use three "next" operations to say that x1 must hold three time steps in the future. The combination of these two parts of the formula makes it an interesting way to define traces.
XXX(x1) refers to x1 being true in three states down the line, and G(F(...)) means that this will always eventually hold.
774
2020So, it always eventually holds that x1 is true in the third state and all subsequent states.G(F(XXX(x1)))
For me the interesting part about the formula is that there is the "always eventually" part of the formula and that I use three "next" operations to say that x1 must hold three time steps in the future. The combination of these two parts of the formula makes it an interesting way to define traces.
From the handout, X(x1) => w^1 satisfies x1. I assume XXX(x1) => w^3 satisfies x1. So, x1 is true in w^3 and later.
775
2020It always holds that eventually the next, next, next x1 holds true.G(F(XXX(x1)))
For me the interesting part about the formula is that there is the "always eventually" part of the formula and that I use three "next" operations to say that x1 must hold three time steps in the future. The combination of these two parts of the formula makes it an interesting way to define traces.
776
2020Always, x1 next next next finally is trueG(F(XXX(x1)))
For me the interesting part about the formula is that there is the "always eventually" part of the formula and that I use three "next" operations to say that x1 must hold three time steps in the future. The combination of these two parts of the formula makes it an interesting way to define traces.
777
2020It always eventually holds that x1 holds at the next next next state (i.e. three states forward).G(F(XXX(x1)))
For me the interesting part about the formula is that there is the "always eventually" part of the formula and that I use three "next" operations to say that x1 must hold three time steps in the future. The combination of these two parts of the formula makes it an interesting way to define traces.
The formula is wrapped in a G and then an F, which means that something "always eventually" holds. On the inside there are three X\'s (X meaning that the variable holds at the next state) -- XX(x1) means that X(x1) holds at the next state, and XXX(x1) means that XX(x1) holds at the next state, so therefore XXX(x1) means that x1 holds three states ahead.
778
2020It always holds that x1 does not equal x2.
G((x1 => not x2) and (x2 => not x1))
This is interesting because it implies that the values of x1 and x2 must always be opposite, however it doesn't enforce values for x1 and x2.
It is globally true that either x1 is true and x2 is false or x2 is true and x1 is false.
779
2020x1 is true could always imply x2 is false, and x2 could sometimes impliy x1 is false
G((x1 => not x2) and (x2 => not x1))
This is interesting because it implies that the values of x1 and x2 must always be opposite, however it doesn't enforce values for x1 and x2.
G means the always, and the second formula has no G so it's not always
780
2020It always holds that if x1 is true, x2 is false; and if x2 is true, x1 is false.
G((x1 => not x2) and (x2 => not x1))
This is interesting because it implies that the values of x1 and x2 must always be opposite, however it doesn't enforce values for x1 and x2.
G means always, and => means implication
781
2020It always holds that x1 implies not x2 and x2 implies not x1.
G((x1 => not x2) and (x2 => not x1))
This is interesting because it implies that the values of x1 and x2 must always be opposite, however it doesn't enforce values for x1 and x2.
The G symbols signifies that the formula accepts traces where the given predicate is always true. So the formula accepts traces where x1 implies not x2 and x2 implies not x1 is always true.
782
2020It always holds that x2 is false if x1 is true and x1 is false if x2 is true.
G((x1 => not x2) and (x2 => not x1))
This is interesting because it implies that the values of x1 and x2 must always be opposite, however it doesn't enforce values for x1 and x2.
(x1 => not x2) - means that x2 is false if x1 is true\n(x2 => not x1) - means that x1 is false if x2 is true \nG is for always
783
2020x1 and x2 are never both true.
G((x1 => not x2) and (x2 => not x1))
This is interesting because it implies that the values of x1 and x2 must always be opposite, however it doesn't enforce values for x1 and x2.
(x1 => not x2) = x2 is false if x1 is true\n(x2 => not x1) = x1 is false if x2 is true\nglobal operator applies this to all states.\nSo x1 and x2 can never both be true in the same state.
784
2020It always holds that x2 is true if x1 is false.
G(N(N(N(x1))) => x2)
Its always true that if there is a state where x1 is true three time steps later, then x2 must be true in that state
I was not sure what N means here. I assume that it means not.
785
2020It is always true that if in 3 time steps from now x1 is true then x2 is true now.
G(N(N(N(x1))) => x2)
Its always true that if there is a state where x1 is true three time steps later, then x2 must be true in that state
x1 being true 3 steps from now implies that x2 is true now (assuming N is the same as X?)
786
2020if it is always true that, 4 steps after the current step, x1 is true, then x2 is true
G(N(N(N(x1))) => x2)
Its always true that if there is a state where x1 is true three time steps later, then x2 must be true in that state
I haven\'t seen "N" before but I\'m guessing it means "X". This is saying that if it always the case that x1 is true in the 4th step ahead, then x2 is true in the current step
787
2020Unsure
G(N(N(N(x1))) => x2)
Its always true that if there is a state where x1 is true three time steps later, then x2 must be true in that state
I don't believe that we learned about N, and it's not on the PDF, so I'm not really sure how to answer this unfortunately.
788
2020It always holds the third sequence following the current subtrace evaluating true for x1 implies that x2 is true.
G(N(N(N(x1))) => x2)
Its always true that if there is a state where x1 is true three time steps later, then x2 must be true in that state
I'm guessing the N means X, n for next.
789
2020it eventually always eventually holds that x1 is trueFGF(x1)
This is very interesting because it is very hard to determine the actual semantics of such a statement without thinking. To me, it means finally there is a state that all states after it would finally make x1 to be true. However, this is redundant, due to that it is true if and only if GF(x1), which means it is always true that you'll see an x1 state to be true.
the combination of F and Gs are reflected in the combo of eventually and always
790
2020It's eventually true that it always eventually holds that x1 is TrueFGF(x1)
This is very interesting because it is very hard to determine the actual semantics of such a statement without thinking. To me, it means finally there is a state that all states after it would finally make x1 to be true. However, this is redundant, due to that it is true if and only if GF(x1), which means it is always true that you'll see an x1 state to be true.
The F means eventually, while G(F( means always eventually, so eventually it also eventually holds
791
2020eventually x1 is always eventually trueFGF(x1)
This is very interesting because it is very hard to determine the actual semantics of such a statement without thinking. To me, it means finally there is a state that all states after it would finally make x1 to be true. However, this is redundant, due to that it is true if and only if GF(x1), which means it is always true that you'll see an x1 state to be true.
FGF() means eventually, it is always eventually x. However, eventually always eventually doesn't make much intuitive sense to me -- is eventually not the same thing as eventually always eventually?
792
2020it always eventually holds that x1 is trueFGF(x1)
This is very interesting because it is very hard to determine the actual semantics of such a statement without thinking. To me, it means finally there is a state that all states after it would finally make x1 to be true. However, this is redundant, due to that it is true if and only if GF(x1), which means it is always true that you'll see an x1 state to be true.
F(x1) means that x1 will eventually be true. G(F(x1)) means that it always eventually holds that x1 is true. F(G(F(x1))) means that G(F(x1)) will eventually hold; but since this is global, if it eventually holds, it must always hold.
793
2020It always eventually holds that x1 is true.FGF(x1)
This is very interesting because it is very hard to determine the actual semantics of such a statement without thinking. To me, it means finally there is a state that all states after it would finally make x1 to be true. However, this is redundant, due to that it is true if and only if GF(x1), which means it is always true that you'll see an x1 state to be true.
If x1 eventually always eventually holds true, then it always will eventually be true.
794
2020x1 is always eventually trueFGF(x1)
This is very interesting because it is very hard to determine the actual semantics of such a statement without thinking. To me, it means finally there is a state that all states after it would finally make x1 to be true. However, this is redundant, due to that it is true if and only if GF(x1), which means it is always true that you'll see an x1 state to be true.
A direct translation could be "there is a state after which for all future states, x1 is true at some later state"
795
2020It always holds true that if x1 is false in the next state, then it is true in the current stateG(X(!x1) => x1)
I think this question is interesting because it demonstrates how the next of x1 implies a fact about x1's value, always.
pretty straight forward
796
2020It globally holds that in the next state x1 will be false if and only if x1 is true.G(X(!x1) => x1)
I think this question is interesting because it demonstrates how the next of x1 implies a fact about x1's value, always.
x1 is false in the next state if x1 is true in the current state. This is always true.
797
2020Its always true that if x1 is false, then it was true in the previous stateG(X(!x1) => x1)
I think this question is interesting because it demonstrates how the next of x1 implies a fact about x1's value, always.
G means that its always true. X(!x1) means that if x1 is false in the next state, and x1 is just x1 being true in the current state\n
798
2020x1 is always true immediately before it isn'tG(X(!x1) => x1)
I think this question is interesting because it demonstrates how the next of x1 implies a fact about x1's value, always.
The G makes it hold always, and the clause inside says that x1 being false in the next state implies x1 is true now
799
2020it always holds that not x1 implies x1 in the next stateG(X(!x1) => x1)
I think this question is interesting because it demonstrates how the next of x1 implies a fact about x1's value, always.
G is always true, X is true in the next state
800
2020For every state if the next state of x1 being false is true, then x1 is true.G(X(!x1) => x1)
I think this question is interesting because it demonstrates how the next of x1 implies a fact about x1's value, always.
I'm wondering if it's a paradox, if for some state x1 is false then its next state would be false because if it's true, it will make current x1 to be true, which contridicts the assumption, so the sequence might be looks like [true] * [false]*.
801
2020It always holds that x1 implies that not x2 and not x3, and it always holds that x2 union x3 implies that not x1
G(x1=>!x2 and !x3) and G(x2 or x3 => !x1)
it's how read/write lock behaves
The statement seems to describe how x1 influences x2 and x3, as well as vice versa
802
2020It is always true that if x1 is true then x2 is false and x3 is false. Also, it is always true that if x2 is true or x3 is true then x1 is false.
G(x1=>!x2 and !x3) and G(x2 or x3 => !x1)
it's how read/write lock behaves
x1 => !x2 and !x3 means that if x1 is true, then !x2 and !x3 are false (so x2 and x3 are true). x2 or x3 => !x1 means that if "x2 or x3" is true then !x1 is true so x1 is false. Wrapping these with G and using "and" we get that it is always true that if x1 is true then x2 is false and x3 is false. Also, it is always true that if x2 is true or x3 is true then x1 is false.
803
2020always if x1 true, then x2 is false and x3 is false, or always if x2 or x3 are true, then x1 is false
G(x1=>!x2 and !x3) and G(x2 or x3 => !x1)
it's how read/write lock behaves
G enforces always, while (x1 => ...) and (x2 or x3 => ...) mean that the left implies the right. Altogether we have that either of two situations are always true
804
2020Its always true that if x1 is true that x2 and x3 are false, and its always true that if x2 or x3 are true that x1 is false
G(x1=>!x2 and !x3) and G(x2 or x3 => !x1)
it's how read/write lock behaves
The and means that both are true. The G(x1 => !x2 and !x3) means that its always true that if x1 is true that x2 and x3 are false. While G(x2 or x3 => !x1) means that if x2 or x3 are true that x1 is false
805
2020It always holds that if x1 is true then x2 must be false and x3 is false and if x3 is true then x1 must be false or x2 is true.
G(x1=>!x2 and !x3) and G(x2 or x3 => !x1)
it's how read/write lock behaves
#ERROR!
806
2020it is always the case that x2 and x3 are both false if x1 is true, and it is always the case that x2 is true or x1 is false if x3 is true
G(x1=>!x2 and !x3) and G(x2 or x3 => !x1)
it's how read/write lock behaves
translating each part of formula, it might be possible to simplify English representation
807
2020it always holds that x1 implies x2 in the next state and not x3 in the following state
G(x1 -> (X(x2) and XX(!x3)))
I think the notion of being true in future states is pretty neat. For me, understanding those concepts was challenging at first, so translating from LTL to English and vice versa has helped a lot!
Example traces: [x1, x2, !x3, ...]
808
2020It is always true that x1 implies that x2 is true in the i=1th entry, and x3 is false in the following entry)
G(x1 -> (X(x2) and XX(!x3)))
I think the notion of being true in future states is pretty neat. For me, understanding those concepts was challenging at first, so translating from LTL to English and vice versa has helped a lot!
G means always, X means the i=1 entry.
809
2020it is always true that x1 being true implies that both at the next timestep x2 is true and at the timestep after the next timestep x3 is false
G(x1 -> (X(x2) and XX(!x3)))
I think the notion of being true in future states is pretty neat. For me, understanding those concepts was challenging at first, so translating from LTL to English and vice versa has helped a lot!
this establishes a statement about how x1 being true implies these things about the next and next next timestep (/the following things after you eliminate t0 for the first X(x2) and eliminating t1 and t2 for XX(!x3))
810
2020It always holds true that x1 being true in the first sequence implies that in the next sequence x2 is true and the next next sequence x3 does not.
G(x1 -> (X(x2) and XX(!x3)))
I think the notion of being true in future states is pretty neat. For me, understanding those concepts was challenging at first, so translating from LTL to English and vice versa has helped a lot!
G has a definition of subtraces including all sequences following, so in my mind, I\'m imagining the only trace that can satisfy this is if x1 and x2 are true for all sequences, and x3 is false for all sequences. Only then could it be "always" true, for all subtraces.
811
2020It is always true that x1 implies that x2 will be true in the next state and x3 will be false in the state after that.
G(x1 -> (X(x2) and XX(!x3)))
I think the notion of being true in future states is pretty neat. For me, understanding those concepts was challenging at first, so translating from LTL to English and vice versa has helped a lot!
This one seems pretty straightforward - G being Always, and the inner formula is just the sentence used.
812
2020It always holds that eventually x1 implies that the next sequence has x1 as false and that x2 implies that x2 is false.
G(F(x1 -> X(!x1)) and x2 -> !x2)
It always holds that eventually x1 implies that the next state of x1 is false will hold and x2 implies that x2 is false
I think that the second imply statement is applying to the same sequence, so as I understand, both statements can't evaluate to true. This follows the same structure with the implies on either side of the and.
813
2020There will be always that eventually if x1 is true and then it will be false, and if x2 is true, x2 is false
G(F(x1 -> X(!x1)) and x2 -> !x2)
It always holds that eventually x1 implies that the next state of x1 is false will hold and x2 implies that x2 is false
The last part of the formula means it will never be true
814
2020always, eventually x1 being true will imply x1 is false in the next state, and x2 being true implies not x2
G(F(x1 -> X(!x1)) and x2 -> !x2)
It always holds that eventually x1 implies that the next state of x1 is false will hold and x2 implies that x2 is false
unsat
815
2020it always holds that x1 can only be true at t0 and x2 is false
G(F(x1 -> X(!x1)) and x2 -> !x2)
It always holds that eventually x1 implies that the next state of x1 is false will hold and x2 implies that x2 is false
F(x1 -> X(!x1)) equals F(!x1 or X(!x1), which means x1 can only be true at t0; x2 -> !x2 equals !x2 or !x2, which means x2 is false
816
2020x1 will always eventually go false the state after it is true, and x2 is always false
G(F(x1 -> X(!x1)) and x2 -> !x2)
It always holds that eventually x1 implies that the next state of x1 is false will hold and x2 implies that x2 is false
F(x1->X(!x1)) means that eventually, if x1 is true, it must be false in its next state. x2->!x2 means x2 must be false. These both hold globally.
817
2020if x1 is never true, that implies that either x1 is always true or x2 is eventually true
G(!x1) => (G(x1) or F(x2))
I think that this is an interesting formula because it includes a contradiction between G(!x1) or G(x1). Therefore, if x1 is always false, then x2 must be true eventually (because x1 can never be true)
first clause of the or operator is meaningless since we know that x1 is never true (from the condition of the implies operator)
818
2020if x1 is always false, x1 will always be true or x2 will eventually be false
G(!x1) => (G(x1) or F(x2))
I think that this is an interesting formula because it includes a contradiction between G(!x1) or G(x1). Therefore, if x1 is always false, then x2 must be true eventually (because x1 can never be true)
G means always, and F means eventually, so it's easy to figure out
819
2020If x1 is never true then x1 is always true or eventually x2 is true.
G(!x1) => (G(x1) or F(x2))
I think that this is an interesting formula because it includes a contradiction between G(!x1) or G(x1). Therefore, if x1 is always false, then x2 must be true eventually (because x1 can never be true)
F(x2) says that eventually x2 is true. G(x1) says that x1 is always true. G(!x1) says that !x1 is always true, so x1 is always false, meaning x1 is never true. Combining these all with implication and or gives us "if x1 is never true then x1 is always true or x2 is eventually true."
820
2020It is always true that x1 is always true if and only if x2 is eventually trueG(G(x1) => F(x2))
Most prompts that I received used "=>" with only variables like "x1" and "x2", and never with any of the more interesting LTL things
The G(G(x1)) throws me off a bit (idk the difference between that and G(x1)). But in order for x1 to be globally true, x2 has to eventually be true.
821
2020If x1 is always true, then x2 is always eventually true.G(G(x1) => F(x2))
Most prompts that I received used "=>" with only variables like "x1" and "x2", and never with any of the more interesting LTL things
G(x1) => F(x2) = If x1 is true, then x2 is eventually true.\nG(G(x1) = G(x1) because if x1 is always true, then it is always the case that x1 is always true.\nG(F(x2) = x2 is always eventually true \nso G(G(x1) => F(x2)) = G(x1) => G(F(x2))
822
2020It always holds that if x1 is globally true, then x2 is eventually true.G(G(x1) => F(x2))
Most prompts that I received used "=>" with only variables like "x1" and "x2", and never with any of the more interesting LTL things
This means that if the trace has x1 true in every state, then eventually x2 will stay true in the trace. If the trace does not hold G(x1), then this is an arbitrary constraint that doesn't describe anything. If the trace does hold G(x1), then the formula is the same as just saying F(x2).
823
2020if x1 is set to true in all states, then x2 is eventually trueG(G(x1) => F(x2))
Most prompts that I received used "=>" with only variables like "x1" and "x2", and never with any of the more interesting LTL things
the formula suggests whenever x1 is true for all states, x2 will eventually become true
824
2020It will always hold that if x1 always holds true, then it eventually holds that x2 is true.G(G(x1) => F(x2))
Most prompts that I received used "=>" with only variables like "x1" and "x2", and never with any of the more interesting LTL things
G means globally/always, and F means eventually.
825
2020x1 and x2 are both eventually true where x1 becomes true before x2, or it always holds that if x1 is true then x1 is false in the next state.
(x1 U x2) OR G(x1 => X(!x1))
I think using the OR to combine these two clauses is interesting since they both use x1 as an initial condition. Either x1 is always true until x2 is true, or when x1 is true it will always be false in the next state.
I'm not sure if my understanding of U is correct! The second clause is always true because of the G, and the implication shows that x1 switches to false in the next state.
826
2020x1 is true until x2 is true or it is always true that if x1 is true that implies that at the next time step x1 will be false
(x1 U x2) OR G(x1 => X(!x1))
I think using the OR to combine these two clauses is interesting since they both use x1 as an initial condition. Either x1 is always true until x2 is true, or when x1 is true it will always be false in the next state.
we have an or statement with one until statement for x1's trueness being true until x2 is true and one global statement which says that x1 being true implies that x1 is false at the next time step
827
2020x1 is true until x2 is true, or whenever x1 is true, it will be false in the next state.
(x1 U x2) OR G(x1 => X(!x1))
I think using the OR to combine these two clauses is interesting since they both use x1 as an initial condition. Either x1 is always true until x2 is true, or when x1 is true it will always be false in the next state.
Direct translation.
828
2020x1 is true until x2 is true, or it is always true that when x1 is true, x1 at the next time step is False
(x1 U x2) OR G(x1 => X(!x1))
I think using the OR to combine these two clauses is interesting since they both use x1 as an initial condition. Either x1 is always true until x2 is true, or when x1 is true it will always be false in the next state.
G(x1 => X(!x1)) means it is always true that when x1 is true, x1 at the next time step is False\n(x1 U x2) means x1 is true until x2 is true
829
2020it always holds that in the next state x1 is true and that eventually, x1 will imply x2
G(X(x1) and F(x1 => x2))
I think this is an interesting formula because the G(X()) construct is in a sense a more specific F, where i is just 1. Then, side by side with F shows the difference, where x1=>x2 doesn't need to happen until much later.
830
2020x1 is guaranteed to be true in all states but the first, and it always eventually holds that x2 is true.
G(X(x1) and F(x1 => x2))
I think this is an interesting formula because the G(X()) construct is in a sense a more specific F, where i is just 1. Then, side by side with F shows the difference, where x1=>x2 doesn't need to happen until much later.
G(X(x1)) says that x1 is true in the next state of every state, making x1 true in every state but the first. Since x1 is already always true, G(F(x1 => x2)) becomes just G(F(x2)) which means it always eventually holds that x2 is true.
831
2020It always holds that in the second state x1 is true and eventually x1 implies x2; that is if x1 is true then x2 has to be true.
G(X(x1) and F(x1 => x2))
I think this is an interesting formula because the G(X()) construct is in a sense a more specific F, where i is just 1. Then, side by side with F shows the difference, where x1=>x2 doesn't need to happen until much later.
Since the global symbol is applied to both sides they both always hold. X(x1) holds if in the second state the proposition x1 is correct, and F(x1 => x2) if eventually there is a state where x1 implies x2.
832
2020If x1 is false until the next state, where it becomes true, then x1 is eventually false
X(!x1 U x1) => F(!x1)
My rationale is that it holds that x1 is false until x1 is true in the next state implies that it eventually holds that x1 is false.
X(!x1 U x1) was confusing, at first I thought it was a contradiction. But all it means is that x1 is false until some state, then x1 is true. Thus, we know that the next state satisfying this structure (x starting false and going to true) implies the second part of the formula, F(!x1), which is simply x1 eventually being false.
833
2020if at the next(following, so at any timestop thats not t0) time step x1 is false until x1 is true then this implies that eventually x1 will be false
X(!x1 U x1) => F(!x1)
My rationale is that it holds that x1 is false until x1 is true in the next state implies that it eventually holds that x1 is false.
this one was kind of confusing, but that in the time steps following the present one, the until statement being true implies this eventual statement about the falseness of x1
834
2020This evaluates to true for any trace in which x1 is eventually false.
X(!x1 U x1) => F(!x1)
My rationale is that it holds that x1 is false until x1 is true in the next state implies that it eventually holds that x1 is false.
This says: Starting from the "next" state, x1 is false until it is true implies that x1 is eventually false. This should just be equivalent to F(!x1) because (!x1 U x1) is true by definition.
835
2020x1 is always true.!F(!x1)
This formula would evaluate to G(x1) and I think it is interesting to write G as a function of F. This is because "always x1" is the same as "they will never be not x1"
x1 will never be false after this.
836
2020This formula accepts traces where x1 is always true.!F(!x1)
This formula would evaluate to G(x1) and I think it is interesting to write G as a function of F. This is because "always x1" is the same as "they will never be not x1"
It is not the case that sometimes x1 is false.
837
2020not x1 does not eventually have to hold!F(!x1)
This formula would evaluate to G(x1) and I think it is interesting to write G as a function of F. This is because "always x1" is the same as "they will never be not x1"
the not kinda cancels out the F
838
2020if eventually x1 and x2 are true, either x1 holds until x2 is true or x2 is true in the next state
F(x2 and x1) => (x1 U x2) or (X x2)
I thing the relationships between the global operators and the next operator is interesting - like what does next actually mean here?
i wrote this question
839
2020If x2 and x1 are ever true at the same time, then x1 is true until x2 is true or x2 is true in the second state.
F(x2 and x1) => (x1 U x2) or (X x2)
I thing the relationships between the global operators and the next operator is interesting - like what does next actually mean here?
x2 and x1 is sometimes true implies x1 is true until x2 is true or x2 is true in the next state.
840
2020Eventually x2 and x1 being true implies that either x1 and x2 is true or x2 initially being true.
F(x2 and x1) => (x1 U x2) or (X x2)
I thing the relationships between the global operators and the next operator is interesting - like what does next actually mean here?
F means eventually. Arrow means imply and X1 U x2 ask both have to be true.
841
2020if x1 is true in the state 0, then eventually always x1 is truex1 -> F(G(x1))
If x1 is ever true, eventually it will take over and always be true, inspired by coronavirus.
Using rule 1) in the handout, x1 is set to true. Then, x1 is set to true implies eventually globally x1
842
2020If x1 is true, then eventually x1 will always be true.x1 -> F(G(x1))
If x1 is ever true, eventually it will take over and always be true, inspired by coronavirus.
G(x1) says that x2 is always true. F(G(x1)) says that there exists a state where G(x1) is true, so eventually there is a state where G(x1) is always true. Adding the implication to this we get that if x1 is true, then eventually x1 will always be true.
843
2020if x1 is true, then it always eventually holds that x1 is true.x1 -> F(G(x1))
If x1 is ever true, eventually it will take over and always be true, inspired by coronavirus.
F means eventually and G means always.
844
2020eventually x1 is true and eventually x2 is true and eventually x3 is true implies that x4 is true AND x1 never being true or x2 never being true or n3 never being true implies that x4 is false
((F(x1) and F(x2) and F(x3)) => x4) and (!F(x1) or !F(x2) or !F(x3) => !x4)
This is expressing if (and only if) x1, x2, and x3 are "switched on" x4 turns on. It is interesting as it seems very easy to map to a real life example.
several eventually statements which implies something about x4, i think both of these guarantee the same thing?
845
2020x1 if and only if x2, and eventually x1 and x2 hold
G(((!x1) => (!x2)) and ((!x2)=>(!x1))) and F(x1 and x2)
F(x1 and x2) means that x1 and x2 will eventually be true at the same time, but not that will *become* true at the same time.
The first part says that it is global that !x1 implies !x2 and vice versa. We can simplify the negations leading to x1 iff x2. The last part states that finally, x1 and x2 hold.
846
2020It always holds that if x1 is false then x2 is false, and if x2 is false then x1 is false (i.e. x1 is false iff x2 is false). Also, it eventually holds that x1 and x2 are both true.
G(((!x1) => (!x2)) and ((!x2)=>(!x1))) and F(x1 and x2)
F(x1 and x2) means that x1 and x2 will eventually be true at the same time, but not that will *become* true at the same time.
The first part of the expression is wrapped in a G, meaning that the contents are always true. Inside we have ((!x1) => (!x2)), which means that x1 being false implies that x2 is also false. We also have ((!x2) => (!x1)), which means the reverse: x2 being false implies that x1 is also false. Taken together, this means that x1 is false iff x2 is false. We also have F(x1 and x2), which means that eventually x1 and x2 are both true.
847
2020If x1 is eventually true, x2 is always trueF(x1) -> G(x2)
It represents the state transformation, which is very interesting in the language.
848
2020x1 eventually being true implies that x2 is always trueF(x1) -> G(x2)
It represents the state transformation, which is very interesting in the language.
an eventually statement wrapping x1 implying a global statement about x2
849
2020x1 is true in the next state, and x2 is true three states laterX(x1)->XXX(x2)
help student understand next
850
2020If x1 is true in the next state, then x2 is true three states after the current stateX(x1)->XXX(x2)
help student understand next
X(x1) = x1 is true in the next state \nXXX(x2) = x2 is true in the state after the state after the next state
851
2020x1 is always true, and x2 is false and if x1 is not equal to x2, x2 will always being false.
G(x1) and !x2 and x1!=x2 => G(!x2)
I think it's interesting that the left side appears more verbose, but it can mean the same thing (x1 and x2 are never equal, specifically that x1 = T, and x2 = F). This is because the G on the left will propagate to every instance in the trace.
Its logic is clear and it can be followed by the formula to imply it's meaning
852
2020For every state x1 is true, if x2 is false and x1 is not equals x2, then for every state x2 is false
G(x1) and !x2 and x1!=x2 => G(!x2)
I think it's interesting that the left side appears more verbose, but it can mean the same thing (x1 and x2 are never equal, specifically that x1 = T, and x2 = F). This is because the G on the left will propagate to every instance in the trace.
853
2020It eventually holds that always x1 is true or always eventually x1 is true.
F(G(x1)) or G(F(x1))
I think this is an interesting formula because while "eventually x is always true" and "always, x is eventually true" sound very similar, they are slightly different - in the first one, there will be a state after which x is always true, while in the second one, no matter what state you're in you will eventually reach a state where x1 is true.
854
2020It eventually holds that x1 is always true or always holds that x1 is eventually true.
F(G(x1)) or G(F(x1))
I think this is an interesting formula because while "eventually x is always true" and "always, x is eventually true" sound very similar, they are slightly different - in the first one, there will be a state after which x is always true, while in the second one, no matter what state you're in you will eventually reach a state where x1 is true.
The first half expresses that it eventually holds that x1 is always true and the second half expresses that x1 is eventually true always.
855
2020If that x1 starts as true, it will eventually hold that it is false, whereas if x1 starts as false, it will eventually hold as true.
x1 -> F(!x1) & !x1 -> F(x1)
my question describes the relation between x1's current value and its future value, which I think is interesting!
x1 implies eventually finally not x1 and not x1 implies eventually finally x1
856
2020If x1 is true then eventually x1 is false and if x1 is false then eventually x1 is true.
x1 -> F(!x1) & !x1 -> F(x1)
my question describes the relation between x1's current value and its future value, which I think is interesting!
F(x1) indicates that there is some state in which x1 is true, so x1 is eventually true. F(!x1) thus means !x1 is eventually true, so x1 is eventually not true. Adding in the implications and the & symbol, we get the x1 implies that eventually x1 is false and !x1 implies that eventually x1 is true.
857
2020x1 if and only if x2 until x1, until x3
(x1 <=> x2) U (x1 U x3)
The double until should be tricky. Is it possible for an until statement to be false and then become true? Or, to occur without having occurred previously?
The <=> symbol is if and only if, and the U's are until. It means that x1 and x2 imply each other until x1 is invalid, which happens only when x3 becomes invalid.
858
2020x1 xor x2 will always be true (i.e. one of x1,x2 will always be true, and the other false)
G(!(x1 and x2)) and G(x1 or x2)
This is essentially saying G(x1 Xor x2). I think it could be interesting to think about a truth function like this that applies over all states even though it's separated into two clauses.
G(!(x1 and x2)) means that x1 and x2 can never both be true, but G(x1 or x2) means that at least one of x1 and x2 will always be true. So, one will always be true, and the other cannot be.
859
2020It is always true that x1 or x2 is true, and in the future it is always true that x2 is true.
G(x1 or x2) and X(G(x2))
Either x1 or x2 must always be true, but starting at the next state x2 must always be true, so starting at the next state the second clause being true will mean that the first clause is already also true no matter what x1 is.
G(x1 or x2) tells us that x1 or x2 is always true. X(G(x2)) tells in the next state, G(x2) is true, which means that starting from the next state, x2 is always true, so x2 is always true in the future. Combining them with "and" we get that it is always true that x1 or x2 is true, and in the future it is always true that x2 is true.
860
2020It always holds that x1 or x2 is true and that in the next sequence, it always holds that x2 holds as true.
G(x1 or x2) and X(G(x2))
Either x1 or x2 must always be true, but starting at the next state x2 must always be true, so starting at the next state the second clause being true will mean that the first clause is already also true no matter what x1 is.
The "next sequence" might also mean the "next sequence on" for the G to evaluate correctly, although I\'m not sure.
861
2020For every state, if x1 is false in the current state and true in the next state, then in the next state x2 is eventually true
G((!x1 and X(x1)) => X(F(x2))
I like this because it's the idea of causation - like if I flick a switch, the light will eventually turn on after the switch is flicked.
!x1 and X(x1) = x1 is false in current in true in next\nX(F(x2) = in the next state, x2 is eventually true. \nThe global operator makes this apply to every state
862
2020It is always true that eventually x1 or x2 is true. Seems invalid...G(F(x1) or F(x2))
I think having the or within the always is interesting
The outer G indicates that the inner clauses must always hold at any point in a trace, while the inner ones only make guarantees about things _eventually_ happening. This seems like a contradiction.
863
2020eventually x2 is true, and whenever x1 is true, it will become false in the next state
x1 -> X(!x1) and F(x2)
x1 -> X(!x1) is interesting because implication actually adds more constraints to X
864
2020x1 is always trueF(x1) => G(x1)
This is interesting because it shows how if something is eventually true then it is always true.
If x1 being eventually true implies that x1 is always true then the traces that satisfy this are just traces where x1 is always true.
865
2020x1 always implies that x2 will be valid until x3 is not, from the next state on.
G(F(x1 -> X(x2 U !x3)))
many symbols adding together is tricky
Globally true that eventually x1 implies that in a next state, x2 will be valid until x3 is not (reading logically from left-to-right).
866
2020It always eventually holds that x1 is falseG(F(x1 -> !x1))
The "always eventually" is applying that eventually to subtraces, which is an interesting concept. The x1 implying its own falsity means that it can only evaluate to true when both are not true.
For it to be the case that x1->!x1, x1 must be false (when x1 is true, T->F is false; when x1 is false, F->T is true). F(x1->!x1) means that x1 will eventually be false, and G(F(x1->!x1)) means that x1 will always eventually be false.
867
2020It always eventually holds that x1 is always true and It eventually always holds that x2 is eventually true.
G(F(G(x1))) and F(G(F(x2)))
I think the intersection of F and G is very interesting, especially when they are combined in succession - i.e. how does FGF differ from GFG? The ideas of "always" and "finally" seem very critical in LTL, especially when they are combined like this.
G means always , F means eventually.
868
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