ABCDEFGHIJKLMNOPQRSTUVWXYZ
1
Totals ==>125004021000001132
2
Ok?BPBSIBSQIFIGOIWUEUXXRV
Scope
Answer
Explanation
IDFormulaExpected
3
1
It is always the case that eventually the Red light is on. In other words (applying reasoning) the Red light is on infinitely often.
R_10pfB7n4SRWFEiA
G(F(r))
r infinitely often
4
1The red light is on infinitely often.
R_1d6RMi3t0gkcM5M
G(F(r))
r infinitely often
5
1The red light is on infinitely often.
R_e8QLXoeqrle4xS9
G(F(r))
r infinitely often
6
1Red light is on infinitely often
R_z5PVfNBIPKX5S4F
G(F(r))
r infinitely often
7
1The red light is on infinitely often times.
R_1E0ztwRbCQeEjAn
G(F(r))
r infinitely often
8
1There are infinitely many states in the trace where the red light is on.
R_d6w1dXzacJ9KN0Z
G(F(r))
r infinitely often
9
1
The red light occurs in infinitely many states: at each point in time, there is always a later one for which the light is read. (Always eventually red.)
R_3MuS1ttcj8sdoRv
G(F(r))
r infinitely often
10
1The red light is on infinitely often.
R_1hEGqpvvTZegzQ7
G(F(r))
r infinitely often
11
1The red light is on infinitely often
R_3gO868rLSy0R2Ep
G(F(r))
r infinitely often
12
now vs future1
At all timesteps t>0, there exists a timestep t'>t such that the Red light is on at timestep t'.
R_332amsBlozpU5n8
G(F(r))
r infinitely often
13
1The Red light is on infinitely often.
Repeated reachabilty: \\ From any state we reach on the trace we eventually get a state with Red on.
R_QcBt5TnllV1Ynq9
G(F(r))
r infinitely often
14
1The Red light is on infinitely often.
R_2t50ks4XN0wxwLl
G(F(r))
r infinitely often
15
1Red holds infinitely often.
R_0Vg19zBXa3hIGzf
G(F(r))
r infinitely often
16
1The Red light will always eventually be on.
R_33wl4yU4sEzJKu4
G(F(r))
r infinitely often
17
1The red light is infinitely often on
R_3iWIMn2gWx1V4Lp
G(F(r))
r infinitely often
18
1
At every step, Red will be on after a finite (possibly zero) number of steps - i.e. Red will never be off forever
R_W7oqYTh5ZL9V9vP
G(F(r))
r infinitely often
19
1The Red light will always be on at some point in the future, forever.
R_3e8mdHJtEsNstwp
G(F(r))
r infinitely often
20
1Red appears infinitely often.
R_1i3xVhhSYFfFwDR
G(F(r))
r infinitely often
21
now vs future1There is always a state in the future in which the Red light is on.
R_6F2ykLOE4e1S4IF
G(F(r))
r infinitely often
22
1The red light is on infinitely often.
R_sb9MnVHZ3rBbcuR
G(F(r))
r infinitely often
23
1Red is on infinitely often.
R_3GB8QvZ8WJ7imO7
G(F(r))
r infinitely often
24
1For each state, a red light will be on eventually.
R_3KSFgYqXtEshyWa
G(F(r))
r infinitely often
25
1If the Red light is on now then it will be also on in 3 steps.
R_10pfB7n4SRWFEiA
r => X(X(X(r)))
if red now, then red in 3 steps
26
1If the red light is on on the first stage, it is on on the fourth stage.
R_1d6RMi3t0gkcM5M
r => X(X(X(r)))
if red now, then red in 3 steps
27
1If red is on at the starting state, then red is also on at the fourth state.
R_e8QLXoeqrle4xS9
r => X(X(X(r)))
if red now, then red in 3 steps
28
1If the current light is red, the third state of its following state is red
R_z5PVfNBIPKX5S4F
r => X(X(X(r)))
if red now, then red in 3 steps
29
1
If the red light is on at the first state, then it must be on at the fourth state (3 states after).
R_1E0ztwRbCQeEjAn
r => X(X(X(r)))
if red now, then red in 3 steps
30
1If the first state satisfies that the red light is on, then the third next state is also so.
R_d6w1dXzacJ9KN0Z
r => X(X(X(r)))
if red now, then red in 3 steps
31
1If the light is Red at the first state, then it is red three states later.
R_3MuS1ttcj8sdoRv
r => X(X(X(r)))
if red now, then red in 3 steps
32
1Whenever the red light is on, the red light is on 3 time steps later
R_1hEGqpvvTZegzQ7
r => X(X(X(r)))
if red now, then red in 3 steps
33
1If the red light is on now, then it will be on after 3 transitions.
R_3gO868rLSy0R2Ep
r => X(X(X(r)))
if red now, then red in 3 steps
34
1
From the initial timestep, if it happens that the Red light is turned on, the Red light will be on again in 3 timesteps.
R_332amsBlozpU5n8
r => X(X(X(r)))
if red now, then red in 3 steps
35
1If the Red light is on, then in 3 moments from now it will be Red again
R_QcBt5TnllV1Ynq9
r => X(X(X(r)))
if red now, then red in 3 steps
36
1If the Red light is on in the first state then it must be also on in the fourth state.
R_2t50ks4XN0wxwLl
r => X(X(X(r)))
if red now, then red in 3 steps
37
1
If the first state satisfies red then the state you arrive at in 3 transitions must also satisfy red.
R_0Vg19zBXa3hIGzf
r => X(X(X(r)))
if red now, then red in 3 steps
38
1If Red light is currently on, it will also be on in the 3rd next state.
R_33wl4yU4sEzJKu4
r => X(X(X(r)))
if red now, then red in 3 steps
39
1If the first state is red, then the fourth state is also red
R_3iWIMn2gWx1V4Lp
r => X(X(X(r)))
if red now, then red in 3 steps
40
1If Red is on, it will also be on after 3 steps
R_W7oqYTh5ZL9V9vP
r => X(X(X(r)))
if red now, then red in 3 steps
41
1If the red light is on, it will also be on in 3 states' time
R_3e8mdHJtEsNstwp
r => X(X(X(r)))
if red now, then red in 3 steps
42
1If the 1st state is Red, then the 4th state is Red. Or the first state is not Red.
R_1i3xVhhSYFfFwDR
r => X(X(X(r)))
if red now, then red in 3 steps
43
1In the first state, if Red is on then Red will be on 3 states later.
R_6F2ykLOE4e1S4IF
r => X(X(X(r)))
if red now, then red in 3 steps
44
1If the red light is on at the beginning, it is also on in three time steps.
R_sb9MnVHZ3rBbcuR
r => X(X(X(r)))
if red now, then red in 3 steps
45
1If red is on at the beginning, it is also on in three time steps.
R_3GB8QvZ8WJ7imO7
r => X(X(X(r)))
if red now, then red in 3 steps
46
1If the first state is red, then the 4th state is red.
R_3KSFgYqXtEshyWa
r => X(X(X(r)))
if red now, then red in 3 steps
47
1In two steps it is the case that eventually the Red light will be on next.
R_10pfB7n4SRWFEiA
X(X(F(X(r))))
r in 3 or more steps
48
1The red light will eventually from the fourth stage.
R_1d6RMi3t0gkcM5M
X(X(F(X(r))))
r in 3 or more steps
49
1The red light is on at a state after the 3rd.
R_e8QLXoeqrle4xS9
X(X(F(X(r))))
r in 3 or more steps
50
1
starting from the next state of the next state, its next state is red or it is red after the next state
R_z5PVfNBIPKX5S4F
X(X(F(X(r))))
r in 3 or more steps
51
1The red light is on at some state starting from the fourth state.
R_1E0ztwRbCQeEjAn
X(X(F(X(r))))
r in 3 or more steps
52
1There exists a state where the red light is on from the fourth state of the trace on.
R_d6w1dXzacJ9KN0Z
X(X(F(X(r))))
r in 3 or more steps
53
1
Two steps after the first one, there is a point in time for which the light is red at the next step.
R_3MuS1ttcj8sdoRv
X(X(F(X(r))))
r in 3 or more steps
54
1It holds that starting in three timesteps, the red light will eventually be on.
R_1hEGqpvvTZegzQ7
X(X(F(X(r))))
r in 3 or more steps
55
1After at least 3 transitions, the red light is on.
R_3gO868rLSy0R2Ep
X(X(F(X(r))))
r in 3 or more steps
56
1
In two timesteps from the initial state (t=1), in other words we are at timestep t=3, there will exist a timestep t' more than or equal to 3 such that the Red light will be on at timestep t'+1.
R_332amsBlozpU5n8
X(X(F(X(r))))
r in 3 or more steps
57
1From the third moment, the Red light will eventually be on.
From the second state we eventually get in the next state Red and this is equivalent to the proposed solution
R_QcBt5TnllV1Ynq9
X(X(F(X(r))))
r in 3 or more steps
58
1
Starting from the second state, there will be a state such that its successor has the Red light on.
R_2t50ks4XN0wxwLl
X(X(F(X(r))))
r in 3 or more steps
59
1A state 3 transitions or more from the initial state must satisfy red.
R_0Vg19zBXa3hIGzf
X(X(F(X(r))))
r in 3 or more steps
60
1From the third state onwards, the Red light will be eventually be on.
R_33wl4yU4sEzJKu4
X(X(F(X(r))))
r in 3 or more steps
61
1Some state strictly after the second has red
R_3iWIMn2gWx1V4Lp
X(X(F(X(r))))
r in 3 or more steps
62
1There is a time step t>3 finite when Red is on
In two steps time, Red will eventually be on in the next step, and hence it cannot happen in the 3rd step but will happen after some final time after that
R_W7oqYTh5ZL9V9vP
X(X(F(X(r))))
r in 3 or more steps
63
1In 2 states' time, the red light will eventually be on after at least one state.
R_3e8mdHJtEsNstwp
X(X(F(X(r))))
r in 3 or more steps
64
1There is a Red state from the 4th state or beyond.
R_1i3xVhhSYFfFwDR
X(X(F(X(r))))
r in 3 or more steps
65
1
From the next next state, there will be a state in the future where the state has Red on.
R_6F2ykLOE4e1S4IF
X(X(F(X(r))))
r in 3 or more steps
66
1After three time steps, the right light will eventually be on. \\
R_sb9MnVHZ3rBbcuR
X(X(F(X(r))))
r in 3 or more steps
67
1After three time steps, red will eventually be on.
R_3GB8QvZ8WJ7imO7
X(X(F(X(r))))
r in 3 or more steps
68
1There will be a red state after the 3rd state in the end (excluding the 3rd state).
R_3KSFgYqXtEshyWa
X(X(F(X(r))))
r in 3 or more steps
69
1If the Red light is on eventually, then the Blue light is on always.
R_10pfB7n4SRWFEiA
F(r) => G(b)
if any r, then all b
70
1If the right light will eventually on, the blue light is always on.
R_1d6RMi3t0gkcM5M
F(r) => G(b)
if any r, then all b
71
1If at some point the red light is on, the blue light is always on.
R_e8QLXoeqrle4xS9
F(r) => G(b)
if any r, then all b
72
1If the red light is on eventually, then the Blue light is on in any state
R_z5PVfNBIPKX5S4F
F(r) => G(b)
if any r, then all b
73
1If at some point the red light is on, then the blue light is always on.
R_1E0ztwRbCQeEjAn
F(r) => G(b)
if any r, then all b
74
1
If there is a state where the red light is on, then blue light is on for every state of the trace.
R_d6w1dXzacJ9KN0Z
F(r) => G(b)
if any r, then all b
75
1If the red light is eventually on, then the blue light is always on.
R_3MuS1ttcj8sdoRv
F(r) => G(b)
if any r, then all b
76
1
If there exists a point in time where the red light is on, the blue light must be on anytime.
R_1hEGqpvvTZegzQ7
F(r) => G(b)
if any r, then all b
77
1If the blue light is ever off, then red light is off in every state.
Contrapositive
R_3gO868rLSy0R2Ep
F(r) => G(b)
if any r, then all b
78
11
If it happens that eventually a Red light is turned on, then at that same timestep the Blue light will turn on and will stay turned on forever.
R_332amsBlozpU5n8
F(r) => G(b)
if any r, then all b
79
1Red light is never on or Blue light is always on
This is equivalent to !F(Red) or G(Blue)
R_QcBt5TnllV1Ynq9
F(r) => G(b)
if any r, then all b
80
1If there is a state that has the Red light on, then the Blue light is on in every state.
R_2t50ks4XN0wxwLl
F(r) => G(b)
if any r, then all b
81
1If red ever holds, then all states must satisfy blue.
R_0Vg19zBXa3hIGzf
F(r) => G(b)
if any r, then all b
82
1If the Red light will eventually be on, then the Blue light will always be on.
R_33wl4yU4sEzJKu4
F(r) => G(b)
if any r, then all b
83
1
If the red light is ever on, then the blue light is always on. \\ Similarly, if the blue light is ever off, the red light is never on
R_3iWIMn2gWx1V4Lp
F(r) => G(b)
if any r, then all b
84
1If Red ever turns on, then Blue will always be on
R_W7oqYTh5ZL9V9vP
F(r) => G(b)
if any r, then all b
85
1If the red light will be on at some point in the future, the blue light must always be on.
R_3e8mdHJtEsNstwp
F(r) => G(b)
if any r, then all b
86
1Once there is a Red state, the states from that state on will have Blue.
R_1i3xVhhSYFfFwDR
F(r) => G(b)
if any r, then all b
87
1
From the first state, if there is eventually a state where Red is on, then Blue will always be on (in all states).
R_6F2ykLOE4e1S4IF
F(r) => G(b)
if any r, then all b
88
1If the red is on at some point, the blue is on all along the trace.
R_sb9MnVHZ3rBbcuR
F(r) => G(b)
if any r, then all b
89
1If red is on at some point, then blue is on all along the trace.
R_3GB8QvZ8WJ7imO7
F(r) => G(b)
if any r, then all b
90
1If there eventually has a red state, then all the states are blue.
R_3KSFgYqXtEshyWa
F(r) => G(b)
if any r, then all b
91
1The red light is on forever and the blue light is on eventually.
R_10pfB7n4SRWFEiA
(r U b) & G(r)
always r and eventually b
92
1The red light is always on and the blue light is on at some point.
R_1d6RMi3t0gkcM5M
(r U b) & G(r)
always r and eventually b
93
1Red is always on, and the blue light is on at least one state.
R_e8QLXoeqrle4xS9
(r U b) & G(r)
always r and eventually b
94
1
Red light is always on and blue light is on in some state and any state before that state is red
R_z5PVfNBIPKX5S4F
(r U b) & G(r)
always r and eventually b
95
1The red light is on all the time, and at some state the blue light must be on.
R_1E0ztwRbCQeEjAn
(r U b) & G(r)
always r and eventually b
96
1Red light is on at every state, and there exists a state where blue light is on
R_d6w1dXzacJ9KN0Z
(r U b) & G(r)
always r and eventually b
97
1The red light is always on, and the blue one eventually is on.
R_3MuS1ttcj8sdoRv
(r U b) & G(r)
always r and eventually b
98
1The red light is on anytime and the blue light is on for at least one point in time.
R_1hEGqpvvTZegzQ7
(r U b) & G(r)
always r and eventually b
99
1The red light is always on, and the blue light is eventually on.
Equiv to G(Red) && F(Blue)
R_3gO868rLSy0R2Ep
(r U b) & G(r)
always r and eventually b
100
1
The Red light is always turned on for all timesteps, and at one point the Blue light will be turned on as well.
R_332amsBlozpU5n8
(r U b) & G(r)
always r and eventually b
101
1The Red light is always on and at some point the Blue light will be on
R_QcBt5TnllV1Ynq9
(r U b) & G(r)
always r and eventually b
102
1
The Red light is on in each state that precedes the state where the Blue light is on for the first time, and the Red light is on in every state.
R_2t50ks4XN0wxwLl
(r U b) & G(r)
always r and eventually b
103
1Red must always hold, and eventually blue must also hold.
R_0Vg19zBXa3hIGzf
(r U b) & G(r)
always r and eventually b
104
1The Red light is on until the Blue light is on, and the Red light is always on.
R_33wl4yU4sEzJKu4
(r U b) & G(r)
always r and eventually b
105
1All states have red, and some state has blue
R_3iWIMn2gWx1V4Lp
(r U b) & G(r)
always r and eventually b
106
1Red always is on, and Blue is on after a finite time
The second term forces Red to always be on, and the strength of the Until requires Blue to hold after some finite time.
R_W7oqYTh5ZL9V9vP
(r U b) & G(r)
always r and eventually b
107
1The red light is always on.
R_3e8mdHJtEsNstwp
(r U b) & G(r)
always r and eventually b
108
1Every state will be Red and there will be a state that's Red and Blue simutaneously.
R_1i3xVhhSYFfFwDR
(r U b) & G(r)
always r and eventually b
109
1
Red is on in all states until some state where Blue is on, and Red is always on. \\ --> Red is always on and Blue is on in some state.
R_6F2ykLOE4e1S4IF
(r U b) & G(r)
always r and eventually b
110
1Red is always on and at some point blue is on.
R_sb9MnVHZ3rBbcuR
(r U b) & G(r)
always r and eventually b
111
1Red is always on and blue is on at some point.
R_3GB8QvZ8WJ7imO7
(r U b) & G(r)
always r and eventually b
112
1All the states are red, and there will have a blue state.
R_3KSFgYqXtEshyWa
(r U b) & G(r)
always r and eventually b
113
1
It is always the case that if the Red light is on then at the next step the Red light is off and in two steps the Red light is on again. In other words, (applying reasoning) if the Red light is on now then it will alternate being on and off forever.
R_10pfB7n4SRWFEiA
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
114
1The red light is on and off alternarely.
R_1d6RMi3t0gkcM5M
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
115
1
If red is on at a state, then in the next state it is off, and in the state after that, it is on. I.e., at some point there is a red on state, and then it goes red on, red off, red on, red off, etc.
R_e8QLXoeqrle4xS9
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
116
1
Whenever Red light is on, the next state is not red and the next state of the next state is red
R_z5PVfNBIPKX5S4F
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
117
1
Whenever the red light is on, it is off at the next state, and it is on at the state after the next state.
R_1E0ztwRbCQeEjAn
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
118
1Whenever the red light is on, the next state it's off and the next state it's on again.
R_d6w1dXzacJ9KN0Z
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
119
1
Whenever the red light is on, at the next step it should not be, but it should be on at the one after.
R_3MuS1ttcj8sdoRv
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
120
1
Whenever the red light is on, it is off in the next time step and on again in two timesteps.
R_1hEGqpvvTZegzQ7
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
121
1
If the red light is ever turned on, then it will infinitely toggle off/on in every following transition
Before the red light first turns on there are no limitations, once the light turns on it forces the light to be on after 2 transitions, which forces it to be on after 2 more transitions etc...
R_3gO868rLSy0R2Ep
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
122
1
It is always the case that if the Red light is turned on, then the Red light will be turned off in the next time step, and the Red light will be turned back on in the following timestep.
R_332amsBlozpU5n8
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
123
1Once the Red light is on, it alternates between off and on
If the Red light is on immediately after it won't and after that it will be
R_QcBt5TnllV1Ynq9
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
124
1
Whenever a state has the Red light on, it must be the case that the next state has the Red light off and the second next state has the Red light on.
R_2t50ks4XN0wxwLl
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
125
1
If we ever have a state satisfying red. Then from there onwards, we must have the next state being not red, then the next state being red etc. forever.
R_0Vg19zBXa3hIGzf
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
126
1
It is always true that if the Red light is on in one state, then it will be off in next state, then turn back on in the 2nd next state.
R_33wl4yU4sEzJKu4
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
127
1Either red is never on, or red alternates on and off state to state
R_3iWIMn2gWx1V4Lp
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
128
1Once Red is turned on, it will alternate between being on and off forever
If Red ever holds, then the next step will have Red off and the step after will have Red on, which will repeat the process.
R_W7oqYTh5ZL9V9vP
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
129
1The red light will always alternate between being on and off.
R_3e8mdHJtEsNstwp
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
130
1
If a state is Red then its next state is not Red and the state after the next state is Red.
R_1i3xVhhSYFfFwDR
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
131
1
Whenever Red is on, Red is not on in the next state and Red is on in the state after that.
R_6F2ykLOE4e1S4IF
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
132
1If the red light ever gets on, from then, the red light alternates between on and off.
R_sb9MnVHZ3rBbcuR
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
133
1If red ever gets on, then it alternates between on and off.
R_3GB8QvZ8WJ7imO7
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next
134
1
For all the states, if this state is red, then the next state is not red and the next state of the next state is red.
R_3KSFgYqXtEshyWa
G(r => X(!r) & X(X(r)))
whenever r, off next on next-next