ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
1
Totals ==>1171000000000016000100126
2
Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRVAnswer
Explanation
IDFormulaExpected
3
1True in LTLf
We eventually get a is equivalent to havin true until a is found
R_2PtHv3KdsGqQLoB
F(a) == true U a
TRUE
4
1True in LTLf
R_3079Udc3BOxNquc
F(a) == true U a
TRUE
5
1True in LTLf
R_2U5AqOKS1RFOYwc
F(a) == true U a
TRUE
6
1True in LTLf
R_2yscxjylS1SXLu5
F(a) == true U a
TRUE
7
1True in LTLf
R_oYU5yKiaqPgXzMd
F(a) == true U a
TRUE
8
1True in LTLf
R_5vtXYvxyZKlCUx3
F(a) == true U a
TRUE
9
1True in LTLf
R_2aenS9mNn6Dk4Cz
F(a) == true U a
TRUE
10
1True in LTLf
R_3s0nFuWCbCk5CjP
F(a) == true U a
TRUE
11
1True in LTLf
Yes, that equation is valid in LTLf when the variable a is replaced by any LTLf term.
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F(a) == true U a
TRUE
12
1True in LTLf
The Finally operator requires a to hold in some state, which the Until operator also enforces.
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F(a) == true U a
TRUE
13
1True in LTLf
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F(a) == true U a
TRUE
14
1True in LTLf
This is simply the definition of the F operator from the basic syntax of LTLf. a holds in some state, all states up to that point are not relevant
R_2zeuLOxEmJflmcu
F(a) == true U a
TRUE
15
1True in LTLf
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F(a) == true U a
TRUE
16
1True in LTLf
R_1FRNUwvnkvroHRF
F(a) == true U a
TRUE
17
1True in LTLf
R_3IXnVYuOI9QeI10
F(a) == true U a
TRUE
18
1True in LTLf
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F(a) == true U a
TRUE
19
1True in LTLf
R_2QS5lE5zYMCL1s7
F(a) == true U a
TRUE
20
1True in LTLf
R_31j7Jp3yQbqaZqk
F(a) == true U a
TRUE
21
1False in LTLf
For any formula in the final state !X(a) is vacuously true and X(!a) is vacuously false
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!X(a) == X(!a)FALSE
22
1False in LTLf
R_3079Udc3BOxNquc
!X(a) == X(!a)FALSE
23
1False in LTLf
R_2U5AqOKS1RFOYwc
!X(a) == X(!a)FALSE
24
1False in LTLf
R_2yscxjylS1SXLu5
!X(a) == X(!a)FALSE
25
1False in LTLf
Consider !X(Red) == X(!Red), with a single state transition system. LHS is satisfied as there is no next state, RHS is not
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!X(a) == X(!a)FALSE
26
1False in LTLf
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!X(a) == X(!a)FALSE
27
1False in LTLf
R_2aenS9mNn6Dk4Cz
!X(a) == X(!a)FALSE
28
1False in LTLf
Take the finite trace of a single state. Then since there is no next state, Xa isn't satisfied and hence this trace satisfies !Xa. \\ \\ But then since there is no next state, X!a isn't true. \\ \\ Hence, these two aren't equivalent.
R_3s0nFuWCbCk5CjP
!X(a) == X(!a)FALSE
29
1False in LTLf
No, that equation is not valid, for example !X(Red) is not equivalent to X(!Red), as the first equation can allow for there to be no next state and the equation will automatically hold, but the second equation requires the existence of a next state, and that the next state should not have the Red light
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!X(a) == X(!a)FALSE
30
1False in LTLf
Let a = True, and consider the trace <p>. Then !X(True) holds for this trace as there is no next step, but X(!True) does not as there is no next step.
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!X(a) == X(!a)FALSE
31
1False in LTLf
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!X(a) == X(!a)FALSE
32
1False in LTLf
the first is always satisfied in the last state, the second is never satisfied in the last state
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!X(a) == X(!a)FALSE
33
1False in LTLf
R_1rjpSmO14SmyPCQ
!X(a) == X(!a)FALSE
34
weak X1True in LTLf
R_1FRNUwvnkvroHRF
!X(a) == X(!a)FALSE
35
1False in LTLf
R_3IXnVYuOI9QeI10
!X(a) == X(!a)FALSE
36
1False in LTLf
R_3KGxBOYYe7aGaXI
!X(a) == X(!a)FALSE
37
weak X1True in LTLf
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!X(a) == X(!a)FALSE
38
1False in LTLf
R_31j7Jp3yQbqaZqk
!X(a) == X(!a)FALSE
39
weak X1True in LTLf
The first says that starting from state 2 we always have a whereas the second says that we always have in the next state a. These are always equivalent for traces of size >= 2. For size 1 both of them are vacuously false so the relation holds
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X(G(a)) == G(X(a))
FALSE
40
1False in LTLf
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X(G(a)) == G(X(a))
FALSE
41
1False in LTLf
R_2U5AqOKS1RFOYwc
X(G(a)) == G(X(a))
FALSE
42
1False in LTLf
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X(G(a)) == G(X(a))
FALSE
43
1False in LTLf
Let a:= Red, RHS is unsatisfiable as it requires infinite states, LHS is satisfied by the two state system with trace Red->Red
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X(G(a)) == G(X(a))
FALSE
44
1False in LTLf
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X(G(a)) == G(X(a))
FALSE
45
1False in LTLf
R_2aenS9mNn6Dk4Cz
X(G(a)) == G(X(a))
FALSE
46
1False in LTLf
[] -> [a] satisfies XGa, but doesn't satisfy GXa since from the second state there is no next state and hence state 2 doesn't satisfy Xa.
R_3s0nFuWCbCk5CjP
X(G(a)) == G(X(a))
FALSE
47
1False in LTLf
No, that equation is not valid, as X(G(a)) can be satisfied by finite traces but G(X(a)) cannot be satisfied by finite traces.
R_2uWWZejY6DyOYaH
X(G(a)) == G(X(a))
FALSE
48
1False in LTLf
G(X(True)) enforces that the trace must be infinite as the next state must always exist, whereas <p1, p2> satisfies X(G(True)).
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X(G(a)) == G(X(a))
FALSE
49
1False in LTLf
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X(G(a)) == G(X(a))
FALSE
50
1False in LTLf
the first is satisfied by any trace with at least 2 states, in which all states after the first satisfy a. The second is not satisfied by any finite trace, since all states must satisfy a formula referencing their successor strictly.
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X(G(a)) == G(X(a))
FALSE
51
1False in LTLf
Second can only be satisfied in infinite trace
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X(G(a)) == G(X(a))
FALSE
52
1False in LTLf
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X(G(a)) == G(X(a))
FALSE
53
1False in LTLf
R_3IXnVYuOI9QeI10
X(G(a)) == G(X(a))
FALSE
54
1False in LTLf
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X(G(a)) == G(X(a))
FALSE
55
(maybe) weak X
1True in LTLf
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X(G(a)) == G(X(a))
FALSE
56
1False in LTLf
R_31j7Jp3yQbqaZqk
X(G(a)) == G(X(a))
FALSE
57
1True in LTLf
This is true. The LHS says that from state 2 we have a until b, whereas the RHS says that the next state is a until the next state is b. \\ \\ LHS => RHS as considering a trace satisfying LHS, it should have a in states 2, 3,..., i and in (i+1) b. Thus X(a) is satisfied up to state i-1 and in state i X(b) is satisfied. \\ \\ RHS => LHS as considering a trace satisfying RHS, it should have the next state always a until the next state is b. Hence, from state 2 a holds until b holds.
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X(a U b) == X(a) U X(b)
TRUE
58
1True in LTLf
R_3079Udc3BOxNquc
X(a U b) == X(a) U X(b)
TRUE
59
1True in LTLf
R_2U5AqOKS1RFOYwc
X(a U b) == X(a) U X(b)
TRUE
60
1True in LTLf
R_2yscxjylS1SXLu5
X(a U b) == X(a) U X(b)
TRUE
61
1True in LTLf
Asssume LHS satisfied: b holds in state n, and a holds in states [2, n). Therefore, X(b) holds in state (n-1), and a holds in states [2,n), so X(a) holds in states [1,(n-1)), so RHS satisfied. \\ Assume RHS satisfied, argument symmetric
R_oYU5yKiaqPgXzMd
X(a U b) == X(a) U X(b)
TRUE
62
1True in LTLf
R_5vtXYvxyZKlCUx3
X(a U b) == X(a) U X(b)
TRUE
63
1True in LTLf
R_2aenS9mNn6Dk4Cz
X(a U b) == X(a) U X(b)
TRUE
64
1True in LTLf
R_3s0nFuWCbCk5CjP
X(a U b) == X(a) U X(b)
TRUE
65
1True in LTLf
Yes, that equation is valid in LTLf when the variables a and b is replaced by any LTLf term.
R_2uWWZejY6DyOYaH
X(a U b) == X(a) U X(b)
TRUE
66
1True in LTLf
They both require a to hold in the second step onwards until b holds, after which they can both terminate at any point
R_2WwptK2YPEgUAGJ
X(a U b) == X(a) U X(b)
TRUE
67
1True in LTLf
R_sLorPWGMjcKuGYx
X(a U b) == X(a) U X(b)
TRUE
68
1True in LTLf
They are both satisfied by traces in which some state after the first satisfies b, and all states after the first, and before the one satisfying b satisfy a
R_2zeuLOxEmJflmcu
X(a U b) == X(a) U X(b)
TRUE
69
1True in LTLf
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X(a U b) == X(a) U X(b)
TRUE
70
1True in LTLf
R_1FRNUwvnkvroHRF
X(a U b) == X(a) U X(b)
TRUE
71
1True in LTLf
R_3IXnVYuOI9QeI10
X(a U b) == X(a) U X(b)
TRUE
72
1True in LTLf
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X(a U b) == X(a) U X(b)
TRUE
73
1True in LTLf
R_2QS5lE5zYMCL1s7
X(a U b) == X(a) U X(b)
TRUE
74
???1False in LTLf
R_31j7Jp3yQbqaZqk
X(a U b) == X(a) U X(b)
TRUE
75
1False in LTLf
If we have a trace having only one state satisfying a, LHS is true and RHS is not true as X(G(a)) is vacuously false.
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G(a) == a & X(G(a))
FALSE
76
1False in LTLf
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G(a) == a & X(G(a))
FALSE
77
1False in LTLf
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G(a) == a & X(G(a))
FALSE
78
1False in LTLf
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G(a) == a & X(G(a))
FALSE
79
1False in LTLf
Consider the one state transition system with a := Red and state 1 satisfies Red. G(Red) is satisfied, but X(G(Red)) is not, as there is no second state.
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G(a) == a & X(G(a))
FALSE
80
1False in LTLf
R_5vtXYvxyZKlCUx3
G(a) == a & X(G(a))
FALSE
81
1False in LTLf
R_2aenS9mNn6Dk4Cz
G(a) == a & X(G(a))
FALSE
82
1False in LTLf
A finite trace of [a] will break this since Ga is satisfied by a finite trace of just a. But since there is no next state, XGa isn't satisfied, hence these aren't equivalent.
R_3s0nFuWCbCk5CjP
G(a) == a & X(G(a))
FALSE
83
1False in LTLf
No, that equation is not valid, as G(a) can hold for a finite trace of just one time step, but a & X(G(a)) needs the existence of at least two time steps where both a holds.
R_2uWWZejY6DyOYaH
G(a) == a & X(G(a))
FALSE
84
1False in LTLf
<a> satisfies G(a) but not X(G(a)) as there is no second step.
R_2WwptK2YPEgUAGJ
G(a) == a & X(G(a))
FALSE
85
1False in LTLf
R_sLorPWGMjcKuGYx
G(a) == a & X(G(a))
FALSE
86
1False in LTLf
this is the fixed point definition for G in LTL, in LTLf it would be x & Wx(G(a)) since a only needs to hold in the states that exist. The RHS of the inequality would not hold on a singleton trace that satisfies a, whereas the LHS would
R_2zeuLOxEmJflmcu
G(a) == a & X(G(a))
FALSE
87
1False in LTLf
R_1rjpSmO14SmyPCQ
G(a) == a & X(G(a))
FALSE
88
1False in LTLf
R_1FRNUwvnkvroHRF
G(a) == a & X(G(a))
FALSE
89
1False in LTLf
R_3IXnVYuOI9QeI10
G(a) == a & X(G(a))
FALSE
90
1False in LTLf
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G(a) == a & X(G(a))
FALSE
91
weak X1True in LTLf
R_2QS5lE5zYMCL1s7
G(a) == a & X(G(a))
FALSE
92
1False in LTLf
R_31j7Jp3yQbqaZqk
G(a) == a & X(G(a))
FALSE
93
1True in LTLf
This is true as finally having a is equivalent to having a now or starting from the next state we eventually have a.
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F(a) == a || X(F(a))
TRUE
94
1True in LTLf
R_3079Udc3BOxNquc
F(a) == a || X(F(a))
TRUE
95
1True in LTLf
R_2U5AqOKS1RFOYwc
F(a) == a || X(F(a))
TRUE
96
1True in LTLf
R_2yscxjylS1SXLu5
F(a) == a || X(F(a))
TRUE
97
1True in LTLf
F(a) true iff state 1 satisfies a, or some future state exists which satisfies a. i.e. a || X(F(a))
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F(a) == a || X(F(a))
TRUE
98
1True in LTLf
R_5vtXYvxyZKlCUx3
F(a) == a || X(F(a))
TRUE
99
bad prop?1False in LTLf
R_2aenS9mNn6Dk4Cz
F(a) == a || X(F(a))
TRUE
100
1True in LTLf
R_3s0nFuWCbCk5CjP
F(a) == a || X(F(a))
TRUE
101
1True in LTLf
Yes, that equation is valid in LTLf when the variable a is replaced by any LTLf term.
R_2uWWZejY6DyOYaH
F(a) == a || X(F(a))
TRUE
102
1True in LTLf
The RHS enforces that either a holds, or there is a next state after which a will hold, which is exactly F(a)
R_2WwptK2YPEgUAGJ
F(a) == a || X(F(a))
TRUE
103
1True in LTLf
R_sLorPWGMjcKuGYx
F(a) == a || X(F(a))
TRUE
104
1True in LTLf
Since a must eventually hold, if it doesn't hold in this state then there must be some later state for it to hold in. This means strong next is fine here.
R_2zeuLOxEmJflmcu
F(a) == a || X(F(a))
TRUE
105
1True in LTLf
R_1rjpSmO14SmyPCQ
F(a) == a || X(F(a))
TRUE
106
1True in LTLf
R_1FRNUwvnkvroHRF
F(a) == a || X(F(a))
TRUE
107
1True in LTLf
R_3IXnVYuOI9QeI10
F(a) == a || X(F(a))
TRUE
108
1True in LTLf
R_3KGxBOYYe7aGaXI
F(a) == a || X(F(a))
TRUE
109
1True in LTLf
R_2QS5lE5zYMCL1s7
F(a) == a || X(F(a))
TRUE
110
1True in LTLf
R_31j7Jp3yQbqaZqk
F(a) == a || X(F(a))
TRUE
111
1True in LTLf
RHS says that always we eventually get a whereas LHS says that we eventually always have a. \\ \\ RHS => LHS as a trace satisfying RHS reaches a state from which a is satisfied until the end. In any state of such a trace we have F(a) satisfied, so RHS is satisfied. \\ \\ LHS => RHS as a trace satisfying lhs has the last state satisfying a, meaning that G(a) satisfies the last state, hence RHS is satisfied. \\
R_2PtHv3KdsGqQLoB
G(F(a)) == F(G(a))
TRUE
112
1True in LTLf
R_3079Udc3BOxNquc
G(F(a)) == F(G(a))
TRUE
113
1True in LTLf
R_2U5AqOKS1RFOYwc
G(F(a)) == F(G(a))
TRUE
114
1True in LTLf
R_2yscxjylS1SXLu5
G(F(a)) == F(G(a))
TRUE
115
1True in LTLf
LHS is equivalent to saying the final state satisfies F(a) \\ RHS is equivalent to saying the final state satisfies a. \\ Final state satisfies F(a) iff it satisfies a
R_oYU5yKiaqPgXzMd
G(F(a)) == F(G(a))
TRUE
116
1True in LTLf
R_5vtXYvxyZKlCUx3
G(F(a)) == F(G(a))
TRUE
117
1True in LTLf
R_2aenS9mNn6Dk4Cz
G(F(a)) == F(G(a))
TRUE
118
wrong F starting point?
1False in LTLf
GFa says that in every state, eventually a holds. So from a certain point of time, a must hold for every \\ []->[a]->[]->[a] satisfies GFa but doesn't satisfy FGa. \\ \\ Hence, these aren't the same.
R_3s0nFuWCbCk5CjP
G(F(a)) == F(G(a))
TRUE
119
1True in LTLf
Yes, that equation is valid in LTLf when the variable a is replaced by any LTLf term, since both equations simply need that a holds at the last state.
R_2uWWZejY6DyOYaH
G(F(a)) == F(G(a))
TRUE
120
1True in LTLf
These both just enforce that the final state satisfies a.
R_2WwptK2YPEgUAGJ
G(F(a)) == F(G(a))
TRUE
121
1True in LTLf
R_sLorPWGMjcKuGYx
G(F(a)) == F(G(a))
TRUE
122
1True in LTLf
Both statements just mean that the last state satisfies a. \\ The LHS says that all states are, or are followed by a state that satisfies a. This means the last state must satisfy a since it is followed by nothing, and the last state follows everything so it satisfying a already makes every other state satisfy the formula. \\ the RHS says that eventually all states up to and including the last satisfy a. this means the last state satisfies a, and thus the suffix including only the last state is the string required to satisfy the formula
R_2zeuLOxEmJflmcu
G(F(a)) == F(G(a))
TRUE
123
1True in LTLf
R_1rjpSmO14SmyPCQ
G(F(a)) == F(G(a))
TRUE
124
1True in LTLf
R_1FRNUwvnkvroHRF
G(F(a)) == F(G(a))
TRUE
125
1True in LTLf
R_3IXnVYuOI9QeI10
G(F(a)) == F(G(a))
TRUE
126
1True in LTLf
R_3KGxBOYYe7aGaXI
G(F(a)) == F(G(a))
TRUE
127
1True in LTLf
R_2QS5lE5zYMCL1s7
G(F(a)) == F(G(a))
TRUE
128
?1False in LTLf
R_31j7Jp3yQbqaZqk
G(F(a)) == F(G(a))
TRUE