ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
1
Totals ==>95000000000007140004000120
2
Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRV
Scope
Answer
Explanation
IDFormulaExpected
3
1True in LTLf
R_3MS5InthwHWtaZT
F(a) == true U aTRUE
4
1True in LTLf
R_3G88D8gpZ7v63nz
F(a) == true U aTRUE
5
1True in LTLf
R_3Dw4BW8x0ynmmQm
F(a) == true U aTRUE
6
R_DSlVV5XN50E2Mzn
F(a) == true U aTRUE
7
R_WByy6AuVYHXV65H
F(a) == true U aTRUE
8
1True in LTLf
R_vvG6iiXnXM4vZJv
F(a) == true U aTRUE
9
1True in LTLf
R_9Td5mdNtgffkvJL
F(a) == true U aTRUE
10
1True in LTLf
R_9ukqO8kCEIR4pLb
F(a) == true U aTRUE
11
R_2saTh9ICAgM0BHV
F(a) == true U aTRUE
12
R_3J3NZWyA2hcl5by
F(a) == true U aTRUE
13
R_1mfLnqhYsR97vhv
F(a) == true U aTRUE
14
1True in LTLf
Definition of eventually
R_VHZdlnCi0wPqPGF
F(a) == true U aTRUE
15
R_0ize6N4UFi1irrb
F(a) == true U aTRUE
16
1True in LTLf
R_3kpTXi28NBbr8mO
F(a) == true U aTRUE
17
1True in LTLf
R_3Rt21EZCScgM3oC
F(a) == true U aTRUE
18
R_enENzp0guuxVGwh
F(a) == true U aTRUE
19
1True in LTLf
R_1g8tSbzTTgxijrd
F(a) == true U aTRUE
20
1True in LTLf
R_3M5V4sL4b6COyD2
F(a) == true U aTRUE
21
1True in LTLf
R_23fQaq1kM65aSP7
F(a) == true U aTRUE
22
1True in LTLf
R_1QFcce33iJrOdLO
F(a) == true U aTRUE
23
1True in LTLf
R_2EnQzWu9HQsIjn8
F(a) == true U aTRUE
24
1True in LTLf
R_R9tc8AnNRKwPdkd
F(a) == true U aTRUE
25
1True in LTLf
R_2QPzqLc2iRGU0Jd
F(a) == true U aTRUE
26
1True in LTLf
R_3EO18al85qREqgF
F(a) == true U aTRUE
27
1False in LTLf
R_3MS5InthwHWtaZT
!X(a) == X(!a)FALSE
28
1False in LTLf
R_3G88D8gpZ7v63nz
!X(a) == X(!a)FALSE
29
1False in LTLf
R_3Dw4BW8x0ynmmQm
!X(a) == X(!a)FALSE
30
R_DSlVV5XN50E2Mzn
!X(a) == X(!a)FALSE
31
1False in LTLf
R_WByy6AuVYHXV65H
!X(a) == X(!a)FALSE
32
1False in LTLf
R_vvG6iiXnXM4vZJv
!X(a) == X(!a)FALSE
33
1False in LTLf
R_9Td5mdNtgffkvJL
!X(a) == X(!a)FALSE
34
R_9ukqO8kCEIR4pLb
!X(a) == X(!a)FALSE
35
R_2saTh9ICAgM0BHV
!X(a) == X(!a)FALSE
36
1False in LTLf
R_3J3NZWyA2hcl5by
!X(a) == X(!a)FALSE
37
1False in LTLf
R_1mfLnqhYsR97vhv
!X(a) == X(!a)FALSE
38
1False in LTLf
First formula accepts trace of length 1. second one does not
R_VHZdlnCi0wPqPGF
!X(a) == X(!a)FALSE
39
1False in LTLf
R_0ize6N4UFi1irrb
!X(a) == X(!a)FALSE
40
1False in LTLf
R_3kpTXi28NBbr8mO
!X(a) == X(!a)FALSE
41
1False in LTLf
R_3Rt21EZCScgM3oC
!X(a) == X(!a)FALSE
42
1True in LTLf
R_enENzp0guuxVGwh
!X(a) == X(!a)FALSE
43
1False in LTLf
R_1g8tSbzTTgxijrd
!X(a) == X(!a)FALSE
44
1False in LTLf
R_3M5V4sL4b6COyD2
!X(a) == X(!a)FALSE
45
1False in LTLf
R_23fQaq1kM65aSP7
!X(a) == X(!a)FALSE
46
1True in LTLf
R_1QFcce33iJrOdLO
!X(a) == X(!a)FALSE
47
R_2EnQzWu9HQsIjn8
!X(a) == X(!a)FALSE
48
1True in LTLf
In the last state X(a) is false, so !X(a) is true, and X(!a) is true as well.
R_R9tc8AnNRKwPdkd
!X(a) == X(!a)FALSE
49
1True in LTLf
R_2QPzqLc2iRGU0Jd
!X(a) == X(!a)FALSE
50
R_3EO18al85qREqgF
!X(a) == X(!a)FALSE
51
1False in LTLf
R_3MS5InthwHWtaZT
X(G(a)) == G(X(a))FALSE
52
1False in LTLf
R_3G88D8gpZ7v63nz
X(G(a)) == G(X(a))FALSE
53
1False in LTLf
GXa is false in LTLf
R_3Dw4BW8x0ynmmQm
X(G(a)) == G(X(a))FALSE
54
1False in LTLf
R_DSlVV5XN50E2Mzn
X(G(a)) == G(X(a))FALSE
55
1False in LTLf
R_WByy6AuVYHXV65H
X(G(a)) == G(X(a))FALSE
56
1False in LTLf
R_vvG6iiXnXM4vZJv
X(G(a)) == G(X(a))FALSE
57
R_9Td5mdNtgffkvJL
X(G(a)) == G(X(a))FALSE
58
1False in LTLf
G(X(a)) is only true on infinite traces, while X(G(a)) is true on some finite traces.
R_9ukqO8kCEIR4pLb
X(G(a)) == G(X(a))FALSE
59
1False in LTLf
R_2saTh9ICAgM0BHV
X(G(a)) == G(X(a))FALSE
60
1False in LTLf
R_3J3NZWyA2hcl5by
X(G(a)) == G(X(a))FALSE
61
1False in LTLf
R_1mfLnqhYsR97vhv
X(G(a)) == G(X(a))FALSE
62
1False in LTLf
For a finite trace of length > 1 the last state would always fail X(a) so the formula F( G(X(a)) is not satisfied. However, the formula X(G(a)) would be since the next only applies to the first state.
R_VHZdlnCi0wPqPGF
X(G(a)) == G(X(a))FALSE
63
1False in LTLf
R_0ize6N4UFi1irrb
X(G(a)) == G(X(a))FALSE
64
R_3kpTXi28NBbr8mO
X(G(a)) == G(X(a))FALSE
65
1False in LTLf
R_3Rt21EZCScgM3oC
X(G(a)) == G(X(a))FALSE
66
R_enENzp0guuxVGwh
X(G(a)) == G(X(a))FALSE
67
R_1g8tSbzTTgxijrd
X(G(a)) == G(X(a))FALSE
68
1False in LTLf
R_3M5V4sL4b6COyD2
X(G(a)) == G(X(a))FALSE
69
R_23fQaq1kM65aSP7
X(G(a)) == G(X(a))FALSE
70
R_1QFcce33iJrOdLO
X(G(a)) == G(X(a))FALSE
71
R_2EnQzWu9HQsIjn8
X(G(a)) == G(X(a))FALSE
72
R_R9tc8AnNRKwPdkd
X(G(a)) == G(X(a))FALSE
73
R_2QPzqLc2iRGU0Jd
X(G(a)) == G(X(a))FALSE
74
1False in LTLf
R_3EO18al85qREqgF
X(G(a)) == G(X(a))FALSE
75
R_3MS5InthwHWtaZT
X(a U b) == X(a) U X(b)
TRUE
76
R_3G88D8gpZ7v63nz
X(a U b) == X(a) U X(b)
TRUE
77
R_3Dw4BW8x0ynmmQm
X(a U b) == X(a) U X(b)
TRUE
78
1False in LTLf
R_DSlVV5XN50E2Mzn
X(a U b) == X(a) U X(b)
TRUE
79
1True in LTLf
R_WByy6AuVYHXV65H
X(a U b) == X(a) U X(b)
TRUE
80
R_vvG6iiXnXM4vZJv
X(a U b) == X(a) U X(b)
TRUE
81
1True in LTLf
R_9Td5mdNtgffkvJL
X(a U b) == X(a) U X(b)
TRUE
82
1True in LTLf
R_9ukqO8kCEIR4pLb
X(a U b) == X(a) U X(b)
TRUE
83
1False in LTLf
R_2saTh9ICAgM0BHV
X(a U b) == X(a) U X(b)
TRUE
84
R_3J3NZWyA2hcl5by
X(a U b) == X(a) U X(b)
TRUE
85
R_1mfLnqhYsR97vhv
X(a U b) == X(a) U X(b)
TRUE
86
R_VHZdlnCi0wPqPGF
X(a U b) == X(a) U X(b)
TRUE
87
1True in LTLf
R_0ize6N4UFi1irrb
X(a U b) == X(a) U X(b)
TRUE
88
1True in LTLf
R_3kpTXi28NBbr8mO
X(a U b) == X(a) U X(b)
TRUE
89
1True in LTLf
R_3Rt21EZCScgM3oC
X(a U b) == X(a) U X(b)
TRUE
90
1False in LTLf
R_enENzp0guuxVGwh
X(a U b) == X(a) U X(b)
TRUE
91
R_1g8tSbzTTgxijrd
X(a U b) == X(a) U X(b)
TRUE
92
R_3M5V4sL4b6COyD2
X(a U b) == X(a) U X(b)
TRUE
93
1True in LTLf
R_23fQaq1kM65aSP7
X(a U b) == X(a) U X(b)
TRUE
94
R_1QFcce33iJrOdLO
X(a U b) == X(a) U X(b)
TRUE
95
1False in LTLf
R_2EnQzWu9HQsIjn8
X(a U b) == X(a) U X(b)
TRUE
96
1False in LTLf
I'm quite not able to explain why but I have the impression that the second formula may have differences in the behaviour in the last state.
R_R9tc8AnNRKwPdkd
X(a U b) == X(a) U X(b)
TRUE
97
1False in LTLf
R_2QPzqLc2iRGU0Jd
X(a U b) == X(a) U X(b)
TRUE
98
1False in LTLf
R_3EO18al85qREqgF
X(a U b) == X(a) U X(b)
TRUE
99
R_3MS5InthwHWtaZT
G(a) == a & X(G(a))FALSE
100
1False in LTLf
R_3G88D8gpZ7v63nz
G(a) == a & X(G(a))FALSE
101
R_3Dw4BW8x0ynmmQm
G(a) == a & X(G(a))FALSE
102
1False in LTLf
R_DSlVV5XN50E2Mzn
G(a) == a & X(G(a))FALSE
103
R_WByy6AuVYHXV65H
G(a) == a & X(G(a))FALSE
104
R_vvG6iiXnXM4vZJv
G(a) == a & X(G(a))FALSE
105
1False in LTLf
R_9Td5mdNtgffkvJL
G(a) == a & X(G(a))FALSE
106
1False in LTLf
R_9ukqO8kCEIR4pLb
G(a) == a & X(G(a))FALSE
107
1False in LTLf
R_2saTh9ICAgM0BHV
G(a) == a & X(G(a))FALSE
108
1True in LTLf
R_3J3NZWyA2hcl5by
G(a) == a & X(G(a))FALSE
109
1False in LTLf
R_1mfLnqhYsR97vhv
G(a) == a & X(G(a))FALSE
110
1False in LTLf
G(a) also accepts trace of length 1 with an a in the state. However, it would fail with the next in the other formula
R_VHZdlnCi0wPqPGF
G(a) == a & X(G(a))FALSE
111
1False in LTLf
R_0ize6N4UFi1irrb
G(a) == a & X(G(a))FALSE
112
1False in LTLf
R_3kpTXi28NBbr8mO
G(a) == a & X(G(a))FALSE
113
R_3Rt21EZCScgM3oC
G(a) == a & X(G(a))FALSE
114
1True in LTLf
R_enENzp0guuxVGwh
G(a) == a & X(G(a))FALSE
115
1False in LTLf
R_1g8tSbzTTgxijrd
G(a) == a & X(G(a))FALSE
116
1False in LTLf
R_3M5V4sL4b6COyD2
G(a) == a & X(G(a))FALSE
117
R_23fQaq1kM65aSP7
G(a) == a & X(G(a))FALSE
118
1True in LTLf
R_1QFcce33iJrOdLO
G(a) == a & X(G(a))FALSE
119
1False in LTLf
R_2EnQzWu9HQsIjn8
G(a) == a & X(G(a))FALSE
120
1False in LTLf
I'm not sure. My guess is Xw(G(a)) is needed because X(G(a)) should be false in the last state.
R_R9tc8AnNRKwPdkd
G(a) == a & X(G(a))FALSE
121
R_2QPzqLc2iRGU0Jd
G(a) == a & X(G(a))FALSE
122
R_3EO18al85qREqgF
G(a) == a & X(G(a))FALSE
123
1True in LTLf
R_3MS5InthwHWtaZT
F(a) == a || X(F(a))TRUE
124
R_3G88D8gpZ7v63nz
F(a) == a || X(F(a))TRUE
125
1True in LTLf
R_3Dw4BW8x0ynmmQm
F(a) == a || X(F(a))TRUE
126
1True in LTLf
R_DSlVV5XN50E2Mzn
F(a) == a || X(F(a))TRUE
127
1True in LTLf
R_WByy6AuVYHXV65H
F(a) == a || X(F(a))TRUE
128
1True in LTLf
R_vvG6iiXnXM4vZJv
F(a) == a || X(F(a))TRUE
129
1True in LTLf
R_9Td5mdNtgffkvJL
F(a) == a || X(F(a))TRUE
130
1False in LTLf
R_9ukqO8kCEIR4pLb
F(a) == a || X(F(a))TRUE
131
1True in LTLf
R_2saTh9ICAgM0BHV
F(a) == a || X(F(a))TRUE
132
1True in LTLf
R_3J3NZWyA2hcl5by
F(a) == a || X(F(a))TRUE
133
1True in LTLf
R_1mfLnqhYsR97vhv
F(a) == a || X(F(a))TRUE
134
1True in LTLf
Unfolding of the eventually. Next is not an issue since if a is not true in the last state(and was neve true before) then both formulas are not satisfied.
R_VHZdlnCi0wPqPGF
F(a) == a || X(F(a))TRUE
135
R_0ize6N4UFi1irrb
F(a) == a || X(F(a))TRUE
136
R_3kpTXi28NBbr8mO
F(a) == a || X(F(a))TRUE
137
1True in LTLf
R_3Rt21EZCScgM3oC
F(a) == a || X(F(a))TRUE
138
1True in LTLf
R_enENzp0guuxVGwh
F(a) == a || X(F(a))TRUE
139
1True in LTLf
R_1g8tSbzTTgxijrd
F(a) == a || X(F(a))TRUE
140
1True in LTLf
R_3M5V4sL4b6COyD2
F(a) == a || X(F(a))TRUE
141
1False in LTLf
R_23fQaq1kM65aSP7
F(a) == a || X(F(a))TRUE
142
1False in LTLf
R_1QFcce33iJrOdLO
F(a) == a || X(F(a))TRUE
143
1True in LTLf
R_2EnQzWu9HQsIjn8
F(a) == a || X(F(a))TRUE
144
R_R9tc8AnNRKwPdkd
F(a) == a || X(F(a))TRUE
145
1False in LTLf
R_2QPzqLc2iRGU0Jd
F(a) == a || X(F(a))TRUE
146
1True in LTLf
R_3EO18al85qREqgF
F(a) == a || X(F(a))TRUE
147
how to explain?! LTL intuition says FALSE
1False in LTLf
R_3MS5InthwHWtaZT
G(F(a)) == F(G(a))TRUE
148
1True in LTLf
equivalent to last a
R_3G88D8gpZ7v63nz
G(F(a)) == F(G(a))TRUE
149
1True in LTLf
R_3Dw4BW8x0ynmmQm
G(F(a)) == F(G(a))TRUE
150
1False in LTLf
R_DSlVV5XN50E2Mzn
G(F(a)) == F(G(a))TRUE
151
1True in LTLf
R_WByy6AuVYHXV65H
G(F(a)) == F(G(a))TRUE
152
1True in LTLf
R_vvG6iiXnXM4vZJv
G(F(a)) == F(G(a))TRUE
153
R_9Td5mdNtgffkvJL
G(F(a)) == F(G(a))TRUE
154
R_9ukqO8kCEIR4pLb
G(F(a)) == F(G(a))TRUE
155
1False in LTLf
R_2saTh9ICAgM0BHV
G(F(a)) == F(G(a))TRUE
156
1True in LTLf
R_3J3NZWyA2hcl5by
G(F(a)) == F(G(a))TRUE
157
1True in LTLf
R_1mfLnqhYsR97vhv
G(F(a)) == F(G(a))TRUE
158
R_VHZdlnCi0wPqPGF
G(F(a)) == F(G(a))TRUE
159
1True in LTLf
R_0ize6N4UFi1irrb
G(F(a)) == F(G(a))TRUE
160
1False in LTLf
R_3kpTXi28NBbr8mO
G(F(a)) == F(G(a))TRUE
161
R_3Rt21EZCScgM3oC
G(F(a)) == F(G(a))TRUE
162
1True in LTLf
R_enENzp0guuxVGwh
G(F(a)) == F(G(a))TRUE
163
1True in LTLf
R_1g8tSbzTTgxijrd
G(F(a)) == F(G(a))TRUE
164
R_3M5V4sL4b6COyD2
G(F(a)) == F(G(a))TRUE
165
1False in LTLf
R_23fQaq1kM65aSP7
G(F(a)) == F(G(a))TRUE
166
1False in LTLf
R_1QFcce33iJrOdLO
G(F(a)) == F(G(a))TRUE
167
1False in LTLf
R_2EnQzWu9HQsIjn8
G(F(a)) == F(G(a))TRUE
168
1True in LTLf
Both formulas force a to be true in at least the last step.
R_R9tc8AnNRKwPdkd
G(F(a)) == F(G(a))TRUE
169
1True in LTLf
R_2QPzqLc2iRGU0Jd
G(F(a)) == F(G(a))TRUE
170
1True in LTLf
R_3EO18al85qREqgF
G(F(a)) == F(G(a))TRUE