ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
1
Totals ==>1421302000003481100100020180
2
Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRV
Scope
AnswerExplanationIDFormulaExpected
3
1G(Red => X (!Green))
R_2PtHv3KdsGqQLoB
whenever r, cannot be next with g
G(r => !X(g))
4
1G(Red => !X(Green))
R_3079Udc3BOxNquc
whenever r, cannot be next with g
G(r => !X(g))
5
1G(red=> X(!green))
R_2U5AqOKS1RFOYwc
whenever r, cannot be next with g
G(r => !X(g))
6
1G(Red => X(!Green))
R_2yscxjylS1SXLu5
whenever r, cannot be next with g
G(r => !X(g))
7
1G(Red => !X(Green))
R_oYU5yKiaqPgXzMd
whenever r, cannot be next with g
G(r => !X(g))
8
1G(Red => !X(Green))
R_5vtXYvxyZKlCUx3
whenever r, cannot be next with g
G(r => !X(g))
9
11
G(Red => X(G(!Green))) \\ \\ (Interpreting "there cannot be a next state in which Green is on", as "Green is never on in the future".)
R_2aenS9mNn6Dk4Cz
whenever r, cannot be next with g
G(r => !X(g))
10
1G (Red => !XGreen)
R_3s0nFuWCbCk5CjP
whenever r, cannot be next with g
G(r => !X(g))
11
1G(Red => !X(Green))
R_2uWWZejY6DyOYaH
whenever r, cannot be next with g
G(r => !X(g))
12
1G(Red => X(!Green))
R_2WwptK2YPEgUAGJ
whenever r, cannot be next with g
G(r => !X(g))
13
1G(Red => !X(Green))
R_sLorPWGMjcKuGYx
whenever r, cannot be next with g
G(r => !X(g))
14
1G(Red => !X(Green))
R_2zeuLOxEmJflmcu
whenever r, cannot be next with g
G(r => !X(g))
15
1G(R => !X(G))
R_1rjpSmO14SmyPCQ
whenever r, cannot be next with g
G(r => !X(g))
16
1G(Red =>!X(Green))
R_1FRNUwvnkvroHRF
whenever r, cannot be next with g
G(r => !X(g))
17
1G(red => X(!green))
R_3IXnVYuOI9QeI10
whenever r, cannot be next with g
G(r => !X(g))
18
1G(Red => !(X(Green)))
R_3KGxBOYYe7aGaXI
whenever r, cannot be next with g
G(r => !X(g))
19
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
whenever r, cannot be next with g
G(r => !X(g))
20
1G(Red => !X(Green))
R_31j7Jp3yQbqaZqk
whenever r, cannot be next with g
G(r => !X(g))
21
1F(Blue & X(F(Blue)))
R_2PtHv3KdsGqQLoB
There are at least two states in which the Blue light is on
F(b & X(F(b)))
22
1F(Blue & X(F(Blue)))
R_3079Udc3BOxNquc
There are at least two states in which the Blue light is on
F(b & X(F(b)))
23
1F(blue && X(F blue))
R_2U5AqOKS1RFOYwc
There are at least two states in which the Blue light is on
F(b & X(F(b)))
24
1F(Blue & X(F(Blue)))
R_2yscxjylS1SXLu5
There are at least two states in which the Blue light is on
F(b & X(F(b)))
25
1F(Blue & X(F(Blue))
R_oYU5yKiaqPgXzMd
There are at least two states in which the Blue light is on
F(b & X(F(b)))
26
1F(Blue & X(F(Blue)))
R_5vtXYvxyZKlCUx3
There are at least two states in which the Blue light is on
F(b & X(F(b)))
27
1F(Blue && X(F(Blue)))
R_2aenS9mNn6Dk4Cz
There are at least two states in which the Blue light is on
F(b & X(F(b)))
28
1F (Blue && XF Blue)
R_3s0nFuWCbCk5CjP
There are at least two states in which the Blue light is on
F(b & X(F(b)))
29
1F(Blue & X(F(Blue)))
R_2uWWZejY6DyOYaH
There are at least two states in which the Blue light is on
F(b & X(F(b)))
30
1F(Blue & X(F(Blue)))
R_2WwptK2YPEgUAGJ
There are at least two states in which the Blue light is on
F(b & X(F(b)))
31
1F(Blue & X(F(Blue)))
R_sLorPWGMjcKuGYx
There are at least two states in which the Blue light is on
F(b & X(F(b)))
32
1F(Blue & X(F(Blue))
R_2zeuLOxEmJflmcu
There are at least two states in which the Blue light is on
F(b & X(F(b)))
33
1F(B & X(F(B)))
R_1rjpSmO14SmyPCQ
There are at least two states in which the Blue light is on
F(b & X(F(b)))
34
1F(Blue & X(F(Blue)))
R_1FRNUwvnkvroHRF
There are at least two states in which the Blue light is on
F(b & X(F(b)))
35
1F(blue && X(F blue))
R_3IXnVYuOI9QeI10
There are at least two states in which the Blue light is on
F(b & X(F(b)))
36
1F(Blue & F(Blue))
R_3KGxBOYYe7aGaXI
There are at least two states in which the Blue light is on
F(b & X(F(b)))
37
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
There are at least two states in which the Blue light is on
F(b & X(F(b)))
38
1F(Blue && X(Blue))
R_31j7Jp3yQbqaZqk
There are at least two states in which the Blue light is on
F(b & X(F(b)))
39
1inexpressible
R_2PtHv3KdsGqQLoB
Green is on in the final state.
inexpressible
40
1inexpressible
R_3079Udc3BOxNquc
Green is on in the final state.
inexpressible
41
1Impressible.
R_2U5AqOKS1RFOYwc
Green is on in the final state.
inexpressible
42
1
Inexpressible - No final state. Weak way to enforce could be F(G(Green)).
R_2yscxjylS1SXLu5
Green is on in the final state.
inexpressible
43
1Inexpressible, there is no final state
R_oYU5yKiaqPgXzMd
Green is on in the final state.
inexpressible
44
1Unexpresable, because there is no final state.
R_5vtXYvxyZKlCUx3
Green is on in the final state.
inexpressible
45
1Inexpressible (no final state)
R_2aenS9mNn6Dk4Cz
Green is on in the final state.
inexpressible
46
1Inexpressible.
R_3s0nFuWCbCk5CjP
Green is on in the final state.
inexpressible
47
1Inexpressible.
R_2uWWZejY6DyOYaH
Green is on in the final state.
inexpressible
48
1inexpressible
R_2WwptK2YPEgUAGJ
Green is on in the final state.
inexpressible
49
1inexpressible
R_sLorPWGMjcKuGYx
Green is on in the final state.
inexpressible
50
1
Inexpressible. No notion of "final state". Perhaps mostly closely approximated by F(G(Green))
R_2zeuLOxEmJflmcu
Green is on in the final state.
inexpressible
51
assume G = Green
1F(G(G))
R_1rjpSmO14SmyPCQ
Green is on in the final state.
inexpressible
52
1NO?
R_1FRNUwvnkvroHRF
Green is on in the final state.
inexpressible
53
1Inexpressible
R_3IXnVYuOI9QeI10
Green is on in the final state.
inexpressible
54
1no such a LTL formula
R_3KGxBOYYe7aGaXI
Green is on in the final state.
inexpressible
55
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
Green is on in the final state.
inexpressible
56
1None LTL formula can satisfy
R_31j7Jp3yQbqaZqk
Green is on in the final state.
inexpressible
57
1Blue & G(Blue => (X(!Blue) & X(X(Blue))))
R_2PtHv3KdsGqQLoB
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
58
1Blue & G(Blue => X(!Blue & X(Blue)))
R_3079Udc3BOxNquc
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
59
1blue && G(blue => (X(!blue) && X(X(blue))))
R_2U5AqOKS1RFOYwc
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
60
1Blue & G(Blue => X(!Blue) & X(X(Blue)))
R_2yscxjylS1SXLu5
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
61
1Blue & G(Blue => X(!Blue & X(Blue)))
R_oYU5yKiaqPgXzMd
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
62
1Blue & G(Blue => X(!Blue & X(Blue)))
R_5vtXYvxyZKlCUx3
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
63
1Blue && G(Blue => X(!Blue) && X(X(Blue)))
R_2aenS9mNn6Dk4Cz
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
64
1Blue && (Blue => (X!Blue && XXBlue))
R_3s0nFuWCbCk5CjP
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
65
1Blue & G(Blue => X(!Blue & X(Blue)))
R_2uWWZejY6DyOYaH
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
66
1Blue & G(Blue => X(!Blue) && X(X(Blue)))
R_2WwptK2YPEgUAGJ
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
67
1Blue & X(!Blue) & G(Blue => X(!Blue)) & G(!Blue => X(Blue))
R_sLorPWGMjcKuGYx
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
68
1Blue & G(Blue => X(!Blue & X(Blue)))
R_2zeuLOxEmJflmcu
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
69
?1B & X(!B) &X(G(X(B)=>X(X(!B))|| X(!B)=>X(X(B))))
R_1rjpSmO14SmyPCQ
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
70
?11
Blue & X(!Blue) &(X(X(G(Blue => X(!Blue))&(!Blue => X(Blue)))))
R_1FRNUwvnkvroHRF
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
71
1blue && G(blue => (X(!blue) && XX(blue)))
R_3IXnVYuOI9QeI10
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
72
1Blue & G(Blue => X(! Blue) & X(X(Blue)))
R_3KGxBOYYe7aGaXI
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
73
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
74
1Blue && G((Blue && X(!Blue)) || ((!Blue)&&X(Blue)))
R_31j7Jp3yQbqaZqk
Blue on in 1, off in 2, alts evermore
b & G(b => X(!b & X(b)))
75
1F(Red) & G(Red => X(G(!Red)))
R_2PtHv3KdsGqQLoB
Red is on exactly once.
!r U (r & G(!X(r)))
76
1G(Red => X(G(!Red)))
R_3079Udc3BOxNquc
Red is on exactly once.
!r U (r & G(!X(r)))
77
1F(red) & G(red => X(G(!red)))
R_2U5AqOKS1RFOYwc
Red is on exactly once.
!r U (r & G(!X(r)))
78
1F(Red) & G(Red => X(G(!Red)))
R_2yscxjylS1SXLu5
Red is on exactly once.
!r U (r & G(!X(r)))
79
1!Red U (Red & X(G(!Red)))
R_oYU5yKiaqPgXzMd
Red is on exactly once.
!r U (r & G(!X(r)))
80
1(!Red) U (Red & X(G(!Red)))
R_5vtXYvxyZKlCUx3
Red is on exactly once.
!r U (r & G(!X(r)))
81
1G(Red => X(G(!Red))) && F(Red)
R_2aenS9mNn6Dk4Cz
Red is on exactly once.
!r U (r & G(!X(r)))
82
1F Red && !(F Red && X(F Red))
R_3s0nFuWCbCk5CjP
Red is on exactly once.
!r U (r & G(!X(r)))
83
1F(Red & X(G(!Red)))
R_2uWWZejY6DyOYaH
Red is on exactly once.
!r U (r & G(!X(r)))
84
1F(R && X(G(!Red)))
R_2WwptK2YPEgUAGJ
Red is on exactly once.
!r U (r & G(!X(r)))
85
1F(Red) & !F(Red & X(F(Red)))
R_sLorPWGMjcKuGYx
Red is on exactly once.
!r U (r & G(!X(r)))
86
1!Red U (Red & X(G(!Red)))
R_2zeuLOxEmJflmcu
Red is on exactly once.
!r U (r & G(!X(r)))
87
1F(R) & G(R=>X(G(!R)))
R_1rjpSmO14SmyPCQ
Red is on exactly once.
!r U (r & G(!X(r)))
88
1F(Red & X(G(!Red)))
R_1FRNUwvnkvroHRF
Red is on exactly once.
!r U (r & G(!X(r)))
89
1F(red) & G(red => (XG!red))
R_3IXnVYuOI9QeI10
Red is on exactly once.
!r U (r & G(!X(r)))
90
1F(Red & G(X(! Red)))
R_3KGxBOYYe7aGaXI
Red is on exactly once.
!r U (r & G(!X(r)))
91
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
Red is on exactly once.
!r U (r & G(!X(r)))
92
1None LTL formula can satisfy
R_31j7Jp3yQbqaZqk
Red is on exactly once.
!r U (r & G(!X(r)))
93
1G(Red => Xw(!Green))
Xw instead of X for the end state situation.
R_2PtHv3KdsGqQLoB
whenever r, cannot be next with g
same
94
1same
R_3079Udc3BOxNquc
whenever r, cannot be next with g
same
95
1G(red => Xw(!green))
R_2U5AqOKS1RFOYwc
whenever r, cannot be next with g
same
96
1G(Red => !X(Green))
R_2yscxjylS1SXLu5
whenever r, cannot be next with g
same
97
1Same
R_oYU5yKiaqPgXzMd
whenever r, cannot be next with g
same
98
1G(Red => !X(Green))
R_5vtXYvxyZKlCUx3
whenever r, cannot be next with g
same
99
11G(Red => Xw(G(!Green)))
R_2aenS9mNn6Dk4Cz
whenever r, cannot be next with g
same
100
strong X1G (Red => !Xw Green)
R_3s0nFuWCbCk5CjP
whenever r, cannot be next with g
same
101
1Same.
R_2uWWZejY6DyOYaH
whenever r, cannot be next with g
same
102
1G(Red => Xw(!Green))
R_2WwptK2YPEgUAGJ
whenever r, cannot be next with g
same
103
1same
R_sLorPWGMjcKuGYx
whenever r, cannot be next with g
same
104
1Same
R_2zeuLOxEmJflmcu
whenever r, cannot be next with g
same
105
1Same, G(R => !X(G))
R_1rjpSmO14SmyPCQ
whenever r, cannot be next with g
same
106
1G(Red=>Xw(!Green))
R_1FRNUwvnkvroHRF
whenever r, cannot be next with g
same
107
1G(red => Xw(!green))
R_3IXnVYuOI9QeI10
whenever r, cannot be next with g
same
108
1same
R_3KGxBOYYe7aGaXI
whenever r, cannot be next with g
same
109
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
whenever r, cannot be next with g
same
110
1G(Red => !X(Green))
R_31j7Jp3yQbqaZqk
whenever r, cannot be next with g
same
111
1same
The same this time as we need the 2 states present.
R_2PtHv3KdsGqQLoB
There are at least two states in which the Blue light is on
same
112
1same
R_3079Udc3BOxNquc
There are at least two states in which the Blue light is on
same
113
1same
R_2U5AqOKS1RFOYwc
There are at least two states in which the Blue light is on
same
114
1Same
R_2yscxjylS1SXLu5
There are at least two states in which the Blue light is on
same
115
1Same
R_oYU5yKiaqPgXzMd
There are at least two states in which the Blue light is on
same
116
1F(Blue & X(F(Blue)))
R_5vtXYvxyZKlCUx3
There are at least two states in which the Blue light is on
same
117
1same
R_2aenS9mNn6Dk4Cz
There are at least two states in which the Blue light is on
same
118
1Same
R_3s0nFuWCbCk5CjP
There are at least two states in which the Blue light is on
same
119
1Same.
R_2uWWZejY6DyOYaH
There are at least two states in which the Blue light is on
same
120
1Same
R_2WwptK2YPEgUAGJ
There are at least two states in which the Blue light is on
same
121
1same
R_sLorPWGMjcKuGYx
There are at least two states in which the Blue light is on
same
122
1Same
since there must be two, we are safe to assert there is a successor state to the first
R_2zeuLOxEmJflmcu
There are at least two states in which the Blue light is on
same
123
1Same
R_1rjpSmO14SmyPCQ
There are at least two states in which the Blue light is on
same
124
1F(Blue & X(F(Blue)))
R_1FRNUwvnkvroHRF
There are at least two states in which the Blue light is on
same
125
1F(blue && X(F blue))
R_3IXnVYuOI9QeI10
There are at least two states in which the Blue light is on
same
126
1same
R_3KGxBOYYe7aGaXI
There are at least two states in which the Blue light is on
same
127
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
There are at least two states in which the Blue light is on
same
128
1F(Blue && X(Blue))
R_31j7Jp3yQbqaZqk
There are at least two states in which the Blue light is on
same
129
1G(Xw(false) => Green)
Inexpressible in LTL. In LTLf: if we are at the final state, Green is on.
R_2PtHv3KdsGqQLoB
Green is on in the final state.
F(g & Xw(false))
130
1F(G(Green))
R_3079Udc3BOxNquc
Green is on in the final state.
F(g & Xw(false))
131
1G(Xw(false) => green)
R_2U5AqOKS1RFOYwc
Green is on in the final state.
F(g & Xw(false))
132
1F(G(Green))
R_2yscxjylS1SXLu5
Green is on in the final state.
F(g & Xw(false))
133
1F(G(Green))
R_oYU5yKiaqPgXzMd
Green is on in the final state.
F(g & Xw(false))
134
1F(G(Green))
R_5vtXYvxyZKlCUx3
Green is on in the final state.
F(g & Xw(false))
135
1F(Blue && !X(true))
R_2aenS9mNn6Dk4Cz
Green is on in the final state.
F(g & Xw(false))
136
1GF Green
R_3s0nFuWCbCk5CjP
Green is on in the final state.
F(g & Xw(false))
137
1G(F(Green))
R_2uWWZejY6DyOYaH
Green is on in the final state.
F(g & Xw(false))
138
1G(F(Green))
R_2WwptK2YPEgUAGJ
Green is on in the final state.
F(g & Xw(false))
139
1F(Xw(false) & Green)
R_sLorPWGMjcKuGYx
Green is on in the final state.
F(g & Xw(false))
140
1G(F(Green))
the last equivalence from the last section explained this
R_2zeuLOxEmJflmcu
Green is on in the final state.
F(g & Xw(false))
141
1F(G(G))
R_1rjpSmO14SmyPCQ
Green is on in the final state.
F(g & Xw(false))
142
1F(Green)
R_1FRNUwvnkvroHRF
Green is on in the final state.
F(g & Xw(false))
143
1G(Xw false => green)
R_3IXnVYuOI9QeI10
Green is on in the final state.
F(g & Xw(false))
144
1G(F(Green))
R_3KGxBOYYe7aGaXI
Green is on in the final state.
F(g & Xw(false))
145
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
Green is on in the final state.
F(g & Xw(false))
146
1F(Green && Xw(false))
R_31j7Jp3yQbqaZqk
Green is on in the final state.
F(g & Xw(false))
147
1Blue & X(!Blue) & G(Blue => (Xw(!Blue) & Xw(Xw(Blue))))
Xw instead of X for final states. Also we need to be sure the second state exists.
R_2PtHv3KdsGqQLoB
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
148
allows len=11Blue & G(Blue => Xw(!Blue & Xw(Blue)))
R_3079Udc3BOxNquc
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
149
1blue && X(!blue) && G(blue => (Xw(!blue) && Xw(Xw(blue))))
R_2U5AqOKS1RFOYwc
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
150
1Blue & !Blue & G(!Blue => Xw(Blue) & Xw(Xw(!Blue)))
R_2yscxjylS1SXLu5
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
151
allows len=11Blue & G(Blue => Xw(!Blue & Xw(Blue)))
R_oYU5yKiaqPgXzMd
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
152
1Blue & X(!Blue) & G(Blue => Xw(!Blue & Xw(Blue)))
R_5vtXYvxyZKlCUx3
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
153
1Blue && G(Blue => Xw(!Blue) && Xw(Xw(Blue)))
R_2aenS9mNn6Dk4Cz
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
154
11Blue && (Blue => (Xw !Blue && Xw Xw Blue))
R_3s0nFuWCbCk5CjP
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
155
requires len>21
Blue & X(!Blue) & Xw(X(Blue)) & Xw(X(G(Blue => Xw(!Blue & Xw(Blue)))))
R_2uWWZejY6DyOYaH
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
156
allows len=11Blue & G(Blue => Xw(!Blue) && Xw(Xw(Blue)))
R_2WwptK2YPEgUAGJ
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
157
1Blue & X(!Blue) & G(Blue => Xw(!Blue)) & G(!Blue => Xw(Blue))
R_sLorPWGMjcKuGYx
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
158
allows len=11Blue & G(Blue => Xw (!Blue & Xw (Blue)))
Since only the remaining states are important, the lack of a successor is enough to satisfy the formula
R_2zeuLOxEmJflmcu
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
159
1
B & X(!B) &X(G( \\ Xw(B)=>Xw(Xw(!B))|| \\ Xw(!B)=>Xw(Xw(B)) \\))
R_1rjpSmO14SmyPCQ
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
160
allows len=1, two !blue
111
Blue & Xw(!Blue) &(Xw(Xw(G(Blue => Xw(!Blue))&(!Blue => Xw(Blue)))))
R_1FRNUwvnkvroHRF
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
161
1blue && X(!blue) && G(blue => (Xw(!blue) && XwXw(blue)))
R_3IXnVYuOI9QeI10
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
162
allows len=11Blue & G(Blue => Xw(! Blue) & Xw(Xw(Blue)))
R_3KGxBOYYe7aGaXI
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
163
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
164
1None LTLf can satisfy
R_31j7Jp3yQbqaZqk
Blue on in 1, off in 2, alts evermore
b & G(b => Xw(!b & Xw(b)))
165
1F(Red) & G(Red => Xw(G(!Red)))
We get Red on and once we get Red on it will never be on later on. Xw in LTLf for the final state.
R_2PtHv3KdsGqQLoB
Red is on exactly once.
same
166
1G(Red => Xw(G(!Red)))
R_3079Udc3BOxNquc
Red is on exactly once.
same
167
1F(red) & G(red => Xw(G(!red)))
R_2U5AqOKS1RFOYwc
Red is on exactly once.
same
168
1F(Red) & G(Red => Xw(G(!Red)))
R_2yscxjylS1SXLu5
Red is on exactly once.
same
169
1!Red U (Red & Xw(G(!Red)))
R_oYU5yKiaqPgXzMd
Red is on exactly once.
same
170
1(!Red) U (Red & Xw(G(!Red)))
R_5vtXYvxyZKlCUx3
Red is on exactly once.
same
171
1G(Red => Xw(G(!Red))) && F(Red)
R_2aenS9mNn6Dk4Cz
Red is on exactly once.
same
172
1F Red && !(F Red && Xw(F Red))
R_3s0nFuWCbCk5CjP
Red is on exactly once.
same
173
1F(Red & Xw(G(!Red)))
R_2uWWZejY6DyOYaH
Red is on exactly once.
same
174
1F(R && Xw(G(!Red)))
R_2WwptK2YPEgUAGJ
Red is on exactly once.
same
175
1same
R_sLorPWGMjcKuGYx
Red is on exactly once.
same
176
1!Red U (Red & Xw (G(!Red)))
some state exists which is Red, and is not followed by any state that is Red, and all states before this state are not Red. Since a state cannot be both Red and not Red, this is enough
R_2zeuLOxEmJflmcu
Red is on exactly once.
same
177
1F(R) & G(R=>!X(F(R)))
R_1rjpSmO14SmyPCQ
Red is on exactly once.
same
178
1F(Red & Xw(G(!Red)))
R_1FRNUwvnkvroHRF
Red is on exactly once.
same
179
1F(red) & G(red => (XwG!red))
R_3IXnVYuOI9QeI10
Red is on exactly once.
same
180
1F(Red & G(! X(Red)))
R_3KGxBOYYe7aGaXI
Red is on exactly once.
same
181
noop1
Your LTLf descriptions may ignore the possibility of empty traces.
R_2QS5lE5zYMCL1s7
Red is on exactly once.
same
182
1None LTLf formula can satisfy
R_31j7Jp3yQbqaZqk
Red is on exactly once.
same