ABCDEFGHIJKLMNOPQRSTUVWXYZAA
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Totals ==>106100000010000108
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Ok?BPBSI
BSQ
IFIGOIWUEUXXRV
Scope
AnswerExplanationIDFormulaTraceExpected
3
1Yes
This is true as from any state we start, we eventually get Red (in State 4).
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G(F(r)){r} {} {} {r}Yes
4
1Yes
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G(F(r)){r} {} {} {r}Yes
5
1Yes
R_2U5AqOKS1RFOYwc
G(F(r)){r} {} {} {r}Yes
6
1Yes
R_2yscxjylS1SXLu5
G(F(r)){r} {} {} {r}Yes
7
1Yes
The final state has red turned on
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G(F(r)){r} {} {} {r}Yes
8
1Yes
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G(F(r)){r} {} {} {r}Yes
9
1Yes
R_2aenS9mNn6Dk4Cz
G(F(r)){r} {} {} {r}Yes
10
1Yes
In every state here, red is eventually satisfied.
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G(F(r)){r} {} {} {r}Yes
11
1Yes
Because at every step, either the Red light is currently on, or if there exists next steps, one of the future states have a Red light turned on.
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G(F(r)){r} {} {} {r}Yes
12
1Yes
The red light is on in the final step, and so every step before that satisfies F(Red).
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G(F(r)){r} {} {} {r}Yes
13
1Yes
R_sLorPWGMjcKuGYx
G(F(r)){r} {} {} {r}Yes
14
1Yes
just needs the last state to be red
R_2zeuLOxEmJflmcu
G(F(r)){r} {} {} {r}Yes
15
1Yes
R_1rjpSmO14SmyPCQ
G(F(r)){r} {} {} {r}Yes
16
1Yes
R_1FRNUwvnkvroHRF
G(F(r)){r} {} {} {r}Yes
17
1Yes
R_3IXnVYuOI9QeI10
G(F(r)){r} {} {} {r}Yes
18
1Yes
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G(F(r)){r} {} {} {r}Yes
19
1Yes
R_2QS5lE5zYMCL1s7
G(F(r)){r} {} {} {r}Yes
20
1Yes
R_31j7Jp3yQbqaZqk
G(F(r)){r} {} {} {r}Yes
21
1No
The formula assumes that we have Red on until there is a state with Red off. However, all states have Red on in this trace.
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r U (!r){r} {r} {r}No
22
1No
R_3079Udc3BOxNquc
r U (!r){r} {r} {r}No
23
1No
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r U (!r){r} {r} {r}No
24
1No
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r U (!r){r} {r} {r}No
25
1No
!Red never satisfied
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r U (!r){r} {r} {r}No
26
1No
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r U (!r){r} {r} {r}No
27
1No
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r U (!r){r} {r} {r}No
28
1No
!red is never satisfied.
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r U (!r){r} {r} {r}No
29
1No
The formula needs that !Red eventually holds within the finite trace, which is not true.
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r U (!r){r} {r} {r}No
30
1No
(!Red) never holds as the red light is always on
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r U (!r){r} {r} {r}No
31
1No
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r U (!r){r} {r} {r}No
32
1No
strong until requires a non-red state
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r U (!r){r} {r} {r}No
33
1No
!Red is not satisfied in any of the circle
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r U (!r){r} {r} {r}No
34
1No
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r U (!r){r} {r} {r}No
35
1No
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r U (!r){r} {r} {r}No
36
1No
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r U (!r){r} {r} {r}No
37
1Yes
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r U (!r){r} {r} {r}No
38
1No
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r U (!r){r} {r} {r}No
39
1Yes
This is true as in state 2 X(Red) is False because no next state exists.
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F(!X(r)){r} {r}Yes
40
1Yes
R_3079Udc3BOxNquc
F(!X(r)){r} {r}Yes
41
1Yes
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F(!X(r)){r} {r}Yes
42
1Yes
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F(!X(r)){r} {r}Yes
43
1Yes
State 2 does not satisfy X(Red), so F(!X(Red)) satisfied
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F(!X(r)){r} {r}Yes
44
1Yes
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F(!X(r)){r} {r}Yes
45
1Yes
R_2aenS9mNn6Dk4Cz
F(!X(r)){r} {r}Yes
46
1Yes
In state 2, X red is not satisfied since there is no next state. Hence !X Red is satisfied in state 2. Hence, eventually !X(Red) holds.
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F(!X(r)){r} {r}Yes
47
1Yes
The formula needs there to either eventually exist a state such that the next state does not have the Red light turned on, or there doesn't exist a next state, and this is true at state 2.
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F(!X(r)){r} {r}Yes
48
1Yes
In the second timestep, X(Red) is false as there is no next step
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F(!X(r)){r} {r}Yes
49
1Yes
R_sLorPWGMjcKuGYx
F(!X(r)){r} {r}Yes
50
1Yes
!X (f) is the dual of Xw(!f). since state 2 has no next state, it does not satisfy X(Red) and thus does satisfy it's negation, as required
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F(!X(r)){r} {r}Yes
51
1Yes
2 statisfy !X(Red)
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F(!X(r)){r} {r}Yes
52
1Yes
R_1FRNUwvnkvroHRF
F(!X(r)){r} {r}Yes
53
1Yes
R_3IXnVYuOI9QeI10
F(!X(r)){r} {r}Yes
54
1Yes
R_3KGxBOYYe7aGaXI
F(!X(r)){r} {r}Yes
55
1Yes
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F(!X(r)){r} {r}Yes
56
1Yes
R_31j7Jp3yQbqaZqk
F(!X(r)){r} {r}Yes
57
1No
In state 1 X(!Red) is False because state 2 has Red and in state 2 X(!Red) is False because no state follows it.
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F(X(!r)){r} {r}No
58
1No
R_3079Udc3BOxNquc
F(X(!r)){r} {r}No
59
1No
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F(X(!r)){r} {r}No
60
1No
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F(X(!r)){r} {r}No
61
1No
All states satisfy Red
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F(X(!r)){r} {r}No
62
1No
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F(X(!r)){r} {r}No
63
1No
R_2aenS9mNn6Dk4Cz
F(X(!r)){r} {r}No
64
1No
X (!Red) doesn't hold in state 1 because 2 satisfies red. \\ X (!Red) doesn't hold in state 2 since there is no next state.
R_3s0nFuWCbCk5CjP
F(X(!r)){r} {r}No
65
1No
The formula needs there to eventually exist a state such that the next state does not have the Red light turned on, which is not true.
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F(X(!r)){r} {r}No
66
1No
In the first timestep X(!Red) does not hold as the red light is on in the second timestep. In the second timestep X(!Red) does not hold as there is no next step
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F(X(!r)){r} {r}No
67
1No
R_sLorPWGMjcKuGYx
F(X(!r)){r} {r}No
68
1No
some state would need to satisfy X(!Red) which means the state needs a successor, and that successor cannot have red. The last state has no successor, so cannot satisfy
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F(X(!r)){r} {r}No
69
1No
pai, 1 does not satisfy since 2 is red, pai, 2 does not satisfy as well since there is no next state
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F(X(!r)){r} {r}No
70
1No
R_1FRNUwvnkvroHRF
F(X(!r)){r} {r}No
71
1No
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F(X(!r)){r} {r}No
72
1No
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F(X(!r)){r} {r}No
73
weak X1Yes
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F(X(!r)){r} {r}No
74
1No
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F(X(!r)){r} {r}No
75
1Yes
From state 4 onwards (until state 4) we always have Red on.
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F(G(r)){} {} {} {r}Yes
76
1Yes
R_3079Udc3BOxNquc
F(G(r)){} {} {} {r}Yes
77
1Yes
R_2U5AqOKS1RFOYwc
F(G(r)){} {} {} {r}Yes
78
1Yes
R_2yscxjylS1SXLu5
F(G(r)){} {} {} {r}Yes
79
1Yes
Final state is red, so state 4 satisfies G(Red)
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F(G(r)){} {} {} {r}Yes
80
1Yes
R_5vtXYvxyZKlCUx3
F(G(r)){} {} {} {r}Yes
81
1Yes
R_2aenS9mNn6Dk4Cz
F(G(r)){} {} {} {r}Yes
82
1Yes
Semantics for LTLf state that 4 satisfies G Red.
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F(G(r)){} {} {} {r}Yes
83
1Yes
This formula is satisfied since from time step 4 onwards, the Red light is turned on.
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F(G(r)){} {} {} {r}Yes
84
1Yes
In the fourth timestep, the red light remains on for the remainder of the trace
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F(G(r)){} {} {} {r}Yes
85
1Yes
R_sLorPWGMjcKuGYx
F(G(r)){} {} {} {r}Yes
86
1Yes
Semantically identical to G(F(Red)), just needs last state Red
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F(G(r)){} {} {} {r}Yes
87
1Yes
R_1rjpSmO14SmyPCQ
F(G(r)){} {} {} {r}Yes
88
1Yes
R_1FRNUwvnkvroHRF
F(G(r)){} {} {} {r}Yes
89
1Yes
R_3IXnVYuOI9QeI10
F(G(r)){} {} {} {r}Yes
90
1Yes
R_3KGxBOYYe7aGaXI
F(G(r)){} {} {} {r}Yes
91
1Yes
R_2QS5lE5zYMCL1s7
F(G(r)){} {} {} {r}Yes
92
1Yes
R_31j7Jp3yQbqaZqk
F(G(r)){} {} {} {r}Yes
93
1Yes
In state 3 we have Red and from that point on we eventually have Blue (in state 4).
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F(r & F(b)){b} {r} {r} {b}Yes
94
1Yes
R_3079Udc3BOxNquc
F(r & F(b)){b} {r} {r} {b}Yes
95
1Yes
R_2U5AqOKS1RFOYwc
F(r & F(b)){b} {r} {r} {b}Yes
96
1Yes
R_2yscxjylS1SXLu5
F(r & F(b)){b} {r} {r} {b}Yes
97
1Yes
State 4 satisfies Blue, so state 2 satisfies Red & F(Blue)
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F(r & F(b)){b} {r} {r} {b}Yes
98
1Yes
R_5vtXYvxyZKlCUx3
F(r & F(b)){b} {r} {r} {b}Yes
99
1Yes
R_2aenS9mNn6Dk4Cz
F(r & F(b)){b} {r} {r} {b}Yes
100
1Yes
State 2 and State 3 both satisfy Red. \\ They also both satisfy eventually Blue, since state 4 satisfies blue. \\ \\ Hence, from state 1, eventaully Red and (F Blue) hold.
R_3s0nFuWCbCk5CjP
F(r & F(b)){b} {r} {r} {b}Yes
101
1Yes
The formula holds, as at the 2nd time step the Red light is turned on, and eventually the Blue light is turned on at the 4th time step.
R_2uWWZejY6DyOYaH
F(r & F(b)){b} {r} {r} {b}Yes
102
1Yes
In the second timestep the red light is on and the blue light turns on later in the fourth timestep.
R_2WwptK2YPEgUAGJ
F(r & F(b)){b} {r} {r} {b}Yes
103
1Yes
R_sLorPWGMjcKuGYx
F(r & F(b)){b} {r} {r} {b}Yes
104
1Yes
states 2 and 3 satisfy Red & F(Blue), so state 1 Satisfies F(Red & F(Blue))
R_2zeuLOxEmJflmcu
F(r & F(b)){b} {r} {r} {b}Yes
105
1Yes
2 statisfy Red & F(Blue)
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F(r & F(b)){b} {r} {r} {b}Yes
106
1Yes
R_1FRNUwvnkvroHRF
F(r & F(b)){b} {r} {r} {b}Yes
107
1Yes
R_3IXnVYuOI9QeI10
F(r & F(b)){b} {r} {r} {b}Yes
108
1Yes
R_3KGxBOYYe7aGaXI
F(r & F(b)){b} {r} {r} {b}Yes
109
1Yes
R_2QS5lE5zYMCL1s7
F(r & F(b)){b} {r} {r} {b}Yes
110
1Yes
R_31j7Jp3yQbqaZqk
F(r & F(b)){b} {r} {r} {b}Yes
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