A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z | AA | AB | AC | AD | AE | AF | AG | AH | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1 | Totals ==> | 117 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 6 | 0 | 0 | 0 | 1 | 0 | 0 | 126 | ||||||||||||
2 | Ok | ? | BP | BSI | BSQ | CyG | EU | TSU | BU | IF | IG | IPfx | OI | LenX | SeqX | Last | WU | WF | SG | RV | Answer | Explanation | ID | Formula | Expected | |||||||||
3 | 1 | True in LTLf | We eventually get a is equivalent to havin true until a is found | R_2PtHv3KdsGqQLoB | F(a) == true U a | TRUE | ||||||||||||||||||||||||||||
4 | 1 | True in LTLf | R_3079Udc3BOxNquc | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
5 | 1 | True in LTLf | R_2U5AqOKS1RFOYwc | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
6 | 1 | True in LTLf | R_2yscxjylS1SXLu5 | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
7 | 1 | True in LTLf | R_oYU5yKiaqPgXzMd | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
8 | 1 | True in LTLf | R_5vtXYvxyZKlCUx3 | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
9 | 1 | True in LTLf | R_2aenS9mNn6Dk4Cz | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
10 | 1 | True in LTLf | R_3s0nFuWCbCk5CjP | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
11 | 1 | True in LTLf | Yes, that equation is valid in LTLf when the variable a is replaced by any LTLf term. | R_2uWWZejY6DyOYaH | F(a) == true U a | TRUE | ||||||||||||||||||||||||||||
12 | 1 | True in LTLf | The Finally operator requires a to hold in some state, which the Until operator also enforces. | R_2WwptK2YPEgUAGJ | F(a) == true U a | TRUE | ||||||||||||||||||||||||||||
13 | 1 | True in LTLf | R_sLorPWGMjcKuGYx | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
14 | 1 | True 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 | 1 | True in LTLf | R_1rjpSmO14SmyPCQ | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
16 | 1 | True in LTLf | R_1FRNUwvnkvroHRF | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
17 | 1 | True in LTLf | R_3IXnVYuOI9QeI10 | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
18 | 1 | True in LTLf | R_3KGxBOYYe7aGaXI | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
19 | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
20 | 1 | True in LTLf | R_31j7Jp3yQbqaZqk | F(a) == true U a | TRUE | |||||||||||||||||||||||||||||
21 | 1 | False in LTLf | For any formula in the final state !X(a) is vacuously true and X(!a) is vacuously false | R_2PtHv3KdsGqQLoB | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
22 | 1 | False in LTLf | R_3079Udc3BOxNquc | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
23 | 1 | False in LTLf | R_2U5AqOKS1RFOYwc | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
24 | 1 | False in LTLf | R_2yscxjylS1SXLu5 | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
25 | 1 | False 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 | R_oYU5yKiaqPgXzMd | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
26 | 1 | False in LTLf | R_5vtXYvxyZKlCUx3 | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
27 | 1 | False in LTLf | R_2aenS9mNn6Dk4Cz | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
28 | 1 | False 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 | 1 | False 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 | R_2uWWZejY6DyOYaH | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
30 | 1 | False 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. | R_2WwptK2YPEgUAGJ | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
31 | 1 | False in LTLf | R_sLorPWGMjcKuGYx | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
32 | 1 | False in LTLf | the first is always satisfied in the last state, the second is never satisfied in the last state | R_2zeuLOxEmJflmcu | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
33 | 1 | False in LTLf | R_1rjpSmO14SmyPCQ | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
34 | weak X | 1 | True in LTLf | R_1FRNUwvnkvroHRF | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
35 | 1 | False in LTLf | R_3IXnVYuOI9QeI10 | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
36 | 1 | False in LTLf | R_3KGxBOYYe7aGaXI | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
37 | weak X | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | !X(a) == X(!a) | FALSE | ||||||||||||||||||||||||||||
38 | 1 | False in LTLf | R_31j7Jp3yQbqaZqk | !X(a) == X(!a) | FALSE | |||||||||||||||||||||||||||||
39 | weak X | 1 | True 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 | R_2PtHv3KdsGqQLoB | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||
40 | 1 | False in LTLf | R_3079Udc3BOxNquc | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
41 | 1 | False in LTLf | R_2U5AqOKS1RFOYwc | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
42 | 1 | False in LTLf | R_2yscxjylS1SXLu5 | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
43 | 1 | False 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 | R_oYU5yKiaqPgXzMd | X(G(a)) == G(X(a)) | FALSE | ||||||||||||||||||||||||||||
44 | 1 | False in LTLf | R_5vtXYvxyZKlCUx3 | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
45 | 1 | False in LTLf | R_2aenS9mNn6Dk4Cz | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
46 | 1 | False 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 | 1 | False 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 | 1 | False 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)). | R_2WwptK2YPEgUAGJ | X(G(a)) == G(X(a)) | FALSE | ||||||||||||||||||||||||||||
49 | 1 | False in LTLf | R_sLorPWGMjcKuGYx | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
50 | 1 | False 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. | R_2zeuLOxEmJflmcu | X(G(a)) == G(X(a)) | FALSE | ||||||||||||||||||||||||||||
51 | 1 | False in LTLf | Second can only be satisfied in infinite trace | R_1rjpSmO14SmyPCQ | X(G(a)) == G(X(a)) | FALSE | ||||||||||||||||||||||||||||
52 | 1 | False in LTLf | R_1FRNUwvnkvroHRF | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
53 | 1 | False in LTLf | R_3IXnVYuOI9QeI10 | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
54 | 1 | False in LTLf | R_3KGxBOYYe7aGaXI | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
55 | (maybe) weak X | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | X(G(a)) == G(X(a)) | FALSE | ||||||||||||||||||||||||||||
56 | 1 | False in LTLf | R_31j7Jp3yQbqaZqk | X(G(a)) == G(X(a)) | FALSE | |||||||||||||||||||||||||||||
57 | 1 | True 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. | R_2PtHv3KdsGqQLoB | X(a U b) == X(a) U X(b) | TRUE | ||||||||||||||||||||||||||||
58 | 1 | True in LTLf | R_3079Udc3BOxNquc | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
59 | 1 | True in LTLf | R_2U5AqOKS1RFOYwc | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
60 | 1 | True in LTLf | R_2yscxjylS1SXLu5 | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
61 | 1 | True 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 | 1 | True in LTLf | R_5vtXYvxyZKlCUx3 | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
63 | 1 | True in LTLf | R_2aenS9mNn6Dk4Cz | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
64 | 1 | True in LTLf | R_3s0nFuWCbCk5CjP | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
65 | 1 | True 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 | 1 | True 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 | 1 | True in LTLf | R_sLorPWGMjcKuGYx | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
68 | 1 | True 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 | 1 | True in LTLf | R_1rjpSmO14SmyPCQ | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
70 | 1 | True in LTLf | R_1FRNUwvnkvroHRF | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
71 | 1 | True in LTLf | R_3IXnVYuOI9QeI10 | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
72 | 1 | True in LTLf | R_3KGxBOYYe7aGaXI | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
73 | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | X(a U b) == X(a) U X(b) | TRUE | |||||||||||||||||||||||||||||
74 | ??? | 1 | False in LTLf | R_31j7Jp3yQbqaZqk | X(a U b) == X(a) U X(b) | TRUE | ||||||||||||||||||||||||||||
75 | 1 | False 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. | R_2PtHv3KdsGqQLoB | G(a) == a & X(G(a)) | FALSE | ||||||||||||||||||||||||||||
76 | 1 | False in LTLf | R_3079Udc3BOxNquc | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
77 | 1 | False in LTLf | R_2U5AqOKS1RFOYwc | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
78 | 1 | False in LTLf | R_2yscxjylS1SXLu5 | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
79 | 1 | False 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. | R_oYU5yKiaqPgXzMd | G(a) == a & X(G(a)) | FALSE | ||||||||||||||||||||||||||||
80 | 1 | False in LTLf | R_5vtXYvxyZKlCUx3 | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
81 | 1 | False in LTLf | R_2aenS9mNn6Dk4Cz | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
82 | 1 | False 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 | 1 | False 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 | 1 | False 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 | 1 | False in LTLf | R_sLorPWGMjcKuGYx | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
86 | 1 | False 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 | 1 | False in LTLf | R_1rjpSmO14SmyPCQ | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
88 | 1 | False in LTLf | R_1FRNUwvnkvroHRF | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
89 | 1 | False in LTLf | R_3IXnVYuOI9QeI10 | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
90 | 1 | False in LTLf | R_3KGxBOYYe7aGaXI | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
91 | weak X | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | G(a) == a & X(G(a)) | FALSE | ||||||||||||||||||||||||||||
92 | 1 | False in LTLf | R_31j7Jp3yQbqaZqk | G(a) == a & X(G(a)) | FALSE | |||||||||||||||||||||||||||||
93 | 1 | True 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. | R_2PtHv3KdsGqQLoB | F(a) == a || X(F(a)) | TRUE | ||||||||||||||||||||||||||||
94 | 1 | True in LTLf | R_3079Udc3BOxNquc | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
95 | 1 | True in LTLf | R_2U5AqOKS1RFOYwc | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
96 | 1 | True in LTLf | R_2yscxjylS1SXLu5 | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
97 | 1 | True in LTLf | F(a) true iff state 1 satisfies a, or some future state exists which satisfies a. i.e. a || X(F(a)) | R_oYU5yKiaqPgXzMd | F(a) == a || X(F(a)) | TRUE | ||||||||||||||||||||||||||||
98 | 1 | True in LTLf | R_5vtXYvxyZKlCUx3 | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
99 | bad prop? | 1 | False in LTLf | R_2aenS9mNn6Dk4Cz | F(a) == a || X(F(a)) | TRUE | ||||||||||||||||||||||||||||
100 | 1 | True in LTLf | R_3s0nFuWCbCk5CjP | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
101 | 1 | True 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 | 1 | True 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 | 1 | True in LTLf | R_sLorPWGMjcKuGYx | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
104 | 1 | True 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 | 1 | True in LTLf | R_1rjpSmO14SmyPCQ | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
106 | 1 | True in LTLf | R_1FRNUwvnkvroHRF | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
107 | 1 | True in LTLf | R_3IXnVYuOI9QeI10 | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
108 | 1 | True in LTLf | R_3KGxBOYYe7aGaXI | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
109 | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
110 | 1 | True in LTLf | R_31j7Jp3yQbqaZqk | F(a) == a || X(F(a)) | TRUE | |||||||||||||||||||||||||||||
111 | 1 | True 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 | 1 | True in LTLf | R_3079Udc3BOxNquc | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
113 | 1 | True in LTLf | R_2U5AqOKS1RFOYwc | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
114 | 1 | True in LTLf | R_2yscxjylS1SXLu5 | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
115 | 1 | True 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 | 1 | True in LTLf | R_5vtXYvxyZKlCUx3 | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
117 | 1 | True in LTLf | R_2aenS9mNn6Dk4Cz | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
118 | wrong F starting point? | 1 | False 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 | 1 | True 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 | 1 | True in LTLf | These both just enforce that the final state satisfies a. | R_2WwptK2YPEgUAGJ | G(F(a)) == F(G(a)) | TRUE | ||||||||||||||||||||||||||||
121 | 1 | True in LTLf | R_sLorPWGMjcKuGYx | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
122 | 1 | True 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 | 1 | True in LTLf | R_1rjpSmO14SmyPCQ | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
124 | 1 | True in LTLf | R_1FRNUwvnkvroHRF | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
125 | 1 | True in LTLf | R_3IXnVYuOI9QeI10 | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
126 | 1 | True in LTLf | R_3KGxBOYYe7aGaXI | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
127 | 1 | True in LTLf | R_2QS5lE5zYMCL1s7 | G(F(a)) == F(G(a)) | TRUE | |||||||||||||||||||||||||||||
128 | ? | 1 | False in LTLf | R_31j7Jp3yQbqaZqk | G(F(a)) == F(G(a)) | TRUE |