ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAH
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Totals ==>67500000000000000000072
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Ok?BPBSIBSQCyGEUTSUBUIFIGIPfxOI
LenX
SeqX
LastWUWFSGRVAnswer
Explanation
IDFormulaExpected
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
!X(a) is true if there is no next state, X(!a) is not.
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
X(!a) requires a next state to exist
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
This is clearly true on any trace of length at least 2. On any trace of length 1, however, the right side is false whereas the left side is true.
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1True in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
The first formula is true also if the trace has no second state (i.e. the trace ends after the first state), while the second formula is false in this case, since !a cannot hold in a non-existent state.
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!X(a) == X(!a)FALSE
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1False in LTLf
If there is only one state and "!a" holds, the first formula will return true and the second one false (both because there is no next).
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1True in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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!X(a) == X(!a)FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
G(a) is true if the trace has only one state where a is satisfied, but a & X(G(a)) is not.
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
For trace (a), LHS holds, but RHS does not
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
X(G(a)) requires a next state to exist
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
On any trace of length 1, the right side of the equation will always be false due to the X operator, whereas the left side may be simultaneously true (simply consider any trace whose initial state makes atom a true)
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
The above is false, because, when reaching the last state, X ( G (a)) will always be false, since the trace ends, hence there is no further state where G (a) could hold. The actual correct formula is G(a) == a & Xw (G(a)).
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
If there is only one state and "a" holds, the first formula will return true and the second one false.
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G(a) == a & X(G(a))
FALSE
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1True in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1False in LTLfNeeds Xw
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G(a) == a & X(G(a))
FALSE
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1False in LTLf
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G(a) == a & X(G(a))
FALSE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1False in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
If "a" eventually holds (first formula), then there is definitely a state that satisfies "a". The second formula is equivalent: If the current state does not satisfy "a", then there must be a next state from which "a" will eventually hold.
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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F(a) == a || X(F(a))
TRUE
55
1True in LTLf
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F(a) == a || X(F(a))
TRUE
56
1True in LTLf
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F(a) == a || X(F(a))
TRUE
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1True in LTLf
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G(F(a)) == F(G(a))
TRUE
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1True in LTLf
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G(F(a)) == F(G(a))
TRUE
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missing iff1True in LTLf
For G(F(a)) to be true, a must be true in the last state. Then, in any such formula F(G(a)) is also satisfied.
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G(F(a)) == F(G(a))
TRUE
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1True in LTLf
Both hold iff last state satisfies a. G(F(a)) => for last state F(a) holds => so Last satisfies a Last satisfies a => G(F(a)) F(G(a)) => exists state s.t. in the following states a holds => last satisfies a last satisfies a => F(G(a))
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G(F(a)) == F(G(a))
TRUE
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1True in LTLf
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G(F(a)) == F(G(a))
TRUE
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1True in LTLf
If G(F(a)), then the final state must always have a. If F(G(a)), then the final state must always have a. So for both, they are satisfied only when the final state satisfies a.
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G(F(a)) == F(G(a))
TRUE
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1True in LTLf
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G(F(a)) == F(G(a))
TRUE
64
1True in LTLf
On finite traces, both sides of the equation are equivalent with simply the final state satisfying a.
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G(F(a)) == F(G(a))
TRUE
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1True in LTLf
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G(F(a)) == F(G(a))
TRUE
66
1True in LTLf
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G(F(a)) == F(G(a))
TRUE
67
1True in LTLf
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G(F(a)) == F(G(a))
TRUE
68
1True in LTLf
Both of the above are equivalent to the last state of the trace satisfying a.
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G(F(a)) == F(G(a))
TRUE
69
1True in LTLf
Both become equivalent to whether "a" holds in the last state.
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G(F(a)) == F(G(a))
TRUE
70
1True in LTLf
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G(F(a)) == F(G(a))
TRUE
71
1True in LTLf
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G(F(a)) == F(G(a))
TRUE
72
1True in LTLf
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G(F(a)) == F(G(a))
TRUE
73
1True in LTLf
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G(F(a)) == F(G(a))
TRUE
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1False in LTLf
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G(F(a)) == F(G(a))
TRUE
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