I just come up with this question. As written in the book of Logic in Computer Science, one of the important equivalence of LTL is this: Fp=TUp. And the T means no constraints.
Yet what if I replace the T with (not p)? Does Fp=(not p)Up hold? Since in this case I actually put some constraints (not p) in the formula, but in the meantime there could be no state can satisfy (not p) and p together. And I tried with different LTL formula as p, and as long as p is satisfiable, then for every path with p, it must satisfy Fp and (not p)Up as well. Does it means that I can rewrite F in this way or there is some counter example?
The short answer:
Yes, both formulas are equivalent and you can rewrite Fp also with (¬p)Up.
and a proof:
We can investigate the problem by looking at the definition of pUq (I think it's defined this way in the book Model Checking by Clarke, Grumberg, Peled).
A path s is a model for the formula (written s ⊨ pUq):
s ⊨ pUq <=> ∃k: s^k ⊨ q
∧ ∀i: 0<=i<k => s^i ⊨ q
(With s^i being the path s with the first i steps removed.)
We have (1):
s ⊨ (¬p)Up <=> ∃k: s^k ⊨ p
∧ ∀i: 0<=i<k => s^i ⊨ ¬p
and (2):
s ⊨ TUp <=> ∃k: s^k ⊨ p
∧ ∀i: 0<=i<k => s^i ⊨ true
<=> ∃k: s^k ⊨ p
We want to show (1) <=> (2) (I renamed the ks to k1 and k2 to avoid confusion):
∃k1: s^k1 ⊨ p
∧ ∀i: 0<=i<k1 => s^i ⊨ ¬p
<=>
∃k2: s^k2 ⊨ p
The direction (1) => (2) is trivial.
For (2) => (1) we have to show that from
∃k2: s^k2 ⊨ p
follows
∃k1: s^k1 ⊨ p ∧ ∀i: 0<=i<k1 => s^i ⊨ ¬p
We know that there exists a value for k1 (namely k2) such that s^k1 ⊨ p holds. But what about the second part? We can now just use for k1 the smallest value such that s^i ⊨ p holds. Then the second part is true, because if there would be an i such that s^i ⊨ not p does not hold, we know that s^i |= p holds. But in that case we would have choosen i for k1 because i is strictly smaller then k1.
So both formulas (1) and (2) are equivalent.