A∀ ({q,p}, ¬q∧¬p , (Xq↔(¬q∧¬a))∧(Xp↔(¬p→q)),GFp)
my solution:
∃ q,p. ¬q (0) ∧ ¬p(0) ∧
( ∀ t. ( q(t+1) ↔ ¬ q (t) ∧ ¬a(t) ) ∧ (p(t+1) ↔ ¬p(t) → q (t) ) ) ∧
∀ t1 ∃ t2. t1 < t2 ∧ p(t2)
Questions:
1. can you please explain how to convert temporal to LO2, When I apply the algorithm of the chapter VRS08-PredLogic- slide# 48 it usually explodes the result into a very long formula but when I see the exam solutions they are very short.
1.1. Also I get confuse when calling the method recursively with the new time like in the case of SU and PX where "t1" is passed but the input to the function is always "t0" which also propagates to the internal cases of the algo. For instance [PXφ SU ψ] may lead to the recursive time calls
1.2 In general when to use "≤", "<" for time t1, t2, etc and quantifiers?.
1.3 Is there a better short approach to solve these kinds of questions?
2. Is the solution of automata to LO2 correct to the given automata (In the exam solutions it is explained using state transition diagram )?
3. In the algorithm " ΘLO2(t, Φ) " of the chapter VRS08-PredLogic slide# 25 the acceptance condition always uses "∀t1.∃t2.t1" but in the exam solution i also see it reversed in case of FG. I believe it should be "∀t1 ∃ t2. t1 < t2" for GF and "∃ t1 ∀ t2. t1 <t2" for FG. Please correct me if I'm wrong.
3. In these type of questions when do we use "t1 ≤ t2" and "t1< t2". Is there any rule of thumb?
Link to the paper for the reference (Page 26).
https://es.cs.uni-kl.de/teaching/vrs/exams/2019.08.27.vrs/2019.08.27.vrs.solutions.pdf