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CTL, LTL and CTL∗

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Hello,

I wanted to confirm, Is the below solution correct?. I am trying to violate S1 as S2 is always true. 

in * TF "Emb. Sys. and Rob." by (300 points)

1 Answer

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You are right. The second formula is always true. The first formula needs to be not satisfied. This can be done by having a path that (from the third step on) shows both a at some point and ¬a at some other point. Alternating between these two states is an option.

See also task 8a in https://es.cs.uni-kl.de/teaching/vrs/exams/2018.08.21.vrs/2018.08.21.vrs.solutions.pdf

by (25.6k points)

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