World's most popular travel blog for travel bloggers.

[Answers] Counterexample for LTL - CTL equivalence

, , No Comments
Problem Detail: 

I have to find an example of a model where the LTL-formula $F G p \wedge F q$ is valid and the CTL-formula $EF AG p \wedge AF q$ is not valid. I found this example, but I'm not completely sure whether it's correct:

enter image description here

Asked By : Pieter Verschaffelt

Answered By : Shaull

Consider the following model: you have 3 states, $s_0,s_1,s_2$ with the transitions: $s_0\to s_0$, $s_0\to s_1$, $s_1\to s_2$ and $s_2\to s_2$ and the labels are $L(s_0)=\{p,q\}$, $L(s_1)=\emptyset$ and $L(s_2)=p$.

Then, every computation starts with $q$, so $Fq$ holds, and every infinite computation eventually gets stuck in $s_2$, or it is $s_0^\omega$, and both satisfy $FGp$, so the LTL formula holds.

However, it never holds that $AFq$, so the CTL formula does not hold.

Best Answer from StackOverflow

Question Source :

3.2K people like this

 Download Related Notes/Documents


Post a Comment

Let us know your responses and feedback