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February 16, 2022, 7e

In my eyes, S2 is satisfied in the path s0->s1 where there exists a path where finally (!a & EGa ) holds. Please help me understand how S1 distinguishes S2.
in # Study-Organisation (Master) by (850 points)

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The formula !a&EGa cannot hold in any state since it is a contradiction. You can see that by unrolling EGa into a & EX EG a which then contradicts directly !a. Hence, S2 is simply false and cannot hold in any state.

EGa holds in state s0, but not in state s1. Consider now E(F!a & Era) which is equivalent to (EF!a)&(EGa). This holds in state s0 since the path s0->s1^omega satisfies F!a and the path s0^omega satisfies EGa. Hence, s0 satisfies (EF!a)&(EGa), but s1 does not. Finally, abbreviate this formula by a new variable b which is therefore a label of s0, but not of s1, and consider the formula EGb. It clearly holds in s0, but not in s1, and therefore S1 holds in s0.
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