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Concurrent Systems/notes/11 - LTSs and Bisimulation.md
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@ -67,3 +67,9 @@ However, p0 and q0 are not bisimilar: the transition q0 -> a -> q1 is not bisimu
$S = \{(p0,q0), (p1,q1), (p2,q1), (p0,q2)\}$ $S = \{(p0,q0), (p1,q1), (p2,q1), (p0,q2)\}$
$S^{−1} = \{(q0,p0), (q1,p1), (q1,p2), (q2,p0)\}$ $S^{−1} = \{(q0,p0), (q1,p1), (q1,p2), (q2,p0)\}$
>[!def] Theorem
>Bisimilarity is an equivalence relation
*Proof:*
Reflexivity: we have to show that q ∼ q, for every q. Consider the following relation