vault backup: 2025-04-14 08:54:56
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@ -67,3 +67,9 @@ However, p0 and q0 are not bisimilar: the transition q0 -> a -> q1 is not bisimu
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$S = \{(p0,q0), (p1,q1), (p2,q1), (p0,q2)\}$
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$S^{−1} = \{(q0,p0), (q1,p1), (q1,p2), (q2,p0)\}$
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>[!def] Theorem
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>Bisimilarity is an equivalence relation
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*Proof:*
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Reflexivity: we have to show that q ∼ q, for every q. Consider the following relation
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