vault backup: 2025-04-28 08:47:04

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Marco Realacci 2025-04-28 08:47:04 +02:00
parent 9b50fc51be
commit 1103ed5d9f
3 changed files with 17 additions and 6 deletions

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@ -213,9 +213,10 @@
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@ -36,6 +36,18 @@ $\approx$ is a
*Proof:* *Proof:*
take the symmetric closure of the following relations, that can be easily shown to be weak simulations: take the symmetric closure of the following relations, that can be easily shown to be weak simulations:
1. $S = \{ P,\tau.P \}\cup Id$ 1. $S = \{ (P,\tau.P )\}\cup Id$
2. $S=\{ M+N+\tau.N,M+\tau.N \}\cup Id$ 2. $S=\{ (M+N+\tau.N,M+\tau.N )\}\cup Id$
3. $S=\\{ (M+\alpha.P+) \}$ 3. $S=\{ ((M+\alpha.P+\alpha.(N+\tau.P), M+\alpha.(N+\tau.P)) \} \cup Id$
#### Weak bisimilarity abstracts from any $\tau$
![](../../Pasted%20image%2020250428084340.png)
**There exists no weak bisimulation S that contains (P, Q).**
*Proof:*
By contr. suppose that a bisimulation exists
Since Q −τ→ b.0, there must exist a P such that P ⇒ P and (P,b.0) ∈ S The only P that satisfies P ⇒ P is P itself
hence it should be (P,b.0) ∈ S
Contradiction: P can perform a whereas b.0 cannot !!
Similarly, P/R and Q/R are NOT weakly bisimilar

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