vault backup: 2025-04-14 08:29:56

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Marco Realacci 2025-04-14 08:29:56 +02:00
parent a8ea324614
commit 16a89a0328
5 changed files with 70 additions and 15 deletions

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@ -55,3 +55,9 @@ q0 is simulated by p0; this is shown by the following simulation relation: $$S =
To let p0 be simulated by q0, we should have that p1 is simulated by q1 or q2. To let p0 be simulated by q0, we should have that p1 is simulated by q1 or q2.
If S contained one among (p1,q1) or (p1,q2), then it would not be a simulation: indeed, p1 can perform both a c (whereas q1 cannot) and a b (whereas q2 cannot). If S contained one among (p1,q1) or (p1,q2), then it would not be a simulation: indeed, p1 can perform both a c (whereas q1 cannot) and a b (whereas q2 cannot).
**Remark:** for proving equivalence, it is NOT enough to find a simulation of p by q and a simulation of q by p
![](../../Pasted%20image%2020250414082824.png)
p0 is simulated by q0: $$S = {(p0, q0), (p1, q2), (p2, q3)}$$
q0 is simulated by p0: $$S ={(q0,p0),(q1,p1),(q2,p1),(q3,p2)}$$
However, p0 and q0 are not bisimilar: the transition q0 -> a -> q1 is not bisimulable by any transition from p0

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