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Marco Realacci 2025-04-08 09:51:02 +02:00
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.obsidian/workspace.json vendored
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@ -213,6 +213,7 @@
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Concurrent Systems/notes/11 - LTSs and Bisimulation.md
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@ -48,4 +48,9 @@ A binary relation S ⊆ Q×Q is a simulation if and only if
We say that p is simulated by q if there exists a simulation S such that $$(p,q) ∈ S$$ We say that p is simulated by q if there exists a simulation S such that $$(p,q) ∈ S$$
We say that S is a bisimulation if both S and S−1 are simulations (where $$S^{-1} = \{(p,q) : (q,p) ∈ S\}$$) We say that S is a bisimulation if both S and S−1 are simulations (where $$S^{-1} = \{(p,q) : (q,p) ∈ S\}$$)
Two states q and p are bisimulation equivalent (or, simply, bisimilar) if there exists a bisimulation S such that (p, q) ∈ S; we shall then write p ∼ q. Two states q and p are bisimulation equivalent (or, simply, bisimilar) if there exists a bisimulation S such that (p, q) ∈ S; we shall then write p ∼ q.
![](../../Pasted%20image%2020250408094749.png)
q0 is simulated by p0; this is shown by the following simulation relation: $$S = \{(q0,p0), (q1,p1), (q2,p1), (q3,p2), (q4,p3)\}$$
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)

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