vault backup: 2025-04-09 17:50:41

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Concurrent Systems/notes/11 - LTSs and Bisimulation.md

@@ -12,13 +12,13 @@ Automata Behaviour: language equivalence
 >[!note] Language equivalence
 >M1 and M2 are *language equivalent* if and only if L(M1)=L(M2)
 
-![](../../Pasted%20image%2020250408091924.png)
+![](images/Pasted%20image%2020250408091924.png)
 By considering the starting states as also final, they both generate the same language, i.e.:
 $$(20.(tea + 20.coffee))∗ = (20.tea + 20.20.coffee)∗$$
 
 
 But, do they behave the same from the point of view of an external observer??
-![](../../Pasted%20image%2020250408092853.png)
+![](images/Pasted%20image%2020250408092853.png)
 The essence of the difference is WHEN the decision to branch is taken
 -  language equivalence gets rid of branching points
 	- it is too coarse for our purposes!
@@ -38,7 +38,7 @@ We shall usually write s –a–> s′ instead of ⟨s,a,s′⟩ ∈ T.
 
 ### Bisimulation
 Intuitively, two states are equivalent if they can perform the same actions that lead them in states where this property still holds
-![](../../Pasted%20image%2020250408093840.png)
+![](images/Pasted%20image%2020250408093840.png)
 P0 and Q0 are different because, after an a, the former can decide to do b or c, whereas the latter must decide this before performing a.
 
 Let (Q,T) be an LTS.
@@ -50,7 +50,7 @@ We say that S is a bisimulation if both S and S−1 are simulations (where $$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.
 
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+![](images/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).