Strong Bisimulation

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~ is an Equivalence Relation p ~ p p ~ q implies q ~ p p ~ q and q ~ r imply p ~ r 47
Reflexivity Let Q be a process and Id Q = {(p,p) | p ∈ Q}. For reflexivity, it is enough to show that Id Q is a bisimulation. Proof: Suppose Id Q = {(p,p) | p ∈ Q}. We have to show that for all (p,p) ∈ Id Q , if p ⎯→p' α ⎯ , then there exists q’ such that p α ⎯⎯→q' and (p’,q’) ∈ Id Q. Now, let p Id Q p, if p ⎯→p' α ⎯ , then we have to find a state q’ ∈ Q such that p ⎯→q' α ⎯ and p’ Id Q q’. By assumption, p’ ∈ Q, we take q’ = p’, hence p α ⎯⎯→p' , and by definition of Id Q , we have p’ Id Q p’, as required. Finally, since Id Q = Id Q -1 , Id Q is a bisimulation. q.e.d. 48
Page 1 and 2: Strong Bisimulation Overview • Ac
Page 3 and 4: General Automaton a,b,c b,c q3 a,b
Page 5 and 6: LTS and Automaton An LTS can be tho
Page 7 and 8: Strong Simulation - Definition Let
Page 9 and 10: Defining S If we define S = {(p0, q
Page 11 and 12: Diagrams The condition for S to be
Page 13 and 14: Working With Simulations What do we
Page 15: Linking States b p0 b a a p1 p2 a a
Page 19 and 20: Transitivity S 1 ° S 2 = {(p, r) |
Page 21: Bisimulation - Summary Bisimulation

Strong Bisimulation

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