Strong Bisimulation

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Symmetry For symmetry, we have to show that if S is a bisimulation then so is its converse S -1 . However, this is obvious from the definition of bisimulation. 49
Transitivity S 1 ° S 2 = {(p, r) | q exists with (p, q) ∈ S 1 and (q, r) ∈ S 2 } Proof: Let (p, r) ∈ S 1 ° S 2 . Then there exists a q with (p, q) ∈ S 1 and (q, r) ∈ S 2 . (→) If p α ⎯⎯→ p' , then since (p, q) ∈ S 1 there exists q’ and q α ⎯⎯→ q' and (p’, q’) ∈ S 1 . Furthermore, since (q, r) ∈ S 2 there exists a r’ with r α ⎯⎯→ r' and (q’, r’) ∈ S 2 . Due to the definition of S 1 ° S 2 it holds that (p’, r’) ∈ S 1 ° S 2 as required. (←) similar to (→). 50
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 and 16: Linking States b p0 b a a p1 p2 a a
Page 17: Reflexivity Let Q be a process and
Page 21: Bisimulation - Summary Bisimulation

Strong Bisimulation

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