Fall 2001, CSE 520:
Lecture 19
Properties of bisimulation
Proposition Bisimulation is an equivalence relation. Namely:
 Reflexivity: for every P, P is bisimilar to itself
 Symmetry: for every P and Q, if P is bisimilar to Q then Q is bisimilar to P
 Transitivity: for every P, Q and R, if P is bisimilar to Q, and Q is bisimilar to R, then P is bisimilar to R
Proof These properties can be proved as follows:
 (Reflexivity) The identical relation Id =def= {(P,P)  P in Proc} is a bisimulation.
 (Symmetry) If R is a bisimulation, then R^{1} =def= {(Q,P)  (P,Q) in R} is a bisimulation too.
 (Transitivity) If R_{1}, R_{2} are bisimulations, then
R_{1}R_{2} =def= {(P,R)  exists Q s.t (P,Q) in R_{1} and (Q,R) in R_{2}} is a bisimulation too.
Proposition Bisimulation is a congruence. Namely, if P is strongly bisimilar to Q, then
 for every a, a.P is strongly bisimilar to a.Q
 for every R, P + R is strongly bisimilar to Q + R
 for every R, P  R is strongly bisimilar to Q  R
 for every L, P\L is strongly bisimilar to Q\L
 for every f, P[f] is strongly bisimilar to Q[f]
Proof We prove here only (2). (3) is one exercise in Assignment #5. All the other ones are easy.
Let R be the bisimulation between P and Q.
Then define R'= { (P+R,Q+R) } union R union Id,
where Id is the identity relation.
We need to show that R' is a bisimulation.
The cases of pairs in R' and in Id are easy.
Consider the other possible pair, namely (P+R,Q+R).

Assume there is a transition from P+R, say P+R a> P'. We have two cases:
 If the transition is due to P, then we have also P a> P'.
Then we know that we also have Q a> Q' and hence Q+R a> Q'
for some Q'. Furthermore (P',Q') in R and therefore (P',Q') in R'.
 If the transition is due to R, then we have also R a> P' and we can simulate the same transition from
Q+R, i.e. we have Q+R a> P'. Furthermore (P',P') in Id and therefore (P',P') in R'.

Similarly for the transitions from Q+R.