Fall 2001, CSE 520: Lecture 19

Properties of bisimulation

Proposition Bisimulation is an equivalence relation. Namely:
  1. Reflexivity: for every P, P is bisimilar to itself
  2. Symmetry: for every P and Q, if P is bisimilar to Q then Q is bisimilar to P
  3. 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:
  1. (Reflexivity) The identical relation Id =def= {(P,P) | P in Proc} is a bisimulation.
  2. (Symmetry) If R is a bisimulation, then R-1 =def= {(Q,P) | (P,Q) in R} is a bisimulation too.
  3. (Transitivity) If R1, R2 are bisimulations, then R1R2 =def= {(P,R) | exists Q s.t (P,Q) in R1 and (Q,R) in R2} is a bisimulation too.

Proposition Bisimulation is a congruence. Namely, if P is strongly bisimilar to Q, then

  1. for every a, a.P is strongly bisimilar to a.Q
  2. for every R, P + R is strongly bisimilar to Q + R
  3. for every R, P | R is strongly bisimilar to Q | R
  4. for every L, P\L is strongly bisimilar to Q\L
  5. 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).

  1. Assume there is a transition from P+R, say P+R -a-> P'. We have two cases:

    1. 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'.

    2. 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'.

  2. Similarly for the transitions from Q+R.