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The equivalence studied up to now is quite discriminating, in the sense that it distinguishes, for example, τ.P and τ.τ.P.
- If an external observer can count the number of non-observable actions (i.e., the τ’s), this distinction makes sense.
- If we assume that an observer cannot access any internal information of the system, then this distinction is not acceptable.
The idea of the new equivalence is to ignore (some) τ’s:
- a visible action must be replied to with the same action, possibly together with some internal actions
- an internal action must be replied to by a (possibly empty) sequence of internal actions.
We define the relation \implies
as:
P \implies P'
if and only if there exist P_{0}, P_{1},\dots,P_{k}
(for k \geq 0
) such that P=P_{0} \xrightarrow{\tau} P_{1} \xrightarrow{\tau}\dots\xrightarrow{\tau}Pk=P'
relation \xRightarrow{\hat{\alpha}}
:
- if
\alpha=\tau
then\xRightarrow{\hat{\alpha}}\triangleq\implies
- otherwise
\xRightarrow{\hat{\alpha}}\triangleq\implies\xrightarrow{\alpha}\implies
S is a weak simulation if and only if \forall(p, q) \in S \space \forall p \xrightarrow{\alpha} p' \exists q' \space s.t. \space q\xRightarrow{\hat{\alpha}}q' \space and \space (p', q') \in S
A relation S is called weak bisimulation if both S
and S^{-1}
are weak simulations.
We say that p and q are weakly bisimilar, written p \approx q
, if there exists a weak bisimulation S
such that (p, q) \in S
.
Prop:
\approx
is a
- equivalence
- congruence
- weak bisimulation
\sim \subset \approx
Examples of weakly bisimilar processes
Theorem: given any process P and any sum M, N, then:
P \approx \tau.{P}
M+N+\tau.N \approx M + \tau.N
M+\alpha.P+\alpha.(N+\tau.P) \approx M + \alpha.(N + \tau.P)
Proof: take the symmetric closure of the following relations, that can be easily shown to be weak simulations:
S = \{ (P,\tau.P )\}\cup Id
S=\{ (M+N+\tau.N,M+\tau.N )\}\cup Id
S=\{ ((M+\alpha.P+\alpha.(N+\tau.P), M+\alpha.(N+\tau.P)) \} \cup Id
Weak bisimilarity abstracts from any \tau
There exists no weak bisimulation S that contains (P, Q). Proof: By contr. suppose that a bisimulation exists Since Q −τ→ b.0, there must exist a P’ such that P ⇒ P and (P,b.0) ∈ S The only P that satisfies P ⇒ P’ is P itself hence it should be (P,b.0) ∈ S
Contradiction: P can perform a whereas b.0 cannot !! Similarly, P/R and Q/R are NOT weakly bisimilar