Reactive Systems: Modelling, Specification and Verification - Cs.ioc.ee

3.3. STRONG BISIMILARITY 49
Theorem 3.1 For all LTSs, the relation ∼ is
1. an equivalence relation,
2. the largest strong bisimulation, and
3. satisfies the following property:
s1 ∼ s2 iff for each action α,
- if s1 α → s ′ 1 , then there is a transition s2 α → s ′ 2 such that s′ 1 ∼ s′ 2 ;
- if s2 α → s ′ 2 , then there is a transition s1 α → s ′ 1 such that s′ 1 ∼ s′ 2 .
Proof: Consider an LTS (Proc, Act, { α →| α ∈ Act}). We prove each of the statements
in turn.
1. In order to show that ∼ is an equivalence relation over the set of states Proc,
we need to argue that it is reflexive, symmetric and transitive. (See Definition
3.1.)
To prove that ∼ is reflexive, it suffices only to provide a bisimulation that
contains the pair (s, s), for each state s ∈ Proc. It is not hard to see that the
identity relation
I = {(s, s) | s ∈ Proc}
is such a relation.
We now show that ∼ is symmetric. Assume, to this end, that s1 ∼ s2 for
some states s1 and s2 contained in Proc. We claim that s2 ∼ s1 also holds.
To prove this claim, recall that, since s1 ∼ s2, there is a bisimulation R that
contains the pair of states (s1, s2). Consider now the relation
R −1 = {(s ′ , s) | (s, s ′ ) ∈ R} .
You should now be able to convince yourselves that the pair (s2, s1) is contained
in R −1 , and that this relation is indeed a bisimulation. Therefore
s2 ∼ s1, as claimed.
We are therefore left to argue that ∼ is transitive. Assume, to this end, that
s1 ∼ s2 and s2 ∼ s3 for some states s1, s2 and s3 contained in Proc. We
claim that s1 ∼ s3 also holds. To prove this, recall that, since s1 ∼ s2 and
s2 ∼ s3, there are two bisimulations R and R ′ that contain the pairs of states
(s1, s2) and (s2, s3), respectively. Consider now the relation
S = {(s ′ 1, s ′ 3) | (s ′ 1, s ′ 2) ∈ R and (s ′ 2, s ′ 3) ∈ R ′ , for some s ′ 2} .