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

50 CHAPTER 3. BEHAVIOURAL EQUIVALENCES
The pair (s1, s3) is contained in S. (Why?) Moreover, using that R and
R ′ are bisimulations, you should be able to show that so is S. Therefore
s1 ∼ s3, as claimed.
2. We aim at showing that ∼ is the largest strong bisimulation over the set of
states Proc. To this end, observe, first of all, that the definition of ∼ states
that
∼ = {R | R is a bisimulation} .
This yields immediately that each bisimulation is included in ∼. We are
therefore left to show that the right-hand side of the above equation is itself
a bisimulation. This we now proceed to do.
Since we have already shown that ∼ is symmetric, it is sufficient to prove
that if
(s1, s2) ∈ {R | R is a bisimulation} and s1 α → s ′ 1
then there is a state s ′ 2 such that s2 α → s ′ 2 and
(s ′ 1, s ′ 2) ∈ {R | R is a bisimulation} .
Assume, therefore, that (3.3) holds. Since
(s1, s2) ∈ {R | R is a bisimulation} ,
, (3.3)
there is a bisimulation R that contains the pair (s1, s2). As R is a bisimu-
lation and s1 α → s ′ 1 , we have that there is a state s′ 2 such that s2 α → s ′ 2 and
(s ′ 1 , s′ 2 ) ∈ R. Observe now that the pair (s′ 1 , s′ 2 ) is also contained in
{R | R is a bisimulation} .
Hence, we have argued that there is a state s ′ 2 such that s2 α → s ′ 2 and
which was to be shown.
(s ′ 1 , s′ 2 ) ∈ {R | R is a bisimulation} ,
3. We now aim at proving that ∼ 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 .