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

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

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46 CHAPTER 3. BEHAVIOURAL EQUIVALENCES<br />

This completes the proof that R is a strong bisimulation <strong>and</strong>, since (s, t) ∈ R, we<br />

get that s ∼ t.<br />

In order to prove that, e.g., s1 ∼ s2 we can use the following relation<br />

R = {(s1, s2), (s2, s2)} .<br />

The reader is invited to verify that R is ind<strong>ee</strong>d a strong bisimulation. <br />

Example 3.2 In this example we shall demonstrate that it is possible for the initial<br />

state of a labelled transition system with infinitely many reachable states to be<br />

strongly bisimilar to a state from which only finitely many states are reachable.<br />

Consider the labelled transition system (Proc, Act, { α →| α ∈ Act}) where<br />

• Proc = {si | i ≥ 1} ∪ {t},<br />

• Act = {a}, <strong>and</strong><br />

• a →= {(si, si+1) | i ≥ 1} ∪ {(t, t)}.<br />

Here is a graphical representation of this labelled transition system.<br />

s1<br />

t<br />

<br />

a<br />

a <br />

s2<br />

a <br />

s3<br />

a <br />

s4<br />

We can now observe that s1 ∼ t because the relation<br />

R = {(si, t) | i ≥ 1}<br />

a <br />

. . .<br />

is a strong bisimulation <strong>and</strong> it contains the pair (s1, t). The reader is invited to<br />

verify this simple fact. <br />

Consider now the two coff<strong>ee</strong> <strong>and</strong> tea machines in our running example. We can<br />

argue that CTM <strong>and</strong> CTM ′ are not strongly bisimilar thus. Assume, towards a<br />

contradiction, that CTM <strong>and</strong> CTM ′ are strongly bisimilar. This means that there is<br />

a strong bisimulation R such that<br />

Recall that<br />

CTM R CTM ′<br />

CTM ′ coin<br />

→ tea.CTM ′<br />

.<br />

.

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