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

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46 CHAPTER 3. BEHAVIOURAL EQUIVALENCES This completes the proof that R is a strong bisimulation and, since (s, t) ∈ R, we get that s ∼ t. In order to prove that, e.g., s1 ∼ s2 we can use the following relation R = {(s1, s2), (s2, s2)} . The reader is invited to verify that R is indeed a strong bisimulation. Example 3.2 In this example we shall demonstrate that it is possible for the initial state of a labelled transition system with infinitely many reachable states to be strongly bisimilar to a state from which only finitely many states are reachable. Consider the labelled transition system (Proc, Act, { α →| α ∈ Act}) where • Proc = {si | i ≥ 1} ∪ {t}, • Act = {a}, and • a →= {(si, si+1) | i ≥ 1} ∪ {(t, t)}. Here is a graphical representation of this labelled transition system. s1 t a a s2 a s3 a s4 We can now observe that s1 ∼ t because the relation R = {(si, t) | i ≥ 1} a . . . is a strong bisimulation and it contains the pair (s1, t). The reader is invited to verify this simple fact. Consider now the two coffee and tea machines in our running example. We can argue that CTM and CTM ′ are not strongly bisimilar thus. Assume, towards a contradiction, that CTM and CTM ′ are strongly bisimilar. This means that there is a strong bisimulation R such that Recall that CTM R CTM ′ CTM ′ coin → tea.CTM ′ . .
3.3. STRONG BISIMILARITY 47 So, by the second requirement in Definition 3.2, there must be a transition CTM coin → P for some process P such that P R tea.CTM ′ . A moment of thought should be enough to convince yourselves that the only process that CTM can reach by receiving a coin as input is coffee.CTM + tea.CTM. So we are requiring that (coffee.CTM + tea.CTM) R tea.CTM ′ . However, now a contradiction is immediately reached. In fact, coffee.CTM + tea.CTM coffee → CTM , but tea.CTM ′ cannot output coffee. Thus the first requirement in Definition 3.2 cannot be met. It follows that our assumption that the two machines were strongly bisimilar leads to a contradiction. We may therefore conclude that, as claimed, the processes CTM and CTM ′ are not strongly bisimilar. Example 3.3 Consider the processes P and Q defined thus: and P def = a.P1 + b.P2 P1 P2 def = c.P def = c.P Q def = a.Q1 + b.Q2 def = c.Q3 Q1 Q2 Q3 def = c.Q3 def = a.Q1 + b.Q2 . We claim that P ∼ Q. To prove that this does hold, it suffices to argue that the following relation is a strong bisimulation R = {(P, Q), (P, Q3), (P1, Q1), (P2, Q2)} . We encourage you to check that this is indeed the case.
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Reactive Systems: Modelling, Specif
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iv CONTENTS 5 Hennessy-Milner logic
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List of Figures 2.1 The interface f
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List of Tables 2.1 An alternative f
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xii PREFACE It has been very satisf
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96 CHAPTER 4. THEORY OF FIXED POINT
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98 CHAPTER 4. THEORY OF FIXED POINT
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Chapter 5 Hennessy-Milner logic In
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103 We are interested in using the
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1. Decide whether the following sta
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109 Note that logical negation is n
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111 Another example of a process th
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a s s1 b s2 b a a t b t1 b t
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Chapter 6 Hennessy-Milner logic wit
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CHAPTER 6. HML WITH RECURSION 117 e
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CHAPTER 6. HML WITH RECURSION 119
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6.1. EXAMPLES OF RECURSIVE PROPERTI
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6.2. SYNTAX AND SEMANTICS OF HML WI
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6.2. SYNTAX AND SEMANTICS OF HML WI
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6.5. MUTUALLY RECURSIVE EQUATIONAL
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6.5. MUTUALLY RECURSIVE EQUATIONAL
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6.6. CHARACTERISTIC PROPERTIES 139
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6.7. MIXING LARGEST AND LEAST FIXED
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6.8. FURTHER RESULTS ON MODEL CHECK
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Chapter 7 Modelling and analysis of
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CHAPTER 7. MODELLING MUTUAL EXCLUSI
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CHAPTER 7. MODELLING MUTUAL EXCLUSI
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7.3. TESTING MUTUAL EXCLUSION 169 i
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7.3. TESTING MUTUAL EXCLUSION 171 D
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