vault backup: 2025-04-30 13:27:49
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@ -10,7 +10,7 @@ $$S_{2}^{(2)} \triangleq v \cdot S_{1}^{(2)}$$
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If we consider S(2) as the specification of the expected behavior of a binary semaphore and S(1) | S(1) as its concrete implementation, we can show that $$S^{(1)}|S^{(1)} \space \textasciitilde \space S^{2}$$
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This means that the implementation and the specification do coincide. To show this equivalence, it suffices to show that following relation is a bisimulation:
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## Restrictions
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**Proposition:** $a.P \textbackslash a ∼ 0$
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@ -50,7 +50,7 @@ One of the main aims of an equivalence notion between processes is to make equat
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**This feature on an equivalence makes it a *congruence***
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Not all equivalences are necessarily congruences (even though most of them are).
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To properly define a congruence, we first need to define an execution context, and then what it means to run a process in a context. Intuitively:
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where C is a context (i.e., a process with a hole ☐), P is a process, and $C[P]$ denotes filling the hole with P
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@ -27,7 +27,7 @@ $\approx$ is a
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4. $\sim \subset \approx$
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#### Examples of weakly bisimilar processes
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**Theorem:** given any process P and any sum M, N, then:
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1. $P \approx \tau.{P}$
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@ -41,7 +41,7 @@ take the symmetric closure of the following relations, that can be easily shown
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3. $S=\{ ((M+\alpha.P+\alpha.(N+\tau.P), M+\alpha.(N+\tau.P)) \} \cup Id$
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#### Weak bisimilarity abstracts from any $\tau$
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**There exists no weak bisimulation S that contains (P, Q).**
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*Proof:*
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@ -71,7 +71,7 @@ A possible implementation of this specification is obtained by having two worker
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- For difficult works, they have to use the special machine.
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There is only one special and only one general machine that the workers have to share.
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where rg and rs are used to require the general/special machine, lg and ls are used to leave the general/special machine, and S and G implement a semaphore on the two different machines.
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@ -83,7 +83,7 @@ i.e., that the specification and the implementation of the factory behave the sa
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Let N denote {rg,rs,lg,ls} and x,y ∊ {E,M,D}
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We can prove that the following relation is a weak bisimulation:
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This is a family of relations:
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- 3 pairs of the second form (one for every x)
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@ -101,7 +101,7 @@ We want to model a lottery L where we can select any ball from a bag that contai
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The specification is: $$L \triangleq \tau.\bar{p_{1}}L+\tau.\bar{p_{2}}.L+\dots+\tau.\bar{p_{n}}.L$$
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where $\tau$'s represent ball extractions and $\tilde{p_{i}}$ is the action that communicates with the value of the extracted ball. The LTS resulting from this specification is:
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We now build a system with n components, one for every ball.
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@ -7,19 +7,19 @@ Inference system = axioms + inference rules
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- completeness: whatever is bisimilar, it can be inferred
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#### Axioms & Rules for Strong Bisimilarity
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quite obvious.
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basically we can let the left or the right process evolve, leaving the other unchanged, or they can synchronize.
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- if a process does not perform any action, a restriction won't do anything
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- ...
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$P$ is in standard form if and only if $P \triangleq \sum_{i}\alpha_{i}P_{i}$ and $\forall_{i}P_{i}$ is in standard form.
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**Lemma:** $\forall P \exists P'$* in standard form such that $\vdash P = P'$
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@ -28,15 +28,15 @@ $P$ is in standard form if and only if $P \triangleq \sum_{i}\alpha_{i}P_{i}$ an
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**Base case:** $P \triangleq 0$. It suffices to consider $P' \triangleq 0$ and conclude reflexivity.
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**Inductive step:** we have to consider three cases.
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replacing one by one every continuation with its standard form, obtaining standard form.
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### Axioms & Rules for Weak Bisimilarity
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#### Example
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A server for exchanging messages, in its minimal version, receives a request for sending messages and delivers the confirmation of the reception
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@ -55,9 +55,9 @@ Let us consider the parallel of processes M and R, by using the axiom for parall
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By using the same axiom to the parallel of the three processes, we obtain
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$$\vdash S|(M|R)=send.(\overline{put}|(M|R))+put.(\overline{go}|R|S)+go.(\overline{rcv}|S|M)$$
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By restricting *put* and *go*, and by using the second axiom for restriction, we have that:
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We now apply the third axiom for restriction to the three summands:
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