Stronger Validity Criteria for Encoding Synchrony

Abstract

We analyse two translations from the synchronous into the asynchronous $\pi$-calculus, both without choice, that are often quoted as standard examples of valid encodings, showing that the asynchronous $\pi$-calculus is just as expressive as the synchronous one. We examine which of the quality criteria for encodings from the literature support the validity of these translations. Moreover, we prove their validity according to much stronger criteria than considered previously in the literature.

This work was partially supported by the DFG (German Research Foundation).

A Appendix: Boudol’s Translation is Valid up to

Before we prove validity of Boudol’s translation up to weak barbed bisimulation, we further investigate the protocol steps established by Boudol’s encoding. Let and . Pick u, v not free in P and Q, with . Then

Here structural congruence is applied in omitting parallel components $\varvec{0}$ and empty binders (u) and (v). Now the crucial idea in our proof is that the last two reductions are inert, in that set of the potential behaviours of a process is not diminished by doing (internal) steps of this kind. The first reduction above in general is not inert, as it creates a commitment between a sender and a receiver to communicate, and this commitment goes at the expense of the potential of one of the two parties to do this communication with another partner. We employ a relation that captures these inert reductions in a context.

Definition 14

([18]). Let $\equiv \!\Rrightarrow$ be the smallest relation on $\mathcal {P}_{\mathrm{a\pi }}$ such that

• 1. ,

• 2. if $P\equiv \!\Rrightarrow Q$ then $P|C \equiv \!\Rrightarrow Q|C$,

• 3. if $P\equiv \!\Rrightarrow Q$ then $(w) P\equiv \!\Rrightarrow (w) Q$,

• 4. if $P\mathrel {\equiv }P' \equiv \!\Rrightarrow Q' \mathrel {\equiv }Q$ then $P\equiv \!\Rrightarrow Q$,

where .

First of all observe that whenever two processes are related by $\equiv \!\Rrightarrow$, an actual reduction takes place.

Lemma 3

([18]). If $P \equiv \!\Rrightarrow Q$ then $P \longmapsto Q$.

The next two lemmas confirm that inert reductions do not diminish the potential behaviour of a process.

Lemma 4

([18]). If $P \equiv \!\Rrightarrow Q$ and $P \longmapsto P'$ with then there is a $Q'$ with $Q \longmapsto Q'$ and $P' \equiv \!\Rrightarrow Q'$.

Corollary 3

If $P \equiv \!\Rrightarrow ^* Q$ and $P \longmapsto P'$ then either or there is a $Q'$ with $Q \longmapsto Q'$ and $P' \equiv \!\Rrightarrow ^* Q'$.

Proof

By repeated application of Lemma 4.    $\square$

Lemma 5

If $P \equiv \!\Rrightarrow Q$ and $P {\downarrow _{a}}$ for $a \in \{ x, \bar{x} \,|\, x \in \mathcal N \}$ then $Q {\downarrow _{a}}$.

Proof

Let $(\tilde{w})P$ for with denote $(w_1)\cdots (w_n)P$ for some arbitrary order of the $(w_i)$. Using a trivial variant of Lemma 1.2.20 in [39], there are $\tilde{w} \subseteq \mathcal {N}$, and , such that $x\in \tilde{w}$ and . Since $P{{\downarrow _{a}}}$, it must be that $a{=}u$ or $\bar{u}$ with $u\notin \tilde{w}$, and $C{{\downarrow _{a}}}$. Hence $Q{{\downarrow _{a}}}$.    $\square$

The following lemma states, in terms of Gorla’s framework, operational completeness [22]: if a source term is able to make a step, then its translation is able to simulate that step by protocol steps.

Lemma 6

([18]). Let . If $P \longmapsto P'$ then .

Finally, the next lemma was a crucial step in establishing operational soundness [22].

Lemma 7

([18]). Let and . If then there is a $P'$ with $P \longmapsto P'$ and .

Using these lemmas, we prove the validity of Boudol’s encoding up to weak barbed bisimilarity.

Theorem 4

Boudol’s encoding is valid up to .

Proof

Define the relation $\mathrel {\mathcal {R}}$ by $P\mathrel {\mathcal {R}}Q$ iff . It suffices to show that the symmetric closure of $\mathrel {\mathcal {R}}$ is a weak barbed bisimulation.

To show that $\mathrel {\mathcal {R}}$ satisfies Clause 1 of Definition 8, suppose $P \mathrel {\mathcal {R}}Q$ and $P {\downarrow _{a}}$ for $a \in \{ x, \bar{x} \,|\, x \in \mathcal N \}$. Then by Lemma 1. Since , we obtain by Lemma 3, and thus $Q {\Downarrow _{a}}$.

To show that $\mathrel {\mathcal {R}}$ also satisfies Clause 2, suppose $P \mathrel {\mathcal {R}}Q$ and $P \longmapsto P'$. Since , by Lemmas 3 and 6 we have .

To show that $\mathrel {\mathcal {R}}^{-1}$ satisfies Clause 1, suppose $P \mathrel {\mathcal {R}}Q$ and $Q {\downarrow _{a}}$. Since , Lemma 5 yields , and Lemma 1 gives $P {\downarrow _{a}}$, which implies $P {\Downarrow _{a}}$.

To show that $\mathrel {\mathcal {R}}^{-1}$ satisfies Clause 2, suppose $P \mathrel {\mathcal {R}}Q$ and $Q \longmapsto Q'$. Since , by Corollary 3 either or there is a $Q''$ with and . In the first case $P \mathrel {\mathcal {R}}Q'$, so taking $P':=P$ we are done. In the second case, by Lemma 7 there is a $P'$ with $P \longmapsto P'$ and . We thus have $P' \mathrel {\mathcal {R}}Q'$.    $\square$