In "A Calculus of Mobile Processes, Part 1" [1], Milner et al. give an example for transmitting a pair of values $(u,v)$ from the process $P$ to either $R$ or $Q$ (see page 13). All three processes are assumed to run in parallel.

They present following solution:

$$ \begin{align} P &= (w) (\bar{x}(w).P' | \bar{w}u.\bar{w}v.0) \\ Q &= x(w).w(y).w(z).Q' \\ R &= x(w).w(y).w(z).R' \end{align} $$

They claim that "it is vital to this idea that w (...) is indeed a private name". Why is this a necessity?

Wouldn't following solution without restriction work?

$$ \begin{align*} P &= \bar{x}(w).P' | \bar{w}u.\bar{w}v.0 \\ Q &= x(\alpha).\alpha(y).\alpha(z).Q' \\ R &= x(\beta).\beta(y).\beta(z).R' \end{align*} $$

[1] http://citeseerx.ist.psu.edu/viewdoc/download?doi=


1 Answer 1


Your solution would work only if you consider $P$, $Q$ and $R$ in parallel as a "closed system", i.e. a system which is not supposed to be composed and interact with an external environment but just to perform internal reactions.

In the context of your quotation, the translation of a polyadic action into a monadic one is supposed to be compositional and to yield a bisimilar labelled semantics. The role of the restriction is to avoid exposing the actions on $w$ to the environment: what happens if the environment performs actions on the public channel in your solution? For instance, consider the process $S$ $$ S = w(y).S' $$ If put in parallel with your version of $P$, $Q$ and $R$ it can "steal" the first/second component of the couple from $Q$ or $R$. This is because you do not need to receive $w$ on $x$ to be able to use it if $w$ is a public name, while that is the only way to acquire it when it is restricted.

  • $\begingroup$ Oh yes, that would be nasty indeed! :) Thank you for your clear and enlightening answer. $\endgroup$
    – ftl
    Sep 24, 2015 at 23:16

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