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=10.1.1.159.363&rep=rep1&type=pdf
Pi Calculus: Restriction necessary for molecular (atomic) action?
algorithms;computation models;pi calculus
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.