3 votes

How exactly does a compatible reduction relation change the $\pi$-calculus?

Interesting question. As Damiano says, while syntactically trivial a change, the π-calculus with non-blocking inputs is a different model of computing. (A very different one, and an extremely ...
3 votes

How exactly does a compatible reduction relation change the $\pi$-calculus?

I don't know "exactly" how it changes the calculus, in the sense that I don't have a formal statement measuring the difference (and I am not aware that there exists one), but allowing ...

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