Re: [Haskell-cafe] Proof that Haskell is RT

You can't write a straightforward dynamic semantics (in, say, denotational style) for Haskell. The problem is that with type classes you need to know the types compute the values. You could use a two step approach: first make a static semantics that does type inference/checking and translates Haskell into a different form that has resolved all overloading. And, secondly, you can write a dynamic semantics for that language. But since the semantics has to have the type inference engine inside it, it's going to be a pain. It's possible that there's some more direct approach that represents types as some kind of runtime values, but nobody (to my knowledge) has done that. -- Lennart On Wed, Nov 12, 2008 at 12:39 PM, Luke Palmer wrote:

On Wed, Nov 12, 2008 at 3:21 AM, Jules Bean wrote:

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Andrew Birkett wrote:

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Hi,

Is a formal proof that the Haskell language is referentially transparent? Many people state "haskell is RT" without backing up that claim. I know that, in practice, I can't write any counter-examples but that's a bit handy-wavy. Is there a formal proof that, for all possible haskell programs, we can replace coreferent expressions without changing the meaning of a program?

The (well, a natural approach to a) formal proof would be to give a formal semantics for haskell.

Haskell 98 does not seem that big to me (it's not teeny, but it's nothing like C++), yet we are continually embarrassed about not having any formal semantics. What are the challenges preventing its creation?

Luke _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe