Can I extract a Coq proof as a Haskell function?

Question

• Ever since I learned a little bit of Coq I wanted to learn to write a Coq proof of the so-called division algorithm that is actually a logical proposition: forall n

Answer 1

The current copy of Software Foundations dated July 25, 2012, answers this quite concisely in the late chapter "Extraction2". The answer is that it can certainly be done, much like this:

Extraction Language Haskell
Extraction "divalg.hs" divalg

One more trick is necessary. Instead of a Prop, divalg must be a Type. Otherwise it will be erased in the process of extraction.

Uh oh, @Anthill is correct, I haven't answered the question because I don't know how to explain how Prof. Pierce accomplished that in his NormInType.v variant of his Norm.v and MoreStlc.v.

OK, here's the rest of my partial answer anyway.

Where "divalg" appears above, it will be necessary to provide a space-separated list of all of the propositions (which must each be redefined as a Type rather than a Prop) on which divalg relies. For a thorough, interesting, and working example of a proof extraction, one may consult the chapter Extraction2 mentioned above. That example extracts to OCaml, but adapting it for Haskell is simply a matter of using Extraction Language Haskell as above.

In part, the reason that I spent some time not knowing the above answer is that I have been using the copy of Software Foundations dated October 14, 2010, that I downloaded in 2011.