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Re: RE : Re: [Coq-Club]extensionality and eta-conversion


chronological Thread 
  • From: roconnor AT theorem.ca
  • To: Fabrice Lemercier <nouvid-coq AT yahoo.fr>
  • Cc: coq-club AT pauillac.inria.fr
  • Subject: Re: RE : Re: [Coq-Club]extensionality and eta-conversion
  • Date: Mon, 7 Aug 2006 03:32:14 -0400 (EDT)
  • List-archive: <http://pauillac.inria.fr/pipermail/coq-club/>

On Mon, 7 Aug 2006, Fabrice Lemercier wrote:


You cannot. Eta allows you to identify very similar
functions but not,
say, quicksort and bubble sort.

What about the following argument taken from
Wikipedia?

if  f a==g a  for all lambda expressions a, then in
particular by taking a to be a variable x not
appearing free in f nor g we have  f x == g x  and
hence  ? x. f x == ? x. g x,  and so by eta-conversion
f == g.

You cannot go from f x == g x to \x. f x == \x. g x because in the first case the two x's refer to the same binding, while in the second case the two x's refer to different bindings.

This argument doesn't seem sound in any reasonable lambda calculus, and probably ought to be removed.

--
Russell O'Connor                                      <http://r6.ca/>
``All talk about `theft,''' the general counsel of the American Graphophone
Company wrote, ``is the merest claptrap, for there exists no property in
ideas musical, literary or artistic, except as defined by statute.''





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