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[roconnor AT theorem.ca]

On Sun, 30 May 2004, Roland Zumkeller wrote:

> More generally I'd like to prove
>    (fun x => f x) = (fun x => g x)
> under the hypthesis
>    forall x, f x = g x

I'm pretty sure this is impossible since f and g may not have the same
normal form.  If two expressions do not have the same normal form, they
cannot be proved equal.

-- 
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.''