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[roconnor AT theorem.ca]

Can anyone provide me with a reference that shows that

forall (T U : Set) (f g : T -> U),
(forall x : T, f x = g x) -> (fun y : T => f y) = (fun y : T => g y)

is not provable in Coq?

Thanks.

-- 
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.''