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[Coq-Club] Functional equality with dependent arguments.


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  • From: Daniel Wyckoff <dwyckoff76 AT hotmail.com>
  • To: "coq-club AT inria.fr" <coq-club AT inria.fr>
  • Subject: [Coq-Club] Functional equality with dependent arguments.
  • Date: Thu, 19 Dec 2013 04:08:58 +0000
  • Importance: Normal

  Dear Coq Club,
  I don't know how to prove something like the following  (without unfolding of course) (or if it's possible or what the workaround is). (Of course, I don't mind using proof irrelevance or any consistent
  axiom.) :


Variable T:Type.
Variable x y:T.
Variable P:T->Prop.
Hypothesis equal : x = y.
Variable pfx:(P x).
Variable pfy:(P y).
Definition def (a:T) (pf:(P a)) := 0 (*or whatever arbitrary defined function*).

Goal def x pfx = def y pfy.

--Daniel Wyckoff




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