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Re: [Coq-Club]identical functions up to rewriting


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  • From: "Yevgeniy Makarov" <emakarov AT gmail.com>
  • To: coq-club AT pauillac.inria.fr
  • Subject: Re: [Coq-Club]identical functions up to rewriting
  • Date: Fri, 16 Feb 2007 11:38:36 +0100
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Hi,

Coq does not have an extensionality axiom or theorem, though this
axiom is consistent with the Set-predicative Calculus of Inductive
Constructions (see FAQ at
http://coq.inria.fr/V8.1/faq.html#htoc41). You can prove instead that
the results of application of both functions to the same argument are
equal (if it helps).

Yevgeniy

On 16 Feb 2007 01:17:48 +0000, 
rcp AT cs.nott.ac.uk
 
<rcp AT cs.nott.ac.uk>
 wrote:
Hello,


How can I prove the following two functions are the same (if at all)?

Lemma id_fin_fun: (fun (n:nat) (i: Fin n) => i) = (fun (n:nat) (i:Fin n) =>
rv n (rv n i).

Where rv (n:nat) : Fin n - Fin n, and reverses the order of Fin n. There is
a proof: forall (n:nat) (i: Fin n), rv n (rv n i) = i. When I use
extensionality I get an error - "unbounded reference in environment".


Thanks,

RP.





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