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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.
- [Coq-Club]identical functions up to rewriting, rcp
- Re: [Coq-Club]identical functions up to rewriting, Yevgeniy Makarov
- <Possible follow-ups>
- Re: [Coq-Club]identical functions up to rewriting, rcp
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