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- From: Roland Zumkeller <Roland.Zumkeller AT polytechnique.fr>
- To: robert dockins <robdockins AT fastmail.fm>
- Cc: Coq Club <coq-club AT pauillac.inria.fr>
- Subject: Re: [Coq-Club] decidable eq on a well-specified type
- Date: Mon, 2 May 2005 12:43:14 +0200
- List-archive: <http://pauillac.inria.fr/pipermail/coq-club/>
Hi,
The problem comes down to showing that two proofs of (x<limit) are equivalent.
If you "compile" the the inductive predicate "le" into a computational definition this becomes rather easy.
Fixpoint le (a b : nat) { struct a } : Prop :=
match a, b with
| S a, S b => le a b
| S _, O => False
| O, _ => True
end.
The proof is then:
Goal forall a b (p1 p2 : le a b), p1 = p2.
induction a; simpl; intros.
exact(match p1 as p1', p2 as p2' return p1'=p2' with I,I => refl_equal _ end).
destruct b.
exact (False_ind _ p1).
apply IHa.
Qed.
I would be happy to see practical or philosophical arguments against this approach:
What is the advantage of defining "le" inductively?
Best,
Roland
- Re: [Coq-Club] decidable eq on a well-specified type, Roland Zumkeller
- Re: [Coq-Club] decidable eq on a well-specified type, Pierre Courtieu
- Re: [Coq-Club] decidable eq on a well-specified type, roconnor
- <Possible follow-ups>
- Re: [Coq-Club] decidable eq on a well-specified type, Haixing Hu
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