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- From: Pierre-Marie Pédrot <pierremarie.pedrot AT ens-lyon.fr>
- To: Coq Club <coq-club AT inria.fr>
- Subject: [Coq-Club] Dependent types and equality
- Date: Thu, 30 Sep 2010 22:54:26 +0200
Dear Coq users,
I've fighting for some days with equality and dependent types for very
simple cases. I'm currently trying to fiddle with Fin, which defined as
in CPDT :
Inductive Fin : nat -> Type :=
| Fin0 : forall n, Fin (S n)
| FinS : forall n, Fin n -> Fin (S n).
and I would like to find an axiom-free proof of decidability of equality
for Fin :
Lemma Fin_eq_dec : forall n (x y : Fin n), {x = y} + {x <> y}.
I have been unable to prove it without using JMeq axioms, and I did not
succeed in proving it through hand-fed terms either. Yet, it works out
of the box in Agda.
Is there an axiom-free proof of this property, and what should it be ? I
did not find anything relevant while skimming CPDT. Do you have any
useful pointer ?
PMP
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- [Coq-Club] Estimate resulting error with [round_double], Michael
- [Coq-Club] Dependent types and equality, Pierre-Marie Pédrot
- Re: [Coq-Club] Dependent types and equality, Adam Chlipala
- [Coq-Club] Dependent types and equality, Pierre-Marie Pédrot
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