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Re: [Coq-Club] equality between dependent pairs


Chronological Thread 
  • From: Robbert Krebbers <mailinglists AT robbertkrebbers.nl>
  • To: Dimitri Hendriks <diem AT xs4all.nl>
  • Cc: coq-club AT inria.fr
  • Subject: Re: [Coq-Club] equality between dependent pairs
  • Date: Tue, 08 Oct 2013 13:01:25 +0200

On 10/08/2013 12:07 PM, Dimitri Hendriks wrote:
I have that valq is injective. Now for surjectivity I want to show
that locq is a right-inverse of valq, i.e., forall r:Q, valq (locq r) =
r. So I have to deal with equality between dependent pairs ... Do I need
proof irrelevance, or can I somehow use that > and = are decidable on
nat (as suggested by the answers to faq <http://coq.inria.fr/faq> 38 and
138)? Any alternatives?
All primitive propositions in

let (a,b) := x in a > 0 /\ b > 0 /\ gcd a b = 1

are proof irrelevant, and proof irrelevant propositions are closed under conjunction. So, the whole proposition is proof irrelevant.

* For proof irrelevance of a > 0 and b > 0 use

https://sympa.inria.fr/sympa/arc/coq-club/2013-04/msg00034.html

* For proof irrelevance of gcd a b = 1 use

http://coq.inria.fr/stdlib/Coq.Logic.Eqdep_dec.html

Cheers,

Robbert





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