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- 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 showAll primitive propositions in
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?
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
- [Coq-Club] equality between dependent pairs, Dimitri Hendriks, 10/08/2013
- Re: [Coq-Club] equality between dependent pairs, Arnaud Spiwack, 10/08/2013
- Re: [Coq-Club] equality between dependent pairs, Robbert Krebbers, 10/08/2013
- Re: [Coq-Club] equality between dependent pairs, Dimitri Hendriks, 10/09/2013
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