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Re: [Coq-Club] Excluded middle and decidable equality


Chronological Thread 
  • From: "N. Raghavendra" <raghu AT hri.res.in>
  • To: coq-club AT inria.fr
  • Subject: Re: [Coq-Club] Excluded middle and decidable equality
  • Date: Fri, 27 Sep 2013 11:42:03 +0530
  • Cancel-lock: sha1:zRSTgc10Likr/vVVYrvpiAcuWQo=

At 2013-09-26T17:07:06+02:00, Arnaud Spiwack wrote:

> Coq doesn't have quotients (yet?). So you cannot quite do this
> directly.
>
> The way to go is to use setoids, that is (dependent) pairs of a type
> and an equivalence relation on that type. Then you have to quantify
> on setoids rather than just types. The equality at each setoid is the
> underlying relation. And quotient is obtained just by changing the
> underlying relation of the setoid.

Thanks for the suggestion. I'll try setoids.

> I'm not sure what reading to suggest for a more in-depth view. Maybe
> the first chapter of my thesis can be of help : http://
> pastel.archives-ouvertes.fr/docs/00/60/58/36/PDF/thesis.spiwack.pdf .

Yes, I'll look at that chapter. Your thesis itself looks very
interesting. It seems to discuss a lot of homological algebra in the
context of Coq.

Best regards,
Raghu.

--
N. Raghavendra
<raghu AT hri.res.in>,
http://www.retrotexts.net/
Harish-Chandra Research Institute, http://www.hri.res.in/




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