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Re: [Coq-Club] Parametricity Axiom


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  • From: Thorsten Altenkirch <t.altenkirch AT googlemail.com>
  • To: "coq-club AT inria.fr" <coq-club AT inria.fr>
  • Subject: Re: [Coq-Club] Parametricity Axiom
  • Date: Tue, 23 Feb 2021 13:44:26 +0000
  • Accept-language: en-US
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No it doesn't as long as Prop means HProp (at most one inhabitant). Classical
logic is consistent with univalence.

This is different if you say "forall P : Type, P \/ ~P, because then you can
use Hedberg's theorem to show that all tyoes have a propositional equality
and this does contradict univalence.



On 20/02/2021, 22:06, "coq-club-request AT inria.fr on behalf of Daniel
Schepler" <coq-club-request AT inria.fr on behalf of dschepler AT gmail.com> wrote:

On Fri, Feb 19, 2021 at 4:34 PM Maximilian Wuttke <mwuttke97 AT posteo.de>
wrote:
> > Variable classic_dec : forall P : Prop, {P} + {~P}.
> > Variable UIP_refl : forall (X : Type) (x : X) (H : x = x), H =
eq_refl.

Incidentally, classic : forall P : Prop, P \/ ~P implies proof
irrelevance, and therefore as a corollary classic_dec implies
UIP_refl.
--
Daniel Schepler



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