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- From: richard dapoigny <richard.dapoigny AT univ-savoie.fr>
- To: coq-club AT inria.fr
- Subject: Re: [Coq-Club] Rewrite with setoid equality
- Date: Tue, 21 Jul 2015 16:03:14 +0200
- Organization: LISTIC
Le 21/07/2015 14:29, Matthieu Sozeau a
écrit :
Hi
Richard,
I can't replay you script so it's unclear what happens.
Best, -- Matthieu Hi Matthieu, Just before the rewrite tactic, i.e., with: ================================================ Lemma ProtoTh5 : forall (p q:Prop), (forall f:Morphism, p ≡≡ (forall r, (p ≡≡ f r) ≡≡ forall r, (q ≡≡ f r))) -> q. Proof. intros p q H1. specialize (H1 T). assert (H2:(forall p:Prop, (forall r, p ≡≡ T r) ≡≡ p)). apply ProtoTh4. assert (H3:=H2). specialize (H2 p). specialize (H3 q). ================================================ I have the following demonstration step: ================================================ 1 subgoals p : Prop q : Prop H1 : p ≡≡ (forall r : Prop, (p ≡≡ T r) ≡≡ (forall r0 : Prop, q ≡≡ T r0)) H2 : (forall r : Prop, p ≡≡ T r) ≡≡ p H3 : (forall r : Prop, q ≡≡ T r) ≡≡ q ______________________________________(1/1) q =============================================== The objective is to rewrite (forall r : Prop, p ≡≡ T r) with p in the _expression_ of H1 (and the same job for (forall r : Prop, q ≡≡ T r) in such a way to get the form: H1 : p ≡≡ (p ≡≡ q) Cheers, Richard |
- [Coq-Club] Rewrite with setoid equality, Richard Dapoigny, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, Matthieu Sozeau, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, richard dapoigny, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, Matthieu Sozeau, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, richard dapoigny, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, Matthieu Sozeau, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, richard dapoigny, 07/21/2015
- Re: [Coq-Club] Rewrite with setoid equality, Matthieu Sozeau, 07/21/2015
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