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- From: Bernard Hurley <bernard AT marcade.biz>
- To: Adam Chlipala <adamc AT csail.mit.edu>
- Cc: "coq-club AT inria.fr" <coq-club AT inria.fr>
- Subject: Re: [Coq-Club] Equality of functions.
- Date: Wed, 25 Apr 2012 01:41:02 +0100
On Tue, Apr 24, 2012 at 07:54:17PM -0400, Adam Chlipala wrote:
> On 04/24/2012 07:47 PM, Bernard Hurley wrote:
>> In the following scenario, is it possible to prove the Lemma?
>>
>> +++++++++++++++++++++++++++
>> Section test.
>> Variables A B : Set.
>> Variables f g : A -> B.
>>
>> Hypothesis Hyp1: forall a, f a = g a.
>>
>> Lemma test : f = g.
>>
>> End test.
>> +++++++++++++++++++++++++++
>>
>
> Short answer: no.
>
So if I had code like:
Inductive nat_tree : Set :=
| NLeaf : nat -> nat_tree
| NNode : (nat -> nat_tree) -> nat_tree.
then [forall n, f n = g n] does not imply [NNode f = NNode g].
Thanks that is very helpful indeed.
Bernard.
- [Coq-Club] Equality of functions., Bernard Hurley
- Re: [Coq-Club] Equality of functions.,
Adam Chlipala
- Re: [Coq-Club] Equality of functions., Bernard Hurley
- Re: [Coq-Club] Equality of functions., Jonas Oberhauser
- Re: [Coq-Club] Equality of functions., Bernard Hurley
- Re: [Coq-Club] Equality of functions., Kristopher Micinski
- Re: [Coq-Club] Equality of functions.,
Adam Chlipala
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