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- From: Pierre Casteran <pierre.casteran AT labri.fr>
- To: Thomas Nelson <thomasharrisonnelson AT gmail.com>
- Cc: coq-club AT pauillac.inria.fr
- Subject: Re: [Coq-Club] can't work with eq
- Date: Tue, 23 Sep 2008 15:42:43 +0200
- List-archive: <http://pauillac.inria.fr/pipermail/coq-club/>
Thomas Nelson a écrit :
I'm having trouble with a very basic problem, I don't know what to do.If you print for instance Gt.
How can one prove
not (eq Eq Gt)
?
I cannot find any theorems about equality that will help me, although
I'm very new to coq.
Thanks,
Tom
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Coq < Print Gt.
Inductive comparison : Set :=
Eq : comparison | Lt : comparison | Gt : comparison
You notice that Gt and Eq are distinct constructors of the inductive type
comparison.
Then the tactic discriminate expresses this fact:
Coq < Lemma diff : Eq <> Gt.
1 subgoal
============================
Eq <> Gt
diff < discriminate.
Proof completed.
diff < Qed.
Pierre
- [Coq-Club] can't work with eq, Thomas Nelson
- Re: [Coq-Club] can't work with eq, Pierre Casteran
- Re: [Coq-Club] can't work with eq,
Serge Leblanc
- Re: [Coq-Club] can't work with eq,
Andrew McCreight
- Re: [Coq-Club] can't work with eq,
Yves Bertot
- Re: [Coq-Club] can't work with eq, Andrew McCreight
- Re: [Coq-Club] can't work with eq,
Yves Bertot
- Re: [Coq-Club] can't work with eq,
Andrew McCreight
- Re: [Coq-Club] can't work with eq,
Serge Leblanc
- Re: [Coq-Club] can't work with eq, Stefan Monnier
- Re: [Coq-Club] can't work with eq, Pierre Casteran
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