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- From: S3 <scubed2 AT gmail.com>
- To: Brian Huffman <brianh AT cs.pdx.edu>
- Cc: coq-club AT inria.fr
- Subject: Re: [Coq-Club] Prove inductive type implies false in cycle
- Date: Mon, 01 Aug 2011 22:16:27 -0700
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> Lemma divides_zero :
> forall n m, divides n m -> m = 0 -> n = 0.
Thanks! That work well and surprisingly easily.
My initial theorem is a direct corollary, as you stated.
It was interesting how minor variations of it didn't work.
(Unfortunately, I'm still a novice, so I don't quite
understand why the simpler form doesn't work.)
If I try a simpler form:
Theorem divides_only_0 n: divides n 0 -> n = 0.
inducting on divides gives a useless hypothesis.
I tried to be fancier and make it like the form you gave
by doing:
assert (exists m, m=0).
That seemed to work identically to your version.
So, the moral of the story is that I have to
additionally state explicitly that m is 0,
otherwise the induction doesn't know
if divides might zig-zag changing m.
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- [Coq-Club] Prove inductive type implies false in cycle, S3
- [Coq-Club] Prove inductive type implies false in cycle,
S3
- Re: [Coq-Club] Prove inductive type implies false in cycle, Brian Huffman
- Message not available
- Re: [Coq-Club] Prove inductive type implies false in cycle, S3
- Re: [Coq-Club] Prove inductive type implies false in cycle, Pierre Casteran
- Re: [Coq-Club] Prove inductive type implies false in cycle, AUGER Cedric
- [Coq-Club] Prove inductive type implies false in cycle,
S3
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