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- From: Helge Bahmann <hcb AT chaoticmind.net>
- To: coq-club AT inria.fr
- Subject: [Coq-Club] Equality of "sig" types
- Date: Fri, 29 Jun 2012 10:06:35 +0200
Hello,
I am stuck in a proof with the goal "exist P a prf_a = exist P b prf_b" with
the assumption "a = b" available -- I have read about dependent pairs, but I
have been unable to figure out how to apply this syntactically to my problem
(if it applies at all?).
Could anyone give me a hint how I could proceed?
Regards
Helge
- [Coq-Club] Equality of "sig" types, Helge Bahmann, 06/29/2012
- Re: [Coq-Club] Equality of "sig" types, Jason Gross, 06/29/2012
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