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- From: Daniel Wyckoff <dwyckoff76 AT hotmail.com>
- To: "coq-club AT inria.fr" <coq-club AT inria.fr>
- Subject: [Coq-Club] Functional equality with dependent arguments.
- Date: Thu, 19 Dec 2013 04:08:58 +0000
- Importance: Normal
Dear Coq Club,
I don't know how to prove something like the following (without unfolding of course) (or if it's possible or what the workaround is). (Of course, I don't mind using proof irrelevance or any consistent
axiom.) :
Variable T:Type.
Variable x y:T.
Variable P:T->Prop.
Hypothesis equal : x = y.
Variable pfx:(P x).
Variable pfy:(P y).
Definition def (a:T) (pf:(P a)) := 0 (*or whatever arbitrary defined function*).
Goal def x pfx = def y pfy.
--Daniel Wyckoff
I don't know how to prove something like the following (without unfolding of course) (or if it's possible or what the workaround is). (Of course, I don't mind using proof irrelevance or any consistent
axiom.) :
Variable T:Type.
Variable x y:T.
Variable P:T->Prop.
Hypothesis equal : x = y.
Variable pfx:(P x).
Variable pfy:(P y).
Definition def (a:T) (pf:(P a)) := 0 (*or whatever arbitrary defined function*).
Goal def x pfx = def y pfy.
--Daniel Wyckoff
- [Coq-Club] Functional equality with dependent arguments., Daniel Wyckoff, 12/19/2013
- Re: [Coq-Club] Functional equality with dependent arguments., Jason Gross, 12/19/2013
- RE: [Coq-Club] Functional equality with dependent arguments., Daniel Wyckoff, 12/19/2013
- Re: [Coq-Club] Functional equality with dependent arguments., Jason Gross, 12/19/2013
- RE: [Coq-Club] Functional equality with dependent arguments., Daniel Wyckoff, 12/19/2013
- Re: [Coq-Club] Functional equality with dependent arguments., Jason Gross, 12/19/2013
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