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Re: [Coq-Club] Proof without "auto with arith"


chronological Thread 
  • From: "Lucian M. Patcas" <lucian.patcas AT gmail.com>
  • To: Adam Chlipala <adam AT chlipala.net>
  • Cc: coq-club <coq-club AT inria.fr>
  • Subject: Re: [Coq-Club] Proof without "auto with arith"
  • Date: Wed, 22 Jun 2011 11:11:49 -0400
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Point taken, but why not "P = NP"? :)

Say f is

f (t : nat) := if t>1 and t<2000 then True else False

How can I prove this

forall t : nat, t >=2 -> t <= 1000 -> f t

Pardon my ignorance.

On Wed, Jun 22, 2011 at 10:52, Adam Chlipala <adam AT chlipala.net> wrote:
Lucian M. Patcas wrote:
My question is what's the strategy to prove this lemma

forall t : nat, t >=2 -> t <= 1000 -> f t

in Coq, assuming f t holds over that interval.

I wouldn't say there is a well-defined strategy for all lemmas of that form.  For instance, consider [f] that ignores its argument and always returns "P <> NP".




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