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Re: [Coq-Club] Coq Proof Help


chronological Thread 
  • From: Adam Chlipala <adam AT chlipala.net>
  • To: zaman <zaman.a25 AT gmail.com>
  • Cc: coq-club AT inria.fr
  • Subject: Re: [Coq-Club] Coq Proof Help
  • Date: Sun, 31 Jul 2011 08:22:02 -0400

zaman wrote:
I wish to say that "A component term must be either a client or server of a
connection; but not both." I have come up with the following as a 
representation
of the above in the coq (based on my definition above):

     (forall con:Connector, forall c:Component, In con (MyConnections x) ->
     (c = (client con) /\ c<>  (server con)) \/ (c<>  (client con) /\ c = 
(server
con)))

The form of this theorem is suspicious. You have a quantified implication whose conclusion mentions [c] but whose assumption does not. Thus, in terms of determining the value of [c], the only help the assumption [In con (MyConnections x)] may be is if it is contradictory, in which case any conclusion follows.

Your question seems to be more about first-order logic than about Coq in particular. Do you believe you can formulate everything on paper in traditional first-order logic? (I expect the theorem statement above is not a desirable part of such a formulation.)



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