coq-club AT inria.fr
Subject: The Coq mailing list
List archive
- 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.)
- [Coq-Club] Coq Proof Help, zaman
- Re: [Coq-Club] Coq Proof Help, Adam Chlipala
Archive powered by MhonArc 2.6.16.