```Date: Dec 13, 2012 10:02 AM
Author: Dan Christensen
Subject: Re: A formal counter-example of Ax Ey P(x,y) -> Ey Ax P(x,y)

On Dec 13, 3:44 am, Graham Cooper <grahamcoop...@gmail.com> wrote:> On Dec 13, 1:41 pm, Dan Christensen <Dan_Christen...@sympatico.ca>> wrote:>>>>>>>>>> > On Dec 12, 6:54 pm, Jan Burse <janbu...@fastmail.fm> wrote:> > > Here is an example where the above method gives a false> > > positive:>> > > Your system should be able to prove:>> > >     ~p & (~p -> p) -> ~p>> > Easy.>> >         1       ~P & [~P => P]> >                 Premise>> >         2       ~P> >                 Split, 1>> > 3       ~P & [~P => P] => ~P> >         Conclusion, 1>> > > But we do not have:>> > >     ~p & (~p -> p) |/- p>> > > In fact by modus ponens it is easy to see that p> > > derives from ~p & (~p -> p).>> > Also easy.>> >         1       ~P & [~P => P]> >                 Premise>> >         2       ~P> >                 Split, 1>> >         3       ~P => P> >                 Split, 1>> >         4       P> >                 Detach, 3, 2>> > 5       ~P & [~P => P] => P> >         Conclusion, 1>> > Sorry, I don't see your point, Jan.>>  3       ~P & [~P => P] => ~P>  5       ~P & [~P => P] => P>> This would definitely qualify for G.I.G.O. though!>~P & [~P => P] is a contradiction, and anything follows from acontradiction.You can show, for example that P & ~P => Q for any Q.DanDownload my DC Proof 2.0 software at http://www.dcproof.com
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