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 a
contradiction.

You can show, for example that P & ~P => Q for any Q.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com