Re: What's Wrong With...

From: George Dance (georgedance_at_hotmail.com)
Date: 11/16/04 

Date: 16 Nov 2004 09:39:00 -0800

num num <num@num.num> wrote in message news:<prdgp0thhljf3j8o8f217n40cnume9vg62@4ax.com>...
> What's wrong with this proof?
>
> Ax(Px -> Qx) |- ~Ex(Px & ~Qx)
>
> 1. Ax(Px -> Qx) Premise
> 2. Pa -> Qa UI 1
> 3. Ex(Px -> Qx) EG 2
> 4. Ex(~Px v Qx) Contra 3
> 5. Ex~(Px & ~Qx) DeMorg 4
> 6. ~Ex(Px & ~Qx) ??
>
> While the argument is valid (e.g, 6 can be derived from 1) the move in
> the proof from 5 to 6 must be illegal,

Yes, the move from 5 ("There is something that's not both a P and a
non-Q") to 6 ("There is nothing that is both a P and a non-Q") is
invalid.

> for if not, then we could also
> conclude by a rule of quantifier negation

> 5. Ex~(Px & ~Qx) DeMorg 4
> 6'. ~Ax(Px & ~Qx) QN 5

You can conclude that; your new 6 (which I've renamed 6', just for
better reference) says "It's not true that everything is both a P and
a non-Q"; and that can be validly concluded from "Something is not
both a P and a non-Q".

> since Ex~Px == ~AxPx; yet the two conclusions are not equivalent; so
> num num puzzled.

The source of your confusion may be that you're relying on syntatics
(the rules governing the wff) exclusively, and ignoring the semantics
(what the wff are actually saying).

> This valid proof of the same argument no puzzle num num:
>
> Ax(Px -> Qx) |- ~Ex(Px & ~Qx)
>
> 1. Ax(Px -> Qx) Premise
> 2. Ax(~Px v Qx) Contra 1
> 3. Ax~(Px & ~Qx) DeMorg 2
> 4. ~Ex(Px & ~Qx) QN 3
>
> Yet the UI and EG in the former proof are legal, right? So what rule
> of syntax is violated in the former proof? Please cite a source if
> possible.

It's only the inference from 5 to 6 in your original proof that is
invalid.
5 states that something is not both a P and a non-Q; and that can be
true even if something else *was* both a P and a non-Q (ie, even if
your original 6 was false); so you can't validly infer 6 from 5.