Date: Dec 8, 2012 6:05 PM
Author: Dan Christensen
Subject: Re: A formal counter-example of Ax Ey P(x,y) -> Ey Ax P(x,y)

On Dec 8, 1:58 pm, Graham Cooper <> wrote:
> On Dec 8, 6:14 am, Dan Christensen <> wrote:

> > Let the domain of quantification be U = {x, y} for distinct x and y.
> > Let P be the "is equal to" relation on U.
> > Then Ax Ey P(x,y) would be true since x=x and y=y
> > And Ey Ax P(x,y) would be false since no element of U would be equal
> > to every element of U.

> > See formal proof (in DC Proof 2.0 format) at
> This is a classic Skolem Function example.

This problem is central to predicate calculus. Like Russell's Paradox,
it has spurred various "solutions," Skolem functions being one of
them. My own DC Proof system is another, more natural one (IMHO).

Download my DC Proof 2.0 software at