The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: A formal counter-example of Ax Ey P(x,y) -> Ey Ax P(x,y)
Replies: 4   Last Post: Dec 9, 2012 12:39 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Dan Christensen

Posts: 8,219
Registered: 7/9/08
Re: A formal counter-example of Ax Ey P(x,y) -> Ey Ax P(x,y)
Posted: Dec 8, 2012 6:05 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.