Re: ZFC means?


G. Frege wrote:
> On 20 Dec 2005 12:17:51 -0800, "Charlie-Boo" <chvol@xxxxxxx> wrote:

> > What is your definition of wff and demonstration that x ~e x
> > satisfies that definition?

> How about (just for the sake of the argument):
>
> <variable> ::= u | v | w | x | y | z
>
> <wff> ::= ~(<wff>) | <variable> e <variable>
>
> Now "x" is a variable (since it's a symbol in one of the clauses given
> at the right side of the definition of /variable/), hence "x e x" is a
> wff (according to the definition of /wff/). Hence "~(x e x)" is a wff.
> Applying our definition from above, we get that "x ~e x" is a wff.
>
> WOW!

Then none of your wffs expresses a set and there is no need to give
axioms to specify which wffs do. (You just threw the baby out with the
bath water.)

Still my comments apply and the axioms of ZF are unneeded.

C-B

> F.

.

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