Re: Russell's paradox and ZF(C)

On Dec 31, 2:07 am, LauLuna <laureanol...@xxxxxxxx> wrote:
On Dec 31, 12:41 am, Zaljo...@xxxxxxxxx wrote:

On Dec 30, 7:36 am, G. Frege <nomail@invalid> wrote:

We might derive the theorem:

        Ax e A: x e B <-> x !e x.

What is that? that is not clear at all

do you mean

A B AA Ax ( x e A -> ( x e B <-> x !e x ) )

if you mean that , then this is not a theorem of
ZF minus Regularity.

This is not what G. Frege means. B has been previously defined and is
not a variable.

Take into account the definition of B and you'll get

Exists A forall x ((xeB <-> xeA & x !e x) -> (xeA -> (xeB <-> x !e
x)))

which is a theorem of ZF.

ah, yes.

I think you wish to argue on this thread that there are in ZF no set
theoretic grounds to avoid Russell's paradox. That's wrong. The
requirement that sets only be built from previously given sets (except
for some initially given ones) avoids Russell's paradox by mirroring
the iterative conception of sets, which is quite a philosophy of sets.

hmmm...., yes that is correct. No doubt that this is the philosophy
behind ZF, which is a stronge one indeed, But what I am arguing here
is that ZF-Regularity doesn't prove the negation of Russell's paradox,
it rely on FOL to do the job ,what ZF-Regularity does is as you said
it construct sets wihtout being incriminated in Russell's paradox,
since it works in a zone outside that that could lead to Russell's
paradox, it cannot itself prove the negation of the paradox, it only
shy away from it. While ZF with Regularity actually proves the
negation of Russell's paradox, and not only limit its comprehension,
there is a difference here!


Regards

.

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