Re: Question about set definition

Owen wrote:
MoBlee:

I am not convinced that Zermelo's
1. AzExAy(y e x <-> (y e z & Fy)), is valid for all z.

This is a separate concern now. The fact is that the existence of an
empty set follows from the axiom schema, which is:

AzExAy(yex <-> (yez & phi))
where x does not occur free in phi.

Let's suppose that z={y}, then 1. implies,
ExAy(y e x <-> Fy), because y e {y} is true,

But {y} is not free for z in ExAy(yex <-> (yez & phi)), so of course
you can't instantiate to {y} in ExAy(yex <-> (yez & phi)). But that has
nothing to do with my proof, from the axiom schema of separation, of
the existence of an empty set.

..which leads to
Russell's paradox.

IF you violate the rule of universal instantiation by instantiating to
a term that is not free for the variable in the formula, then that
leads to the existence of a universal set, which leads to Russell's
paradox.

That is, ~AzExAy(y e x <-> (y e z & Fy)).

If x does not occur free in Fy, then the above is the negation of an
instance of the axiom schema of separation.

Can you demonstrate why I am wrong?

You went wrong in overlooking that
{y} is not free for z in ExAy(y e x <-> (y e z & Fy))
and thinking that making such an incorrect instantiation has anything
to do with the proof I gave of the existence of an empty set.

MoeBlee

.

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