E(Y) XeY <-> P(X,Y)

so Russell's Paradox "doesn't hold".

E(Y) Y C Z ^ ( XeY <-> P(X,Y) )

=

E(Y) XeY <-> xeZ ^ P(X,Y)

=

A(z): A(p1,p2..pn):
E(y): A(x): x e y <-> (x e z & P(x,y,p1,p2..pn))
***AXIOM OF SPECIFICATION***

=

which is just
E(Y) XeY -> P(X,Y)

________________________________

So to find a CONTRADICTION IN ZFC

we need to formulate a Superset Z that Russell's Set can be a subset
of.

E(Y) Y C Z ^ ( XeY <-> P(X,Y) )   //ZFC AXIOM OF SPECIFICATION

LET P(X,Y) = X~eX

Y = RUSSELL'S SET

LET Z = { Y }

http://en.wikipedia.org/wiki/Axiom_of_regularity
Russell's paradox does not manifest in ZF because ZF does not prove
that the proposed paradoxical set actually exists (e.g., ZF's axiom of
separation only allows us to construct subsets of some existing set,
and thus cannot be used to construct the desired set).

But if Y contains all sets that don't contain themselves,
does Y contain Z?

Z does not contain itself, so ZeY
Also YeZ

No contradiction, so the singleton { RUSSELLSSET } could exist in ZFC.

This doesn't seem to be prevented by Axiom_Of_Regularity

Indeed AOR seems superfluous, if all it does is prevent:

XeXeXeX...

but cannot prevent

XeYeXeY...

it doesn't solve any class of problematic constructions.

So if RSeZ
Z = { RUSSELLSSET }

How does tacking on the subset_of_z constraint to Naive S.T.
E(Y) XeY <-> xeZ ^ P(X,Y)

how does SPECIFICATION (into a subset) prevent RUSSELL SETS?

Z itself has to be CONSTRUCTED as a subset of some further set.

Specify 2 RUSSELL SETS

RS1 = { RS2, ... }
RS2 = { RS1, ... }

then RS1 = { X | X e RS2 ^ P1(X,Y) }
and  RS2 = { X | X e RS1 ^ P2(X,Y) }

BOTH RS1 and RS2 are RUSSELL SETS so Pn(X,Y)=X~eX

then RS1 = { X | X e RS2 ^ X~eX }
and  RS2 = { X | X e RS1 ^ X~eX }

BOTH RUSSELL SETS specified in ZFC!

***************
*CONTRADICTION*
***************

Herc