ZFC IS INCONSISTENT

ZFC IS INCONSISTENT
http://inconsistentlogic.blogspot.com/

The proof.
Let's designate a symbol x$Y a predicate: x there is an element of
set Y.
Let's designate a symbol x#Y a predicate: (1) x there is an element
countable sets Y which definable by means of some ZFC-formula F (x)
(2) Existence of set Y certain by means of formula F (x), together
with axioms ZFC.(3) (x$Y) it is demonstrable in ZFC.
Let's designate a symbol x~#Y a predicate: (1) x there is no element
countable sets Y which definable by means of some ZFC-formula F (x)
(2) Existence of set Y certain by means of formula F (x), together
with axioms ZFC. (3)(x$Y) it is demonstrable in ZFC.
Let's assume, that the set of all ZFC-formulas can be considered as
usual ZFC-set. Then by virtue of axioms of substitution in ZFC
existence is demonstrable set W which elements will be all ZFC-
definable sets. Then by virtue of an axiom of allocation in ZFC
existence so-called wild Rassel's set RW which is certain as follows
will be deduced: x$RW<->x~#x. For set RW by obvious image it is
received, that RW#RW<->RW~#RW.
Thus for RW in ZFC at a level of a metatheory it will be
demonstrable, that x$Y it is demonstrable in ZFC and it is
simultaneously indemonstrable in ZFC.
http://planetmath.org/?op=getobj&from=papers&id=329

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