ZFC IS INCONSISTENT

http://www.yqcomputer.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://www.yqcomputer.com/

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