Russell's Paradox/Proof 2

Proof
the, that for all predicates $P$ where $S$ is not free:
 * $\exists S : \forall x : \paren {x \in S \iff \map P x}$

Since $x \notin x$ is a predicate where $S$ is not free, it follows that:
 * $\exists S : \forall x : \paren {x \in S \iff x \notin x}$

is an instance of the.

By Existential Instantiation:
 * $\forall x: \paren {x \in S \iff x \notin x}$

By Universal Instantiation:
 * $S \in S \iff S \notin S$

But this contradicts Biconditional of Proposition and its Negation.

Thus, the yields a contradiction.