Russell's Paradox/Proof 2

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Theorem

The Axiom of Abstraction leads to a contradiction.


Proof

Aiming for a contradiction, suppose the Axiom of Abstraction, 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 Axiom of Abstraction.

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 Axiom of Abstraction yields a contradiction.

$\blacksquare$