Russell's Paradox/Corollary/Proof 2

Theorem

$\not \exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$

Proof

Aiming for a contradiction, suppose:

$\exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$

By Existential Instantiation:

$\forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$

By Universal Instantiation:

$\map \RR {x, x} \iff \neg \map \RR {x, x} $

But this contradicts Biconditional of Proposition and its Negation.

We thus conclude:

$\not \exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$

$\blacksquare$