Russell's Paradox/Corollary/Proof 1

Theorem

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

Proof

Aiming for a contradiction, suppose there does exist such an $x$.

Let $\RR$ be such that $\map \RR {x, x}$.

Then $\neg \map \RR {x, x}$.

Hence it cannot be the case that $\map \RR {x, x}$.

Now suppose that $\neg \map \RR {x, x}$.

Then by definition of $x$ it follows that $\map \RR {x, x}$.

In both cases a contradiction results.

Hence there can be no such $x$.

$\blacksquare$