Class has Subclass which is not Element

Theorem
Let $A$ be a class.

Then $A$ has at least one subclass $B$ which is not an element of $A$.

Proof
Let a set $x$ be defined as ordinary $x \notin x$.

Let $\map \phi x$ be the set property defined as:
 * $\map \pi x := \neg \paren {x \in x}$

Then by the axiom of specification there exists a class, which can be denoted $B$, such that:
 * $B = \set {x \in A: \neg \paren {x \in x} }$

That is, $B$ contains as elements of all the ordinary sets of $A$.

By the axiom of extension $B$ is unique.

Thus $B$ is the class of all ordinary sets of $A$.


 * $x \in B \iff \paren {x \in A \land x \notin x}$

Hence, for all $x$ which happen to be elements of $A$, we have:
 * $x \in B \iff x \notin x$

$B \in A$.

Then we could set $A$ for $x$, and so obtain:
 * $B \in B \iff B \notin B$

This is a contradiction.

Hence by Proof by Contradiction $B$ cannot be an element of $A$.

That is:
 * $B \notin V$