Ordinal is Less than Successor

Theorem
Let $x \in \operatorname{On}$.

Let $x^+$ denote the successor of $x$.


 * $x \in x^+$


 * $x \subset x^+$

Proof
$x$ is an ordinal and must also be a set.


 * $x \in ( x \cup \{ x \} ) \land x \subset ( x \cup \{ x \} )$

so by applying the definition of a successor set:


 * $x \in x^+ \land x \subset x^+$