Ordinal is Less than Successor

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

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

Then:


 * $x \in x^+$


 * $x \subset x^+$

Proof
$x$ is an ordinal and so by definition is also a set.


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

so by applying the definition of a successor set:


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