Ordinal is Less than Successor

Theorem

Let $x$ be an ordinal.

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:

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

$\blacksquare$