Successor Set of Ordinary Transitive Set is Ordinary

Theorem
Let $x$ be a transitive set which is also ordinary.

Let $x^+$ denote the successor set of $x$:
 * $x^+ = x \cup \set x$

Then $x^+$ is also an ordinary set.

Proof
By definition of ordinary set:
 * $x \notin x$

$x^+ \in x^+$.

Then we have:


 * Let $x^+ \in x$

We have that $x$ is transitive.

Hence:
 * $x^+ \subseteq x$


 * Let $x^+ = x$

We have that: $x^+ = x \cup \set x$

Hence by definition of union of set of sets:
 * $x^+ \subseteq x$

Hence by Proof by Cases:
 * $x^+ \subseteq x$

But we have:
 * $x \in x^+$

and because $x^+ \subseteq x$ it follows that:
 * $x \in x$

But this contradicts our supposition that $x \notin x$.

Hence by Proof by Counterexample:
 * $x^+ \notin x^+$