Element of Ordinal is Ordinal/Proof 2

Theorem

Let $n$ be an ordinal.

Let $m \in n$.

Then $m$ is also an ordinal.

That is, the class of all ordinals $\On$ is a transitive class.

Proof

By the definition of ordinal, $n$ is transitive.

Thus $m \subseteq n$.

By Subset of Strictly Well-Ordered Set is Strictly Well-Ordered, it follows that $m$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_m}$.

It is now to be shown that $m$ is transitive.

If $m = \O$ then the result follows by Empty Class is Transitive.

If $m \ne \O$, then let $x \in m$.

If $x = \O$, then $x \subseteq m$ by Empty Set is Subset of All Sets.

If $x \ne \O$, then let $y \in x$.

It suffices to show that $y \in m$.

Since $m \subseteq n$, it follows that $x \in n$.

Also, $y \in x \land x \in n \implies y \in n$ because $n$ is transitive.

And so $x \in n$, $y \in n$, and $m \in n$.

A strict well-ordering is transitive by definition.

Therefore:

$y \in x \land x \in m \implies y \in m$

Hence the result.

$\blacksquare$