Successor Set of Ordinal is Ordinal/Proof 2

Proof
From Ordinal is Transitive, it follows by Successor Set of Transitive Set is Transitive that $\alpha^+$ is transitive.

We now have to show that $\alpha^+$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{\alpha^+} }$.

So suppose that a subset $A \subseteq \alpha^+$ is non-empty.

Then:

We need to show that $A$ has a smallest element.

We first consider the case where $A \cap \alpha$ is empty.

By equation $\paren 1$, it follows that $A \cap \set \alpha$ is non-empty (because $A$ is non-empty).

Therefore:
 * $\alpha \in A$

That is:
 * $\set \alpha \subseteq A$

By Union with Empty Set and Intersection with Subset is Subset, equation $\paren 1$ implies that $A \subseteq \set \alpha$.

Therefore, $A = \set \alpha$ by the definition of set equality.

So $\alpha$ is the smallest element of $A$.

We now consider the case where $A \cap \alpha$ is non-empty.

By Intersection is Subset:
 * $A \cap \alpha \subseteq \alpha$

By the definition of a well-ordered set, there exists a smallest element $x$ of $A \cap \alpha$.

Let $y \in A$.

If $y \in \alpha$, then $y \in A \cap \alpha$

Therefore, by the definition of the smallest element, either $x \in y$ or $x = y$.

Otherwise, $y = \alpha$, and so $x \in \alpha = y$.

That is, $x$ is the smallest element of $A$.