Subset is Compatible with Ordinal Successor/Proof 1

Theorem
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.

Let $x \in y$.

Then:
 * $x^+ \in y^+$

Proof
The last part is a contradiction, so $y^+ \notin x^+$.

By Ordinal Membership is Trichotomy:
 * $x^+ \in y^+$