Subset is Compatible with Ordinal Successor/Proof 3

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

Then:


 * $x \in y \implies x^+ \in y^+$.

Proof
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.

By Ordinal Membership is Trichotomy, one of the following must be true:


 * $x^+ = y^+$
 * $y^+ \in x^+$
 * $x^+ \in y^+$

We will show that the first two are both false, so that the third must hold.

Two preliminary facts:

By $(1)$ and Equality of Successors:


 * $x^+ ≠ y^+$

Thus the first of the three possibilities is false.

Suppose for the sake of contradiction that $y^+ \in x^+$.

Then:

But we already know that $y \notin x$ by $(2)$ and $y \ne x$ by $(1)$, so this is a contradiction, and we conclude that $y^+ \notin x^+$.

Thus we have shown that the second possibility is false.

Thus the third and final one must hold: $x^+ \in y^+$.