Subset is Compatible with Ordinal Successor/Proof 3

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:

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:


 * $x^+ \ne y^+$

Thus the first of the three possibilities is false.

$y^+ \in x^+$.

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: $x^+ \in y^+$.