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 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.

Thus the first of the three possibilities is false.

By the definition of an ordinal, every ordinal is transitive.

By Ordinal Membership is Asymmetric, $y \notin x$.

Since $y \notin x$ and $y \ne x$, $y \notin x^+$ by the definition of a successor set.

But we already know that $y \notin x$ and $y \ne x$, 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: $x^+ \in y^+$.