Subset is Compatible with Ordinal Successor/Proof 3

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

Let $x \in y$.


$x^+ \in y^+$


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:

\(\ds x^+\) \(=\) \(\ds y^+\)
\(\ds y^+\) \(\in\) \(\ds x^+\)
\(\ds x^+\) \(\in\) \(\ds y^+\)

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

Two preliminary facts:

\((1)\)   $:$      \(\ds x \)   \(\ds \ne \)   \(\ds y \)      $x \in y$ and Ordinal is not Element of Itself
\((2)\)   $:$      \(\ds y \)   \(\ds \notin \)   \(\ds x \)      $x \in y$ and Ordinal Membership is Asymmetric

By $(1)$ and Equality of Successors:

$x^+ \ne y^+$

Thus the first of the three possibilities is false.

Aiming for a contradiction, suppose $y^+ \in x^+$.


\(\ds y\) \(\in\) \(\ds y^+\) Definition of Successor Set
\(\ds \leadsto \ \ \) \(\ds y\) \(\in\) \(\ds x^+\) $y^+ \in x^+$ and $x^+$ is an ordinal and therefore transitive.
\(\ds \leadsto \ \ \) \(\ds y\) \(\in\) \(\ds x\) Definition of Successor Set
\(\, \ds \lor \, \) \(\ds y\) \(=\) \(\ds x\)

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