Subset is Compatible with Ordinal Successor/Proof 2
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
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x \in y$.
We wish to show that $x^+ \in y^+$.
By Ordinal Membership is Trichotomy:
- Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.
Aiming for a contradiction, suppose $y^+ = x^+$.
Then $y \in x$ or $y = x$ by the definition of successor set.
If $y = x$ then $x \in x$, contradicting the fact that Ordinal is not Element of Itself.
If $y \in x$ then since an ordinal is transitive, $y \in y$, again contradicting Ordinal is not Element of Itself.
Thus $y^+ ≠ x^+$.
Aiming for a contradiction, suppose $y^+ \in x^+$.
By definition of successor set, $y^+ \in x$ or $y^+ = x$.
If $y^+ \in x$, then since $y^+$ and $x$ are both ordinals, $y^+ \subsetneqq x$.
Then $y \in x$.
Since $y$ is transitive, $y \in y$, contradicting Ordinal is not Element of Itself.
If instead $y^+ = x$, then $x \in y \in y^+ = x$, so the same contradiction arises because $x$ is transitive.
Thus $y^+ \notin x^+$.
So the only remaining possibility, that $x^+ \in y^+$, must hold.
$\blacksquare$