Successor is Less than Successor

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

Then, $x \in y \iff x^+ \in y^+$.

Necessary Condition
The last part is a contradiction, so $y^+ \notin x^+$.

By Ordinal Membership Trichotomy, $x^+ \in y^+$.

Sufficient Condition
Suppose $y^+ \in x^+$.

By the definition of successor, $y^+ \in x \lor y^+ = x$.

Suppose $y^+ = x$.

By Ordinal Less than Successor, $y \in x$.

Suppose $y^+ \in x$.

By Ordinal Less than Successor, $y \in y^+$.

By Ordinal is Transitive, $y \in x$.

Proof 2
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.

Forward Implication
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 $x^+ = y^+$.

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

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 $y^+ ≠ x^+$.

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

By the 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.

Reverse implication
Let $x^+ \in y^+$.

Then since $y^+$ is transitive, $x^+ \subseteq y^+$.

Thus $x \in y$ or $x = y$.

If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.

Thus $x \in y$.