Successor is Less than Successor/Sufficient Condition

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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 1

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

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


Suppose $y^+ = x$.

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


Suppose $y^+ \in x$.

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

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

$\blacksquare$


Proof 2

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


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 $x \in y$.

$\blacksquare$