# Successor is Less than Successor

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It has been suggested that this page be renamed.To discuss this page in more detail, feel free to use the talk page. |

This page has been identified as a candidate for refactoring of basic complexity.In particular: This approach is too simple. I feel the necessary and sufficient conditions should be separate pages in their own right. Then this page, aptly renamed, can invoke both theorems proper.Until this has been finished, please leave
`{{Refactor}}` in the code.
Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Refactor}}` from the code. |

## Theorem

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

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

## Proof

\(\ds x \in y\) | \(\implies\) | \(\ds x^+ \in y^+\) | Subset is Compatible with Ordinal Successor | |||||||||||

\(\ds x \in y\) | \(\impliedby\) | \(\ds x^+ \in y^+\) | Sufficient Condition | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x \in y\) | \(\iff\) | \(\ds x^+ \in y^+\) |

$\blacksquare$