# Subset is Compatible with Ordinal Successor/Proof 1

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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

\(\ds x \in y\) | \(\implies\) | \(\ds x \ne y\) | No Membership Loops | |||||||||||

\(\ds \) | \(\implies\) | \(\ds x^+ \ne y^+\) | Equality of Successors | |||||||||||

\(\ds x \in y\) | \(\implies\) | \(\ds y \notin x\) | No Membership Loops | |||||||||||

\(\ds \) | \(\implies\) | \(\ds y \notin x^+\) | $x ≠ y$ and Definition of Successor Set | |||||||||||

\(\ds y^+ \in x^+\) | \(\implies\) | \(\ds y \in x^+\) | Successor Set of Ordinal is Ordinal, Ordinals are Transitive, and Set is Element of Successor |

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

By Ordinal Membership is Trichotomy:

- $x^+ \in y^+$

$\blacksquare$

#### Axiom of Foundation

This proof depends on the Axiom of Foundation, by way of No Membership Loops.

Most mathematicians accept the Axiom of Foundation, but theories that reject it, or negate it, have found applications in Computer Science and Linguistics.