Successor of Element of Ordinal is Subset
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Theorem
Let $x$ and $y$ be ordinals.
Then:
- $x \in y \iff x^+ \subseteq y$
Proof
\(\ds x\) | \(\in\) | \(\ds y\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x^+\) | \(\in\) | \(\ds y^+\) | Successor is Less than Successor | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x^+\) | \(\in\) | \(\ds y\) | Definition of Successor Set | ||||||||||
\(\, \ds \lor \, \) | \(\ds x^+\) | \(=\) | \(\ds y\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x^+\) | \(\subsetneq\) | \(\ds y\) | Transitive Set is Proper Subset of Ordinal iff Element of Ordinal | ||||||||||
\(\, \ds \lor \, \) | \(\ds x^+\) | \(=\) | \(\ds y\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x^+\) | \(\subseteq\) | \(\ds y\) |
$\blacksquare$