# Singleton Equality

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## Theorems

Let $x$ and $y$ be sets.

Then:

- $\set x \subseteq \set y \iff x = y$
- $\set x = \set y \iff x = y$

## Proof

\(\ds \) | \(\) | \(\ds \) | \(\ds \set x \subseteq \set y\) | |||||||||||

\(\ds \) | \(\leadstoandfrom\) | \(\ds \forall z:\) | \(\ds \paren {z \in \set x \implies z \in \set y}\) | Definition of Subset | ||||||||||

\(\ds \) | \(\leadstoandfrom\) | \(\ds \forall z:\) | \(\ds \paren {z = x \implies z = y}\) | Definition of Singleton | ||||||||||

\(\ds \) | \(\leadstoandfrom\) | \(\ds \) | \(\ds x = y\) | Equality implies Substitution |

$\Box$

Then:

\(\ds x\) | \(=\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \set x\) | \(=\) | \(\ds \set y\) | Substitutivity of Equality | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \set x\) | \(\subseteq\) | \(\ds \set y\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | by the first part |

$\blacksquare$

## Sources

- 1963: Willard Van Orman Quine:
*Set Theory and Its Logic*: $\S 7.7$