# Equality of Ordered Pairs/Necessary Condition/Proof from Empty Set Formalization

## Theorem

Let $\tuple {a, b}$ and $\tuple {c, d}$ be ordered pairs such that $\tuple {a, b} = \tuple {c, d}$.

Then $a = c$ and $b = d$.

## Proof

First a lemma:

Let $\set {a, b}$ and $\set {a, d}$ be doubletons such that $\set {a, b} = \set {a, d}$.

Then:

- $b = d$

$\Box$

Let $\tuple {a, b} = \tuple {c, d}$.

From the empty set formalization:

- $\set {\set {\O, a}, \set {\set \O, b} } = \set {\set {\O, c}, \set {\set \O, d} }$

First we note the special case where $a = \set \O$ and $b = \O$.

Then we have:

\(\ds \set {\set {\O, a}, \set {\set \O, b} }\) | \(=\) | \(\ds \set {\set {\O, \set \O}, \set {\set \O, \O} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \set {\set {\O, \set \O}, \set {\O, \set \O} }\) | Axiom of Extension: $\set {\O, \set \O } = \set {\set \O, \O}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \set {\set {\O, \set \O} }\) | Doubleton Class of Equal Sets is Singleton Class | |||||||||||

\(\ds \) | \(=\) | \(\ds \set {\set {\O, c}, \set {\set \O, d} }\) | by definition |

Thus we have that:

- $c = d = \set {\O, \set \O}$

and so:

- $c = \set \O$ and $d = \O$

leading to the result that $a = c$ and $b = d$.

$\Box$

Suppose otherwise, that either $a \ne \set \O$ or $b \ne \O$, or both.

Let $x \in \set {\set {\O, a}, \set {\set \O, b} }$.

Then either:

- $x = \set {\O, a}$

or:

- $x = \set {\set \O, b}$

First suppose $x = \set {\O, a}$.

Then:

- $x = \set {\O, c}$ or $x = \set {\set \O, d}$

Aiming for a contradiction, suppose $x = \set {\set \O, d}$.

That is:

- $\set {\O, a} = \set {\set \O, d}$

which means that $a = \set \O$ and $d = \O$.

But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\set \O, d}$.

Thus $x = \set {\O, c}$.

We have that:

- $x = \set {\O, a}$ and $x = \set {\O, c}$

and so:

- $\set {\O, a} = \set {\O, c}$

It follows from the lemma that:

- $a = c$

Now suppose $x = \set {\set \O, b}$.

Then:

- $x = \set {\O, c} \lor \set {\set \O, d}$

Aiming for a contradiction, suppose $x = \set {\O, c}$.

That is:

- $\set {\set \O, b} = \set {\O, c}$

which means that $b = \O$ and $c = \set \O$.

But as $\O \ne \set \O$ it follows by Proof by Contradiction that $x \ne \set {\O, c}$.

Thus $x = \set {\set \O, d}$.

We have that:

- $x = \set {\set \O, b}$ and $x = \set {\set \O, d}$

and so:

- $\set {\set \O, b} = \set {\set \O, d}$

It follows from the lemma that:

- $b = d$

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Exercise $4.1. \ \text{(a)}$