# Equality of Ordered Pairs/Necessary Condition/Proof from Wiener 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 Wiener formalization:

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

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

Then either:

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

or:

$x = \set {\set b}$

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

Then:

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

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

That is:

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

Then:

$\set d = \set a$

and also:

$\set d = \O$

But that means:

$a \in \O$

which contradicts the definition of the empty set $\O$.

It follows by Proof by Contradiction that $x \ne \set {\set d}$.

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

We have that:

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

and so:

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

It follows from the lemma that:

$\set a = \set c$

Hence from Singleton Equality:

$a = c$

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

Then:

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

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

That is:

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

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

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

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

We have that:

$x = \set {\set b}$ and $x = \set {\set d}$
$\set b = \set d$

and again by Singleton Equality:

$b = d$

$\blacksquare$