# Equality of Ordered Pairs/Necessary Condition

## 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 from Kuratowski Formalization

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 Kuratowski formalization:

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

There are two cases: either $a = b$, or $a \ne b$.

#### Case 1

Suppose $a = b$.

Then:

$\set {\set a, \set {a, b} } = \set {\set a, \set a} = \set {\set a}$

Thus $\set {\set c, \set {c, d} }$ has only one element.

Thus $\set c = \set {c, d}$ and so $c = d$.

So:

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

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

Thus the result holds.

$\Box$

#### Case 2

Now suppose $a \ne b$. By the same argument it follows that $c \ne d$.

So that means that either $\set a = \set c$ or $\set a = \set {c, d}$.

Since $\set {c, d}$ has $2$ distinct elements, $\set a \ne \set {c, d}$.

Thus:

$\set a = \set c$

and so $a = c$.

Then:

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

But as $a = c$ that means we have:

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

It follows from the lemma that:

$b = d$

$\blacksquare$

## Proof from Empty Set Formalization

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:

 $\displaystyle \set {\set {\O, a}, \set {\set \O, b} }$ $=$ $\displaystyle \set {\set {\O, \set \O}, \set {\set \O, \O} }$ $\displaystyle$ $=$ $\displaystyle \set {\set {\O, \set \O}, \set {\O, \set \O} }$ Axiom of Extension: $\set {\O, \set \O } = \set {\set \O, \O}$ $\displaystyle$ $=$ $\displaystyle \set {\set {\O, \set \O} }$ Doubleton Class of Equal Sets is Singleton Class $\displaystyle$ $=$ $\displaystyle \set {\set {\O, c}, \set {\set \O, d} }$ 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:

$b = d$

$\blacksquare$

## Proof from Wiener Formalization

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:

$b = d$

$\blacksquare$