# Equality of Ordered Pairs

## Theorem

Two ordered pairs are equal if and only if corresponding coordinates are equal:

$\tuple {a, b} = \tuple {c, d} \iff a = c \land b = d$

It follows directly that:

$\tuple {a, b} = \tuple {b, a} \iff a = b$

or, equivalently, that:

$a \ne b \iff \tuple {a, b} \ne \tuple {b, a}$

## Proof

### Necessary Condition

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:

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

and so $b = d$.

$\Box$

### Sufficient Condition

Now suppose $a = c$ and $b = d$.

Then:

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

Thus:

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

$\blacksquare$