Equality of Ordered Pairs

Theorem
Two ordered pairs are equal iff corresponding elements are equal:


 * $\left({a, b}\right) = \left({c, d}\right) \iff a = c \land b = d$

It follows directly that:
 * $\left({a, b}\right) = \left({b, a}\right) \iff a = b$

or, equivalently, that:
 * $a \ne b \iff \left({a, b}\right) \ne \left({b, a}\right)$

Proof
Let $\left({a, b}\right) = \left({c, d}\right)$.

From the Kuratowski formalization:
 * $\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\}$

Suppose $a = b$.

Then:
 * $\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{a}\right\}, \left\{{a}\right\}}\right\} = \left\{{\left\{{a}\right\}}\right\}$

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

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

So:
 * $\left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\} = \left\{{\left\{{a}\right\}}\right\}$

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

Thus the result holds.

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

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

Since $\left\{{c, d}\right\}$ has distinct elements, $\left\{{a}\right\} \ne \left\{{c, d}\right\}$.

Thus:
 * $\left\{{a}\right\} = \left\{{c}\right\}$

and so $a = c$.

Then:
 * $\left\{{a, b}\right\} = \left\{{c, d}\right\}$

and so $b = d$.

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

Then:
 * $\left\{{a}\right\} = \left\{{c}\right\}$ and $\left\{{a, b}\right\} = \left\{{c, d}\right\}$

Thus:
 * $\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\}$