Axiom of Pairing from Powers and Replacement

Theorem
The Axiom of Pairing is a consequence of:
 * the Axiom of the Empty Set

and
 * the Axiom of Powers

and also
 * the Axiom of Replacement.

Proof
It is to be shown that $2 = \left\{{\varnothing, \left\{{\varnothing}\right\}}\right\}$ is a set.

By the Axiom of the Empty Set, it is seen that $\varnothing$ is a set.

By the Axiom of Powers, it is seen that $\mathcal P \left({\varnothing}\right)$ is also a set.

It is to be shown that $\mathcal P \left({\varnothing}\right) = \left\{{\varnothing}\right\}$.

To determine the elements of $\mathcal P \left({\varnothing}\right)$, we must find all of the sets $z$ such that:
 * $\forall x: x \in z \implies x \in \varnothing$

From the definition of the empty set, the right hand side of this expression is invariably false.

So by Rule of Transposition, we deduce that:
 * $\forall x: x \notin z$

But this is the definition of the empty set.

So by Empty Set is Unique, it follows that $\mathcal P \left({\varnothing}\right) = \left\{{\varnothing}\right\}$.

It is now to be shown that $\mathcal P \left({\left\{{\varnothing}\right\}}\right) = \left\{{\varnothing, \left\{{\varnothing}\right\}}\right\}$.

By Empty Set is Subset of All Sets, $\varnothing \in \mathcal P \left({\left\{{\varnothing}\right\}}\right)$.

By Set is Subset of Itself, $\left\{{\varnothing}\right\} \in \mathcal P \left({\left\{{\varnothing}\right\}}\right)$.

And so:
 * $(1) \quad \left\{{\varnothing, \left\{{\varnothing}\right\}}\right\} \subseteq \mathcal P \left({\left\{{\varnothing}\right\}}\right)$

It is now to be shown that $\left\{{\varnothing, \left\{{\varnothing}\right\}}\right\} \supseteq \mathcal P \left({\left\{{\varnothing}\right\}}\right)$.

Aiming for contradiction, suppose that there exists a set $S \in \mathcal P \left({\left\{{\varnothing}\right\}}\right)$ such that $S \ne \varnothing$ and $S \ne \left\{{\varnothing}\right\}$.

But then $\exists x \ne \varnothing: x \in \left\{{\varnothing}\right\}$, which is lunacy.

Therefore by contradiction:
 * $(2) \quad \mathcal P \left({\left\{{\varnothing}\right\}}\right) \subseteq \left\{{\varnothing, \left\{{\varnothing}\right\}}\right\}$.

Putting $(1)$ and $(2)$ together yields:
 * $\mathcal P \left({\left\{{\varnothing}\right\}}\right) = \left\{{\varnothing, \left\{{\varnothing}\right\}}\right\}$

The set $2 = \left\{\varnothing, \left\{\varnothing\right\}\right\}$ is used with the Axiom of Replacement as the domain for a mapping whose image is $\left\{A, B \right\}$.

A suitable mapping would be:
 * $\left({y = \varnothing \land z = A}\right) \lor \left({y = \left\{{\varnothing}\right\} \land z = B}\right)$

Usually this mapping would be written as:
 * $f \left({x}\right) = \begin{cases}

A & : x = \varnothing \\ B & : x = \left\{{\varnothing}\right\} \end{cases}$