# Equivalence of Formulations of Axiom of Pairing

## Theorem

The following formulations of the **Axiom of Pairing** in the context of **axiomatic set theory** are equivalent:

### Strong Form

For any two sets, there exists a set to which only those two sets are elements:

- $\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \iff z \in c}$

### Weak Form

For any two sets, there exists a set to which those two sets are elements:

- $\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \implies z \in c}$

## Proof

### Strong Form implies Weak Form

Let the strong form of the Axiom of Pairing be assumed:

For any two sets, there exists a set to which only those two sets are elements:

- $\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \iff z \in c}$

By definition of the biconditional, this can be expressed as:

- $\forall a: \forall b: \exists c: \forall z: \paren {\paren {z = a \lor z = b \implies z \in c} \land \paren {z \in c \implies z = a \lor z = b} }$

from which, by the Rule of Simplification:

- $\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \implies z \in c}$

Thus the weak form of the Axiom of Pairing is seen to hold.

$\Box$

### Weak Form implies Strong Form

Let the weak form of the Axiom of Pairing be assumed:

For any two sets, there exists a set to which those two sets are elements:

- $\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \implies z \in c}$

By the Axiom of Specification, let us create the set $c'$ as:

- $c' = \set {z: \paren {z \in c} \land \paren {z = a \lor z = b} }$

By the Axiom of Extensionality, it follows that:

- $c' = \set {a, b}$

Thus we have:

- $\forall a: \forall b: \exists c': \forall z: \paren {z = a \lor z = b \implies z \in c'}$

and:

- $\forall a: \forall b: \exists c': \forall z: \paren {z \in c' \implies z = a \lor z = b}$

That is:

- $\forall a: \forall b: \exists c': \forall z: \paren {z = a \lor z = b \iff z \in c'}$

Thus the strong form of the Axiom of Pairing is seen to hold.

$\blacksquare$