Equivalence of Formulations of Axiom of Unions/Proof

Proof
It is assumed throughout that the axiom of extensionality and the axiom of specification both hold.

Formulation $1$ implies Formulation $2$
Let formulation $1$ be axiomatic:

Thus it is posited that for a given set of sets $A$ the union of $A$ exists:


 * $x := \ds \bigcup A := \set {y: \exists z \in A: y \in z}$

and that this is a set.

Formulation $2$ asserts that given the existence of $\ds \bigcup A$, it is axiomatic that $\ds \bigcup A$ is itself a set.

Hence it follows that the truth of formulation $2$ follows from acceptance of the truth of formulation $1$.

From Class Union Exists and is Unique, which depends on:
 * the axiom of extensionality
 * the axiom of specification

$\ds \bigcup A$ is unique for a given $A$.

Formulation $2$ implies Formulation $1$
Let formulation $2$ be axiomatic:

Let $A$ be a set of sets.

Let us create the class:
 * $x := \ds \bigcup A := \set {y: \exists z \in A: y \in z}$

From Class Union Exists and is Unique, which depends on:
 * the axiom of extensionality
 * the axiom of specification

$\ds \bigcup A$ exists and is unique for a given $A$.

We have asserted the truth of formulation $2$.

That is, $\ds \bigcup A$ is a set.

As $A$ is arbitrary, it follows that $\ds \bigcup A$ exists and is unique for all sets of sets $A$.

That is, the truth of formulation $1$ follows from acceptance of the truth of formulation $2$.