Existence of Set is Equivalent to Existence of Empty Set

Theorem
Let $V$ be a basic universe.

Let $P$ be the axiom:


 * $V$ has at least one element.

Then $P$ is equivalent to the axiom of the empty set:


 * The empty class $\O$ is a set.

Necessary Condition
Let the axiom of the empty set hold.

That is:
 * $\O$ is a set.

By definition of a basic universe, $V$ is a universal class.

Hence, by definition, every set is an element of $V$.

We have that $\O$ is a set.

Thus:
 * $\O \in V$

and by definition of empty class it is seen that $V$ is not empty.

Sufficient Condition
Let $V$ have at least one element.

Let $x \in V$ be an element of $V$.

Then by the axiom of transitivity $x$ is also a class.

By Empty Class is Subclass of All Classes:
 * $\O \subseteq x$

By the axiom of swelledness, every subclass of $x$ is an element of $V$.

That includes $\O$.

Thus:
 * $\O \in V$

as we were to show.