Axiom of Subsets Equivalents

Theorem
The Axiom of Subsets states that:


 * $\forall z: \forall P \left({y}\right): \exists x: \forall y: \left({y \in x \iff \left({y \in z \land P \left({y}\right)}\right)}\right)$

We will prove that this statement is equivalent to the following statements:


 * $\forall z: \forall A: \left({\left({z \cap A}\right) \in U}\right)$
 * $\forall z: \forall A: \left({A \subseteq z \implies A \in U}\right)$

In the above statements, the universe is $U$.

Proof of the First Statement
The Axiom of Subsets states:


 * $\forall z: \forall P \left({y}\right): \exists x: \forall y: \left({y \in x \iff \left({y \in z \land P \left({y}\right)}\right)}\right)$

$y \in A$ is substituted for the propositional function $P \left({y}\right)$.

This leads to the statement:


 * $\forall z: \forall A: \exists x: \forall y: \left({y \in x \iff \left({y \in z \land y \in A}\right)}\right)$

By definition of intersection:


 * $\forall z: \forall A: \exists x: \forall y: \left({y \in x \iff y \in \left({z \cap A}\right)}\right)$

By definition of class equality:


 * $\forall z: \forall A: \exists x = \left({z \cap A}\right)$

This is equivalent to:


 * $\forall z: \forall A: \left({z \cap A}\right) \in U$

because $A \in U \iff \exists x = A$.

Re-derivation of the Axiom of Subsets
Only bi-conditional ($\iff$) statements were used to prove the first result, so it is possible to reverse the step order and arrive at the original Axiom of Subsets by Biconditional is Commutative.

Although this statement is shorter, it uses defined terms, and is thus unsuitable as an axiom.

Proof of the Second Statement
We will take the result of the first statement:


 * $\forall z: \forall A: \left({\left({z \cap A}\right) \in U}\right)$

We will now take the definition of the subset:


 * $A \subseteq B \iff \forall x: \left({x \in A \implies x \in B}\right)$

From Intersection with Subset is Subset:
 * $A \subseteq B \iff \left({A \cap B}\right) = A$

Thus:


 * $A \subseteq B \implies \left({\left({A \cap B}\right) \in U \implies A \in U}\right)$

We will take the result of the first statement:


 * $\forall z: \forall A: \left({\left({z \cap A}\right) \in U}\right)$

Using the above two statements, substituting $z$ for $B$:


 * $\forall z: \forall A: \left({A \subseteq z \implies A \in U}\right)$

Re-derivation of the Axiom of Subsets
Because $\left({A \cap z}\right) \subseteq z$, the antecedent of $\forall z: \forall A: \left({A \subseteq z \implies A \in U}\right)$ is satisfied.

We now arrive at the first statement (above), which in turn can prove the Axiom of Subsets:


 * $\forall z: \forall A: \left({A \cap z}\right) \in U$