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:


 * $\forall z: \forall A: ( ( z \cap A ) \in U )$
 * $\forall z: \forall A: ( A \subseteq z \implies A \in U )$

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

Proof of the First Statement
The Axiom of Subsets states:


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

We will substitute $y \in A$ for Predicate $P(y)$. We arrive at the statement:


 * $\forall z: \forall A: \exists x: \forall y: ( y \in x \iff ( y \in z \land y \in A ) )$

We apply the definition for intersection:


 * $\forall z: \forall A: \exists x: \forall y: ( y \in x \iff y \in ( z \cap A ) )$

We now apply the definition for set equality:


 * $\forall z: \forall A: \exists x = ( z \cap A ) )$

This is equivalent to:


 * $\forall z: \forall A: ( z \cap A ) \in U$

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.

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:


 * $\forall z: \forall A: ( ( z \cap A ) \in U )$

We will now take the definition of the subset:


 * $A \subseteq B \iff \forall x: ( x \in A \implies x \in B )$

A simple propositional calculus manipulation yields:


 * $A \subseteq B \iff \forall x: ( ( x \in A \land x \in B ) \iff x \in A )$

Applying the definition of definition for set equality and intersection, we arrive at:


 * $A \subseteq B \iff ( A \cap B ) = A$

Thus,


 * $A \subseteq B \implies ( ( A \cap B ) \in U -> A \in U )$

We will take the result of the first statement:


 * $\forall z: \forall A: ( ( z \cap A ) \in U )$

Using the above two statements, substituting $z$ for $B$, we arrive at the desired statement:


 * $\forall z: \forall A: ( A \subseteq z \implies A \in U )$

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

We now arrive at the first statement (above):


 * $\forall z: \forall A: ( A \cap z ) \in U$