# Axiom of Specification from Replacement and Empty Set

## Theorem

The Axiom of Specification is a consequence of:

the Axiom of Replacement

and

the Axiom of the Empty Set.

## Outline of Proof

Take some set $A$ and some propositional function $\map P x$.

We obtain the set:

$B = \set {x \in A : \map P x}$

by defining the mapping:

$\map f x = \begin{cases} x & : \map P x \\ w & : \text{otherwise} \end{cases}$

on $A$, where we choose some fixed $w \in A : \map P w$.

We obtain its image with the Axiom of Replacement.

This mapping maps all of the $x \in A$ that fulfill $\map P x$ to themselves, ensuring that they are in the image.

It redirects the elements that don't fulfill $\map P x$ to some fixed $w$ that does fulfill $\map P x$.

Thus, the image we have obtained is $B$.

We must also deal with the special case where no elements in $A$ fulfill $\map P x$.

We must map all elements to something to fulfill the mapping definition.

All elements need to be redirected to a $w$ that fulfills $\map P x$, but here, there is no such $w$.

However, the Axiom of the Empty Set provides us with the desired set.

Thus, we can produce the set $B$ both when it is empty and when it is non-empty.

While this proof outline would suffice as a proof, the construction of set-theoretic mappings relies on Cartesian Product Exists and is Unique, which relies on Axiom of Specification, making such a proof circular in this context.

We thus must define a propositional function that is not set theoretic to act as the above $f$, leading to a proof that is conceptually the same, but more tedious.

## Proof

Let $A$ be an arbitrary set.

Let $\map P x$ be an arbitrary propositional function.

It is to be shown that there exists a set $B$ consisting of exactly the $y \in A$ such that $\map P y$.

That is:

$\forall A: \exists B: \forall y: \paren {y \in B \iff \paren {y \in A \land \map P y} }$

By Law of Excluded Middle, there are two choices:

$\exists y \in A : \map P y$

or:

$\not \exists y \in A : \map P y$

Suppose $\not \exists y \in A : \map P x$.

By Axiom of the Empty Set, take $B = \O$.

Take arbitrary $A$ and $y$.

Assume $y \in B$.

This contradicts the empty set definition.

By Rule of Explosion, we have:

$y \in A \land \map P y$

giving:

$y \in B \implies \paren {y \in A \land \map P y}$

Now assume:

$y \in A \land \map P y$

$\not \exists y \in A : \map P y$

By Rule of Explosion, we have:

$y \in B$

giving:

$\paren {y \in A \land \map P y} \implies y \in B$

Thus:

$\paren {y \in A \land \map P y} \iff y \in B$
$\forall A: \exists B: \forall y: \paren {y \in B \iff \paren {y \in A \land \map P y} }$

This shows that the Axiom of Specification holds when:

$\not \exists y \in A : \map P y$

$\Box$

Suppose $\exists y \in A : \map P y$.

Take some fixed $w \in A : \map P w$.

Define the propositional function $\map Q {x, z}$ as follows:

$\paren {\map P x \land z = x} \lor \paren {\neg \map P x \land z = w}$

Usually this would be written as the mapping:

$\map f x = \begin {cases} x & : \map P x \\ w & : \text{otherwise} \end {cases}$

It is to be shown that $\map Q {x, z}$ determines a mapping.

That is:

$\forall x: \exists a: \forall z: \paren {\map Q {x, z} \iff a = z}$

Take arbitrary $x$ and $z$.

Assume that $\map Q {x, z}$, giving:

$\paren {\map P x \land z = x} \lor \paren {\neg \map P x \land z = w}$

By Law of Excluded Middle, there are two choices:

$\map P x$

or:

$\neg \map P x$

Suppose $\map P x$.

Aiming for a contradiction, suppose that:

$\paren {\neg \map P x \land z = w}$

$\map P x \land \neg \map P x$

Thus we have:

$\neg \paren{\neg \map P x \land z = w}$

and by Modus Tollendo Ponens:

$\paren {\map P x \land z = x}$

Take $a=x$.

Thus, for all $z$ such that $\map Q {x, z}$, we have $a = x = z$, giving:

$\map Q {x, z} \implies a = z$

Suppose $\neg \map P x$.

Aiming for a contradiction, suppose that:

$\paren {\map P x \land z = x}$

$\map P x \land \neg \map P x$

Thus we have:

$\neg \paren {\map P x \land z = x}$

and by Modus Tollendo Ponens:

$\paren {\neg \map P x \land z = w}$

Take $a=w$.

Thus, for all $z$ such that $\map Q {x, z}$, we have $a = w = z$, giving:

$\map Q {x, z} \implies a = z$

Thus:

$\map Q {x, z} \implies a = z$

follows from both $\map P x$ and $\neg \map P x$.

$\forall x: \exists a: \forall z: \paren {\map Q {x, z} \iff a = z}$

and $\map Q {x, y}$ determines a mapping.

$\Box$

We now have a propositional function $\map Q {x, y}$ satisfying the premise of Axiom of Replacement, giving:

$\forall A: \exists B: \forall y: \paren {y \in B \iff \exists x \in A : \map Q {x, y} }$

By definition of $\map Q {x, y}$, we have:

$\forall A: \exists B: \forall y: \paren {y \in B \iff \exists x \in A : \paren {\paren {\map P x \land y = x} \lor \paren {\neg \map P x \land y = w} } }$

where $w \in A$ and $\map P w$.

We must show that the Axiom of Specification holds:

$\forall A: \exists B: \forall y: \paren {y \in B \iff \paren {y \in A \land \map P y} }$

Assume $y \in B$.

By the Axiom of Replacement, we have:

$\paren {\map P x \land y = x} \lor \paren {\neg \map P x \land y = w}$

for some $x \in A$.

$\map P x \land y = x$ and $x \in A$ imply $y \in A \land \map P y$.

Recalling that $w \in A$ and $\map P w$, $\neg \map P x \land y = w$ implies $y \in A \land \map P y$.

Thus:

$y \in B \implies \paren {y \in A \land \map P y}$

Now assume $y \in A \land \map P y$.

Then there is an $x$, namely $x = y$, such that:

$\exists x \in A : \paren {\map P x \land y = x}$
$\exists x \in A : \paren {\paren {\map P x \land y = x} \lor \paren {\neg \map P x \land y = w} }$

Thus, by the Axiom of Replacement we have $y \in B$.

Thus:

$\paren {y \in A \land \map P y} \implies y \in B$

This completes the biconditional:

$y \in B \iff \paren {y \in A \land \map P y}$
$\forall A: \exists B: \forall y: \paren {y \in B \iff \paren {y \in A \land \map P y} }$

This shows that the Axiom of Specification holds when:

$\exists y \in A : \map P y$

$\Box$

Thus, the Axiom of Specification holds both when:

$\exists y \in A : \map P y$

and when:

$\not \exists y \in A : \map P y$

completing the proof.

$\blacksquare$