Axiom of Specification from Replacement and Empty Set

Theorem
The is a consequence of:
 * the {{axiom-link|Replacement}

and
 * the.

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.

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 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, 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, 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:
 * $y \in A \land \map P y$

This contradicts our assumption that:
 * $\not \exists y \in A : \map P y$

By Rule of Explosion, we have:
 * $y \in B$

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

Thus:
 * $\paren {y \in A \land \map P y} \iff y \in B$

and by Universal Generalisation and Existential Generalisation:
 * $\forall A: \exists B: \forall y: \paren {y \in B \iff \paren {y \in A \land \map P y} }$

This shows that the holds when:
 * $\not \exists y \in A : \map P y$

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 $\map P x$.

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

Then we have the contradiction:
 * $\map P x \land \neg \map P x$

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 $\neg \map P x$.

that:
 * $\paren {\map P x \land z = x}$

Then we have the contradiction:
 * $\map P x \land \neg \map P x$

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 $\neg \map P x$.

By Universal Generalisation and Existential Generalisation:
 * $\forall x: \exists a: \forall z: \paren {\map Q {x, z} \iff a = z}$

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

We now have a propositional function $\map Q {x, y}$ satisfying the premise of, 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 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, 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:
 * $\exists x \in A : \paren {\map P x \land y = x}$

By Rule of Addition:
 * $\exists x \in A : \paren {\paren {\map P x \land y = x} \lor \paren {\neg \map P x \land y = w} }$

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

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

This completes the biconditional:
 * $y \in B \iff \paren {y \in A \land \map P y}$

By Universal Generalisation and Existential Generalisation:
 * $\forall A: \exists B: \forall y: \paren {y \in B \iff \paren {y \in A \land \map P y} }$

This shows that the holds when:
 * $\exists y \in A : \map P y$

Thus, the holds both when:
 * $\exists y \in A : \map P y$

and when:
 * $\not \exists y \in A : \map P y$

completing the proof.