Show that the set of all $x$ such that $x \in P$ and $x \notin Q$ exists

Question

I have to show that set of all $x$ such that $x \in P$ and $x \notin Q$ exists. So, I think I have to assume that sets $P$ and $Q$ already exist. So, with that, I can use axiom schema of specification . This axiom is

$$ \forall w_1,\cdots, w_n \, \forall A \,\exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in A \wedge \varphi(x,w_1, \cdots , w_n, A)] \right) $$

So, I will let $\varphi(x, Q, P) = x \in P \text{ and } x \notin Q $. Also, I will let $A = P$ and $w_1 = Q $. So using universal instantiation, I get the following

$$ \exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge \varphi(x, Q, P)] \right) $$

And using, existential instantiation, there exists a set $B$ such that

$$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge \varphi(x, Q, P)] \right) $$

Using the definition of $\varphi(x, Q, P)$, this can be simplified as following

$$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge x \in P \wedge x \notin Q] \right) $$

$$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge x \notin Q] \right)$$

$$ \forall x \left( x \in B \Longleftrightarrow x \in \{ x \,| x \in P \text{ and } x \notin Q \} \right)$$

Now, using axiom of extensionality, it will follow that

$$ B = \{ x \,| x \in P \text{ and } x \notin Q \} $$

And, since $B$ exists, this means that $\{ x \,| x \in P \text{ and } x \notin Q \} $ also exists. Is the proof good ?

Answer 1

The proof is correct.

As you can already see in your proof, the bit "$x \in P$" is unnecessary in your $\varphi$ because this is already in the axiom of specification (at least, in the formulation you use). There is nothing really wrong with this, but it might trick you into believing that $\varphi$ can contain parameters. Really $\varphi$ is just a formula in the language of set theory. So it would be more precise to say that $\varphi(x, w_1, z)$ is $x \not \in w_1$ (or $x \in z \wedge x \not \in w_1$). Then after the step of universal instantiation the sets $Q$ and $P$ come in the place of $w_1$ and $z$ respectively.