Let $A$ and $B$ be sets. Show there exists a unique set $C$ such that $x \in C$ iff $x \in A$ and $x \notin B$ or $x \in B$ and $x \notin A$

Question

I am working through Hrbacek and Jech's Introduction to Set Theory and am not yet feeling totally confident in my proof writing in this more rigorous setting, so I figured I would post solutions on here to get feedback. Is this a valid proof in ZF? (I'm trying to stay as close to the axioms as possible).

Proof

Existence

Let $A$ and $B$ be sets and consider properties $P(x,B): x \notin B$ and $P(x,A): x \notin A$. By the axiom of separation there is a set X such that $x \in X$ iff $x \in A$ and $x \notin B$. Similarly, there is a set Y such that $x \in Y$ iff $x \in B$ and $x \notin A$.

Now, by the axiom of pairing, there exists a set $S$ such that $x \in S$ iff $x = X$ or $x = Y$; that is, there is a set $S = \{X, Y\}$. By the axiom of union, there is a set $C$ such that $x \in C$ iff $x \in U$ for some $U \in S$, that is, $x \in C$ iff $x \in X$ or $x \in Y$.

But $x \in X$ iff $x \in A$ and $x \notin B$. Also $x \in Y$ iff $x \in B$ and $x \notin A$. Therefore $x \in C$ iff $x \in A$ and $x \notin B$ or $x \in B$ and $x\notin A$, as desired.

Uniqueness

Suppose there exist sets $C$ and $C'$ such that $x \in C$ iff $x \in A$ and $x \notin B$ or $x \in B$ and $x\notin A$ and $x \in C'$ iff $x \in A$ and $x \notin B$ or $x \in B$ and $x\notin A$. Well if $x \in C$ it follows that $x \in C'$, and conversely if $x \in C'$ then $x \in C$. So $x \in C$ iff $x \in C'$ and, by axiom of extensionality, $C = C'$. Therefore C is unique.

Questions

I am somewhat confused by the notation for properties $P(x,y,z...)$ I know that I only include variables in the parentheses if they are parameters of the property. But what exactly does it mean for a variable to be a parameter? How do I distinguish a variable that is a parameter from a variable that is not a parameter?

Answer 1

I would suggest this approach:

The Axiom of Pairing tells you that the set $\{A, B \}$ exists, and the Axiom of Union then tells you that the set $D= A \cup B$ exists. Both of these sets are easily shown to be unique via the Axiom of Extensionality.

Using the Axiom of Separation, define $C= \{ x \in D \mid \lnot (x \in A \land x \in B) \}$. Then $C$ is the desired set.