Prove that for all sets $A,B,C$ $$A \times ( B - C) = (A \times B) - (A \times C).$$
My proof:
$$ \begin{align} A \times (B-C) &= \{(a,b) \mid a \in A \land b \in B-C\} \\ &= \{(a,b) \mid a \in A \land b \in B \land b \notin C\} \end{align} $$
$$ \begin{align} (A\times B) - (A\times C) &= \{(a,b) \mid a \in A \land b \in B\} \cap \{(a,b) \mid a \in A \land b \in C\}^{c} \\ &= \{(a,b) \mid a \in A \land b \in B\} \cap \{(a,b) \mid a \notin A \lor b\notin C\} \\ &= \{(a,b) \mid (a \in A \land b \in B) \land (a \notin A \lor b\notin C)\} \\ &= \{(a,b) \mid b \in B \land [(a \in A \land a \notin A) \lor(a \in A \land b\notin C)]\} \\ &= \{(a,b) \mid b \in B \land a \in A \land b\notin C\} \\ &= \{(a,b) \mid a \in A \land b \in B \land b \notin C\} = A \times (B-C) \end{align} $$ I am especially wondering about the validity of combining the intersection via set builder notation
Your proof is correct. However, it is simpler to do it as follows. Given $(x,y)$,\begin{align}(x,y)\in A\times(B\setminus C)&\iff x\in A\wedge y\in B\setminus C\\&\iff x\in A\wedge y\in B\wedge y\notin C\end{align}and\begin{align}(x,y)\in(A\times B)\setminus(A\times C)&\iff(x,y)\in A\times B\wedge(x,y)\notin A\times C\\&\iff x\in A\wedge y\in B\wedge(x\notin A\vee y\notin C)\\&\iff x\in A\wedge y\in B\wedge y\notin C.\end{align}It's the same idea, but I think that it is easier to read.