Prove that $(A - B) \times X = (A \times X) - (B \times X)$ I tried:
\begin{align} (A - B) \times X & = \{(i, j): i \in (A-B) \wedge j \in X \} \\ & = \{(i, j): i \in A \wedge i \notin B \wedge j \in X \} \\ & = \{(i, j): i \in A \wedge i \notin B \wedge j \in X \wedge j \in X\} \\ & = \{(i,j): (i \in A \wedge j \in X) \wedge j \notin B \wedge j \in X \}
\end{align}
There at the end, $(i \in A \wedge j \in X) \wedge not.something$ would be the $(A \times X) - something$ part, if I managed to get that $not.something$ to be $i \notin B \vee j \notin X$, that is, the negation of $B \times X$.
I'm not sure how to get there. Any hints?
Thank you.
 A: Instead of a string of $=$ signs with manipulations of logical formulae or set expressions at each step, this would be much easier if you used the definition of set equality directly. By this, I mean you should prove that an object is an element of $(A - B) \times X$ if and only if it is an element of $(A \times X) - (B \times X)$.
Here's one direction:
Let $p \in (A - B) \times X$. Then $p=(a,x)$ for some $a \in A-B$ and $x \in X$. In particular, we have


*

*$a \in A$ and $x \in X$, so that $p = (a,x) \in A \times X$;

*$a \not\in B$, so that $p=(a,x) \not\in B \times X$;


It follows that $p \in (A \times X) - (B \times X)$. Hence $(A-B) \times X \subseteq (A \times X) - (B \times X)$.
I'll leave the converse to you.
A: Take any $(i,j)\in (A-B)\times X$. Then $i\in A$, $i\not\in B$ and $j\in X$. Since $i\in A, j\in X$, $(i,j)\in A\times X$. Since $i\not\in B$, we have $(i,j)\not\in B\times X$. 
Thus, $(A-B)\times X \subseteq (A\times X)-(B\times X)$.
In the other direction, take any $(i,j)\in (A\times X)-(B\times X)$. Then since $(i,j)\in A\times X$, we have $i\in A$ and $j\in X$. Since $(i,j)\not\in B\times X$, either $i\not\in B$ or $j\not\in X$. But we already know $j\in X$. So it must be the case that $i\not\in B$. Thus, $(i,j)\in (A-B)\times X$.
Therefore, $(A\times X)-(B\times X)\subseteq (A-B)\times X$.
