What would be a good way to understand (A x A) - (B x C) = (A - B) x (A - C)? I'm studying Cartesian Products and am tasked with proving or disproving (A x A) - (B x C) = (A-B)x(A-C). However, I'm having difficulties understanding what, exactly, is being stated, and hence that prevents me from starting any type of proof. I think that what looms over my lack of understanding in this is how to work with the Cartesian product and the complement set, especially how those two concepts overlap in the problem. Could someone please provide some clarity? 
 A: Take it one step at a time .  It may also help to use $\neg~z\in Z$ rather than $z\notin Z$.
$\begin{align}&\qquad(A\times A)\smallsetminus(B\times C)\\&=\{\langle x,y\rangle: (\langle x,y\rangle\in A\times A)\wedge\neg(\langle x,y\rangle\in B\times C)\}&&\text{def'n of set difference}\\&=\{\langle x,y\rangle: (x\in A\wedge y\in A)\wedge\neg(x\in B\wedge y\in C)\}&&\text{def'n of cartesian product}\\&=\{\langle x,y\rangle: (x\in A\wedge y\in A)\wedge(\neg~x\in B\color{red}\vee\neg~ y\in C)\}&&\text{deMorgan's Rules}\\[3ex]&\qquad(A\smallsetminus B)\times(A\smallsetminus C)\\&=\{\langle x,y\rangle:(x\in A\smallsetminus B)\wedge (y\in A\smallsetminus C)\}&&\text{def'n cartesian product}\\&=\{\langle x,y\rangle:(x\in A\wedge\neg~ x\in B)\wedge(y\in A\wedge\neg~ y\in C)\}&&\text{def'n set difference}\\&=\{\langle x,y\rangle:(x\in A\wedge y\in A)\wedge(\neg~x\in B\color{red}\wedge\neg~ y\in C)\}&&\text{Commutativity and Association}\end{align}$
As you can plainly see, these set builders do not have equivalent conditions.
You shall then be able to find a counter example which will demonstrate this.
A: I would propose using one Cartesian product that you definitely know — the coordinate plane — to help you visualize the problem. After all, the usual coordinate plane
$$\mathbb{R}^2=\color{blue}{\mathbb{R}}\times\color{magenta}{\mathbb{R}}=\{(x,y)\mid x\in\color{blue}{\mathbb{R}},y\in\color{magenta}{\mathbb{R}}\}$$
is an example of the Cartesian product of two sets, isn't it?
If $F\subseteq\color{blue}{\mathbb{R}}$ and $G\subseteq\color{magenta}{\mathbb{R}}$ are, for example, intervals on the $x$- and $y$-axes, respectively, then their Cartesian product is a rectangle formed by the intersection of the two stripes. Say, in the example below, $F=[2,6]$ and $G=[3,5]$, and then $F\times G$ is the rectangle in the middle.
     
Now, just as an example, take some interval $A$, for example $A=[1,10]$. Then $A\times A$ is a square. Consider $B=[1,2]$ on the $x$-axis and $C=[1,3]$ on the $y$-axis. And now draw two pictures: what is $(A\times A)-(B\times C)$ and what is $(A-B)\times(A-C)$?
Of course, this is just an example. You still need to learn to work with Cartesian products in general. But I hope that visualizing can help you develop some initial understanding.
A: The Cartesian product of two sets has an ordered pair between each pair of elements straddling both sets. For example, $\{1,2\}×\{1,2\}=\{(1,1),(1,2),(2,1),(2,2)\}$.
The set difference $A-B$ is a subset of $A$; it contains those elements that are in $A$ but not in $B$. For example, $\{1,2,3,4\}-\{2,3,5\}=\{1,4\}$.
Now set $A=\{1,2\},B=C=\{2\}$ and compare both sides of the claim at hand. See what you get.
A: Consider we can partition $A$ into two distinct parts: All the elements that are also in $B$, and all the elements that are not in $B$.  That is  $A = (A\cap B) \cup (A\setminus B)$.  (And, $(A\cap B)\cap (A\setminus B) =\emptyset$.
Likewise we can partition $A$ into the elements that are in $C$ and the elements that are not in $C$.  That is $A = (A\cap C)\cup (A\setminus C)$.
So that cuts up $A\times A$ into four distinct parts:
$A\times A = [(A\cap B)\cup (A\setminus B)]\times [(A\cap C)\cup (A\setminus C)]=$
1) $[A\cap B]\times [A\cap C] \cup$
2) $[A\cap B]\times [A\setminus C] \cup$
3) $[A\setminus B] \times [A\cap C] \cup$
4) $[A\setminus B]\times [A\setminus C]$
....Now part 1) will be a subset of $B\times C$.  That is $[A\cap B]\times [A\cap C]\subset B\times C$.  But parts 2), 3), and 4) will have no part of $B\times C$.
So $(A\times A)\setminus (B\times C)$ will be parts 2)3) and 4) combined but not 1).
That is $(A\times A)\setminus (B\times C)=[A\cap B]\times [A\setminus C] \cup [A\setminus B] \times [A\cap C] \cup [A\setminus B]\times [A\setminus C]$.
Now $(A\setminus B)\times (A\setminus C)$ is .... simply part 4).  So no, they are not equal
(unless 2) and 3) are empty...)
A: $A\times B:=\{(a,b):a\in A,b\in B\}$ and $A-B:=\{a\in A:a\notin B\}$ BUT 
the claim $$(A\times A) -(B\times C)=(A-B)\times (A-C)$$ is NOT universally true. One counterexample is the case $A=B=\{1,2\}\;\text{with}\;B\cap C=\emptyset$, say $C=\{3,4\}$, so that the claim becomes 
$$\{1,2\}\times\{1,2\}-\{1,2\}\times\{3,4\}=(\{1,2\}-\{1,2\})\times(\{1,2\}-\{3,4\})$$
$$\{1,2\}\times\{1,2\}=\emptyset\times\{1,2\}=\emptyset$$ which is obviously false. 
