Prove or disprove $ \ (A \times A) - (B \times B) = (A-B) \times (A-B)$ Question: 
Let $ A,B$ be sets.
Prove or disprove: $ \ (A \times A) - (B \times B) = (A-B) \times (A-B)$
My attempt:
Let $ \ (x,y) \in (A \times A) - (B \times B) \implies (x,y) \in (A \times A)$ and $ \ (x,y) \notin (B \times B) \implies x \in A $ and $ \ x\notin B$ and $ \ y \in A$ and $ \ y \notin B \implies (x,y) \in (A-B) \times (A-B)$
Let $ \ (x,y) \in (A-B) \times (A-B) \implies x \in A$ and $ x \notin B$ and $ \ y \in A$ and $ \ y \notin B \implies (x,y) \in (A \times A)$ and $ \ (x,y) \notin (B \times B) \implies (x,y) \in (A \times A) - (B \times B)$.
Is this approach correct?
 A: No, because the statement is false. Let $A$ be $\mathbb{R}$ and let $B = \{0\}$. Then, $(A \times A) - (B \times B)$ is just $\mathbb{R}^2$ minus the origin, while $(A - B) \times (A - B)$ is the plane minus both axes.
A: Your proof that $(A-B)\times(A-B) \subseteq (A\times A)-(B\times B)$ is correct. However, the reverse inclusion fails. Consider $A:=\{1,2\}$ and $B:=\{2\}$. Then $$(A\times A)-(B\times B)=\{(1,1),(1,2),(2,1)\}$$ is not a subset of $$(A-B)\times(A-B)=\{(1,1)\}.$$
A: This is incorrect. For example, if $A = \{0,1\}$ and $B = \{1\}$, then $A \times A - B \times B = \{(0,1),(1,0),(0,0)\}$,while $(A - B) \times (A - B) = \{ (0,0)\}$.
The reason why your answer is incorrect, is because of the following logic : If $(x,y) \notin A \times B$, then it is not true that $x \in A ,y \in B$ must both be false, it is enough if one of them is false. In the example above, you can see this clearly.
Even more obviously : Suppose $A$ and $B$ are finite sets, and  $B \subset A$. Then, the cardinality  of $A \times A - B \times B$ is $|A|^2 - |B|^2$, while the cardinality of $(A-B) \times (A - B)$ is $(|A| - |B|)^2$. If the sets are equal, then their cardinalities are equal, but when do you think this happens? Certainly not all the time.
A: Suppose $A$ has $a$ elements, and $B$ is a $b$-element subset of $A.$ Then the left side of your set equation has $a^2-b^2$ elements, while the right side has $(a-b)^2$ elements. For the sets to be equal, you must have $a^2-b^2=(a-b)^2.$ Is that always true?
A: No. An example is $A=[1,3]$ and $B=[2,4]$ so sketch it in he plane you find it!
A: This is incorrect.
The set $A \times A$, is the sets where both coordinates belong in $A$.
The set $B \times B$ is the set where both coordinates belong in $B$.
So, $A \times A-B \times B$ are the points where both coordinates belong in $A$ except the ones that both coordinates belong in $B$. In other words, the first coordinates or the second coordinates does not belong in $B$.
Without words,
$$ A \times A-B \times B= \left ((A-B)\times A \right ) \cup \left (A\times (A-B)  \right )$$
A: 
Let $ \ (x,y) \in (A \times A) - (B \times B) \implies (x,y) \in (A \times A)$ and $ \ (x,y) \notin (B \times B) \implies x \in A $ and $ \ x\notin B$ and $ \ y \in A$ and $ \ y \notin B $...

No.
Indeed $~(x,y)\in A{\times}A ~\iff ~\big((x\in A)~\wedge~ (y\in A)\big)$.


*

*Any $(x,y)$ will be in $A^2$ exactly when both $x$ and $y$ are in $A$.


However $(x,y)\notin B{\times} B ~\iff~ \big((x\notin B)~{\large\vee}~(y\notin B)\big)$. 


*

*Any $(x,y)$ is not in $B^2$ exactly when at least one from $x$ or $y$ is not in $B$.



Which is to say: 
$$(A{\times}A)\setminus (B{\times}B) ~{=~ (A{\times}A)\cap (B{\times}B)^\complement \\=~(A{\times}A)\cap(B^\complement{\times} B\cup B^\complement{\times} B^\complement\cup B{\times}B^\complement) \\ = ((A{\times}A)\cap(B^\complement{\times} B))~\cup~((A{\times}A)\cap(B^\complement{\times} B^\complement))~\cup~((A{\times}A)\cap( B{\times}B^\complement))\\ =~ (A\setminus B){\times}(A\cap B)~\cup~(A\setminus B){\times}(A\setminus B)~\cup~(A\cap B){\times}(A\setminus B)}$$
