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?


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