Proving $(A\times B) \cap (C\times D) = (A\cap C) \times (B\cap D)$ So there is a similiar question in the archives which I looked at after I attempted my proof: Proving that for any sets $A,B,C$, and $D$, if $(A\times B)\cap (C\times D)=\emptyset $, then $A \cap C = \emptyset $ or $B \cap D = \emptyset $
But it is not exactly the same, so I wanted to write my proof out from start to finish to see if my thought process was correct.
1) First I just experimented with some sets to see if anything came about: let $A = \{1\}$, $B = \{2\}$, $C = \{3\}$, $D = \{4\}$:
$(A\times B) \cap (C\times D)$ side:
$$(1,2)\cap (3,4) = (\emptyset,\emptyset)$$ [not sure if i could state this, but it is what I said in my solution]
$(A\cap C)\ \times (B\cap D)$ side:
$$\emptyset\ \times \emptyset = (\emptyset, \emptyset)$$
Ok so I established what appears to be equality, so now I have to prove it.
let $(x,y) \in (A\ X\ B) \cap (C\ X\ D)$
--> $(x\in A \cap y\in B) \cap (x\in C \cap y\in D)$
--> $(x\in A \cap x\in C) \cap (y\in B \cap y\in D)$
--> $x\in (A\cap C)\ X\ y\in (B\cap D)$
--> $ (x,y)\in (A\cap C)\ X\ (B\cap D)$
 Done.
Then I would have to do the other way as well but it would amount to a similar argument.
P.S: How to get lines of my proof to line up with arrows?
 A: $$(x,y)\in (A\times B)\cap (C\times D) \iff  \left\{\begin{array}{c}(x,y)\in A\times B \\ (x,y) \in C\times D \end{array}\right. \iff  \left\{\begin{array}{c}x \in A,\ y\in  B \\   x \in C,\ y\in  D \end{array}\right. \iff$$
$$\iff  \left\{\begin{array}{c}x \in A,\ x\in  C \\   y \in B,\ y\in  D \end{array}\right. \iff \left\{\begin{array}{c}x \in A\cap C \\   y \in B\cap  D \end{array}\right. \iff (x,y) \in (A\cap C)\times (B\cap D)$$
A: $$(A\times B)\cap(C\times D)=(A\cap C)\times (B\cap D) ?$$
Def.:
$$ X\cap Y= \left\{x|x\in X\wedge y\in Y\right\}.$$
$$ X\times Y=\left\{<x,y>|x\in A\wedge y\in B\right\}. $$
$$ <x, y> = \{ \{x\}, \{x, y\} \}$$
Proof:
\begin{align*}
       (A\times B) \cap(C\times D)&= \left\{<x,y>|x\in A\wedge y\in B\right\}\cap  \left\{<x,y>|x\in C\wedge y\in D\right\} \\
       &=   \left\{<x,y>|(x\in A\wedge y\in B)\wedge (x\in C\wedge y\in D)\right\} \\
       &= \left\{<x,y>|(x\in A\wedge x\in C)\wedge (y\in B\wedge y\in D)\right\}\\
       &=(A\cap C)\times (B\cap D).
       \end{align*}
$ \dashv $
