Attempting to prove $(A\setminus B) × (C\setminus D) = (A × C) \setminus [(A × D) ∪ (B × C)]$. I am trying to prove the following equality where A, B, C and D are sets.
(A \ B) × (C \ D) = (A × C) \ [(A × D) ∪ (B × C)]
x stands for the cartesian product.
As I am trying to prove via inclusion equality this is my attempt so far, I am trying to construct the forward implication but I am not sure how. I understand that to prove set equality you must prove one set is a subset of the other and vice versa but this question has me stumped. Find below my thought process so far and thanks in advance for your help!
Foward Implication: (A \ B) × (C \ D) ⊆ (A × C) \ [(A × D) ∪ (B × C)]

*

*(A \ B) × (C \ D) = (A\C)

Reverse Implication: (A × C) \ [(A × D) ∪ (B × C)] ⊆ (A \ B) × (C \ D)
I am unsure how to progress from here any help would be highly appreciated!
 A: This might look a little messy, but it shouldn't be too difficult to follow:
(AxC)\[(A × D) ∪ (B × C)]=(AxC)$\cap$[(AxD) ∪ (BxC) $]^c$=(AxC)$\cap$(AxD$)^c\cap$(BxC$)^c$=[(AxC)$\cap$(AxD$)^c$]$\cap$[(AxC)$\cap$(BxC$)^c$].
Now, (AxC)$\cap$(BxC$)^c$=(AxC)$\cap$[($B^c$xC) ∪ (Bx$C^c$)]=(AxC)$\cap$($B^c$xC)=(A\B)xC. Similarily, (AxC)$\cap$(AxD$)^c$=Ax(C\D). So, the expression becomes Ax(C\D) $\cap$ (A\B)xC=(A\B)x(C\D)
A: We start from the left hand side. 

$(A\times C)$ \ $[(A\times D)\cup(B\times C)]$=$(A\times C)\cap [(A\times D)\cup (B\times C)]^c$=$(A\times C)\cap (A\times D)^c\cap (B\times C)^c$

But right hand side=$(A$ \ $B)$$\times$$(C$ \ $D)$= [$A\times (C$ \ $D)$] \ [$B\times(C$ \ $D)$].

Since $C$ \ $D\subseteq C$, [$A\times (C$ \ $D)$] \ [$B\times(C$ \ $D)$]=[$A\times (C$ \ $D)$] \ [$B\times C$].

(This step is a little tricky, and I will explain it further. If you totally understand what I mean you can skip this paragraph. Since $B\times C=[B\times (C$ \ $D)]\cup [B\times (C\cap D)]$, so [$A\times (C$ \ $D)$] \ [$B\times C$]={[$A\times (C$ \ $D)$] \ $[B\times (C\cap D)]$} \ $[B\times (C$ \ $D)]$. If we look at the second index of $(u,v)\in A\times (C$ \ $D)$, we notice that $v$ cannot be in $D$. But if we look at the second index of $(u',v')\in B\times (C\cap D)$, we notice that $v'$ must stay in $D$. Hence [$A\times (C$ \ $D)$] $\cap$ $[B\times (C\cap D)]=\emptyset$, and excluding one from the other will keep the other unchanged, i.e. [$A\times (C$ \ $D)$] \ $[B\times (C\cap D)]$=$A\times (C$ \ $D)$. Therefore [$A\times (C$ \ $D)$] \ [$B\times C$]={[$A\times (C$ \ $D)$] \ $[B\times (C\cap D)]$} \ $[B\times (C$ \ $D)]$=[$A\times (C$ \ $D)$] \ [$B\times(C$ \ $D)$].)

Now we continue. [$A\times (C$ \ $D)$] \ [$B\times C$]=[$(A\times C)$ \ $(A\times D)$]\ $(B\times C)$=$(A\times C)\cap (A\times D)^c\cap (B\times C)^c$. Therefore, the right hand side equals to the left hand side.
A: We can also use logic to argue.

Denote an element in $(A\backslash B)\times(C\backslash D)$ as $(u,v)$. So $(u,v)\in(A\backslash B)\times(C\backslash D)\Leftrightarrow (u\in A)\land\neg(u\in B)\land(v\in C)\land\neg(v\in D)$.

Denote an element in $(A\times C)\backslash[(A\times D)\cup(B\times C)]$ as $(u',v')$. So $(u',v')\in(A\times C)\backslash[(A\times D)\cup(B\times C)]\Leftrightarrow[(u'\in A)\land(v'\in C)]\land\neg\left\{[(u'\in A)\land(v'\in D)]\vee[(u'\in B)\land(v'\in C)]\right\}$
Since the negation of $[(u'\in A)\land(v'\in D)]\vee[(u'\in B)\land(v'\in C)]$ is true, we must have $[(u'\in A)\land(v'\in D)]$ and $[(u'\in B)\land(v'\in C)]$ both false, so $\neg (u'\in A)\vee \neg(v'\in D)$ true and $\neg(u'\in B)\vee\neg(v'\in C)$true. So the original statement can be rewritten as $[(u'\in A)\land(v'\in C)]\land[\neg (u'\in A)\vee \neg(v'\in D)]\land[\neg(u'\in B)\vee\neg(v'\in C)]$=$[(u'\in A)\land(v'\in C)]\land \neg(v'\in D)\land \neg(u'\in B)$.

Now compare the left hand side and the right hand side we find that they are the same.
A: We have :
$$\begin{array}{lcl}
(x, y) \in (A \times C) \setminus \left[(A \times D) \cup (B \times C)\right] & \iff & (x, y) \in A \times C \\
& & \quad \wedge (x , y) \notin \left[(A \times D) \cup (B \times C)\right] \\
& \iff & x \in (A \times C) \wedge (x, y) \notin A \times D \\
& & \quad \wedge (x, y) \notin B \times C \\
& \iff & x \in A \wedge y \in C \wedge (x \notin A \vee y \notin D) \\
& & \quad \wedge (x \notin B \vee y \notin C) \\
& \iff & \left[x \in A \wedge (x \notin A \vee y \notin D) \right] \\
& & \wedge \left[y \in C \wedge (x \notin B \vee y \notin C)\right] \\
& \iff & (x \in A \wedge y \notin D) \wedge (y \in C \wedge x \notin B) \\
& \iff & (x \in A \wedge x \notin B) \wedge (y \in C \wedge y \notin D) \\
& \iff & x \in A \setminus B \wedge (y \in C \setminus D \\
& \iff & (x, y) \in (A \setminus B) \times (C \setminus D)
\end{array}$$
We deduce that :
$$(A \setminus B) \times (C \setminus D) = (A \times C) \setminus \left[(A \times D) \cup (B \times C)\right]$$
A: Let $(x,y)\in (A \times C) \setminus \left[(A \times D) \cup (B \times C)\right]$.
Then $(x,y)\in (A \times C)$ and $(x,y)\notin \left[(A \times D) \cup (B \times C)\right]$.
Then $(x,y)\in (A \times C)$ and $(x,y)\notin (A \times D) $ and $(x,y)\notin (B \times C)$.
Then, since $x\in A$ and $y\in C$,$$(x,y)\in (A \times C)\text{ and }y\notin D \text{ and } x\notin B,$$
otherwise written $(x,y)\in (A\setminus B)\times(C \setminus D)$.
For the reverse inclusion, just repeat the reasoning in the other direction.
