Prove that $A=\emptyset$ iff the equality $\bigl(((U \backslash A) \cap B) \cup (A \cap (U \backslash B))\bigr)=B$ holds. Let $A$, $B$ and $U$ be sets such that  $A \subseteq U$ and $B \subseteq U$.
Prove that $A=\emptyset$ iff the equality $\bigl(((U \backslash A) \cap B) \cup (A \cap (U \backslash B))\bigr)=B$ holds.

Given $A \subseteq U$, i.e. $\forall x, x \in A \implies x \in U$
and $B \subseteq U$, i.e. $\forall x, x \in B \implies x \in U$,
Prove $A=\emptyset$ iff the equality $\bigl(((U \backslash A) \cap B) \cup (A \cap (U \backslash B))\bigr)=B$
Prove ($\implies$):
Suppose $A = \emptyset$, i.e. $\forall x, x \not \in A$
$\bigl(((U \backslash A) \cap B) \cup (A \cap (U \backslash B))\bigr)\\
i.e. \bigl(((\forall x, x \in U, x \not \in A) \land (x \in B)) \cup ((x \in A) \land (\forall x, x \in U, x \not \in B))\bigr)\\ $

I know that since $A = \emptyset$, the right side of the union is automatically false. However, I have no idea how I am supposed to proceed. Please guide in the correct direction.
 A: Note that 
$$\bigl(((U \backslash A) \cap B) \cup (A \cap (U \backslash B))\bigr) = \big(U\cap A^c\cap B\big) \cup\big(A\cap U\cap B^c\big) = (A^c\cap B)\cup(A\cap B^c).$$
A: For the $\Rightarrow$ direction, instead of starting to look at individual element, it is easier just to work at the set algebra level: When $A=\varnothing$ plug it into the left-hand side to get
$$ ((U\setminus \varnothing)\cap B) \cup (\varnothing\cap(U\setminus B)) =
(U\cap B)\cup \varnothing =
U \cap B $$
which is $B$ becasue $B\subseteq U$.
For the $\Leftarrow$ direction, the easiest way appears to prove the contraposed form:

If $A\ne \varnothing$ then $((U\setminus A)\cap B) \cup (A\cap(U\setminus B))\ne B$.

For this you would start by looking at an element $x\in A$ (and you're assuming exactly that such an element exists). Then split into cases based on whether $x\in B$ or $x\notin B$.

Your proposed proof start is, I'm afraid to say, misusing notation quite a bit. When you write

$\bigl(((\forall \color{blue}x, \color{blue}x \in U, \color{blue}x \not \in A) \land (\color{red}x \in B)) \cup ((\color{red}x \in A) \land (\forall \color{green}x, \color{green}x \in U, \color{green}x \not \in B))\bigr)$

Given the context I can see that you've probably tried to translate the set expression $((U\setminus A)\cap B) \cup (A\cap(U\setminus B))\ne B$ into a logical statement about elements, but you have done it only halfway and arrived at something ill-formed that definitely doesn't express what you want to say.
First, you're mixing up notation for sets (the $\cup$ must come between sets) with notation for propositions (the $\land$ creates a proposition).
Second, the two occurrences of $x$ I've marked in red are not in the scope of any quantifier, whereas the blue and green ones are.
Third, where do you have quantifiers, you're writing some commas whose meaning can only be understood if one already has an idea what you're probably trying to say.
