Prove that following logical implication is true 
I tried to approach this problem but I can't get anywhere near the answer. Can someone help me to the right direction?
This is what i tried,
Suppose $x \in A$, then $x \in A \implies x \in A \cup C \implies x \in B \cup C$, if $x \in B$ then $A=B$. I don't know if this is correct and I have to do a case for when $x \in C$, I think.
 A: Edit: In response to the comment, I have rewritten this to explicitly show that we are proving an implication.
We want to prove:
For sets $A,B,C \subset U$, the following implication holds:
$[(A\cap C = B \cap C) \land (A \cup C = B \cup C)]\implies A=B$.
Proof:
Let $A,B,C \subset U$ be arbitrary sets. Assume the following property holds:
$[(A\cap C = B \cap C) \land (A \cup C = B \cup C)]$.
We will show that $A=B$.
Set equality is almost always a 2-part problem: show the inclusion one way, then the other way.
Let $x \in A$ be arbitrary. Either $x \in C$ or $x \notin C$. (This is a proof by cases.)
Case 1: If $x \in C$ then $x \in A\cap C = B\cap C$ so $x \in B$, and we are done.
Case 2: If $x \notin C$ then we still have $x \in A \cup C = B \cup C$. Since $x \notin C$ and $x \in B \cup C$, $x \in B \cup C -C = B$. So $x \in B$.
By cases, $A \subset B$.
By symmetry (or by doing the same exact proof with $A$ and $B$ reversed), we see that $B \subset A$. Thus $A=B$.
A: If x does not belong to C, xEA => xEA or C => xEB or C. Since x is not in C, xEB.
If x does belong to C, xEA and xEC = xEB and xEC, so xEB when xEC.
In either case, xEB, therefore A is a subset of B
If y belongs to B, then yEB => yEB or C = A or C, so yEA or yEC. 
If yEA, then yEB => yEA
If y belongs to C, then yeB and C = A and C. 
In either case, yeA therefore B is a subset of A
Therefore, A=B
A: Let's assume that $A$ is not empty. Why don't you just take $C:=\{x\}$ for any $x\in A$? The premise is true for this special $C$ and so you have
$$
(A \cap \{x\} = B \cap \{x\}) \wedge (A \cup \{x\}= B \cup \{x\})
$$
which means
$$
(\{x\} = B \cap \{x\}) \wedge (A = B \cup \{x\})
$$
and the first clause means $x\in B$, and the second implies $B\subseteq A$ directly.
A: You are trying to prove an implication. So you need to assume the antecedent
$$(A \cap C = B \cap C) \wedge (A \cup C = B \cup C)$$
to prove that
$$A = B$$
So how do you prove that two sets are equal?
Addendum:
I think you are starting in the right direction. (Of course, there are more than one way to approach a proof like this.)
First note that to be formally correct, you need to deduce two statements from the single hypothesis that I wrote above before continuing with the proof as you started it.
Now let's explore what happens if $x \in C$. We still want to show that $x \in B$. Can you use the other half of our assumption to show this? (Hint: you already assumed that $x \in A$.)
