Proving equality of 3 sets $A, B, C$ are sets. I have to prove equality of this:
$$A \cap C = B \cap C \land A \cup C = B \cup C ⇒ A = B$$
I did this, but I don't know what to do next and whether I even did the right thing:
$A \cap C = x \in A \land x \in C$
$B \cap C ∧ A \cup C = (x \in B \land x \in C) \land (x \in A \lor x \in C)$
$B \cup C \implies A = x \in B \lor x \in C ⇒ A $
$B$
Any ideas what to do next? Or maybe how to solve this somehow else? 
 A: Partition $A$ into the two sets $A'=A\cap C$ and $A''=A\backslash (A\cap C)$, and partition $B$ into the two sets $B'=B\cap C$ and $B''=B\backslash (B\cap C)$. The first equality
$$A\cap C=B\cap C$$
tells us that $A'=B'$. Since $(A\cup C)\backslash C=A''$ and $(B\cup C)\backslash C=B''$, the second equality
$$A\cup C=B\cup C$$
tells us that $A''=B''$. Thus, since the corresponding parts of the partitions of $A$ and $B$ are equal, $A=B$.
I leave it to you to make this more rigorous/formal as needed.
A: Consider for any $x$ in the universal set you have four possibilities for whether $x$ is in $A$ or $C$.
1) $x \in A$ and $x \in C$.
2) $x \in A$ and $x \not \in C$.
3) $x \not \in A$ and $x  \in C$.
4) $x \not \in A$ and $ x\not \in C$
For each of those cases we ask: Is $x\in B$?
If we discover that $x \in B \iff x \in A$ the we determine that $A = B$.
If we discover anything else or any ambiguous situation then we determine $A \ne B$ or that $A$ need not equal $B$
So
1) $x \in A$ and $x \in C$.
So $x \in A\cap C = B\cap C$ so $x\in B$.
2) $x \in A$ and $x \not \in C$.
So $x\in A\cup C= B\cup C$ so $x$ is in either $B$ or $C$.  But $x \not \in C$ so $x \in B$.
3) $x \not \in A$ and $x  \in C$.
So $x \not \in A\cap C = B\cap C$.  So $x$ is not in both $B$ and $C$.  But $x \in C$ so $x$ can't be in $B$.
4) $x \not \in A$ and $ x\not \in C$
$x\not \in A\cup C = B\cup C$ so $x$ is in neither $B$ nor $C$.  So $x\not \in B$.
So $x\in B$ precisely when $x \in A$ and $x \not \in B$ precisely what $x \not \in A$.  So $A$ an $B$ have precisely the same elements and $A = B$.
