$A\cap B = A\cap C$ and $A\cup B = A\cup C$. Show $B=C$. I couldn't gather an idea on how prove that. I tried to form union of $B$ on both sides of equation one just to try if it takes me anywhere:
$$B \cup (A\cap B) = B \cup(A\cap C)$$
$$(B\cup A) \cap (B\cup B) = (B\cup A) \cap (B\cup C) $$
but it didn't. any ideas? 
 A: $B = B \cap (B \cup A) = B \cap (C \cup A) = (B \cap C) \cup (B \cap A) = (C \cap B) \cup (C \cap A) = C \cap (B \cup A) = C \cap (C \cup A) = C$
A: Hint
Suppose $B\neq C$
$\implies \exists x\in B : \; x\notin C$
$$\implies x\in A\cup B$$
$$\implies x\in  A\cup C$$
$$\implies x\in A$$
$$\implies x\in A\cap B$$
$$\implies x\in A\cap C$$
$$\implies x\in C$$
Contradiction, so...
A: We have $(i) \ A\cap B=A\cap C$ and $(ii) \ A\cup B=A\cup C$. Suppose that $b\in B$ but $b\notin C$. Then, if $b\in A$, we have a contradiction on $(i)$. If $b\notin A$, we got a contradiction on $(ii)$. We have the same for any $c\in C$ such that $c\notin B$. So, $B=C$.
A: Assuming you know the symmetric difference $A \Delta B = (A \cup B) \setminus (A \cap B)$ it follows that:
$$A \Delta B = (A \cup B) \setminus (A \cap B) = (A \cup C) \setminus (A \cap C) = A \Delta C$$
Then, using the associativity of $\Delta$ and the facts that $A \Delta \emptyset = A$ and $A \Delta A = \emptyset$:
$$B = \emptyset \Delta B = (A \Delta A) \Delta B = A \Delta (A \Delta B) = A \Delta (A \Delta C) = (A \Delta A) \Delta C = \emptyset \Delta C = C$$
