Let $ A $, $ B $ and $ C $ be sets. If $ A \cup B = A \cup C $ and $ A \cap B = A \cap C $, then show that $ B = C $. I’m stuck on this one problem in my textbook regarding proofs in set theory. I’ve done the following so far:

Let $ x \in B $. As $ B \subseteq A \cup B $, we have $ x \in A \cup B $, so $ x \in A \cup C $ because $ A \cup B = A \cup C $. Hence, $ x \in (A \cup B) \cap (A \cup C) $, which yields $ x \in A \cup (B \cap C) $.
Now, either $ x \in A $ or $ x \in B \cap C $. If $ x \in B \cap C $, then $ x \in C $ because $ B \cap C \subseteq C $, so $ B \subseteq C $.

This is only half of the proof, though, as I have to prove that $ C \subseteq B $ as well; and I also haven’t even considered the case $ x \in A $ for the last step in what I have so far.
I’ve drawn many different Venn diagrams, and although I see why the statement is true, I just can’t formalize it. Any pointers, help or guidance is very much appreciated.
Thanks!
 A: Using the symmetric difference property that $A \Delta B = (A \cup B) \setminus (A \cap B)$, it follows from the given equalities that:
$$A \Delta B = (A \cup B) \setminus (A \cap B) = (A \cup C) \setminus (A \cap C) = A \Delta C$$
Using the associativity of the symmetric difference, and given that $A \Delta A = \emptyset$, and $\emptyset \Delta X = X$:
$$B = (A \Delta A) \Delta B = A \Delta (A \Delta B) = A \Delta (A \Delta C) = (A \Delta A) \Delta C = C$$
A: I'll assume that there is some $x\in B$ that it is not in $C$. If this leads to a contradiction, we'll have $B\subseteq C$. To prove $C\subseteq B$ simply swap $B$ and $C$ in this proof.
First, $x\in A\cup B$, so $x\in A\cup C$. Since $x\notin C$, we have that $x\in A$.
Since $x$ is also in $B$, then $x\in A\cap B$ and therefore, $x\in A\cap C$. This contradicts that $x\notin C$.
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: $A \cup B = A \cup C$ implies $B \setminus A = C \setminus A$, which is to say that the parts of $B$ and $C$ not in $A$ are identical.
$A \cap B = A \cap C$ says that the parts of $B$ and $C$ in $A$ are identical.
So $B=C$.
A: First,Let $x \in B$, by conjunction property
$$x \in B \lor x\in A$$
$\implies$ by definition of union
$$x\in (A \cup B)$$
$\implies$By hypothesis
$$x \in (A \cup C)$$
$\implies$By definition of union, and adition property
$$(x\in A \lor x\in C) \land x\in B$$
$\implies$Distributing
$$(x\in A \land x\in B) \lor (x\in C \land x \in B)$$
$\implies$ By Hypothesis
$$(x \in A \land x\in C) \lor (x\in C \land x \in B)$$
$\implies$
$$x\in C \land (x\in A \lor x \in B) $$
$\implies$ Symplifing
$$x \in C$$
Then, $B \subseteq C$. Remember, you now need proof $C \subseteq B$. Let $x \in C$...
A: 
This means $x  \in  A \lor  x  \in  B\cap C$. If $x  \in  B\cap C$, since $B\cap C  \subseteq  C$, $x  \in  C$ and so $B  \subseteq  C$.

That is wrong. If $x  \in  B\cap C$ then $x  \in  C$, but what if $x  \in  A$ but not $x  \in  B\cap C$?
So how to proof the statement?
Hints:


*

*A set $A$ splits another set $B$ into two disjoint parts:
$$B=(B\cap A) \cup (B\setminus  A)\tag{1}$$
$$\emptyset =(B\cap A) \cap (B\setminus  A)\tag{2}$$
If$A^c$ is the complement of $A$ then $B \setminus A$ is defined as 
$$B \setminus A:=B \cap A^c\tag{3}$$.
Can you roof of $(1)$ and $(2)$? (Annotation: $(2)$ isn't really needed further. I mentioned it only or the sake of completeness)

*If $$(B\cap A)=(C\cap A) \tag{4}$$ and $$(B\cup A)=(C\cup A) \tag{5}$$ then 
$$B \setminus A= C \setminus A \tag{6}$$

*From $(1)$, $(5)$ and $(6)$ immediately follows $B=C$.

