Use direct proof to prove: If $A \cap B = A \cap C$ and $A \cup B = A \cup C$, then $B = C$ I'm interested in knowing if the method I used is correct. I've been teaching myself proofs lately and I am having difficulties with how to approach a problem so any general tips would be awesome as well! Here is what I have:

Assume $A \cap B = A \cap C$ and $A \cup B = A \cup C$.
Let $x \in B$.
Let $x \in A$.
Then, $x \in A \cap B$.
Since $A \cap B = A \cap C$, $x$ must also be an element of $C$.
That is, $x \in C$.

Using the same thinking we can prove the $A \cup B = A \cup C$ case. Ie.

Assume $A \cap B = A \cap C$ and $A \cup B = A \cup C$.
Let $x \in B$.
Let $x \in A$.
Then, $x \in A \cap B$.
Since $A \cup B = A \cup C$, $x$ must also be an element of $C$.
That is, $x \in C$.

Finally, we see that for any $x \in B$, it follows that $x \in C$ as well. We can show that the converse is true by letting $x \in C$. Thus, $B = C$.

I believe the proof should go like this:
Assume $x\in B$. Then $x\in A \cup B=A\cup C$. If $x\in C$ then good. If $x\in A$ then it is in $A\cap B=A\cap C$ so it is in $C$.
Then assume $x\in C$. By a symmetric argument to the one just given, it is in $B$.
