Proving $A \cap (B - C) = (A \cap B) - (A \cap C)$

Question

I am trying to prove that $A \cap (B - C) = (A \cap B) - (A \cap C)$ using biconditionals, but I cannot seem to get the proof to work out. \begin{align*} x \in A \cap (B - C) & \iff x \in A \text{ and } x \in B - C \\ & \iff x \in A \text{ and } (x \in B \text{ and } x \not \in C) \\ & \iff (x \in A \text{ and } x \in B) \text{ and } (x \in A \text{ and } x \not \in C) \\ & \iff x \in A \cap B \text{ and } x \in A - C \\ & \iff x \in (A \cap B) \cap (A-C). \end{align*} I cannot figure out how to get from $(A \cap B) \cap (A - C)$ to $(A \cap B) - (A \cap C)$.

Answer 1

HINT: $(x \in A \text{ and } x \in B) \text{ and } (x \in A \text{ and } x \not \in C)\\ \iff (x \in A \text{ and } x \in B) \text{ and } (x\in A \text{ and }x \not\in A \cap C)$

Answer 2

A small observation is that: if $x \notin C \implies x \notin A\cap C$ because $A\cap C \subseteq C$. Using this you can justify the step above.