Conditional independence intersection I'm given that: 
$$
\mathbb{P}(A \cap B \ |\ C) = \mathbb{P}(A\ |\ C) \cdot \mathbb{P}(B\ |\ C)
$$
Also, $B$ and $C$ are independent. I want to prove that $A$ and $B$ are independent. 
To prove $A$ and $B$ are independent, I know $\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)$. 
I started off with a contradiction by assuming $\mathbb{P}(A \cap B) \neq \mathbb{P}(A) \cdot \mathbb{P}(B)$, but I always end up needing the information that $A$ and $C$ are independent too, which is not given. 
Does anyone have any tips about how to get started initially? 
 A: With this type of question, I would usually just go for a direct proof instead of a proof by contradiction.
Start with the equation $\Bbb P(A \cap B|C)=\Bbb P(A|C) \cdot \Bbb P(B|C)$ and see if you can turn it into an equation for $\Bbb P(A \cap B)$.

 \begin{align} \Bbb P(A \cap B|C) & =\frac{\Bbb P(A\cap B \cap C)}{\Bbb P(C)} \\ & =\frac{\Bbb P(B\cap C|A)\Bbb P(A)}{\Bbb P(C)} \\ & =\frac{\Bbb P(B|A)\Bbb P(C|A) \Bbb P(A)}{\Bbb P(C)} && \text{by independence of $B$ and $C$} \\ & =\frac{\Bbb P(B \cap A)\Bbb P(C|A)}{\Bbb P(C)}\\ & =\frac{\Bbb P(A \cap B)\Bbb P(A|C)}{\Bbb P(A)} && \text{by Bayes' Theorem} \end{align} Thus \begin{align} & \Bbb P(A \cap B|C)=\Bbb P(A|C) \cdot \Bbb P(B|C) \\ \implies & \frac{\Bbb P(A \cap B)\Bbb P(A|C)}{\Bbb P(A)}= \Bbb P(A|C) \cdot \Bbb P(B|C) \\ \implies & \frac{\Bbb P(A \cap B)}{\Bbb P(A)}= \Bbb P(B|C) \\ \implies & \frac{\Bbb P(A \cap B)}{\Bbb P(A)}= \Bbb P(B) && \text{by independence of $B$ and $C$} \\ \implies & \Bbb P(A \cap B)= \Bbb P(A) \Bbb P(B)\end{align}

