If $P(A\cap B\cap C)=P(A\cap B)P(C)$, does it follow that $P(A\cap C)=P(A)P(C)$ and $P(B\cap C)=P(B)P(C)$? 
If $P(A\cap B\cap C)=P(A\cap B)P(C)$, does it follow that $P(A\cap C)=P(A)P(C)$ and $P(B\cap C)=P(B)P(C)$?

In other words, if the conjunction of two events $A$ and $B$ is independent of a third event $C$, is it always true that event $C$ is independent from events $A$ and $B$ separately?
I've gone back and forth between believing that it is likely true and not true. I can't see how to prove it from the basic axioms of probability, but it also seems challenging to think of a counterexample where the conjunction of $A$ and $B$ is independent of $C$, but either $A$ or $B$ are dependent on $C$ (or vice versa).

Here is another counter-example: Consider the sample space $\Omega=\{1,2,3,4,5,6\}$, and suppose $\mathbf{P}(\{\omega\})=\frac{1}{6}$ for each $\omega \in \Omega$ (i.e., each of $1,\ldots,6$ is equally likely to occur):

Now consider the events
\begin{align*}
A &= \{1, 2, 5\}, &
B &= \{2, 3, 5\}, &
C &= \{1, 2, 3\}.
\end{align*}

Then we easily check that

*

*$\mathbf{P}(A\cap B \mid C) = \frac{1}{3} = \mathbf{P}(A\cap B)$, and so, $A\cap B$ and $C$ are independent.

*$\mathbf{P}(A \mid C) = \frac{2}{3} \neq \frac{1}{2} = \mathbf{P}(A)$, hence $A$ and $C$ are not independent.

*$\mathbf{P}(B \mid C) = \frac{2}{3} \neq \frac{1}{2} = \mathbf{P}(B)$, hence $C$ and $C$ are not independent.

