# How to prove that conditional independence does not imply independence?

I am trying to prove that conditional independence does not imply independence, ie that $$P(A|C)P(B|C)=P(A \cap B|C) \nRightarrow P(A \cap B)=P(A)P(B)$$

I guess I need a counter-example but I am struggling to find a way of homing in onto one.

So far I have tried drawing Venn diagrams, and I can see that there is no reason why the sizes of the relevant intersections should multiply as implied by the above, but I am not sure how to proceed from there.

Let $$A,B$$ be events that are not independent.
Observe that $$P(A\mid A)P(B\mid A)=P(B\mid A)=P(A\cap B\mid A)$$ showing that there is conditional independence wrt $$A$$.
If $$C = A^c,$$ then as long as the $$P(A) \neq 1,$$ then the conditional independence equation will just be $$0=0$$ while we've learned nothing about whether $$A$$ and $$B$$ are independent.
That is, take any two $$A, B$$ that aren't independent, then $$P(A) \neq 1$$ (why?) and setting $$C=A^c$$ implies $$A|C, B|C$$ are independent.