Suppose that $A$ is conditionally independent of $(B,C)$ given $D$ (that is $P(A, B, C | D) = P(A|D)P(B,C|D))$.
I want to show the following is true:
$P(A | B, C, D) = P(A | B, D)$
I am not sure how to show this. I have tried beginning with transformations like the product rule and Bayes' Rule, but I am not sure how I can remove $C$ from the equations.
$\frac{P(ABCD)}{P(D)}=\frac{P(AD)}{P(D)}\frac{P(BCD)}{P(D)}$ (by definition of conditional probability) $\Rightarrow P(ABCD)P(D)=P(AD)P(BCD)$ $\Rightarrow \frac{P(ABCD)}{P(BCD)}=\frac{P(AD)}{P(D)}$ $\Rightarrow P(A|BCD)=P(A|D)=P(A|BD)$, since $A$ and $B$ are independent in the conditional universe given $D$, we have $P(A|BD) = P(A|D)$