The wikipedia article on conditional independence states that the following two statements are equivalent, but I can't see how one implies the other and vice versa:
$$\text{Pr}(R\cap{B}\rvert{Y})=\text{Pr}(R\rvert{Y})\;\text{Pr}(B\rvert{Y})\tag{1}$$
$$\text{Pr}(R\rvert{B\cap{Y}})=\text{Pr}(R\rvert{Y})\tag{2}$$
Using Bayes's rule, I can rewrite equation 1 as follows:
$$\text{Pr}(R\cap{B}\rvert{Y})=\frac{\text{Pr}(Y\rvert{R\cap{B}})\;\text{Pr}(R\cap{B})}{\text{Pr}(Y)}$$
but that doesn't help me prove equation 2 -- I don't see a pathway to getting a $B\cap{Y}$ and we have no information whether $R$ and $B$ are independent, so I can't rewrite $\text{Pr}(R\cap{B})$ as $\text{Pr}(R)\;\text{Pr}(B)$. Likewise, expanding equation 2 via Bayes's doesn't help me prove equation 1.
I'm sure it's something obvious, but I just don't see it.
Assuming $\Pr(B \mid Y) \not = 0$, the following statements are equivalent:
$$\Pr(R\cap{B}\mid {Y})=\Pr(R\mid{Y})\;\Pr(B\mid {Y})$$
$$\Pr(R\mid {B \cap Y})\;\Pr(B \mid Y) =\Pr(R\mid{Y})\;\Pr(B\mid {Y})$$
$$\Pr(R\mid {B \cap Y})=\Pr(R\mid{Y})$$