Consider three r.v. X,Y and Z. The r.v. X is independent of Z given Y (i.e. $X \perp Z \ | \ Y$)
then the following is true (for some reason):
$Pr[x,y,z] = \frac{Pr[x,y]Pr[y,z]}{Pr[y]} = Pr[x,y]Pr[z|y]$
The proof of this statement would obvious if all three variables were independent of each other, since we could just cancel the "extra" Pr[y] in the numerator. But without this condition, how can you show that the above relation is true?
(I am only concerned about any of the right equalities being compared to P[x,y,z]. The last equality is obviously true given the definition of conditional probs).
\begin{align*} P(x,y,z) &= P(x,z|y)P(y) & \mbox{Axioms of Probability} \\ &= P(x|y)P(z|y)P(y) & \mbox{Conditional Independency} \\ &= \frac{P(x,y)}{P(y)}\frac{P(z,y)}{P(y)}P(y) & \mbox{Axioms of Probability} \\ &= \frac{P(x,y)P(z,y)}{P(y)} \end{align*}