How do you multiply two conditional probabilities? I have to multiply this:
$P(a|b,c)·P(b|c)$
How do you multiply those two expressions?
It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.
 A: If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) = \frac{P(a,b,c)}{P(c)}
$$
so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$. 
A: Those two expressions simply mean: 
(the Probability of event $a$ given the events $b$ and $c$) $\times$ (the Probability of event $b$ given event $c$).  
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1).  So you simply multiply the two values together as you would any two numbers.  So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:
$P(a|b,c)·P(b|c)=0.40\times0.70=0.28$
UPDATED BASED ON YOUR EDITED QUESTION:
The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $\frac{P(DE)}{P(E)}$
\begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)\frac{P(bc)}{P(c)}\\
 & = & \frac{P(abc)}{P(bc)}\frac{P(bc)}{P(c)}\\
 & = & \frac{P(abc)}{P(c)}\\
 & = & \frac{P[(ab)c]}{P(c)}\\
 & = & P(ab|c)
\end{eqnarray*}
So, in all, we have:
\begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
\end{eqnarray*}
