Which probability law? It may be a basic probability law in another form, but I cannot figure it out. Why can we say the following:
$P(A∩B|C) = $$P(A|B∩C)P(B|C)$
Thank you.
 A: You should simply expand upon the definition:
$P(A \wedge B | C) = \frac{P(A \wedge B \wedge C)}{P(C)}$. 
$P(A | B \wedge C) = \frac {P( A \wedge B \wedge C)}{P( B \wedge C )}$.
$P(B|C)=\frac{P( B \wedge C)}{P(C)}$.
Now, we solve for $P(A \wedge B \wedge C)$ in the first two above and equate them:
$P(A \wedge B | C) \cdot P(C)=P(A | B \wedge C) \cdot P( B \wedge C )$.
Solve for $P(B \wedge C)$ in the third line and plug it in to the expression above.
A: user76553, in your most recent comment you asked "If I did not know the rhs of the equation, how could I derive it from the lhs? Specifically, what probability equations would I use?" 
I'm assuming you mean if you just had $P(A\cap B \mid C)$ sitting in front of you by itself without knowing that it was indeed equivalent to $P(A\mid B \cap C)P(B \mid C)$, and you wanted to see if you could find equivalent expressions. 
In that case, you still make use of the various equalities given by marchimedes in his/her answer above, but you could proceed like this: 
Step 1: use definition of conditional probability to write
$$P(A \cap B \mid C) = \frac{P(A \cap B \cap C)}{P(C)} $$
Step 2: use the definition of conditional probability again on the numerator to express $P(A \cap B \cap C)$ as $P(A \mid B \cap C)P(B \cap C)$. Then the entire expression is now
$$\frac{P(A \mid B \cap C)P(B \cap C)}{P(C)} $$
Step 3: similarly we then express $P(B \cap C)$ as $P(B \mid C)P(C)$ and the entire expression is now
$$\frac{P(A \mid B \cap C)P(B \mid C)P(C)}{P(C)} $$
Step 4: cancel the $P(C)$ terms in the numerator and denominator and then you have
$$P(A \mid B \cap C)P(B \mid C).$$
So really you are just using the definition of conditional probability several times and then one cancelation at the end. 
Hope that helps!
A: Starting from the right and using just the definition of conditional probability,
$$\begin{align}
P(A\mid B\cap C)P(B\mid C) 
&= \frac{P(A\cap B \cap C)}{P(B\cap C)}\times \frac{P(B\cap C)}{P(C)}
&\scriptstyle{\text{substitute using the definition of conditional probability}}\\
&= \frac{P(A\cap B \cap C)}{P(C)}
&\scriptstyle{\text{cancel}}~P(B\cap C)~\scriptstyle{\text{from numerator and denominator}}\\
&= P(A\cap B \mid C).  &\scriptstyle{\text{identify the ratio as a conditional probability}}
\end{align}$$
Starting from the left, we can just reverse the steps above.
$$\begin{align}
P(A\cap B \mid C) &= \frac{P(A\cap B \cap C)}{P(C)}\\
&= \frac{P(A\cap B \cap C)}{P(B\cap C)}\times \frac{P(B\cap C)}{P(C)}\\
&= P(A\mid B\cap C)P(B\mid C).
\end{align}$$

But you might ask: given the expression $P(A\cap B \mid C)$, how would
I even begin to think of ways to get to $P(A\mid B\cap C)P(B\mid C)$ which expression
I have never seen before and have no reason to suspect might
equal $P(A\cap B \mid C)$?  Well, one reason to suspect that something
like this might be true is to consider that conditional probabilities
conditioned on the occurrence of the same event are a valid probability measure, 
that is,
every theorem of probability, e.g. $$P(A\cup B) = P(A)+P(B)-P(A\cap B),$$
holds if all the probabilities in question are changed
to conditional probabilities
given that event $C$ occurred. Thus,
$$P(A\cup B\mid C) = P(A\mid C)+P(B\mid C)-P(A\cap B\mid C),$$
and similarly,
$$P(A\cap B) = P(A\mid B)P(B) ~\Rightarrow ~
P(A\cap B\mid C) = P(A\mid B\cap C)P(B\mid C).$$
So, once you have the right intuition, you can, if you wish,
formally verify the
correctness of the result using the steps outlined above.
