Factoring of conditional probability I have been watching Tom Mitchell's lecture on Bayes Nets: 
http://cc-web.isri.cmu.edu/CourseCast/Viewer/Default.aspx?id=bc507778-7a18-4121-b345-
83d9bab72f55 
He states the following by way of the chain rule. I'm being dense I am sure but don't recognize this quickly. Is it a straight application of the chain rule?
$$P(x_3,x_4\ |\ Y) = P(x_3\  |\  Y, X_4 )P(X_4\  |\  Y)$$
 A: I like to think of the chain rule as:
$$P(A\cap B) = P(B)P(A\mid B) = P(A)P(B \mid A)$$
instead of $P(A\cap B) = P(B)P(A\mid B)$ which is the way G. Jay Kerns expressed
it in his comment on the question.
There is no mathematical difference, of course, but
we can think of $P(A\cap B) = P(A)P(B \mid A)$ as encapsulating the
following heuristic way of thinking.  

We want to find $P(A \cap B)$, the probability that both $A$ and $B$ 
  occurred.  Clearly, $A$ must have occurred (which has probability $P(A)$),
  and if we assume that $A$ has occurred, then in order for $A \cap B$ to
  occur, $B$ must occur too, which has probability $P(B\mid A)$,
  (note: not $P(B)$
  since we have already assumed that $A$ has occurred), and so
  $P(A \cap B) = P(A)P(B \mid A)$.

Generalizing this argument (which fortunately can be backed up
by straight mathematical calculations from the definition of
conditional probability),
$$P(A\cap B \cap C) = P(A)P(B\mid A)P(C\mid (A \cap B))$$
or, proceeding in the opposite direction,
$$P(A\cap B \cap C) = P(C)P(B\mid C)P(A\mid (B \cap C)).$$
Dividing this last equation on both sides by $P(C)$ gives
$$\begin{align*}
\frac{P(A\cap B \cap C)}{P(C)} 
&= \frac{P(C)P(B\mid C)P(A\mid B \cap C)}{P(C)}\\
P(A\cap B \mid C) &= P(B\mid C)P(A\mid (B \cap C))\\
&= P(A\mid (B \cap C))P(B\mid C)
\end{align*}$$
So the OP should accept that the equality follows from 
 the chain rule (plus an extra step).
