Is there a more general form of the conditional probability? Say you have $P(A_1, A_2, \dots, A_n | B_1, B_2, \dots, B_m)$.  Is there a general way to break this down into combinations of $P(X|Y)$'s and $P(Z)$'s?
I understand that $P(X|Y) = \frac{P(X,Y)}{P(Y)}$ and $P(X_1, \dots, X_n) = P(X_1)P(X_2 | X_1)P(X_3 | X_2, X_1)\dots P(X_n|X_{n-1}, \dots, X_1)$, but I'm not sure if there's a way to reduce this using these formulas.
Does anyone have any ideas?
 A: My interpretation of your question: You are looking for a general procedure to break  down $P(A_1, A_2, \dots, A_n | B_1, B_2, \dots, B_m)$ into a combination of terms of the form $P(X|Y)$ and $P(Z)$ where $X,Y,Z$ are the "singleton" events like $A_i, B_j$.  (In particular, $Y$ cannot be a combo like $B_1 \cap B_2$.)
Define the collection of singleton events as $\mathcal S = \{A_1, \dots, A_n, B_1 \dots B_m\}$.
I.e. you seek a function $f_{n,m}$ s.t. for any probability space, 
$$
P(A_1, A_2, \dots, A_n | B_1, B_2, \dots, B_m) = f_{n,m}(\dots P(X) \dots\,\,\,, \,\,\, \dots P(X|Y) \dots)
$$
where the function $f_{n,m}$ takes the following inputs:


*

*$(n+m)$ inputs of the form $P(X)$ where $X \in \mathcal S,$ and,

*$(n+m)(n+m-1)$ inputs of the form $P(X|Y)$ where $X, Y \in \mathcal S$ and $X \neq Y$.

If my interpretation is correct, then that is not possible in general when $m>1$, i.e. no such $f$ exists.
You are asking for "credible or official sources."  Here I provide a counter-example that can be easily verified.  Is that credible enough?  :)  My counter-example is for the simplest case of $n=1, m=2$.
Probability space #$1$: Flip a fair coin twice and let:


*

*$B_1 =$ first coin shows Head

*$B_2 =$ second coin shows Head

*$A =$ two coins are the same
It is obvious that $P(X) = 1/2$ for all $X \in \mathcal S$.  It is also obvious that $P(X|Y) = 1/2$ for all distinct $X,Y \in \mathcal S$ since the three events are pairwise independent.
In this case $P(A|B_1, B_2) = 1$.
Probability space #$2$: Same as above, except instead of $A$ we have $A' =$ two coins are different.  Again all $P(X) = 1/2$ and all $P(X|Y) = 1/2$.  However in this case $P(A'|B_1, B_2) = 0$.
If there exists a function $f$ which can break down $P(X|Y,Z)$, then this function is given identical inputs in both probability spaces above (all its input values are $1/2$), and yet this function must return $1$ for probability space #$1$ and $0$ for #$2$, which is impossible.  Therefore, no such function exists.
A: Any probability of the form P(.|X) forms a probability law. They obey all the axioms.
So, $(_1,_2,…,_n|_1,_2,…,_) = P(A_1|_1,_2,…,_) . P(A_2 | _1,_2,…,_,A_1) . P(A_3 | _1,_2,…,_, A_1, A_2) ... P(A_n| _1,_2,…,_, A_1, A_2 ... A_n)$
.
