Necessary conditions for $P(A|B)=P(A|C)P(C|B)$? Suppose we know $P(A|C)$ and $P(C|B)$ and we want to find $P(A|B)$. What are the necessary conditions under which the $C$ "cancels out" and we have the equality $P(A|B)=P(A|C)P(C|B)$? I have found the following sufficient conditions for which the equality holds true:

*

*$P(AB|C)=P(A|C)P(B|C)\hspace{40 pt}$ ($A$ and $B$ are conditionally independent on $C$)

*$P(A|C^c)=0\quad\mbox{or}\quad{}P(B|C^c)=0\hspace{18 pt}$ (Either $A$ or $B$ is a subset of $C$)

Proof:
$$\begin{align}P(A|B)&=\frac{P(AB)}{P(B)}=\frac{P(AB|C)P(C)+P(AB|C^c)P(C^c)}{P(B)}=\frac{P(A|C)P(B|C)P(C)}{P(B)}&\\&=P(A|C)*\frac{P(BC)}{P(B)}=P(A|C)P(C|B)&\end{align}$$
Are my two conditions necessary conditions for the equality to hold true or are there more general conditions under which it holds?
EDIT:
Second condition should be: $AB$ is a subset of $C$, so $P(AB|C^c)=0$
 A: Consider $A=\{1,2\}, B=\{2,3\}, C=\{2\}, C^{\complement}=\{1,3,4\}$.  In this case, neither $A\subseteq C$ nor $B\subseteq C$, but $A\cap B\cap C^\complement=\emptyset$.
Rather, we have that $\mathsf P(A,B,C^\complement)=0$ when $A\cap B\subseteq C$ almost certainly.

Also:
$$\begin{align}\mathsf P(A,B,C)&=\mathsf P(A\mid B,C)\,\mathsf P(C\mid B)\,\mathsf P(B)&&\text{so if }A\perp B\mid C\text{ then }\ldots\\&=\mathsf P(A\mid C)\,\mathsf P(C\mid A,B)\,\mathsf P(B)&&\text{so if instead }A\perp C\mid B\text{ then }\ldots\end{align}$$
A: 
Necessary conditions for P(A|B)=P(A|C)P(C|B)?

If $p(A|C) = 0$, then you must also have $p(A|B) = 0.$
Without loss of generality, assume that $p(A|C) \neq 0.$ 
Implicit in this assumption is that $p(A) \neq 0 \neq p(C).$
Under this assumption, the LHS of the assertion will equal $0$ if and only if $p(C|B) = 0.$
So, now make the further assumption, without loss of generality, that neither $p(A|B)$ nor $p(C|B)$ is equal to $0$.  This means that you are (also) assuming that $p(B) \neq 0.$

Let $(R,S)$ denote the event that $R$ and $S$ both occur.
Then
$$p(A|B) = \frac{p(A,B)}{p(B)} = \frac{p(A,B)}{p(A,C)} \times \frac{p(A,C)}{p(B)}. \tag1 $$
$$p(A|C) \times p(C|B) = \frac{p(A,C)}{p(C)} \times  \frac{p(B,C)}{p(B)}. \tag2 $$
Using (1) and (2) above, and assuming that none of $p(A|B), p(A|C),$ or $p(C|B)$ are $= 0$, the original assertion will be true if and only if
$$\frac{p(B|A)}{p(C|A)} = \frac{p(A,B)}{p(A,C)} = \frac{p(B,C)}{p(C)} = p(B|C). \tag3 $$
I don't see how you can draw any further conclusions than these.
Edit
(3) above can be equivalently manipulated to :
$$\frac{p(A,B)}{p(B,C)} = p(A|C). \tag4 $$
However, you can also reach (4) above, directly from the original assertion, via
$$p(A|C) = \frac{p(A|B))}{p(C|B)} = \frac{p(A,B)/p(B)}{p(C,B)/p(B)}.$$
