Conditional independence identity If A and B are independent and conditionally independent given C, but A and C and B and C are not necessarily independent, then
$ P(A,B | C) = P(A | C) P(B | C) $
Is it also true that
$ P(C | A, B) = P(C | A)P(C | B) $ ?

When I worked out the algebra with the same independence assumptions, I got the following:
$ P(A,B,C) = P(A,B | C)P(C) $
$ P(A,B,C) = P(A,C | B)P(B) $
Setting these identities equal gives:


*

*$ P(A,C | B)P(B) = P(A,B | C)P(C) $  

*$ P(A | B)P(C | A,B)P(B) = P(A|C)P(B|C)P(C) $

*$ P(A | B)P(C | A,B) = P(A|C)P(C|B) $ by the application of Bayes' rule

*$ P(C | A,B) = \frac{P(A|C)P(C | B)}{P(A|B)} $


Which is the correct answer? Or, if neither, where did I go wrong?
 A: Your conditions are that $\mathbb{P}(A,B) = \mathbb{P}(A) \cdot \mathbb{P}(B)$ and $\mathbb{P}(A,B|C) = \mathbb{P}(A|C) \cdot \mathbb{P}(B|C)$.  From these stipuatled conditions you have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(C|A) \cdot \mathbb{P}(C|B) 
&= \frac{\mathbb{P}(A,C)}{\mathbb{P}(A)} \cdot \frac{\mathbb{P}(B,C)}{\mathbb{P}(B)}  \\[6pt]
&= \frac{\mathbb{P}(A,C) \cdot \mathbb{P}(B,C)}{\mathbb{P}(A) \cdot \mathbb{P}(B)} \\[6pt]
&= \frac{\mathbb{P}(A|C) \cdot \mathbb{P}(B|C) \cdot \mathbb{P}(C)^2}{\mathbb{P}(A) \cdot \mathbb{P}(B)} \\[6pt]
&= \frac{\mathbb{P}(A,B|C) \cdot \mathbb{P}(C)}{\mathbb{P}(A) \cdot \mathbb{P}(B)} \cdot \mathbb{P}(C) \\[6pt]
&= \frac{\mathbb{P}(A,B,C)}{\mathbb{P}(A) \cdot \mathbb{P}(B)} \cdot \mathbb{P}(C) \\[6pt]
&= \mathbb{P}(C|A,B) \cdot \mathbb{P}(C). \\[6pt]
\end{aligned} \end{equation}$$
Thus, we have $\mathbb{P}(C|A,B) = \mathbb{P}(C|A) \cdot \mathbb{P}(C|B)$ if and only if $\mathbb{P}(C) = 1$ or $\mathbb{P}(C|A,B)=0$.  Thus, your result does not hold in general, but requires at least one of these additional conditions.
A: Needn't always be true.
Consider a counter-example where $A$, $B$, $C$ are mutually independent, that is, $p(A, B, C) = p(A)\;p(B)\;p(C)$.  This distribution satisfies your givens.
However, your assertion $p(C|A,B) = p(C|A) \; p(C|B)$ boils down to saying $p(C) = p(C)\;p(C)$, which needn't always hold true.
