Second-order Bayesian inference We have three events $A$, $B$ and $C$ in question, and given appropriate priors, we derive the posterior $\Pr(A|B)$. Now we want to derive a 'second-order' posterior $\Pr(A|B,C)$ by using the 'first-order' posterior $\Pr(A|B)$ as the prior.
First of all, does this mean that $\Pr(A|B,C)$ is the same as $\Pr((A|B)|C)$? If so, is the following correct:
$$\Pr(A|B,C)=\frac{\Pr(C|(A|B))\cdot\Pr(A|B)}{\Pr(C)}$$
and how do we derive $\Pr(C|(A|B))$?
 A: If what you're interested in is the probability of $A$ given $B$ and $C$ occur,  this is just $\Pr(A|B \wedge C)$ (Or $\Pr(A|B\cap C)$ if you prefer that way of talking).
As was mentioned in the comments, $A|B$ isn't an event in the algebra of events, so it's meaningless to say $\Pr((A|B)|C)$.
There's nothing "second order" about this. It's just iterated conditionalisation.
So, using the definition of conditionalisation (And using $XY$ to mean $X\wedge Y$):
$$\Pr(A|BC) = \frac{\Pr(ABC)}{\Pr(BC)}$$
Using the fact that $\Pr(XY)=\Pr(X|Y)\Pr(Y)$ twice, we find:
$$\Pr(A|BC) = \frac{\Pr(AB|C)\Pr(C)}{\Pr(B|C)\Pr(C)} = \frac{\Pr(AB|C)}{\Pr(B|C)}$$
Depending on what probabilities are known, many other manipulations are possible. See the wikipedia page on Bayes' theorem
A: From what you know: 

Pr(A|B), Pr(C|B), Pr(B|A), Pr(C|A) and Pr(A)

You can derive the joint distribution from the factors:
P(A, B, C) = P(C | B) P(B | A) P(A). This set of equations indicates that the structure of your conditional independence assumptions is as follows:

A -> B -> C

Which means that C is conditionally independent of of A given B (once we know the value of B, C provides no additional information about A). In this case P(A | B, C) = P(A | B). 
