what i know is,
conditional probability: P(X|Y) = P(X,Y) / P(Y)
total probability: P(X) = sum Y [P(X|y) P(y)]
bayes rule: p(X|Y) = p(Y|X) P(X)
Is it possible to express P(A,B|C) in term of P(A,C|B) and joint distribution of two random variable, i.e. P(A,C). P(A,B), P(B,C)
Is it possible to express P(A,B|C,D) in term of P(A,C|B, D) and joint distribution of two random variable, i.e. P(A,C). P(A,B), P(B,C)
I think for $P(AB|C)$, the answer is no.
$$P(AB|C)=\frac{P(ABC)}{P(C)}=\frac{P(AC|B)P(B)}{P(C)}=P(AC|B)\frac{P(B)}{P(C)} =P(AC|B)\frac{P(B)}{P(C)}$$
$$\frac{P(AB|C)}{P(AC|B)} =\frac{P(B)}{P(C)}$$
Now $$\frac{P(B)}{P(C)}=\frac{P(CB)/P(C|B)}{P(BC)/P(B|C)}=\frac{P(B|C)}{P(C|B)}$$
$$P(AB|C)=P(AC|B)\frac{P(B)}{P(C)}=P(AC|B)\frac{P(B|C)}{P(C|B)}$$
$P(B|C)$ and $P(C|B)$ include $P(C)$ and $P(B)$.
I think for $P(AB|CD)$, the answer is yes. $$P(AB|CD)=\frac{P(ABCD)}{P(CD)}=\frac{P(AC|BD)P(BD)}{P(CD)}$$