Is it possible to decompose a conditional probability with three or more elements (i.e. number of events $n>=3$, where $n$ is the number of elements or events) into conditional probabilities of only two elements or the marginal probabilities of one element? Knowing this decomposition, it would help to solve higher order Markov Chain mathematically. I also know that this decomposition can be solved if we add assumption of conditional independent.
To make it concrete here is a negative example:
Notice that the RHS still contains a conditional probability with three elements $Pr(a,b│c)$.
Assuming conditional independent on $c$, we have $Pr(a,b│c)=Pr(a│c)∙Pr(b│c)$. Thus, the conditional probability decomposition becomes
My question is whether this type of conditional probability decomposition into one or two element is possible without making assumption. If it is really unsolvable problem, then at least we know that the assumption of conditional independent is a must.