The following statement is true for deductive logic (e.g. in a boolean algebra): $$a\wedge b \Rightarrow \neg c \quad \text{if and only if }\quad a \Rightarrow \neg(b\wedge c).$$
It seems like you can transform deductive inference into inductive inference by replacing implication $(a\Rightarrow b)$ with plausibility $(a|b)$, so my question is whether a similar statement holds in probability theory, i.e. is it true in general that
$$p(AB|\bar C) = p(A|\overline{BC})$$
provided $p(\bar C) \neq 0$ and $p(\overline{BC})\neq 0$?
I'm unfamiliar with axiomatic probability theory, so I'm not sure how to approach such a statement from first principles (rather than just making algebraic manipulations I've been taught are valid.)