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We know that $P(B|A)$ and $P(C|B)$ don't generally constraint $P(C|A)$. But do $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$ jointly constrain $P(C|A)$ and $P(neg A|neg C)$? In particular, let $t > frac{1}{2}$ be such that

$t leq$ $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$.

What does this tell us about $P(C|A)$ and $P(neg A|neg C)$? In particular, does it imply $P(C|A)$, $P(neg A|neg C)$ $leq 1 - t$ can't both hold?

(-----Answer to the old question-----)

Consider the probability function $P$ that assigns the following probabilities to atomic events:
$$P(neg A wedge B wedge C) = frac{2}{3}$$
$$P(A wedge B wedge neg C) = frac{1}{12}$$
$$P(neg A wedge neg B wedge C) = frac{1}{12}$$
$$P(neg A wedge neg B wedge neg C) = frac{1}{6}$$
The other atomic events get null probability by the function $P$.

With such assignment we have that $P(B|A)=1$, $P(C|B)=frac{8}{9}$, $P(neg A|neg B)=1$ and $P(neg B|neg C)=frac{2}{3}$. Note that $P(C|A)=0$ under the distribution $P$. Therefore, the implication you suggest does not hold (take $t=frac{2}{3}$, for example).