If $P(B|A) > P(B)$ and $P(C|B)>P(C)$, then prove or disprove $P(C|A) > P(C)$

Question

I have this question in intro to probability:

If $P(B|A) > P(B)$ and $P(C|B)>P(C)$ then prove or disprove $P(C|A) > P(C)$

It is also given that $A, B,$ and $C$ are events with probability greater than $0$. I have tried to do the following:

$$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)} > P(B)$$

by Bayes

$$\implies P(A|B) > P(A) \text{ because } P(B)>0$$

I did the same for the other expression, but I can't seem to be able to connect the formulas.

Answer 1

Take a roll of a six-sided die, and define the events $$ A = \{ 1\}, C = \{2\}, B = \{1,2\}.$$ Then $P(B) = 1/3, P(B|A) = 1, P(C) = 1/6, P(C|B) = 1/2.$ But $P(C|A) = 0 < P(C)$.

Of course, the opposite can happen too - take $A = \{1,2,3\}, B = \{1,2\}, C = \{1\}.$