Using the law of total probability to prove an inequality The total law of probability says to assume that $B_1, B_2, ..., B_k$ is a partition of S such that $P(B_i)$ > 0, for $i = 1, 2, ..., k.$
$$P(A) = \sum_{i=1}^k P(A|B_i)P(B_i)$$
I am trying to prove that $P(A) > P(B)$, assuming $P(A|C) > P(B|C)$ and $P(A|C^c) > P(B|C^c)$, using the law of total probability.
I know how to visualize an event A as it lies on partitions of a "background" B ($B_1$ might cover a portion of A, $B_2$ might cover another, etc.), but I don't know how to picture this situation as a probability diagram.
All I have so far is:
$$P(A) = P(A|C)P(C) + P(A|C^c)P(C^c)\,\,;\,\,P(B) = P(B|C)P(C) + P(B|C^c)P(C^c)$$
and I'm not sure where to go from here. Any help would be appreciated.

$P(A|C)>P(B|C)$ implies $P(A,C)>P(B,C)$ (multiply $P(C)$ to both sides). Also, $P(A|C^c)>P(B|C^c)$ implies $P(A,C^c)>P(B,C^c)$ (multiply ($P(C^c)$ to both sides). Add the inequalities to get $P(A,C)+P(A,C^c)>P(B,C)+P(B,C^c)$. This is just $P(A)>P(B)$, from the law of total probability if you wish.
