# Proof about an inequality in probability theory

Suppose we have the following inequalities

$$P(A\mid C) > P(B\mid C) \text{ and } P(A\mid \bar{C})>P(B\mid \bar{C})$$

Can we make a conclusion about the following inequality

$$P(A)>P(B)$$

I have been wondering this question for a while and I tried to apply the Total Average Probability law to it but it didn't get me anywhere.

-
$P(A\mid C) > P(B\mid C)$ implies $P(A\cap C)> P(B\cap C)$, and similarly $P(A\mid \overline{C})>P(B\mid \overline{C})$ implies $P(A\cap\overline C)>P(B\cap\overline C)$. Adding the inequalities, $P(A)=P(A\cap C)+P(A\cap\overline C)>P(B\cap C)+P(B\cap\overline C)=P(B)$.