Let A and B be events in a sample space with positive probabilities. Prove that $\displaystyle P(B|A)>P(B)\iff P(A|B)>P(A)$

The seems relatively easy.

1st direction: $\displaystyle P\Rightarrow Q$

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

$\displaystyle \displaystyle\Rightarrow\frac{P(A\cap B)}{P(A)}\cdot\frac{P(A)}{P(B)}>P(B)\cdot\frac{P(A )}{P(B)}$

$\displaystyle \displaystyle\frac{P(A\cap B)}{P(B)}>P(A)\Rightarrow P(A|B)>P(A)$

The second direction is identical so I am not going to post it.

Is this how it is proven though?