Using Bayes Law leads me to a seemingly nonsensical result

Question

I am trying to prove that, given events $A$ and $B$, if $P(A) > P(B) > 0$, then

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

Now, here is my proof:

$P(A|B) = \dfrac{P(A)P(A|B)}{P(B)} \tag*{(by Bayes' Theorem)}$

$P(A|B) > \dfrac{P(B)P(A|B)}{P(B)} \tag*{(since P(A) > P(B) > 0)}$

$P(A|B) > P(B|A) \tag*{}$

This is fine, but then I considered the case where $A$ and $B$ are mutually exclusive. In this case, $P(A|B) = P(B|A) = 0$, so what I just proved doesn't hold. Have I applied Bayes' Theorem incorrectly here?

Answer 1

Your second line does not follow from your first in the case $P(A|B) = P(B|A) = 0$.

Specifically, $a > b > 0$ only implies that $\lambda a > \lambda b > 0$ if $\lambda > 0$.