# Show that if $P(H | E) > P(H)$, then $P (H' | E) < P(H')$ using only probability rules and not using frequentist way of proving this

I am having trouble showing this proof since I got from $P(H | E) > P(H)$, \begin{align} \frac{P (H \cap E)}{P (E)} &> P (H) \\[5pt] P (H \cap E) &> P (H) P(E) \\ P(H \cap E) &> P(E) - P(E)P (H') \end{align} However I get stuck at this point and I am not sure how to continue. Can someone explain in detail the steps I need to do to get to $P (H' | E) < P(H')$?

This is fairly straightforward if you use the rule that $P(X) = 1 - P(X')$
\begin{aligned} P(H | E)&> P(H) \\ 1 - P(H' | E) &> 1 - P(H') \\ -P(H' | E) &> - P(H') \\ \therefore P(H' | E) &< P(H') \end{aligned}
Note that when we multiply by $-1$ in the last step, we switch the direction of the inequality