Is is true that if $H$ entails $H'$, then $P(H\mid E)\le P(H'\mid E)$ for any $E$?

Question

Is it a theorem in probability calculus that if $H$ entails $H'$, then $P(H\mid E)\le P(H'\mid E)$ for any $E$? If yes, how do I prove this?

Thanks!

Answer 1

$H \subseteq H'$

$\implies H \cap E \subseteq H' \cap E$

$\implies \mathbb{P}(H \cap E) \leq \mathbb{P}(H' \cap E)$

$\implies \frac{\mathbb{P}(H \cap E)}{\mathbb{P}(E)} \leq \frac{\mathbb{P}(H' \cap E)}{\mathbb{P}(E)}$ (Provided that $\mathbb{P}(E) \neq 0$)

$\implies \mathbb{P}(H | E) \leq \mathbb{P}(H' | E)$