Since $\mathbb{P}(F) > 0$, then $F \neq \emptyset$. If $F = \Omega$, then result R4338: Conditional probability of almost surely true event yields the claim, so we may further assume that $F \neq \Omega$. Under this assumption, the pair $F, \Omega \setminus F$ is a disjoint partition of $\Omega$ and applying results

(i)	R2089: Unsigned basic expectation is compatible with probability measure
(ii)	R4333: Binary product of indicator functions equals indicator of intersection
(iii)	R3401: Probability calculus expression for conditional expectation given disjoint non-null partition

we have \begin{equation} \begin{split} \frac{\mathbb{P}(E \cap F)}{\mathbb{P}(F)} = \frac{\mathbb{E}(I_{E \cap F})}{\mathbb{P}(F)} & = \frac{\mathbb{E}(I_E I_F)}{\mathbb{P}(F)} \\ & = \mathbb{E}(I_E \mid \sigma \langle F \rangle) & = \mathbb{P}(E \mid \sigma \langle F \rangle) = \mathbb{P}(E \mid F) \end{split} \end{equation} where the last two equalities hold by definition. This is what was required to be shown. $\square$