Applying results

(i)	R4145: Binary union is an upper bound to both sets in the union
(ii)	R2090: Isotonicity of probability measure

we have \begin{equation} 1 = \mathbb{P}(F) \leq \mathbb{P}(E \cup F) \leq 1 \end{equation} Thus, $\mathbb{P}(E \cup F) = 1$. Next, using result R4445: Inclusion-exclusion principle for probability of binary union, we have \begin{equation} \mathbb{P}(E \cap F) = \mathbb{P}(E) + \mathbb{P}(F) - \mathbb{P}(E \cup F) = \mathbb{P}(E) + 1 - 1 = \mathbb{P}(E) \end{equation} This completes the proof. $\square$