Result
R3649: Expectation of conditional probability shows that
\begin{equation}
\mathbb{E}(\mathbb{P}(E \mid \mathcal{G}))
= \mathbb{P}(E) = 1
\end{equation}
Since the random basic real number $\mathbb{P}(E \mid \mathcal{G})$ takes values in the interval $[0, 1]$, result
R4398: then guarantees that
\begin{equation}
\mathbb{P}( \mathbb{P}(E \mid \mathcal{G}) = 1)
= 1
\end{equation}
$\square$
