We proceed by induction on \(\cgG\text{.}\) If \(|\cgG|=1\) and \(\cgG=\{E\}\text{,}\) we are simply asserting that \(P(\overline{E})\ge 1-x(E)\text{,}\) which is true since \(P(E)\le x(E)\text{.}\) Now suppose that \(|\cgG|=k\ge2\) and that the lemma holds whenever \(1\le |\cgG|\lt k\text{.}\) Let \(\cgG =\{E_1,E_2,\dots,E_k\}\text{.}\) Then
\begin{equation*}
P(\prod_{i\ge1}^k \overline{E_i})=P(\overline{E_1}|\prod_{i=2}^k\overline{E_i})
P(\overline{E_2}|\prod_{i=3}^k\overline{E_i})P(E_3|\prod_{i=4}^k\overline{E_i})\dots
\end{equation*}
Now each term in the product on the right has the following form:
\begin{equation*}
P(\overline{E}|\prod_{F\in\cgF_E}\overline{F})
\end{equation*}
where \(|\cgF_E|\lt k\text{.}\)
So, we done if we can show that
\begin{equation*}
P(\overline{E}|\prod_{F\in\cgF_E}\overline{F})\ge 1- x(E)
\end{equation*}
This is equivalent to showing that
\begin{equation*}
P(E|\prod_{F\in\cgF_E}\overline{F})\le x(E)
\end{equation*}
Suppose first that \(\cgF_E\cap\cgN(E)=\emptyset\text{.}\) Then
\begin{equation*}
P(E|\prod_{F\in\cgF_E}\overline{F})=P(E)\le x(E).
\end{equation*}
So we may assume that \(\cgF_E\cap\cgN(E)\neq\emptyset\text{.}\) Let \(\cgF_E=\{F_1,F_2,F_r,F_{r+1},F_{r+2},\dots, F_t\}\text{,}\) with \(F_i\in\cgN_E\) if and only if \(r+1\le i\le t\text{.}\) Then
\begin{equation*}
PE|\prod_{F\in\cgF_E}\overline{F})=
\frac{P(E\prod_{F\in\cgF_E\cap\cgN(E)}\overline{F}|\prod_{F\in\cgF_E-\cgN(E)}\overline{F})}
{P(\prod_{F\in\cgF_E\cap \cgN(E)}\overline{F})}
\end{equation*}
Consider first the numerator in this last expression. Note that
\begin{equation*}
P(E\prod_{F\in\cgF_E\cap\cgN(E)}\overline{F}|\prod_{F\in\cgF_E-\cgN(E)}\overline{F})\le
P(E|\prod_{F\in\cgF_E\cap\cgN(E)}\overline{F})\le x(E)\prod_{F\in\cgF_E\cap\cgN(E)}(1-x(F))
\end{equation*}
Next, consider the denominator. By the inductive hypothesis, we have
\begin{equation*}
P(\prod_F\in\cgF_E\cap\cgN(E)\overline{F}\ge \prod_{F\in\cgF_E\cap\cgN(E)}(1-x(F)).
\end{equation*}
Combining these last two inequalities, we have
\begin{equation*}
P(E|\prod_{F\in\cgF_E}\overline{F})\le x(E)\prod_{\cgN(E)-\cgF_E}(1-x(F))\le x(E),
\end{equation*}
and the proof is complete.
