Every subset $U$ of $N$ that satisfies $U \sups S$ and $U \sups T$ is automatically nonempty (since $U$ contains the nonempty set $S$ as a subset), and thus belongs to $P$. Hence, the $U \in P$ that satisfy $U \sups S$ and $U \sups T$ are precisely the subsets $U$ of $N$ that satisfy $U \sups S$ and $U \sups T$. Thus, we have
\begin{align*}
& \set{ U \in P \mid U \sups S \text{ and } U \sups T } \\
= & \set{ U \subs N \mid U \sups S \text{ and } U \sups T } \\
= & \set{ U \subs N \mid U \sups \underbrace{S \cap T}_{=G} } \\
= & \set{ U \subs N \mid U \sups G } \\
= & \set{ U \subs N \mid N \setminus U \subs N \setminus G }
\end{align*}\begin{align*}
&\ \set{ U \in P \mid U \sups S \text{ and } U \sups T } \\
= &\ \set{ U \subs N \mid U \sups S \text{ and } U \sups T } \\
= &\ \set{ U \subs N \mid U \sups \underbrace{S \cup T}_{=G} } \\
= &\ \set{ U \subs N \mid U \sups G } \\
= &\ \set{ U \subs N \mid N \setminus U \subs N \setminus G }
\end{align*}
(because the condition "$U \sups G$" on a subset $U$ of $N$ is equivalent to the condition "$N \setminus U \subs N \setminus G$").
Hence, the summation sign "$\suml_{\substack{U \in P; \\ U \sups S \text{ and } U \sups T}}$" in \eqref{darij1.pf.l6.1} can be rewritten as "$\suml_{\substack{U \subs N; \\ N \setminus U \subs N \setminus G}}$". Thus, \eqref{darij1.pf.l6.1} rewrites as
\begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \subs N; \\ N \setminus U \subs N \setminus G}}
\tup{q-1}^{n-\abs{U}} .
\label{darij1.pf.l6.2}
\tag{3}
\end{align}
Every subset $U$ of $N$ satisfies $n - \abs{U} = \abs{N \setminus U}$ (since $U \subs N$ yields $\abs{N \setminus U} = \underbrace{\abs{N}}_{=n} - \abs{U} = n - \abs{U}$). Thus, \eqref{darij1.pf.l6.2} becomes
\begin{align}
& \suml_{U \in P} \tup{q-1}^{n-\abs{U}} q^{\abs{S}-n} \ive{ U \sups S } q^{\abs{T}} \ive{ U \sups T } \\
&= q^{\abs{S}-n} q^{\abs{T}} \underbrace{\suml_{\substack{U \subs N; \\ N \setminus U \subs N \setminus G}}}_{\substack{= \suml_{\substack{U \subs N; \\ N \setminus U \subs H}} \\ \tup{ \text{since } N \setminus G = H } }}
\underbrace{\tup{q-1}^{n-\abs{U}}}_{\substack{= \tup{q-1}^{\abs{N \setminus U}} \\ \tup{ \text{since } n - \abs{U} = \abs{N \setminus U} } }} \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \subs N; \\ N \setminus U \subs H}}
\tup{q-1}^{\abs{N \setminus U}} \\
&= q^{\abs{S}-n} q^{\abs{T}} \suml_{\substack{U \subs N; \\ U \subs H}}
\tup{q-1}^{\abs{U}}
\label{darij1.pf.l6.3}
\tag{4}
\end{align}
(here, we have substituted $U$ for $N \setminus U$ in the sum, since the map
\begin{align}
\set{U \subs N} \to \set{U \subs N}, \qquad
U \mapsto N \setminus U
\end{align}
is a bijection).
Recall that $\abs{P} = 2^n - 1$ (as we proved during our proof of Lemma 4 (b)). Also, $\sumS \abs{S} = n 2^{n-1}$ (by Lemma 4 (a)). Thus,
\begin{align}
\sumS \tup{n - \abs{S}}
&= \underbrace{\sumS n}_{= \abs{P} \cdot n} - \underbrace{\sumS \abs{S}}_{= n 2^{n-1}}
\\
&= \underbrace{\abs{P}}_{= 2^n - 1} \cdot n - n 2^{n-1} = \tup{2^n - 1} \cdot n - n 2^{n-1} \\
&= n \tup{2^n - 1 - 2^{n-1}} = n \tup{2^{n-1} - 1}
\end{align}
(since $\underbrace{2^n}_{= 2 \cdot 2^{n-1}} - 1 - 2^{n-1} = 2 \cdot 2^{n-1} - 1 - 2^{n-1} = 2^{n-1} - 1$).
Endow our set $P$ with the same total ordering that we used in the proof of Theorem 1. Now, it is easy to see that the $P \times P$-matrix $E$ is upper-triangular. Since the determinant of an upper-triangular $P \times P$-matrix equals the product of its diagonal entries, we thus conclude that
\begin{align}
\det E
&= \prodS \tup{ \underbrace{\tup{q-1}^{n-\abs{S}} q^{\abs{S}-n}}_{= \tup{\dfrac{q-1}{q}}^n \tup{\dfrac{q}{q-1}}^{\abs{S}} }
\underbrace{\ive{ S \sups S }}_{= 1} }
= \prodS \tup{ \tup{\dfrac{q-1}{q}}^n \tup{\dfrac{q}{q-1}}^{\abs{S}} } \\
&= \underbrace{ \tup{ \prodS \tup{\dfrac{q-1}{q}}^n } }_{= \tup{\tup{\dfrac{q-1}{q}}^n}^{\abs{P}} }
\underbrace{ \tup{ \prodS \tup{\dfrac{q}{q-1}}^{\abs{S}} } }_{= \tup{\dfrac{q}{q-1}}^{\sumS \abs{S}} } \\
&= \tup{\tup{\dfrac{q-1}{q}}^n}^{\abs{P}} \tup{\dfrac{q}{q-1}}^{\sumS \abs{S}} \\
&= \underbrace{ \tup{\tup{\dfrac{q-1}{q}}^n}^{2^n - 1} }_{= \tup{\dfrac{q-1}{q}}^{n \tup{2^n - 1}} }
\underbrace{ \tup{\dfrac{q}{q-1}}^{n 2^{n-1}} }_{= \tup{\dfrac{q-1}{q}}^{-n 2^{n-1}}}\\
& \qquad \tup{ \text{since } \abs{P} = 2^n - 1
\text{ and } \sumS \abs{S} = n 2^{n-1} } \\
&= \tup{\dfrac{q-1}{q}}^{n \tup{2^n - 1}} \tup{\dfrac{q-1}{q}}^{-n 2^{n-1}} \\
&= \tup{\dfrac{q-1}{q}}^{n \tup{2^n - 1} - n 2^{n-1}}
= \tup{\dfrac{q-1}{q}}^{n 2^{n-1} - n}
\end{align}\begin{align}
\det E
&= \prodS \tup{ \underbrace{\tup{q-1}^{n-\abs{S}} q^{\abs{S}-n}}_{= \tup{\dfrac{q-1}{q}}^{n-\abs{S}} } \underbrace{\ive{ S \sups S }}_{= 1} }
\\
&= \prodS \tup{\dfrac{q-1}{q}}^{n-\abs{S}}
= \tup{\dfrac{q-1}{q}}^{\sumS \tup{n - \abs{S}}}
= \tup{\dfrac{q-1}{q}}^{n 2^{n-1} - n}
\end{align}
(since $n \tup{2^n - 1} - n 2^{n-1} = n 2^{n-1} - n$$\sumS \tup{n - \abs{S}} = n \tup{2^{n-1} - 1} = n 2^{n-1} - n$).
