\[It:NoENoPath\] $e \notin H$, and both $a \not\succeq b$ and $b
    \not\succeq a$ in $H$.

4.  \[It:EAtoB\] $e \in H$, and $a \succeq b$ in $H \setminus \{e\}$.

5.  \[It:ENoPath\] $e \in H$, and $a \not\succeq b$ in $H \setminus
    \{e\}$.

Obviously, $H \in {\mathrm{\ref{It:NoEAtoB}}}$ if and only if $H \cup \{e\} \in
{\mathrm{\ref{It:EAtoB}}}$. The number of edges of $H \cup \{e\}$ is the number of edges of $H$ plus $1$, so $$\label{Eq:1Plus4}
\SumSub(\alpha, {\mathrm{\ref{It:NoEAtoB}}} \cup {\mathrm{\ref{It:EAtoB}}}) = \sum_{H \in
\mathrm{\ref{It:NoEAtoB}}} (-1)^{\#\text{ of edges of $H$}} \,(1-1) = 0.$$ Also, $H \in \mathrm{\ref{It:NoENoPath}}$ if and only if $H \cup \{e\} \in
\mathrm{\ref{It:ENoPath}}$, and similar to one has $\SumSub(\alpha,{\mathrm{\ref{It:NoENoPath}}} \cup {\mathrm{\ref{It:ENoPath}}}) = 0$, and therefore $$\label{Eq:AllEq2}
\SumSub(\alpha,G) = \SumSub(\alpha,{\mathrm{\ref{It:NoEAtoB}}} \cup
{\mathrm{\ref{It:NoEBtoA}}} \cup {\mathrm{\ref{It:NoENoPath}}} \cup {\mathrm{\ref{It:EAtoB}}} \cup
{\mathrm{\ref{It:ENoPath}}}) = \SumSub(\alpha,{\mathrm{\ref{It:NoEBtoA}}}).$$ Like in Clause \[SSec:SSCCrucial\] if $H \in {\mathrm{\ref{It:ENoPath}}}$ then $H/e \subset G/e$ is acyclic, and vice versa, if $e \in H$ and $H/e \subset
G/e$ is acyclic then $H \in {\mathrm{\ref{It:ENoPath}}}$. The graph $G/e$ is strongly connected, so by the induction hypothesis $\SumSub(\alpha,{\mathrm{\ref{It:ENoPath}}}) = -\SumSub(\alpha,G/e) = -(-1)^{k-1}
\sigma(G/e) = (-1)^k \sigma(G)$, hence $\SumSub(\alpha,{\mathrm{\ref{It:NoENoPath}}})
= -(-1)^k\sigma(G)$.

If $e \notin H$ and $H$ is acyclic, then $H$ is an acyclic subgraph of the strongly 