connected graph $G \setminus \{e\}$. The graph $G$ is strongly connected, too, so $e$ enters a cycle, and $\beta_1(G \setminus \{e\}) =
\beta_1(G)-1$, which implies $\sigma(G \setminus \{e\}) = -\sigma(G)$. The graph $G \setminus \{e\}$ contains $k-1 < k$ edges, so by the induction hypothesis $$\SumSub(\alpha, {\mathrm{\ref{It:NoEAtoB}}} \cup {\mathrm{\ref{It:NoEBtoA}}} \cup
{\mathrm{\ref{It:NoENoPath}}}) = \SumSub(\alpha,G \setminus \{e\}) = (-1)^{k-1}
\sigma(G \setminus \{e\}) = (-1)^k \sigma(G),$$ and therefore $$\label{Eq:Sum12}
\SumSub(\alpha, {\mathrm{\ref{It:NoEAtoB}}} \cup {\mathrm{\ref{It:NoEBtoA}}}) = 2(-1)^k
\sigma(G).$$ A subgraph $H \subset G$ of class \[It:NoEAtoB\] is at the same time a subgraph $H \subset G_e^{\vee}$ of class \[It:NoEBtoA\]. So, applied to $G_e^{\vee}$ gives $\SumSub(\alpha,
{\mathrm{\ref{It:NoEAtoB}}}) = \SumSub(\alpha, G_e^{\vee})$. If follows now from and that $$\SumSub(\alpha,G) + \SumSub(\alpha,G_e^{\vee}) = 2(-1)^k \sigma(G) =
(-1)^k (\sigma(G) + \sigma(G_e^{\vee})),$$ which proves the lemma.

To complete the proof of Theorem \[Th:Direct\] let $a$ be a vertex of $G$, and let $e_1 {, \dots ,} e_m$ be the complete list of edges finishing at $a$. Consider the sequence of graphs $G_0 = G$, $G_1 = G_{e_1}^{\vee}$, $G_2 = (G_1)_{e_2}^{\vee}$, …, $G_m = (G_{m-1})_{e_m}^{\vee}$. The graphs $G_0$ and $G_1$ are strongly connected; the graph $G_m$ is not, because $a \not\succeq b$ for any $b \ne a$ in it. Take the maximal 