\dots ,} i_s$ and without loops, and let $\Phi \in \SSC_{n,k}^I$ (a strongly semiconnected graph with the isolated vertices $i_1 {, \dots ,} i_s$). Denote by $\widehat{\Phi}$ the graph $\Phi$ with the loops deleted. A contribution of $\Phi$ into $x_G$ is equal to $\frac{1}{k!} (-1)^{\beta_0(\Phi) - \#\text{ of loops in $\Phi$} + n}$ if $\widehat{\Phi} \subset G$ and is $0$ otherwise.

The number of edges of $\widehat{\Phi}$ is $k - \#\text{ of loops of
$\Phi$}$. The graph $\widehat{\Phi}$ is strongly semiconnected if and only if $\Phi$ is. The Euler characteristics of $\widehat{\Phi}$ is $$\beta_0(\widehat{\Phi}) - \beta_1(\widehat{\Phi}) = n - \#\text{ of edges
of $\widehat{\Phi}$} = n-k + \#\text{ of loops of $\Phi$}$$ and $\beta_0(\widehat{\Phi}) = \beta_0(\Phi)$. Therefore, the contribution of $\Phi$ into $x_G$ is $$(-1)^{n+k+\beta_0(\widehat{\Phi})+\#\text{ of edges of
$\widehat{\Phi}$}}\frac{1}{k!} =
(-1)^{k+\beta_1(\widehat{\Phi})}\frac{1}{k!}$$ if $\widehat{\Phi} \subset G$ is strongly semiconnected and $0$ otherwise. Summing up, $$x_G = \frac{(-1)^k}{k!}\SumSub(\sigma;G) = \frac{1}{k!} \alpha(G)$$ by Theorem \[Th:Direct\]’. This proves Theorem \[Th:Diag\].

Proof of Theorem \[Th:Codim1\] {#SSec:PrCodim1}
------------------------------

Note that $\det_{n,k}^{i/i} = \det_{n,k}^{\varnothing}+ \det_{n,k}^{\{i\}}$. Applying the operator $\Lapl$ to equation and using Theorem \[Th:Diag\] with $I = {\varnothing}$ and $I = \{i\}$ one obtains $$\begin{aligned}
0 &= \sum_{i,j=1}^n \Lapl(([ij]) * \det_{n,k}^{i/j}) = \sum_{i=1}^n
\Lapl ([ii]) * 