& 0 & -q_{2} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & 0 & 0 & \cdots & -q_{n} & q_{n} \\ n & 0 & 0 & \cdots & 0 & 0}.\end{aligned}$$

We will proceed in three steps.

We start by showing that $$\begin{aligned}
\label{eq:TildeGtoGN}
\sum_{j_{2}\in\mathbf{E}}\left({\widetilde}{\mG}^{k}\right)\left((n_{1},j_{1}),(n_{2},j_{2}) \right)=\left({\widetilde}{\mG}_{N}^{k}\right)(n_{1},n_{2}),\end{aligned}$$ for any $j_{1}\in\mathbf{E}$, $k\in\mathbb{N}$, and $0\leq n_{1},n_{2}\leq n$, In particular, note that the left-hand-side of does not depend on $j_{1}$.

We will prove by induction in $k$. Clearly holds true for $k=1$. Next, assume that holds for some $k=\ell\in\mathbb{N}$. Now, for $\ell+1$, $$\begin{aligned}
\sum_{j_{2}\in\mathbf{E}}\left({\widetilde}{\mG}^{\ell+1}\right)\left((n_{1},j_{1}),(n_{2},j_{2})\right)&=\sum_{j_{2}\in\mathbf{E}}\sum_{m=0}^{n}\sum_{j\in\mathbf{E}}\left({\widetilde}{\mG}^{\ell}\right)\left((n_{1},j_{1}),(m,j)\right){\widetilde}{\mG}\left((m,j),(n_{2},j_{2})\right)\\
&=\sum_{m=0}^{n}\sum_{j\in\mathbf{E}}\left({\widetilde}{\mG}^{\ell}\right)\left((n_{1},j_{1}),(m,j)\right)\sum_{j_{2}\in\mathbf{E}}{\widetilde}{\mG}\left((m,j),(n_{2},j_{2})\right)\\
&=\sum_{m=0}^{n}\sum_{j\in\mathbf{E}}\left({\widetilde}{\mG}^{\ell}\right)\left((n_{1},j_{1}),(m,j)\right){\widetilde}{\mG}_{N}(m,n_{2})\\
&=\sum_{m=0}^{n}\left({\widetilde}{\mG}_{N}^{\ell}\right)(n_{1},m){\widetilde}{\mG}_{N}(m,n_{2})=\left({\widetilde}{\mG}_{N}^{\ell+1}\right)(n_{1},n_{2}),\end{aligned}$$ where we used the inductive assumptions for $k=1$ and $k=\ell$ in the third and the fourth equalities, respectively. Hence is established.

We will show that $$\begin{aligned}
\label{eq:NNotDependonY}
{\widetilde}{\bP}^{i}\left(\left.N_{t+s}=n_{2}\,\right|N_{t}=n_{1}\right)={\widetilde}{\bP}^{i}\!
\left(\left.N_{t+s}=n_{2}\,\right|N_{t}=n_{1},Y_{t}=j\right)=e^{s{\widetilde}{\mG}_{N}}(n_{1},n_{2}),\end{aligned}$$ for any $t,s\geq 0$, $j\in\mathbf{E}$, and $0\leq n_{1}\leq n_{2}\leq n$. In particular, note that the left-hand side of , and thus ${\widetilde}{\bP}^{i}(N_{t+s}=n_{2}|N_{t}=n_{1})$, does not depend on $t$. We start by checking the second equality in . For any $t,s\geq 0$, $j\in\mathbf{E}$, and $0\leq n_{1}\leq n_{2}\leq n$, $$\begin{aligned}
{\widetilde}{\bP}^{i}\left(\left.N_{t+s}=n_{2}\,\right|N_{t}=n_{1},Y_{t}=j\right)&=\sum_{k\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(\left.N_{t+s}=n_{2},Y_{t+s}=k\,\right|N_{t}=n_{1},Y_{t}=j\right)\\
&=\sum_{k\in\mathbf{E}}e^{s\,{\widetilde}{\mG}}\left((n_{1},j),(n_{2},k)\right)=\sum_{k\in\mathbf{E}}\sum_{\ell=0}^{\infty}\frac{s^{\ell}}{\ell!}\,{\widetilde}{\mG}^{\ell}\left((n_{1},j),(n_{2},k)\right)\\
&=\sum_{\ell=0}^{\infty}\frac{s^{\ell}}{\ell!}\sum_{k\in\mathbf{E}}{\widetilde}{\mG}^{\ell}\left((n_{1},j),(n_{2},k)\right)\\
&=\sum_{\ell=0}^{\infty}\frac{s^{\ell}}{\ell!}\,{\widetilde}{\mG}^{\ell}_{N}(n_{1},n_{2})=e^{s{\widetilde}{\mG}_{N}}(n_{1},n_{2}),\end{aligned}$$ where we used the result of Step 1 in the last two 