# G-Networks with Adders
## 4 Proof of the Theorem  
Proof. : We now write the Chapman-Kolmogorov equations in steady-state, with the term \(\gamma_{ij}(x)\equiv lim_{t\rightarrow\infty}\gamma_{ij}(x,t)\), so that:  
\[ \begin{array}{l}\sum_{i=1}^{N}\left[\lambda_{i}^{+}+(\ \mu_{i}+\lambda_{i}^{-})1_{\{k_{i}>0\}}\right]p(\vec{k})\\ \quad=\sum_{i=1}^{N}[\lambda_{i}^{+}p(\vec{k}-\vec{e}_{i})1_{\{k_{i}>0\}}+(\ \mu_{i}d_{i}+\lambda_{i}^{-}\ )p(\vec{k}+\vec{e}_{i})]+\sum_{i=1}^{N}\sum_{j=1}^{N}\mu_{i}p(\vec{k}+\vec{e}_{ i}-\vec{e}_{j})P_{ij}^{+}1_{\{k_{j}>0\}}\\ \quad+\sum_{i=1}^{N}\sum_{j=1}^{N}[\mu_{i}p(\vec{k}+\vec{e}_{i}+\vec{e}_{j})P_{ ij}^{-}+\mu_{i}p(\vec{k}+\vec{e}_{i})P_{ij}^{-}1_{\{k_{j}>0\}}]+\sum_{i=1}^{N} \sum_{j=1}^{N}\sum_{m=1}^{N}\ \mu_{i}\ p(\vec{k}+\vec{e}_{i}+\vec{e}_{m}-\vec{e}_{j})Q_{imj}1_{\{k_{j}>0\}}\\ \quad+\sum_{i=1}^{N}\sum_{j=1}^{N}\sum_{m=1}^{N}\ \mu_{i}\ p(\vec{k}+\vec{e}_{i})\ Q_{ lim}1_{\{k_{j}>0\}}+\sum_{i=1}^{N}\sum_{j=1}^{N}\sum_{y=0}^{\infty}\ \mu_{i}\ p_{ij}^{A}\ \gamma_{ij}(k_{j})p(\vec{k}+\vec{e}_{i}+\vec{e}_{j}(y-k_{j})),\end{array}  \tag{7}\]  
into which we substitute (5):  
\[ \begin{array}{l}\sum_{i=1}^{N}\left[\lambda_{i}^{+}+(\mu_{i}+\lambda_{i}^{-})1 _{\{k_{i}>0\}}\right]\\ \quad=\ \sum_{i=1}^{N}\frac{\lambda_{i}^{+}1_{\{k_{i}>0\}}}{\rho_{i}}+\sum_{i=1}^{N }(\mu_{i}d_{i}+\lambda_{i}^{-})\rho_{i}+\sum_{i=1}^{N}\mu_{i}\frac{\rho_{i}}{ \rho_{j}}P_{ij}^{+}1_{\{k_{j}>0\}}\\ \quad+\sum_{i=1}^{N}\sum_{j=1}^{N}\mu_{i}\rho_{i}\rho_{j}P_{ij}^{-}+\sum_{i=1}^ {N}\sum_{j=1}^{N}\mu_{i}\rho_{i}P_{ij}^{-}1_{\{k_{j}=0\}}\\ \quad+\sum_{i=1}^{N}\sum_{j=1}^{N}\sum_{m=1}^{N}u_{i}\frac{\rho_{i}\rho_{m}}{ \rho_{j}}Q_{imj}1_{\{k_{j}>0\}}\\ \quad+\sum_{i=1}^{N}\sum_{j=1}^{N}\sum_{m=1}^{N}u_{i}\rho_{i}Q_{ijm}1_{\{k_{j }=0\}}+\sum_{i=1}^{N}\sum_{j=1}^{N}u_{i}\rho_{i}1_{\{k_{j}=0\}}P_{ij}^{A}\frac{ 1-\rho_{i}W_{ij}(\rho_{i})}{1-\rho_{j}}\\ \quad+\sum_{i=1}^{N}\sum_{j=1}^{N}\mu_{i}\frac{\rho_{i}\rho_{j}}{\rho_{j}}1_{\{ k_{j}>0\}}P_{ij}^{A}W_{ij}(\rho_{j}).\end{array}  \tag{8}\]  
Using \(1_{\{k_{j}=0\}}=1-1_{\{k_{j}>0\}}\), we regroup terms on the RHS to the LHS (left-hand-side) changing sign, exchange some \(j\)'s and \(i\)'s, and use \(\rho_{i}\) in (4), and (3), yielding:  
\[\begin{array}{l}\sum_{i=1}^{N}\left[\lambda_{i}^{+}+\frac{\lambda_{i}^{+}}{\rho _{i}}1_{\{k_{i}>0\}}\right]\\ \quad=\ \sum_{i=1}^{N}\frac{\lambda_{i}^{+}1_{\{k_{i}>0\}}}{\rho_{i}}+\sum_{i=1}^{N }(\mu_{i}d_{i}+\lambda_{i}^{-})\rho_{i}+\sum_{i=1}^{N}\mu_{i}\frac{\rho_{i}}{ \rho_{j}}P_{ij}^{+}1_{\{k_{j}>0\}}\\ \quad+\ \sum_{i=1}^{N}\sum_{j=1}^{N}\mu_{i}\rho_{i}P_{ij}^{-}+\sum_{i=1}^{N}\sum_{j=1 }^{N}\sum_{m=1}^{N}\mu_{i}\rho_{i}P_{ij}^{-}+\sum_{i=1}^{N}\sum_{j=1 With (2) we get:  
\[\sum_{i=1}^{N}\lambda_{i}^{+}=\sum_{i=1}^{N}\rho_{i}[\lambda_{i}^{-}+\mu_{i}[d_{i }+\sum_{j=1}^{N}(\rho_{j}P_{ij}^{-}+P_{ij}^{-}+P_{ij}^{A}\frac{1-\rho_{j}W_{ij}( \rho_{j})}{1-\rho_{j}}+\sum_{m=1}^{N}\ \mu_{i}\ Q_{ijm})]]. \tag{10}\]  
To show that (10) is the same as (4) we write \(\rho_{i}\) as: \(\Lambda_{i}^{+}=\rho_{i}(\mu_{i}+\Lambda_{i}^{-})\), in the LHS we use (2) and the RHS (3), so that summing over \(i\) the result is:  
\[ \begin{split}&\sum_{i=1}^{N}\big{[}\lambda_{i}^{+}+\sum_{j=1}^{N}[ \mu_{j}\rho_{j}P_{ji}^{+}+\mu_{j}\rho_{ji}^{A}\rho_{i}W_{ji}(\rho_{i})+\sum_{m= 1}^{N}\ \mu_{j}\rho_{j}\rho_{m}Q_{jmi}\ ]\big{]}\\ &=\sum_{i=1}^{N}\big{[}\rho_{i}\mu_{i}+\lambda_{i}^{-}+\sum_{j=1} ^{N}[\mu_{j}\rho_{j}P_{ji}^{-}+\sum_{m=1}^{N}\ \mu_{j}\rho_{j}Q_{jim}+\mu_{j}\rho_{j}P_{ji}^{A}\frac{1-\rho_{i}W_{ji}(\rho_{i })}{1-\rho_{i}}]\ \big{]}.\end{split}  \tag{11}\]  
Since \(1=d_{i}+\sum_{j=1}^{N}[P_{ij}^{A}+P_{ij}^{+}+P_{ij}^{-}+\sum_{m=1}^{N}Q_{ijm}]\) we have:  
\[