exists.

    -   If $\mathcal N_0^2(p,\mathbf{p},p_{n+1})>1$, a unique strongly endemic equilibrium $(\tilde{S}_h, \tilde{E}_h,\tilde{\mathbf{I}}_h,\tilde{S}_v,\tilde{E}_v,\tilde{I}_v)$ exists.

An equilibrium $(S_h^\ast,E_h^\ast,I_h^\ast,\dots,I_n^\ast,S_v^\ast,E_v^\ast,I_v^\ast)$ for Model (\[ModelAD2Compact\]) satisfies the following relations. $$\label{ModelAD2Endemic1}
\left\{\begin{array}{llll}\displaystyle \Lambda_h=a\,\beta_{vh}\,S_{h}^\ast\,\dfrac{I_{v}^\ast}{N_{h}}+\mu_h S_h^\ast\\
\displaystyle (\mu_h+\nu_h+\eta)E_h^\ast=p\pi_h+ \,\beta_{vh}\,S_{h}^\ast\,\dfrac{I_{v}^\ast}{N_{h}}\\
\displaystyle (\textrm{diag}(\alpha)-M)\mathbf{I}_h^\ast=\pi_h \mathbf{p}+\nu_hE_h^\ast e_1\\
\displaystyle \pi_v=
 a\dfrac{ S_{v}^\ast}{N_h}\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, \mathbf{I}_h^\ast   
\bigr\rangle
+(\mu_v+\delta_v)S_{v}^\ast\\
\displaystyle (\mu_v+\nu_v+\delta_v)E_{v}^\ast=a\dfrac{ S_{v}^\ast}{N_h}\bigl\langle
      \Beta    
 \, \mbox{\Large $\mid$} \, \mathbf{I}_h^\ast   
\bigr\rangle\\
\displaystyle (\mu_v+\delta_v)I_{v}^\ast=\nu_vE_v^\ast,
\end{array}\right.$$ where $\displaystyle\Lambda_h=\pi_h\left( 1-p-\mathds{1}^t\mathbf{p} -p_{n+1}\right)$. Using these relationship and $N_v=S_v-E_v-I_v$, one could express $I_v^\ast$ in terms of $\mathbf{I}_h^\ast$, as follows: $$\label{IVstarEE}I_v^\ast\left(\mu_v+\delta_v+\frac{a}{N_h}\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, \mathbf{I}_h^\ast   
\bigr\rangle\right)=\frac{a\nu_v}{(\mu_v+\nu_v+\delta_v)}\frac{N_v}{N_h}\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, \mathbf{I}_h^\ast   
\bigr\rangle.$$ Moreover, the first equation of (\[ModelAD2Endemic1\]) leads to: $$S_h^\ast=\frac{\Lambda_hN_h}{\mu_hN_h+a\beta_{vh}I_v^\ast}>0.$$ Furthermore, since the matrix $\textrm{diag}(\alpha)-M$ is strictly diagonally dominant and thus invertible, we obtain: $$\mathbf{I}_h^\ast=\pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}+\nu_hE_h^\ast(\textrm{diag}(\alpha)-M)^{-1}e_1,$$ and $$\begin{aligned}
E_h^\ast 
&=& \frac{1}{(\mu_h+\nu_h+\eta)}\left(p\pi_h+\,\dfrac{\Lambda_ha\,\beta_{vh}I_{v}^\ast}{\mu_hN_h+a\beta_{vh}I_v^\ast}\right).\end{aligned}$$ Therefore, $\mathbf{I}_h^\ast$ could be written as: $$\label{Ihast}
\mathbf{I}_h^\ast=\pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}+\frac{\nu_h}{(\mu_h+\nu_h+\eta)}\left(p\pi_h+\,\dfrac{\Lambda_ha\,\beta_{vh}I_{v}^\ast}{\mu_hN_h+a\beta_{vh}I_v^\ast}\right)(\textrm{diag}(\alpha)-M)^{-1}e_1.$$ The relation (\[Ihast\]) is key for the remaining of the proof, as we will use it to compute $\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, \mathbf{I}_h^\ast   
\bigr\rangle$ and obtain a quadratic equation in $I_v^\ast$ using . The latter equation leads to:

$$\begin{gathered}
\label{IVstarEEQuadratic}
0=I_v^\ast\left(\mu_v+\delta_v+\frac{a}{N_h}\bigl\langle
      \Beta   \, \mbox{\Large $\mid$} \, \pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle+\right.\\
      \left.+\frac{a}{N_h}\frac{\nu_h}{(\mu_h+\nu_h+\eta)}\left(p\pi_h+\,\dfrac{\Lambda_ha\,\beta_{vh}I_{v}^\ast}{\mu_hN_h+a\beta_{vh}I_v^\ast}\right)\bigl\langle\Beta   \, \mbox{\Large $\mid$}   (\textrm{diag}(\alpha)-M)^{-1}e_1  
\bigr\rangle\right)-\\
\frac{a\nu_v}{(\mu_v+\nu_v+\delta_v)}\frac{N_v}{N_h}\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, \pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle-\\
 \frac{\nu_h}{(\mu_h+\nu_h+\eta)}\left(p\pi_h+\,\dfrac{\Lambda_ha\,\beta_{vh}I_{v}^\ast}{\mu_hN_h+a\beta_{vh}I_v^\ast}\right)\frac{a\nu_v}{(\mu_v+\nu_v+\delta_v)}\frac{N_v}{N_h}\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \,(\textrm{diag}(\alpha)-M)^{-1}e_1  
\bigr\rangle\end{gathered}$$

After 