# Competing Reverse Channels' Performance with Sustainable Recycle Innovation Input
## Appendix A Proofs for Theorems, Lemmas and Propositions
### Proof for Theorem 6  
**Proof for the Theorem 6.**  
\[\Pi_{M}^{B}(p,b)=\frac{H\left(\alpha-\beta c_{m}\right)^{2}}{\beta\left[4H- \beta(\Delta-A)^{2}\right]}\]  
\[\Pi_{M}^{BE*}(p,b)=\frac{RH\left(\alpha-\beta c_{m}\right)^{2}}{\beta\left(4M -4\beta RN\right)}=\frac{H\left(\alpha-\beta c_{m}\right)^{2}}{\beta\left[4H -\beta*\frac{\beta R}{\beta R-K^{2}}(\Delta-A)^{2}\right]}\]\[\Pi_{M}^{H*}(p,b_{1},b_{2})=\frac{RH\left(\alpha-\beta c_{m}\right)^{2}}{\left[4M-4 \beta RN-\beta R(\Delta-A)^{2}\right]\beta}=\frac{H\left(\alpha-\beta c_{m} \right)^{2}}{\beta\left[4H-\beta*(\frac{\beta R}{\beta R-K^{2}}+1)(\Delta-A)^{ 2}\right]}\]  
Because \(4H-\beta*(\frac{\beta R}{\beta R-K^{2}}+1)(\Delta-A)^{2}>4H-\beta*\frac{\beta R }{\beta R-K^{2}}(\Delta-A)^{2}>4H-\beta(\Delta-A)^{2}\), so we can get \(\Pi_{M}^{B*}<\Pi_{M}^{BEx}<\Pi_{M}^{H*}\).  
_Appendix A.12_.: Proof for Theorem 7  
**Proof for the Theorem 7.**  
\[\Pi_{\mathbb{C}}^{B}(\tau)=\frac{H}{2}\left[\frac{\left(\Delta-A\right)( \alpha-\beta c_{m})}{4H-\beta(\Delta-A)^{2}}\right]^{2}\]  
\[\Pi_{\mathbb{C}}^{BEx*}(p,b)=\left[\frac{H}{2}-\frac{R}{8}\left(\frac{\beta K (\Delta-A)}{\left(\beta R-K^{2}\right)}\right)^{2}\right]\left(\frac{\beta R }{\beta R-K^{2}}*\frac{R(\Delta-A)\left(\alpha-\beta c_{m}\right)}{4M-4\beta RN }\right)^{2}\]  
\[\Pi_{\mathbb{C}_{1}}^{H*}\left(\tau_{1},e\right)=\left[\frac{H}{2}-\frac{R}{8} \left(\frac{\beta K(\Delta-A)}{\left(\beta R-K^{2}\right)}\right)^{2}\right] \left(\frac{\beta R}{\beta R-K^{2}}*\frac{R(\Delta-A)\left(\alpha-\beta c_{m} \right)}{4M-4\beta RN-\beta R(\Delta-A)^{2}}\right)^{2}\]  
\[\Pi_{\mathbb{C}_{2}}^{H*}\left(\tau_{2}\right)=\frac{H}{2}\left[\frac{R( \Delta-A)\left(\alpha-\beta c_{m}\right)}{4M-4\beta RN-\beta R(\Delta-A)^{2}} \right]^{2}\]  
Because \(4H-\beta*(\frac{\beta R}{\beta R-K^{2}}+1)(\Delta-A)^{2}>4H-\beta*\frac{\beta R }{\beta R-K^{2}}(\Delta-A)^{2}>4H-\beta(\Delta-A)^{2}\), so we can get \(\Pi_{\mathbb{C}_{1}}^{H*}>\Pi_{\mathbb{C}}^{BEx*}\), \(\Pi_{\mathbb{C}_{2}}^{H*}>\Pi_{\mathbb{C}}^{B*}\).  
There's an intersection point between \(\Pi_{\mathbb{C}}^{B*}\) and \(\Pi_{\mathbb{C}}^{BEx*}\).  
_Appendix A.13_.: Proof for Theorem 8  
**Proof for the Theorem 8.**  
\[\Pi_{\mathbb{C}}^{B*}+\Pi_{M}^{B*}=\frac{H\left(\alpha-\beta c_{m}\right)^{2} \left[8H-\beta(\Delta-A)^{2}\right]}{2\beta\left[4H-\beta(\Delta-A)^{2}\right] ^{2}}\]  
\[\Pi_{\mathbb{C}}^{BEx*}+\Pi_{M}^{BEx*}=\frac{H(\alpha-\beta c_{m})^{2}}{\beta \left[4H-\beta*\frac{\beta R}{\beta R-K^{2}}(\Delta-A)^{2}\right]}+\left[ \frac{H}{2}-\frac{R}{8}\left(\frac{\beta K(\Delta-A)}{\left(\beta R-K^{2} \right)}\right)^{2}\right]\left(\frac{\beta R}{\beta R-K^{2}}*\frac{R(\Delta- A)\left(\alpha-\beta c_{m}\right)}{4M-4\beta RN}\right)^{2}\]  
Because \(4H-\beta*\frac{\beta R}{\beta R-K^{2}}(\Delta-A)^{2}>4H-\beta(\Delta-A)^{2}\), then we can get \(\Pi_{T}^{B*}<\Pi_{T}^{BEx*}\).  
\[\Pi_{\mathbb{C}_{1}}^{H*}>\Pi_{\mathbb{C}}^{BEx*},\Pi_{\mathbb{C}_{2}}^{H*}>\Pi _{\mathbb{C}}^{B*},\Pi_{M}^{B*}<\Pi_{M}^{BEx*}<\Pi_{M}^{H*}\]  
Thus, \(\Pi_{T}^{B*}<\Pi_{T}^{BEx*}<\Pi_{T}^{H*}\).