## 6 Conclusions
\begin{array}{ll}P^{00}=Pr\Big{\{}R_{SU_{1}}^{x_{2}}<R_{2},R_{SU_{1}}^{x_{1} }<R_{1}\Big{\}}=1+\frac{\sigma^{2}}{\lambda_{TU_{1}}P_{T}}exp\Big{(}\frac{ \sigma^{2}}{\lambda_{TU_{1}}P_{T}}\Big{)}\cdot(Q_{4}-Q_{3}).\\ \end{array}  \tag{11}\]  
which completes the derivation of \(P^{\Theta}\). Then, let us focus on the derivation of \(P^{\Theta}_{out,2}\), which can be expressed as  
\[ \begin{array}{ll}P^{\Theta}_{out,2}=\left\{\begin{array}{ll}Pr\Big{\{}R_{ U_{1}U_{2}}^{x_{2}}<R_{2}\Big{\}}\cdot\prod\limits_{n=1}^{N}\Big{[}1-P^{\Theta}_{n,2 }\Big{]},&\Theta=11,\\ \prod\limits_{n=1}^{N}\Big{[}1-P^{\Theta}_{n,2}\Big{]},&\Theta\in\{10,01,00\}, \end{array}\right.\end{array}  \tag{12}\]  
where \(Pr\Big{\{}R_{U_{1}U_{2}}^{x_{2}}<R_{2}\Big{\}}\) is the probability that U\({}_{1}\) fails to help U\({}_{2}\) decode \(x_{2}\). Based on Equation (8), we have  
\[ \begin{array}{ll}Pr\Big{\{}R_{U_{1}U_{2}}^{x_{2}}<R_{2}\Big{\}}&=Pr\Big{\{} \rho_{R}\big{|}h_{U_{1}U_{2}}\big{|}^{2}<\gamma_{2}\Big{\}}=F_{\rho_{R}|h_{U_{1 }U_{2}}|^{2}}(\gamma_{2})\\ &=1-exp\Big{(}-\frac{\sigma^{2}\gamma_{2}}{\lambda_{U_{1}U_{2}}P_{2}}\Big{)} \Big{(}1-\frac{\lambda_{U_{1}R^{\Theta}}}{\lambda_{U_{1}U_{2}}I_{0t}+\lambda_{ U_{1}R^{\Theta}}}\sigma^{2}\gamma_{2}exp\Big{(}-\frac{I_{0t}}{\lambda_{U_{1}R }P_{2}}\Big{)}\Big{)}.\end{array}  \tag{13}\]  
and \(P^{\Theta}_{n,2}\) in (12) denotes the probability that the optimal relay successfully assists U\({}_{2}\) to decode \(x_{2}\) in the case of \(\Theta\); that is, all relays in the candidate set have ability to make their achievable data rate reach the target data rate, i.e.,  
\[ \begin{array}{ll}P^{\Theta}_{n,2}=\left\{\begin{array}{ll}Pr\Big{\{}R_{SR_{ \alpha}}^{x_{2}}\geq R_{2}\Big{\}}Pr\Big{\{}R_{R_{\alpha}U_{2}}^{x_{2}}\Theta =R_{2}\Big{\}},&\Theta\in\{11,01\},\\ Pr\Big{\{}R_{SR_{\alpha}}^{x_{2}}\geq R_{2},R_{SR_{\alpha}}^{x_{1}}\geq R_{1} \Big{\}}Pr\Big{\{}R_{SR_{\alpha}U_{2}}^{x_{2}}\geq R_{2}\Big{\}},&\Theta\in\{10, 00\}.\end{array}\right.\end{array}  \tag{14}\]  
Based on Equation (2), Proposition 1 and Proposition 2, we can get  
\[ \begin{array}{ll}Pr\Big{\{}R_{SR_{\alpha}}^{x_{2}}\geq R_{2}\Big{\}}&=Pr\Big{\{} \frac{a_{2}\rho_{S}|h_{SR_{\alpha}}|^{2}}{\rho_{T}|h_{TR_{\alpha}}|^{2}+a_{1} \rho_{S}|h_{SR_{\alpha}}|^{2}+1}\geq\gamma_{2}\Big{\}}\\ &=Pr\Big{\{}\rho_{S}|h_{SR_{\alpha}}|^{2}\geq\Big{(}\rho_{T}|h_{TR_{\alpha}}|^{2}+1 \Big{)}\gamma_{\alpha}\Big{\}}\\ &=\int_{0}^{\infty}\Big{[}1-F_{\rho_{S}|h_{SR_{\alpha}}|^{2}}(\gamma_{a}y)\Big{]} \cdot f_{\rho_{T}|h_{TR_{\alpha}}|^{2}+1}(y)dy\\ &=\frac{\sigma^{2}}{\lambda_{TR_{\alpha}}P_{T}}exp\Big{(}\frac{\sigma^{2}}{ \lambda_{TR_{\alpha}}P_{T}}\Big{)}\cdot(Q_{5}-Q_{6}).\end{array}  \tag{15}\]where \(Q_{5}=1/a_{3}exp(-a_{3}),Q_{6}=exp\Big{(}-\frac{I_{0}}{(\lambda_{SR}R_{1})}\Big{)} exp(a_{3}b_{3})\Big{[}\frac{1}{a_{3}}exp(-a_{3}-a_{3}b_{3})-b_{3}E_{1}(a_{3}+a_{3}b_{3}) \Big{]}\). In \(Q_{5}\) and \(Q_{6}\), \(a_{3}=\frac{\sigma^{2}\big{(}\gamma_{a}\lambda_{TR_{a}}P_{T}+\lambda_{SR_{a}}P_{ 1}\big{)}}{\big{(}\lambda_{SR_{a}}P_{1}\lambda_{TR_{a}}P_{T}\big{)}}\) and \(b_{3}=\frac{\big{(}\lambda_{SR_{a}}I_{0}\big{)}}{\big{(}\lambda_{SR}R^{2} \gamma_{a}\big{)}}\). On the other hand, we can obtain \(Pr\Big{\{}R_{SR_{a}}^{x_{2}}\geq R_{2},R_{SR_{a}}^{x_{1}}\geq R_{1}\Big{\}}\) \(=Pr\Big{\{}\frac{a_{3}\rho_{5}|h_{SR_{a}}|^{2}}{\rho_{T}|h_{TR_{a}}|^{2}+a_{1} \rho_{5}|h_{SR_{a}}|^{2}+1}\geq\gamma_{2},\frac{a_{1}\rho_{5}|h_{SR_{a}}|^{2}}{ \rho_{T}|h_{TR_{a}}|^{2}+1}\geq\gamma_{1}\Big{\}}\) \(=Pr\Big{\{}\rho_{5}|h_{SR_{a}}|^{2}\geq\Big{(}\rho_{T}|h_{TR_{a}}|^{2}+1\Big{)} \,\rho_{1}\Big{\}}\) (A10) \(=\int_{0}^{\infty}\!\!\Big{[}1-F_{\rho_{5}|h_{SR_{a}}|^{2}}(\rho_{1}y)\Big{]} \cdot f_{\rho_{T}|h_{TR_{a}}|^{2}+1}(y)dy\) \(=\frac{\sigma^{2}}{\lambda_{TR_{a}}P_{T}}exp\Big{(}\frac{\sigma^{2}}{\lambda_ {TR_{a}}P_{T}}\Big{)}\cdot(Q_{7}-Q_{8})\), where \(Q_{7}=1/a_{4}exp(-a_{4})\), \(Q_{8}=exp\Big{(}-\frac{I_{0}}{(\lambda_{SR}P_{1})}\Big{)}exp(a_{4}b_{4})\Big{[} \frac{1}{a_{4}}exp(-a_{4}-a_{4}b_{4})-b_{4}E_{1}(a_{4}+a_{4}b_{4})\Big{]}\).