## 4 Security Analysis
\begin{split}\hat{e}&(D_{1,i},K_{c,\rho(i)}(i)=\hat{ e}(g^{\frac{\tau_{i}}{\tau_{i}}}(g^{\frac{\hat{e}\tau_{i}}{\tau_{j}}}t_{jk} \cdot(g^{\rho(i)}\cdot H(\rho(i)))^{\tau_{k}\hat{\gamma}_{k}\xi_{j}}))\\ &=\hat{e}(g^{\frac{\tau_{i}}{\tau_{k}}},g^{\frac{\hat{e}\tau_{k}}{ \tau_{i}}}t_{jk})\cdot\hat{e}(g^{\frac{\tau_{i}}{\tau_{k}}}(g^{\rho(i)}\cdot H (\rho(i)))^{\gamma_{k}\hat{\gamma}_{k}\xi_{j}})\\ &=\hat{e}(g,g)^{\frac{\tau_{i}\cdot\hat{\gamma}_{k}}{\tau_{j}}}t_ {jj}\cdot\hat{e}(g^{\frac{\tau_{i}}{\tau_{k}}},g^{\rho(i)\cdot\gamma_{k}\xi_{ j}})\cdot\hat{e}(g^{\frac{\tau_{i}}{\tau_{k}}},(H(\rho(i)))^{\tau_{k}\hat{\gamma}_{k} \xi_{j}})\\ &=\hat{e}(g,g)^{\frac{\tau_{i}\cdot\hat{\gamma}_{k}}{\tau_{j}}}t_ {jk}\cdot\hat{e}(g,g)^{\tau_{i}\cdot\hat{\gamma}_{k}\cdot\hat{c}_{j}}\cdot\hat{ e}(g,H(\rho(i)))^{\tau_{i}\cdot\hat{\gamma}_{k}\cdot\hat{c}_{j}},\end{split}  \tag{5}\]  
In addition, obtaining the second part of the denominator for Equation (2) is hard because it belongs to the DL problem and square problem. \[ \begin{split}\hat{e}(D_{2,i},L_{c_{j,k}})&=\hat{e} (g^{-\frac{\tau_{i}}{\tau_{k}}r_{i}},g^{\frac{\hat{e}_{k}}{\tau_{j}}}t_{jk})\\ &=\hat{e}(g,g)^{\frac{-\tau_{i}\cdot\hat{\gamma}_{k}}{\tau_{j}}} t_{jk}=\frac{1}{\hat{e}(g,g)^{\frac{t_{i}\cdot\hat{\gamma}_{k}}{\tau_{j}}}t_{jk}} \end{split}  \tag{6}\]  
After that, the denominator of Equation (2) can be calculated using Equations (4)-(6):  
\[ \begin{split}\hat{e}&(C_{i},GPK_{c_{j}})\cdot\hat{ e}(D_{1,i},K_{c_{j}\rho(i)})\cdot\hat{e}(D_{2,i},L_{c_{j,k}})\\ &=\frac{\hat{e}(g,g)^{a\cdot\lambda_{i}\cdot c_{j}}\cdot\hat{e}( g,g)^{\frac{\tau_{i}\cdot\hat{\gamma}_{k}}{\tau_{j}}}t_{jk}\cdot\hat{e}(g,g)^{ \tau_{i}\cdot\hat{\gamma}_{k}\cdot c_{j}}\cdot\hat{e}(g,H(\rho(i)))^{\tau_{i} \cdot\hat{\gamma}_{k}\cdot c_{j}}}{\hat{e}(g,g)^{\tau_{i}\cdot\hat{\gamma}_{ i}\cdot\hat{\gamma}_{i}\cdot\hat{\gamma}_{i}}\cdot\hat{e}(g,g)^{\frac{t_{i}\cdot \hat{\gamma}_{k}}{\tau_{j}}}t_{jk}}\\ &\hat{e}(C_{i},GPK_{c_{j}})\cdot\hat{e}(D_{1,i},K_{c_{j}\rho(i)}) \cdot\hat{e}(D_{2,i},L_{c_{j},k})\\ &=\frac{\hat{e}(g,g)^{a\cdot\lambda_{i}\cdot c_{j}}\cdot\hat{e}( g,g)^{\tau_{i}\cdot\hat{\gamma}_{k}\cdot c_{j}}}{\hat{e}(g,g)^{\tau_{i}\cdot \hat{\gamma}_{k}\cdot\hat{\gamma}_{i}}\cdot\hat{e}(g)^{\tau_{i}\cdot\hat{ \gamma}_{i}\cdot\hat{\gamma}_{i}}}\\ &=\frac{\hat{e}(g,g)^{a\cdot\lambda_{i}\cdot c_{j}}\cdot\hat{ \gamma}_{k}}{\hat{e}(g,g)^{\overline{\rho}(i)}}\end{split}  \tag{8}\]  
\(a,\gamma_{k},\vec{\rho}(i)\in\mathbb{Z}_{q}^{*}\) are three randomly chosen constants,  
\[\hat{e}(C_{i},GPK_{c_{j}})\cdot\hat{e}(D_{1,i},K_{c_{j}\rho(i)})\cdot\hat{e}( D_{2,i},L_{c_{j},k})=\hat{e}(g,g)^{a\cdot\lambda_{i}\cdot c_{j}} \tag{9}\]  
By substituting in Equation (2) using Equations (3) and (9):  
\[\begin{split} DT&=\frac{\hat{e}(g,g)^{f\cdot a\cdot c _{j}\cdot\hat{\gamma}_{A}}\underset{i=1}{\overset{k=1}{\overset{k=1}{\overset{k =1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{ \overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{ \overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{\overset{k=1}{ \overset{k=1}{\overset{k=1}{\overset{k=1=1}  
We can find that \(DT=\prod_{k=1}^{N_{A}}\delta(g,g)^{\frac{\alpha_{k}}{j}f}\) is successfully generated. It should be mentioned that if the user does not satisfy the previously mentioned two conditions, once the \(DT\) is successfully generated, the decryption algorithm can generate the original data securely. This is because recovering the original data is firstly based on generating the decryption token. Then, \(DT\) is used to generate (\(ck_{k}=\frac{CT}{DT^{\prime}}\)) for each authority. This is based on two secure factors, \(CT\) and \(DT\). Since \(CT\) and \(DT\) are secure, \(ck_{k}\) is secure and cannot be generated by an attacker who failed to successfully generate the decryption token \(DT\).