## Appendix A
(3) Actually, \(A\cap C=\{(x,\min\{\mu_{A}(x),\mu_{C}(x)\},\max\{\
u_{A}(x),\
u_{C}(x)\})|x\in U\}\), and \(B\cap C=\{(x,\min\{\mu_{B}(x),\mu_{C}(x)\},\max\{\
u_{B}(x),\
u_{C}(x)\})|x\in U\}\). According to Theorem 1, we can obtain  
\[D(A\cap C,B\cap C)=\max_{x\in U}\left\{\frac{1}{2}\left( \begin{array}{c}\left| \min\{\mu_{A}^{2}(x),\mu_{C}^{2}(x)\}-\min\{\mu_{B}^{2}(x),\mu_{C}^{2}(x)\} \right|\\ +\left|\max\{\
u_{A}^{2}(x),\
u_{C}^{2}(x)\}-\max\{\
u_{B}^{2}(x),\
u_{C}^{2}( x)\}\right|\end{array} \right)\right\}\]  
Therefore, \(D(A\cap C,B\cap C)\leq D(A,B)\). (4) Similarly, the proof of \(D(A\cup C,B\cup C)\leq D(A,B)\) is the same as \(D(A\cup C,B\cup C)\leq D(A,B)\). \(\square\)  
**Proof of Property 3. ** (1) From Definition 7, \(\theta_{\alpha}^{\beta}=\theta_{\alpha}\cap\theta^{\beta}\). (2) For \(\forall(x,y)\in\theta_{\alpha_{2}}^{\beta_{2}}\), where \(x,y\in U\), then we can obtain \(\sqrt{S_{\theta}(x,y)}\geq\alpha_{2}\) and \(\sqrt{D_{\theta}(x,y)}\leq\beta_{2}\). Because \(0\leq\alpha_{1}\leq\alpha_{2}\leq 1\) and \(0\leq\beta_{2}\leq\beta_{1}\leq 1\), then \(\sqrt{S_{\theta}(x,y)}\geq\alpha_{1}\) and \(\sqrt{D_{\theta}(x,y)}\leq\beta_{1}\). \((x,y)\in\theta_{\alpha_{1}}^{\beta_{1}}\), namely, \(\theta_{\alpha_{2}}^{\beta_{2}}\subseteq\theta_{\alpha_{1}}^{\beta_{1}}\). (3) For \(\forall(x,y)\in\theta_{\alpha}^{\beta}\), \(\sqrt{S_{\theta}(x,y)}\geq\alpha\) and \(\sqrt{D_{\theta}(x,y)}\leq\beta\). Because \(\theta\subseteq\theta\), \(\sqrt{S_{\theta}(x,y)}\leq\sqrt{S_{\theta}(x,y)}\) and \(\sqrt{D_{\theta}(x,y)}\geq\sqrt{D_{\theta}(x,y)}\). Then \(\sqrt{S_{\theta}(x,y)}\geq\alpha\) and \(\sqrt{D_{\theta}(x,y)}\leq\beta\), \((x,y)\in\theta_{\alpha}^{\beta}\). Therefore, \(\theta_{\alpha}^{\beta}\subseteq\theta_{\alpha}^{\beta}\). (4) Because  
\[(\vartheta\cap\theta)_{\alpha}=\left\{(x,y)\in U\times U\Big{|} \sqrt{S_{(\vartheta\cap\theta)}(x,y)}=\min\Bigl{\{}\sqrt{S_{\theta}(x,y)}, \sqrt{S_{\theta}(x,y)}\Bigr{\}}\geq\alpha\,\right\}\] \[=\left\{(x,y)\in U\times U\Big{|}\sqrt{S_{\theta}(x,y)}\geq \alpha\,\wedge\,\sqrt{S_{\theta}(x,y)}\geq\alpha\,\right\}\] \[=\left\{(x,y)\in U\times U\Big{|}\sqrt{S_{\theta}(x,y)}\geq \alpha\,\right\}\wedge\left\{(x,y)\in U\times U\Big{|}\sqrt{S_{\theta}(x,y)} \geq\alpha\,\right\}\] \[=\vartheta_{\alpha}\cap\theta_{\alpha}\]  
Similarly, it is easy to prove \((\vartheta\cap\theta)^{\beta}=\vartheta^{\beta}\cap\theta^{\beta}\). Otherwise,  
\[(\vartheta\cap\theta)_{\alpha}^{\beta}=\left\{(x,y)\in U\times U \Big{|}\sqrt{S_{(\vartheta\cap\theta)}(x,y)}\geq\alpha\,\wedge\,\sqrt{D_{( \vartheta\cap\theta)}(x,y)}\leq\beta\,\right\}\] \[=\left\{(x,y)\in U\times U\Big{|}\min\Bigl{\{}\sqrt{S_{\theta}(x,y )},\sqrt{S_{\theta}(x,y)}\Bigr{\}}\geq\alpha\,\wedge\,\max\Bigl{\{}\sqrt{D_{ \theta}(x,y)},\sqrt{D_{\theta}(x,y)}\Bigr{\}}\leq\beta\,\right\}\] \[=\left\{(x,y)\in U\times U\Big{|}\sqrt{S_{\theta}(x,y)}\geq\alpha \,\wedge\,\sqrt{S_{\theta}(x,y)}\geq\alpha\,\wedge\,\sqrt{D_{\theta}(x,y)} \leq\beta\,\wedge\,\sqrt{D_{\theta}(x,y)}\leq\beta\,\right\}\] \[=\left\{(x,y)\in U\times U\Big{|}\left(\sqrt{S_{\theta}(x,y)}\geq \alpha\,\wedge\,\sqrt{D_{\theta}(x,y)}\leq\beta\,\right)\wedge\Bigl{(}\sqrt{S_{ \theta}(x,y)}\geq\alpha\,\wedge\,\sqrt{D_{\theta}(x,y)}\leq\beta\,\Bigr{)} \,\right\}\] \[=\vartheta_{\alpha}^{\beta}\cap\theta_{\alpha}^{\beta}\]  
(5) The proof of \((\vartheta\cup\theta)_{\alpha}=\vartheta_{\alpha}\cup\theta_{\alpha_{\alpha}}( \vartheta\cup\theta)^{\beta}=\vartheta^{\beta}\cup\vartheta^{\beta},(\vartheta \cup\theta)_{\alpha}^{\beta}\supseteq\vartheta_{\alpha}^{\beta}\cup\theta_{ \alpha}^{\beta}\) is similar to \((\vartheta\cap\theta)_{\alpha}=\vartheta_{\alpha}\cap\theta_{\alpha_{\alpha}}( \vartheta\cap\theta)^{\beta}=\vartheta^{\beta}\cap\theta^{\beta},(\vartheta\cap \theta)_{\alpha}^{\beta}=\vartheta_{\alpha}^{\beta}\cap\theta_{\alpha}^{\beta}\). \(\square\)