A New Representation Method for Type-2 Fuzzy Sets and Its Application to Multiple Criteria Decision Making

Abstract

Type-2 fuzzy sets (T2FSs) have the advantage of representing higher-order uncertainty, they have shown excellent performance in system modeling. However, complexity in mathematical expressions and fundamental operations restricts their application in multi-criteria decision-making. To address this issue, a new representation method for T2FSs is first proposed, and the concepts of double, single discrete T2FSs and single continuous T2FSs are introduced, respectively. Further, the fundamental operations, ranking method and decision-making interpretations for double discrete T2FSs are studied, and the axiomatic framework of fuzzy entropy and distance measure for double discrete T2FSs are also defined. Finally, the technique for order preference by similarity to ideal solution method in type-2 fuzzy environment is developed, and the practicability and effectiveness of the proposed definition and decision-making method are illustrated by validity tests and comparative analyses.

Appendix

1.1 Proof of Theorem 2

Proof

Verifying that (1), (3) and (5) in Theorem 2 hold true, the remaining (2), (4) and (6) can be similarly proved. According to Definitions 7 and 8, the following results can be obtained:

(1) $(\tilde{A}_{1}^{c})^{\lambda }=\{\langle x_{i},((\mu _{\tilde{A}_{1}}(x_{i}))^{c})^{\lambda }\rangle |x_{i}\in X, ((\mu _{\tilde{A}_{1}}(x_{i}))^{c})^{\lambda }$

$=\{\langle \tau _{j},(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }\rangle |\tau _{j}\in [0,1],$

$(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }\in [0,1]\}\}$,

$(\lambda \tilde{A}_{1})^{c}$ $=\{\langle x_{i},(\lambda \mu _{\tilde{A}_{1}}(x_{i}))^{c}\rangle |x_{i}\in X, (\lambda \mu _{\tilde{A}_{1}}(x_{i}))^{c}$

$=\{\langle \tau _{j},1-(1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda })\rangle |\tau _{j}$

$\in [0,1],(1-(1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }))$

$\in [0,1]\}\}$,

and since

$(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }=1 -(1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda })$,

so $(\tilde{A}_{1}^{c})^{\lambda }=(\lambda \tilde{A}_{1})^{c}$.

(3) $\tilde{A}_{1}^{c}\cup \tilde{A}_{2}^{c}$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i}))^{c}\cup (\mu _{\tilde{A}_{2}}(x_{i}))^{c}\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i}))^{c}\cup (\mu _{\tilde{A}_{2}}(x_{i}))^{c}=\{\langle \tau _{j},\max (1-$

$\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\rangle |\tau _{j}\in [0,1],$

$\max (1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))$

$\in [0,1]\}\}$,

$(\tilde{A}_{1}\cap \tilde{A}_{2})^{c}$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\cap \mu _{\tilde{A}_{2}}(x_{i}))^{c}\rangle |x_{i}\in X, (\mu _{\tilde{A}_{1}}(x_{i})$

$\cap \mu _{\tilde{A}_{2}}(x_{i}))^{c}=\{\langle \tau _{j},1-\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\rangle |\tau _{j}\in [0,1], (1-\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})))\in [0,1]\}\}$.

Since $1-\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}} (x_{i})}(\tau _{j}))=\max (1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))$, hence $\tilde{A}_{1}^{c}\cup \tilde{A}_{2}^{c}=(\tilde{A}_{1}\cap \tilde{A}_{2})^{c}$.

(5) $\tilde{A}_{1}^{\lambda }\cup \tilde{A}_{2}^{\lambda }$

$~~=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i}))^{\lambda }\cup (\mu _{\tilde{A}_{2}}(x_{i}))^{\lambda }\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i}))^{\lambda }\cup (\mu _{\tilde{A}_{2}}(x_{i}))^{\lambda } =\{\langle \tau _{j},$

$\max ((\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda },(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda })\rangle |$

$\tau _{j}\in [0,1],\max ((\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda },(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda })$

$\in [0,1]\}\}$,

$(\tilde{A}_{1}\cup \tilde{A}_{2})^{\lambda }$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\cup \mu _{\tilde{A}_{2}}(x_{i}))^{\lambda }\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i})\cup \mu _{\tilde{A}_{2}}(x_{i}))^{\lambda }=\{\langle \tau _{j},$

$(\max (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})))^{\lambda }\rangle |\tau _{j}\in [0,1],$

$(\max (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})))^{\lambda }\in [0,1]\}\}$,

and since

$\max ((\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda },(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda })$

$=(\max (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})))^{\lambda }$,

thus, $\tilde{A}_{1}^{\lambda }\cup \tilde{A}_{2}^{\lambda }=(\tilde{A}_{1}\cup \tilde{A}_{2})^{\lambda }$.

Proof of Theorem 3

Proof

Verifying that (1), (3) and (5) in Theorem 3 are valid, the rest (2), (4) and (6) can be proved in a similar manner. By Definitions 7 and 8, we can obtain

(1) $\tilde{A}_{1}^{c}\otimes \tilde{A}_{2}^{c}$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i}))^{c}\otimes (\mu _{\tilde{A}_{1}}(x_{i}))^{c}\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i}))^{c}\otimes (\mu _{\tilde{A}_{1}}(x_{i}))^{c}=\{\langle \tau _{j},(1-$

$\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))(1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\rangle |\tau _{j}\in [0,1],$

$(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))(1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\in [0,1]\}\}$,

$(\tilde{A}_{1}\oplus \tilde{A}_{2})^{c}$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\oplus \mu _{\tilde{A}_{2}}(x_{i}))^{c}\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i})\oplus \mu _{\tilde{A}_{2}}(x_{i}))^{c}=\{\langle \tau _{j},1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})$

$-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})+\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})\rangle |\tau _{j}$

$\in [0,1],(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})+$

$\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\in [0,1]\}\}$.

Because $(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))(1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))= 1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})+ \mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})$,

$\tilde{A}_{1}^{c}\otimes \tilde{A}_{2}^{c}=(\tilde{A}_{1}\oplus \tilde{A}_{2})^{c}$.

(3) $\tilde{A}_{1}^{\lambda }\otimes \tilde{A}_{2}^{\lambda }$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i}))^{\lambda }\otimes (\mu _{\tilde{A}_{1}}(x_{i}))^{\lambda }\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i}))^{\lambda }\otimes (\mu _{\tilde{A}_{1}}(x_{i}))^{\lambda }=\{\langle \tau _{j},$

$(\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }\rangle |\tau _{j}\in [0,1],$

$(\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }\in [0,1]\}\}$,

$(\tilde{A}_{1}\otimes \tilde{A}_{2})^{\lambda }$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\otimes \mu _{\tilde{A}_{2}}(x_{i}))^{\lambda }\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i})\otimes \mu _{\tilde{A}_{2}}(x_{i}))^{\lambda }=\{\langle \tau _{j},$

$(\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }\rangle |\tau _{j}\in [0,1],$

$(\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }\in [0,1]\}\}$.

Here, $(\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }= (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }$, so $\tilde{A}_{1}^{\lambda }\otimes \tilde{A}_{2}^{\lambda }=(\tilde{A}_{1}\otimes \tilde{A}_{2})^{\lambda }$.

(5) $\lambda \tilde{A}_{1}\oplus \lambda \tilde{A}_{2}$

$=\{\langle x_{i},\lambda \mu _{\tilde{A}_{1}}(x_{i})\oplus \lambda \mu _{\tilde{A}_{2}}(x_{i})\rangle |x_{i}\in X,\lambda \mu _{\tilde{A}_{1}}(x_{i})$

$\oplus \lambda \mu _{\tilde{A}_{2}}(x_{i})=\{\langle \tau _{j}, 1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }$

$(1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }\rangle |\tau _{j} \in [0,1],(1-(1-$

$\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda } (1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda })\in [0,1]\}\}$,

$\lambda (\tilde{A}_{1}\oplus \tilde{A}_{2})$

$=\{\langle x_{i},\lambda (\mu _{\tilde{A}_{1}}(x_{i})\oplus \mu _{\tilde{A}_{2}}(x_{i}))\rangle |x_{i} \in X,\lambda (\mu _{\tilde{A}_{1}}(x_{i})$

$\oplus \mu _{\tilde{A}_{2}}(x_{i}))=\{\langle \tau _{j},1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})+\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}) \mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }\rangle$

$|\tau _{j}\in [0,1],(1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})$

$+\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda })\in [0,1]\}\}$,

since $1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda }(1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }= 1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}) +\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}) \mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda }$, so $\lambda \tilde{A}_{1}\oplus \lambda \tilde{A}_{2}=\lambda (\tilde{A}_{1}\oplus \tilde{A}_{2})$.

Proof of Theorem 4

Proof

It is proved that (6), (8) and (10) in Theorem 4 hold, the rest (1)-(5), (7) and (9) can be proved similarly. Through Definitions 7 and 8, we can get

(6) $(\tilde{A}_{1}\cup \tilde{A}_{2})\otimes (\tilde{A}_{1}\cap \tilde{A}_{2})$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\cup \mu _{\tilde{A}_{2}}(x_{i}))\otimes (\mu _{\tilde{A}_{1}}(x_{i})\cap \mu _{\tilde{A}_{2}}(x_{i}))\rangle$

$|x_{i}\in X, (\mu _{\tilde{A}_{1}}(x_{i})\cup \mu _{\tilde{A}_{2}}(x_{i}))\otimes (\mu _{\tilde{A}_{1}}(x_{i})\cap$

$\mu _{\tilde{A}_{2}}(x_{i}))=\{\langle \tau _{j},\max (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))$

$\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\rangle |\tau _{j}\in [0,1],$

$\max (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))$

$\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\in [0,1]\}\}$,

and since

$\max (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})) \min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))=\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}) \mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})$,

so $(\tilde{A}_{1}\cup \tilde{A}_{2})\otimes (\tilde{A}_{1}\cap \tilde{A}_{2})=\tilde{A}_{1}\otimes \tilde{A}_{2}$.

(8) $(\tilde{A}_{1}\cap \tilde{A}_{2})\oplus \tilde{A}_{3}$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\cap \mu _{\tilde{A}_{2}}(x_{i}))\oplus \mu _{\tilde{A}_{3}}(x_{i})\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i})\cap \mu _{\tilde{A}_{2}}(x_{i}))\oplus \mu _{\tilde{A}_{3}}(x_{i})=\{\langle \tau _{j},\tilde{\mu }\rangle |\tau _{j}$

$\in [0,1],\tilde{\mu }\in [0,1]\}\}$,

where $\tilde{\mu }=\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})} (\tau _{j}))+\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})$

$-\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}), \mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})$.

$(\tilde{A}_{1}\oplus \tilde{A}_{3})\cap (\tilde{A}_{2}\oplus \tilde{A}_{3})$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\oplus \mu _{\tilde{A}_{3}}(x_{i})) \cap (\mu _{\tilde{A}_{2}}(x_{i})\oplus \mu _{\tilde{A}_{3}}(x_{i}))\rangle$

$|x_{i}\in X, (\mu _{\tilde{A}_{1}}(x_{i})\oplus \mu _{\tilde{A}_{3}}(x_{i})) \cap (\mu _{\tilde{A}_{2}}(x_{i})\oplus$

$\mu _{\tilde{A}_{3}}(x_{i}))=\{\langle \tau _{j},\breve{\mu } \rangle |\tau _{j}\in [0,1],\breve{\mu }\in [0,1]\}\}$,

where $\breve{\mu }=\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})+\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})- \mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})$

$\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})} (\tau _{j})+\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})- \mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})$ $\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))$.

Since $\tilde{\mu }=\breve{\mu }$, hence $(\tilde{A}_{1}\cap \tilde{A}_{2})\oplus \tilde{A}_{3}=(\tilde{A}_{1}\oplus \tilde{A}_{3})\cap (\tilde{A}_{2}\oplus \tilde{A}_{3})$.

(10) $(\tilde{A}_{1}\cap \tilde{A}_{2})\otimes \tilde{A}_{3}$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\cap \mu _{\tilde{A}_{2}}(x_{i}))\otimes \mu _{\tilde{A}_{3}}(x_{i})\rangle |x_{i}\in X,$

$(\mu _{\tilde{A}_{1}}(x_{i})\cap \mu _{\tilde{A}_{2}}(x_{i}))\otimes \mu _{\tilde{A}_{3}}(x_{i})=\{\langle \tau _{j},$

$\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})\rangle |\tau _{j}$

$\in [0,1],\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))$

$\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})\in [0,1]\}\}$,

$(\tilde{A}_{1}\otimes \tilde{A}_{3})\cap (\tilde{A}_{2}\otimes \tilde{A}_{3})$

$=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i})\otimes \mu _{\tilde{A}_{3}}(x_{i}))\cap (\mu _{\tilde{A}_{2}}(x_{i})\otimes \mu _{\tilde{A}_{3}}(x_{i}))\rangle$

$|x_{i}\in X,(\mu _{\tilde{A}_{1}}(x_{i})\otimes \mu _{\tilde{A}_{3}}(x_{i}))\cap (\mu _{\tilde{A}_{2}}(x_{i})\otimes$

$\mu _{\tilde{A}_{3}}(x_{i}))=\{\langle \tau _{j},\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}),$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))\rangle |\tau _{j}\in [0,1],$

$\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}),$

$\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))\in [0,1]\}\}$.

Since $\min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}),\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j})= \min (\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}), \mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))$,

hence $(\tilde{A}_{1}\cap \tilde{A}_{2})\otimes \tilde{A}_{3}=(\tilde{A}_{1}\otimes \tilde{A}_{3})\cap (\tilde{A}_{2}\otimes \tilde{A}_{3})$.

Proof of Theorem 6

Proof

Here, we provide the proof process of (2) and (4), (1) and (3) can be proved in a similar manner.

(2) If $\tilde{A}_{1}\succeq \tilde{A}_{2}$, then $S(\tilde{A}_{1})\ge S(\tilde{A}_{2})$, that is

$\sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{1}}(x_{i})|}_{j=1}\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\ge \sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{2}}(x_{i})|}_{j=1}\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})$,

so there exists

$\sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{1}}(x_{i})|}_{j=1}(1-(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))^{\lambda })\ge$

$\sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{2}}(x_{i})|}_{j=1}(1-(1-\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j}))^{\lambda })$,

i.e., $\lambda \tilde{A}_{1}\succeq \lambda \tilde{A}_{2}$.

(4) If $\tilde{A}_{1}\succeq \tilde{A}_{2}$, then $S(\tilde{A}_{1})\ge S(\tilde{A}_{2})$, i.e.,

$\sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{1}}(x_{i})|}_{j=1}\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\ge \sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{2}}(x_{i})|}_{j=1}\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})$,

hence we can obtain

$\sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{1}}(x_{i})|}_{j=1}(\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})(1- \mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))+$

$\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))\ge \sum \limits ^{|X|}_{i=1}\sum \limits ^{|\mu _{\tilde{A}_{2}}(x_{i})|}_{j=1}(\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})(1-$

$\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))+\mu _{\mu _{\tilde{A}_{3}}(x_{i})}(\tau _{j}))$,

that is, $\tilde{A}_{1}\oplus \tilde{A}_{3}\succeq \tilde{A}_{2}\oplus \tilde{A}_{3}$.

Proof of Theorem 7

Proof

Verifying that $E(\tilde{A})=F(f(\tilde{A}(x_{1})),f(\tilde{A}(x_{2})),..., f(\tilde{A}(x_{n})))$ satisfies axioms (E1)-(E4) in Definition 11. For any $\tilde{A}, \tilde{A}_{1},\tilde{A}_{2}\in T2FS(X)$, we can obtain

(E1) Since $\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})\in [0,1]$, $E(\tilde{A})=F(f(\tilde{A}(x_{1})), f(\tilde{A}(x_{2})),\ldots ,f(\tilde{A}(x_{n})))=0\Leftrightarrow f(\tilde{A}(x_{i}))=g(a_{i1},a_{i2}, \ldots ,a_{i|\tilde{A}(x_{i})|})=0\Leftrightarrow a_{ij}=|\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})-0.5|=0.5$, so $\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})=0$ or $\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})=1$.

(E2) Since $\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})\in [0,1]$, $E(\tilde{A})=F(f(\tilde{A}(x_{1})), f(\tilde{A}(x_{2})),\ldots ,f(\tilde{A}(x_{n})))=1\Leftrightarrow f(\tilde{A}(x_{i}))=g(a_{i1},a_{i2}, \ldots ,a_{i|\tilde{A}(x_{i})|})=1\Leftrightarrow a_{ij}=|\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})-0.5|=0$, hence $\mu _{\mu _{\tilde{A}}(x_{i})}(\tau _{j})=0.5$.

(E3) Since $\tilde{A}_{1}^{c}=\{\langle x_{i},(\mu _{\tilde{A}_{1}}(x_{i}))^{c}\rangle |x_{i}\in X,(\mu _{\tilde{A}_{1}}(x_{i}))^{c} =\{\langle \tau _{j},1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})\rangle |\tau _{j}\in [0,1],(1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j}))\in [0,1]\}\}$, and $a_{ij}=|1-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-0.5|=|0.5-\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})|$, thus, $E(\tilde{A}_{1})=E(\tilde{A}_{1}^{c})$.

(E4) Because $g(a_{i1},a_{i2},\ldots ,a_{i|\tilde{A}(x_{i})|})$ is strictly monotonically decreasing with respect to each $a_{ij}(j=1,2,\ldots ,|\tilde{A}(x_{i})|)$, $F(f(\tilde{A}(x_{1})),f(\tilde{A}(x_{2})),\ldots ,f(\tilde{A}(x_{n})))$ is strictly monotonically increasing with respect to each $f(\tilde{A}(x_{i}))(i=1,2,\ldots ,n)$, when $|\mu _{\mu _{\tilde{A}_{1}}(x_{i})}(\tau _{j})-0.5|\ge |\mu _{\mu _{\tilde{A}_{2}}(x_{i})}(\tau _{j})-0.5|$, then we can obtain $f(\tilde{A}_{1}(x_{i}))\le f(\tilde{A}_{2}(x_{i}))$, and $F(f(\tilde{A}_{1}(x_{1}),f(\tilde{A}_{1}(x_{2})),\ldots ,f(\tilde{A}_{1}(x_{n}))) \le F(f(\tilde{A}_{2}(x_{1})),f(\tilde{A}_{2}(x_{2})),\ldots ,f(\tilde{A}_{2}(x_{n})))$. Therefore, $E(\tilde{A}_{1})\le E(\tilde{A}_{2})$.