A Multi-attribute Decision-Making Approach Based on Spherical Fuzzy Sets for Yunnan Baiyao’s R&D Project Selection Problem

Abstract

As we all know, the research and development (R&D) is crucial for enterprises. How to choose the R&D project is an important research topic, and at the same time, it is also a typical multi-attribute decision-making (MADM) problem. In this paper, we propose a novel MADM method for selecting the Yunnan Baiyao’s R&D project of toothpastes. Firstly, we use T-spherical fuzzy sets (T-SFSs) to express the evaluation information of the toothpastes from decision makers to overcome the shortcomings existing that the traditional information form may cause information distortion. Secondly, we extend the generalized Maclaurin symmetric mean (GMSM) operator to T-spherical fuzzy environment and propose the T-spherical fuzzy GMSM operator (T-SFGMSM) and the T-spherical fuzzy weighted GMSM operator (T-SFWGMSM), where these operators are suitable for the situation which the input information is interrelated. Further, we put forward a novel MADM approach for selecting toothpaste product based upon the T-SFWGMSM operator, which can not only handle a more extensive scope of fuzzy information, but also process the interrelationships between multiple attributes. Lastly, we solve a R&D project selection problem for Yunnan Baiyao Co., Ltd. by the proposed method and make a detailed comparative analysis with other MADM methods to demonstrate the effectiveness and superiority of the novel approach.

Appendices

Appendix 1

Suppose $T = \left\langle {s,i,d} \right\rangle$, $T_{A} = \left\langle {s_{A} ,i_{A} ,d_{A} } \right\rangle$, $T_{B} = \left\langle {s_{B} ,i_{B} ,d_{B} } \right\rangle$are any three T-SFNs, and $\lambda ,\lambda_{1,} \lambda_{2} > 0$, Then they have the following operational properties.

$$({\text{i}})\;T_{A} \oplus T_{B} = T_{B} \oplus T_{A}$$

(8)

$$({\text{ii}})\;T_{A} \otimes T_{B} = T_{B} \otimes T_{A}$$

(9)

$$({\text{iii}})\;\lambda \left( {T_{A} \oplus T_{B} } \right) = \lambda T_{A} \oplus \lambda T_{B}$$

(10)

$$({\text{iv}})\;\lambda_{1} T \oplus \lambda_{2} T = \left( {\lambda_{1} \oplus \lambda_{2} } \right)T$$

(11)

$$({\text{v}})\;\left( {T_{A} \otimes T_{B} } \right)^{\lambda } = T_{A}^{\lambda } \otimes T_{B}^{\lambda }$$

(12)

$$({\text{vi}})\;T^{{\lambda_{1} }} \otimes T^{{\lambda_{2} }} = T^{{\left( {\lambda_{1} + \lambda_{2} } \right)}}$$

(13)

Proof

• 1.
  $$\begin{array}{*{20}l} {T_{A} \oplus T_{B} = \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{1/n} ,i_{A} i_{B} ,d_{A} d_{B} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - s_{B}^{n} } \right)\left( {1{ - }s_{A}^{n} } \right)} \right)^{1/n} ,i_{B} i_{A} ,d_{B} d_{A} } \right\rangle = T_{B} \oplus T_{A} } \hfill \\ \end{array}$$
• 2.
  $$\begin{array}{*{20}l} {T_{A} \otimes T_{B} = \left\langle {s_{A} s_{B} ,\left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{1/n} ,\left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s_{B} s_{A} ,\left( {1 - \left( {1 - i_{B}^{n} } \right)\left( {1{ - }i_{A}^{n} } \right)} \right)^{1/n} ,\left( {1 - \left( {1 - d_{B}^{n} } \right)\left( {1{ - }d_{A}^{n} } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = T_{B} \otimes T_{A} .} \hfill \\ \end{array}$$
• 3.
  $$\begin{array}{*{20}l} {\lambda \left( {T_{A} \oplus T_{B} } \right) = \lambda \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{1/n} ,i_{A} i_{B} ,d_{A} d_{B} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{1/n} } \right)^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{A}^{\lambda } ,d_{A}^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {\left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ \end{array}$$

Because $\lambda T_{A} = \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{A}^{\lambda } ,d_{A}^{\lambda } } \right\rangle ,$$\lambda T_{B} = \left\langle {\left( {1 - \left( {1 - s_{B}^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{B}^{\lambda } ,d_{B}^{\lambda } } \right\rangle$

$$\begin{array}{*{20}l} {\lambda T_{A} \oplus \lambda T_{B} = \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{A}^{\lambda } ,d_{A}^{\lambda } } \right\rangle \oplus \left\langle {\left( {1 - \left( {1 - s_{B}^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{B}^{\lambda } ,d_{B}^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - s_{A}^{n} } \right)^{\lambda } } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - s_{B}^{n} } \right)^{\lambda } } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {1 - \left( {1 - s_{A}^{n} } \right)^{\lambda } } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - s_{B}^{n} } \right)^{\lambda } } \right)} \right)} \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {\left( {1 - s_{A}^{n} } \right)^{\lambda } } \right)\left( {\left( {1 - s_{B}^{n} } \right)^{\lambda } } \right)} \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {\left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ \end{array}$$

So, $\lambda \left( {T_{A} \oplus T_{B} } \right) = \lambda T_{A} \oplus \lambda T_{B}$.

• 4.
  $$\begin{array}{*{20}l} {\lambda_{1} T \oplus \lambda_{2} T{ = }\left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} ,i^{{\lambda_{1} }} ,d^{{\lambda_{1} }} } \right\rangle \oplus \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{2} }} ,d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} ,i^{{\lambda_{1} }} i^{{\lambda_{2} }} ,d^{{\lambda_{1} }} d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)} \right)} \right)^{1/n} ,i^{{\lambda_{1} }} i^{{\lambda_{2} }} ,d^{{\lambda_{1} }} d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{1} }} i^{{\lambda_{2} }} ,d^{{\lambda_{1} }} d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} { + }\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{1} { + }\lambda_{2} }} ,d^{{\lambda_{1} { + }\lambda_{2} }} } \right\rangle } \hfill \\ \end{array}$$

and $\left( {\lambda_{1} \oplus \lambda_{2} } \right)T = \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} { + }\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{1} { + }\lambda_{2} }} ,d^{{\lambda_{1} { + }\lambda_{2} }} } \right\rangle$So $\lambda_{1} T \oplus \lambda_{2} T = \left( {\lambda_{1} \oplus \lambda_{2} } \right)T$

• 5. $\begin{array}{*{20}l} {\left( {T_{A} \otimes T_{B} } \right)^{\lambda } { = }\left\langle {s_{A} s_{B} ,\left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{1/n} ,\left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{1/n} } \right\rangle^{\lambda } } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{1/n} } \right)^{n} } \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{1/n} } \right)^{n} } \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)} \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {\left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {\left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ \end{array}$

and

$$\begin{array}{*{20}l} {T_{A}^{\lambda } \otimes T_{B}^{\lambda } = \left\langle {s_{A}^{\lambda } ,\left( {1 - \left( {1 - i_{A}^{n} } \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {1 - d_{A}^{n} } \right)^{\lambda } } \right)^{1/n} } \right\rangle \otimes \left\langle {s_{B}^{\lambda } ,\left( {1 - \left( {1 - i_{B}^{n} } \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {1 - d_{B}^{n} } \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle \begin{aligned} s_{A}^{\lambda } s_{B}^{\lambda } ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - i_{A}^{n} } \right)^{\lambda } } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - i_{B}^{n} } \right)^{\lambda } } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - d_{A}^{n} } \right)^{\lambda } } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - d_{B}^{n} } \right)^{\lambda } } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} \hfill \\ \end{aligned} \right\rangle } \hfill \\ { = \left\langle {s_{A}^{\lambda } s_{B}^{\lambda } ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - i_{A}^{n} } \right)^{\lambda } } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - i_{B}^{n} } \right)^{\lambda } } \right)} \right)} \right)^{1/n} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - d_{A}^{n} } \right)^{\lambda } } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - d_{B}^{n} } \right)^{\lambda } } \right)} \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s_{A}^{\lambda } s_{B}^{\lambda } ,\left( {1 - \left( {\left( {1 - i_{A}^{n} } \right)^{\lambda } } \right)\left( {\left( {1 - i_{B}^{n} } \right)^{\lambda } } \right)} \right)^{1/n} ,\left( {1 - \left( {\left( {1 - d_{A}^{n} } \right)^{\lambda } } \right)\left( {\left( {1 - d_{B}^{n} } \right)^{\lambda } } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {\left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {\left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ \end{array}$$

So $\left( {T_{A} \otimes T_{B} } \right)^{\lambda } = T_{A}^{\lambda } \otimes T_{B}^{\lambda }$

• 6.
  $$\begin{array}{*{20}l} {T^{{\lambda_{1} }} \otimes T^{{\lambda_{2} }} = \left\langle {s^{{\lambda_{1} }} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right\rangle \otimes \left\langle {s^{{\lambda_{2} }} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle \begin{aligned} s^{{\lambda_{1} }} s^{{\lambda_{2} }} ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} \hfill \\ \end{aligned} \right\rangle } \hfill \\ { = \left\langle {s^{{\lambda_{1} }} s^{{\lambda_{2} }} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)} \right)} \right)^{1/n} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)} \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s^{{\lambda_{1} }} s^{{\lambda_{2} }} ,\left( {1 - \left( {\left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)\left( {\left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)} \right)^{1/n} ,\left( {1 - \left( {\left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)\left( {\left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} } \right\rangle } \hfill \\ \end{array}$$

and $T^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} = \left\langle {s,i,d} \right\rangle^{{\left( {\lambda_{1} + \lambda_{2} } \right)}}$$= \left\langle {s^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} } \right\rangle$

So $T^{{\lambda_{1} }} \otimes T^{{\lambda_{2} }} = T^{{\left( {\lambda_{1} + \lambda_{2} } \right)}}$.□

Appendix 2

Theorem 1

Suppose$T_{1} ,T_{2} , \ldots$and$T_{k}$are T-SFNs, where$T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$(m = 1,2, \ldots ,k)$. The synthesized result of the T-SFGMSM operator of the T-SFNs$T_{1} ,T_{2} , \ldots$and$T_{k}$is still a T-SFN, shown as follows:

$$\begin{aligned} & {\hbox{T-SFGMSM}^{{\left( {r,\lambda_{1},\lambda_{2} ,\ldots ,\lambda_{r} } \right)}} \left( {T_{1} ,T_{2}, \ldots ,T_{k} } \right) = \left\langle {\left( {\left( {1 - \left({\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1- \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.} \hfill \\ & {\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle } \hfill \\ \end{aligned}$$

(16)

Proof

According to the operational rules of T-SFNs, we can obtain

$$T_{{m_{j} }}^{{\lambda_{j} }} = \left\langle {s_{{m_{j} }}^{{\lambda_{j} }} ,\left( {1 - \left( {1 - i_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } \right)^{1/n} ,\left( {1 - \left( {1 - d_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } \right)^{1/n} } \right\rangle$$

and

$$\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} = \left\langle {\prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} }} } ,\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - i_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} ,\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - d_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} } \right\rangle .$$

Then, we can calculate

$$\mathop \oplus \limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} \left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right) = \left\langle \begin{aligned} \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{1/n} , \hfill \\ \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} ,} \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } \hfill \\ \end{aligned} \right\rangle$$

and

$$\begin{array}{*{20}l} {\frac{{\mathop \oplus \limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} \left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right)}}{{\left( \begin{aligned} k \hfill \\ r \hfill \\ \end{aligned} \right)}} = \left\langle {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} ,} \right.} \hfill \\ {\left. {\left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} ,\left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right\rangle } \hfill \\ \end{array}$$

So the comprehensive result of the T-SFGMSM operator is

$$\left( {\frac{{\mathop \oplus \limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} \left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right)}}{{\left( \begin{aligned} k \hfill \\ r \hfill \\ \end{aligned} \right)}}} \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} = \left\langle {\left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle$$

That proves Eq. (16) is right.□

Appendix 3

(1) Idempotency: If $T_{1} ,T_{2} , \ldots ,$ and $T_{k}$ are T-SFNs meeting the condition with$T_{m} = \left\{ {s_{m} ,i_{m} ,d_{m} } \right\} = T = \left\{ {s,i,d} \right\}$$(m = 1,2, \ldots ,k)$. Then, $\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T$.

Proof.

According to the above Theorem 1, we can obtain that $T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle = T = \left\langle {s,i,d} \right\rangle$ for any $m = 1,2, \ldots ,k$.

$$\begin{array}{*{20}l} {\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = \left\langle {\left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.} \hfill \\ {\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,} \hfill \\ { \left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - s^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right) \times n}} } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.} \hfill \\ {\left. \begin{aligned} \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle } \hfill \\ { = \left\langle {\left( {\left( {1 - \left( {1 - s^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right) \times n}} } \right)} \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.} \hfill \\ {\left( {1 - \left( {1 - \left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)} \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left. {\left( {1 - \left( {1 - \left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)} \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s,} \right.i,\left. d \right\rangle } \hfill \\ \end{array}$$

□

Appendix 4

(2) Monotonicity: If $T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$\left( {m = 1,2, \ldots ,k} \right)$ and $\hat{T}_{m} = \left\langle {\hat{s}_{m} ,\hat{i}_{m} ,\hat{d}_{m} } \right\rangle$$\left( {m = 1,2, \ldots ,k} \right)$ are any two sets of T-SFNs. They accord the prerequisite “$\hat{s}_{m} \le s_{m}$, $\hat{i}_{m} \le i_{m}$ and $\hat{d}_{m} \ge d_{m}$,” then$\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).$

Proof

Suppose that $\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) = \hat{T} = \left\langle {\hat{s},\hat{i},\hat{d}} \right\rangle$ and $\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T = \left\langle {s,i,d} \right\rangle$, then

$$\hat{s} = \left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {\hat{s}_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,$$

$$s = \left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ;$$

$$\hat{i} = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{i}_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,$$

$$i = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ;$$

$$\hat{d} = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{d}_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,$$

$$d = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} .$$

In order to prove this property, we need to compute their score function values $S(\hat{T})$ and $S(T)$, and their accuracy function values $A(\hat{T})$ and $A(T)$ to compare their synthesized results, i.e., $\hat{T} \le T$. Firstly, based on the conditions $\hat{s}_{m} \le s_{m}$, $\hat{i}_{m} \le i_{m}$ and $\hat{d}_{m} \ge d_{m}$, we can get the compared results of their MD, AMD and NMD.

• 1. For the membership degree

Based on $\hat{s}_{m} \le s_{m}$, we can get

$$\left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {\hat{s}_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} \le \left( {\left( {1 - \left( {\prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} \times n}} } } \right)} } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}}$$

That is $\hat{s}_{m} \le s_{m}$

• 2. For the abstinence degree

Based on $\hat{i}_{m} \le i_{m}$ , we can obtain

$$\begin{aligned} \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{i}_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \le \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned}$$

That is $\hat{i}_{m} \le i_{m}$

• 3. For the non-membership degree

Based on $\hat{d}_{m} \ge d_{m}$ , we can obtain

$$\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{d}_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n}$$

$$\ge \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{1/\left( \begin{array}{l} k \\ r \end{array} \right)}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n}$$

That is $\hat{d}_{m} \ge d_{m}$Thus, it can be obtained that $S(\hat{T}) = \hat{s}^{n} - \hat{d}^{n} \le S(T) = s^{n} - d^{n}$. Next, we explore two cases.

• 1. If $S(\hat{T}) < S(T)$, then $\hat{T} < T$ according to Definition 4.

• 2. If $S(\hat{T}) = S(T)$, then $\hat{s} = s$, $\hat{d} = d$ Because $\hat{i} \le i$, so $A(\hat{T}) = \hat{s}^{n} + \hat{i}^{n} { + }\hat{d}^{n} = A(T) = s^{n} + i^{n} { + }d^{n}$, which testifies $\hat{T} \le T$.

In conclusion, the synthesized result $\hat{T} \le T$, which explains$\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right)$□

Appendix 5

(3) Boundedness: Suppose $T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$\left( {m = 1,2, \ldots ,k} \right)$ is any a set of T-SFNs, $T^{ - } = \mathop { \hbox{min} }\limits_{m} T_{m}$ and $T^{ + } = \mathop { \hbox{max} }\limits_{m} T_{m}$, then $T^{ - } \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \cdots ,T_{k} } \right) \le T^{ + }$.

Proof

Owing to $T_{m} \ge T^{ - } = \mathop { \hbox{min} }\limits_{m} T_{m}$, based on the Monotonicity and Idempotency of the novel $\hbox{T-SFGMSM}$ operator, the following results can be obtained:

$$\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) \ge \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T^{ - } ,T^{ - } , \ldots ,T^{ - } } \right) = T^{ - }$$

comparably, the corresponding result for $T^{ + } = \mathop { \hbox{max} }\limits_{m} T_{m}$ can be obtained:

$$\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T^{ + } ,T^{ + } , \ldots ,T^{ + } } \right) = T^{ + }$$

Therefore, $T^{ - } \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) \le T^{ + }$.□

Appendix 6

Theorem 3

Let$T_{1} ,T_{2} , \cdots$and$T_{k}$be T-SFNs, where$T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$m = \left( {1,2, \ldots ,k} \right)$, and let the weight of input argument$T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$m = \left( {1,2, \ldots ,k} \right)$be$\omega_{m}$, where$\omega_{m} \in \left[ {0,1} \right]$and$\sum\nolimits_{m = 1}^{k} {\omega_{m} } = 1$. Then, the synthesized result of the$\hbox{T-SFGMSM}$operator of the T-SFNs$T_{1} ,T_{2} , \ldots$and$T_{k}$is still a T-SFN, shown as follows:

$$\begin{array}{*{20}l} {{\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = } \hfill \\ {\left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,} \right.} \hfill \\ {\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle \left( {1 \le r < k} \right)} \hfill \\ {\left\langle {\left( {\prod\limits_{m = 1}^{r} {s_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,\left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - i_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - d_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle \left( {r = k} \right)} \hfill \\ \end{array}$$

(18)

Proof

Obtained by the operational laws of T-SFNs defined in Definition 5, Theorem 3 can be easily proved.

• 1. For $1 \le r < k$, we can have $\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} = \left\langle {\prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} }} } ,\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - i_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} ,\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - d_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} } \right\rangle$

Obtained by the operational laws of T-SFNs, we can derive the below equations:

$$\left( {1 - \sum\limits_{j= 1}^{r} {\omega_{{m_{j} }} } } \right)\left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right) = \left\langle \begin{aligned} \left( {1 - \left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } }} } \right)^{1/n} , \hfill \\ \left( {\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - i_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} } \right)^{{1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } }} ,\left( {\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - d_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} } \right)^{{1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } }} \hfill \\ \end{aligned} \right\rangle$$

and

$$\begin{array}{*{20}l} {\mathop \oplus \limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} \left( {1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)\left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right) = } \hfill \\ {\left\langle {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } }} } } \right)^{1/n} , \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } \right)^{{1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } }} ,} \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } \right)^{{1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } }} } } \right\rangle } \hfill \\ \end{array}$$

then

$$\begin{array}{*{20}l} {\frac{{\mathop \oplus \limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} \left( {1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)\left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right)}}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}} = } \hfill \\ {\left\langle \begin{aligned} \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} , \hfill \\ \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} ,} \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{1/n} } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } \hfill \\ \end{aligned} \right\rangle } \hfill \\ {\left( {\frac{{\mathop \oplus \limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} \left( {1 - \sum\nolimits_{j= 1}^{r} {\omega_{{m_{j} }} } } \right)\left( {\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} } \right)}}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}} \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} = } \hfill \\ {\left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.} \hfill \\ {\left. \begin{aligned} \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j= 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j= 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle } \hfill \\ \end{array}$$

So formula (18) is right when $1 \le r < n$.

• 2. For $r = k$, we have known that

  $$\mathop \otimes \limits_{m = 1}^{r} T_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} = \left\langle {\prod\limits_{m = 1}^{r} {s_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } ,\left( {1 - \prod\limits_{m = 1}^{r} {(1 - i_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{1/n} ,\left( {1 - \prod\limits_{m = 1}^{r} {(1 - d_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{1/n} } \right\rangle$$

Based on the proof of the first case when $1 \le r < n$, we can derive the below equations:

$$\begin{array}{*{20}l} {\left( {\mathop \otimes \limits_{m = 1}^{r} T_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} = } \hfill \\ {\left\langle \begin{aligned}& \left( {\prod\limits_{m = 1}^{r} {s_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\& \left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - i_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - d_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle } \hfill \\ \end{array}$$

□

Appendix 7

(1) Idempotency: If $T_{1} ,T_{2} , \ldots ,$ and $T_{k}$ are T-SFNs and meet $T_{m} = \left\langle {s_{m} , i_{m} ,d_{m} } \right\rangle = T = \left\langle {s, i,d} \right\rangle$$(m = 1,2, \ldots ,k)$. Then, ${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T$.

Proof

According to the above Theorem 3, we can obtain that

• 1. For $1 \le r < k$, we can get

$${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = \left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - s^{{q \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,$$

$$\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \left( {1 - s^{{n \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\sum\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)} }} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\sum\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\sum\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \left( {1 - s^{{n \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}} \left( {\left( \begin{array}{*{20}l} k\\ r \\ \end{array}\right) -\sum\limits_{{1 \le m_{1} \cdots < m_{r} \le k}}^{k} {\left( {\sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)} }\right)}}} \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \sum\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {\sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \sum\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {\sum\limits_{j = 1}^{r} {\omega_{{m_{j} }} } } \right)} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \left( {1 - s^{{q \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \sum\limits_{{m_{j} = 1}}^{k} {\left( {\left( \begin{array}{l} k - 1 \\ r - 1 \end{array} \right)\omega_{{m_{j} }} } \right)} } \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \sum\limits_{{m_{j} = 1}}^{k} {\left( {\left( \begin{array}{l} k - 1 \\ r - 1 \end{array} \right)\omega_{{m_{j} }} } \right)} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \sum\limits_{{m_{j} = 1}}^{k} {\left( {\left( \begin{array}{l} k - 1 \\ r - 1 \end{array} \right)\omega_{{m_{j} }} } \right)} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \left( {1 - s^{{n \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \left( \begin{array}{l} k - 1 \\ r - 1 \end{array} \right)} \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \left( \begin{array}{l} k - 1 \\ r - 1 \end{array} \right)} \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}\left( {\left( \begin{array}{l} k \\ r \end{array} \right) - \left( \begin{array}{l} k - 1 \\ r - 1 \end{array} \right)} \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle \begin{aligned} \left( {\left( {1 - \left( {1 - s^{{n \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right) } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\ \left( {1 - \left( {1 - \left( {1 - (1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)} \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {1 - \left( {1 - (1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)} \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle \begin{aligned} \left( {\left( {s^{{n \times \left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\ \left( {1 - \left( {(1 - i^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {(1 - d^{n} )^{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle {u ,v} \right\rangle$$

• 2. For $r = k$, we can get

$${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) =$$

$$\left\langle \begin{aligned} \left( {\prod\limits_{m = 1}^{r} {s^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\ \left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - i^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - d^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$${ = }\left\langle {\left( {s^{{\sum\nolimits_{m = 1}^{r} {\left( {\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}} \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,\left( {1 - \left( {(1 - i^{n} )^{{\sum\nolimits_{m = 1}^{r} {\left( {\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}} \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {(1 - d^{n} )^{{\sum\nolimits_{m = 1}^{r} {\left( {\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}} \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle$$

$$= \left\langle \begin{aligned} \left( {s^{{\sum\nolimits_{m = 1}^{r} {\left( {\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}} \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\ \left( {1 - \left( {(1 - i^{n} )^{{\sum\nolimits_{m = 1}^{r} {\left( {\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}} \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {(1 - d^{n} )^{{\sum\nolimits_{m = 1}^{r} {\left( {\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}} \right)} }} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle \begin{aligned} \left( {s^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right) + \frac{{\sum\nolimits_{m = 1}^{r} {\left( {1 - k\omega_{m} } \right)} }}{k - 1}}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\ \left( {1 - \left( {(1 - i^{n} )^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right) + \frac{{\sum\nolimits_{m = 1}^{r} {\left( {1 - k\omega_{m} } \right)} }}{k - 1}}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {(1 - d^{n} )^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right) + \frac{{\sum\nolimits_{m = 1}^{r} {\left( {1 - k\omega_{m} } \right)} }}{k - 1}}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

Because $r = k$ and $\sum\nolimits_{m = 1}^{k} {\omega_{m} } = 1$, we have $\sum\nolimits_{m = 1}^{r} {\left( {1 - k\omega_{m} } \right)} = k - k\sum\nolimits_{m = 1}^{k} {\omega_{m} } = 0$ and the upper equality becomes the following form:

$$\begin{aligned} \left\langle {\left( {s^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,\left( {1 - \left( {(1 - i^{n} )^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {(1 - d^{n} )^{{\left( {\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} } \right)}} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle \hfill \\ = \left\langle {s,\left( {1 - (1 - i^{n} )} \right)^{1/n} ,\left( {1 - (1 - d^{n} )} \right)^{1/n} } \right\rangle = \left\langle {s,i,d} \right\rangle \hfill \\ \end{aligned}$$

Therefore, the ${\hbox{T-SFWGMSM}}$ operator has the Idempotency property.□

Appendix 8

(2) Monotonicity: If $T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$\left( {m = 1,2, \ldots ,k} \right)$ and $\hat{T}_{m} = \left\langle {\hat{s}_{m} ,\hat{i}_{m} ,\hat{d}_{m} } \right\rangle$$\left( {m = 1,2, \ldots ,k} \right)$ are any two sets about T-SFNs. They accord the prerequisite $\hat{s}_{m} \le s_{m}$, $\hat{i}_{m} \le i_{m}$ and $\hat{d}_{m} \ge d_{m}$, then ${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le {\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).$

Proof

Similar to the monotonicity property of the ${\hbox{T-SFWGMSM}}$ operator, the following formulas can be assumed:${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) = \hat{T} = \left\langle {\hat{s},\hat{i},\hat{d}} \right\rangle$ and ${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T = \left\langle {s,i,d} \right\rangle$. Since formula (18) is divided into two cases, we discuss each case separately.

• 1. For $1 \le r < k$, we can get

$${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = \left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. {\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} } \right\rangle$$

$$\hat{s} = \left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {\hat{s}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,$$

$$s = \left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,$$

$$\hat{i} = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{i}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,$$

$$i = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,$$

$$\hat{d} = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{d}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,$$

$$d = \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} .$$

Because $\hat{s}_{m} \le s_{m}$, $\hat{i}_{m} \le i_{m}$ and $\hat{d}_{m} \ge d_{m}$, it is easy to obtain the following inequalities:

$$\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {\hat{s}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} \le \left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}}$$

$$\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{i}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \le \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n}$$

and

$$\left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - \hat{d}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n}$$

$$\ge \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\omega_{{m_{j} }} } }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n}$$

i.e., $\hat{s} \le s$ and $\hat{i} \le i$.Thus, it can be obtained that $S(\hat{T}) = \hat{s}^{n} - \hat{d}^{n} \le S(T) = s^{n} - d^{n}$. Next, we discuss two cases.

• (a) If $S(\hat{T}) < S(T)$, then $\hat{T} < T$ according to Definition 4.

• (b) If $S(\hat{T}) = S(T)$, then $\hat{s} = s$, $\hat{d} = d$ Because $\hat{i} \le i$, so $A(\hat{T}) = \hat{s}^{n} + \hat{i}^{n} { + }\hat{d}^{n} = A(T) = s^{n} + i^{n} { + }d^{n}$, which testifies $\hat{T} < T$.

To sum up, the synthesized result is $\hat{T} < T$.

• 2. For $r = k$, its proof process is similar to the one of the first case where $1 \le r < k$, so we omit it here.

According to (i) and (ii), we can get $\hat{T} \le T$, which explains ${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le {\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).$□

Appendix 9

(3) Boundedness: Let $T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle$$\left( {m = 1,2, \ldots ,k} \right)$ be any a set of T-SFNs, $T^{ - } = \mathop { \hbox{min} }\limits_{m} T_{m}$ and$T^{ + } = \mathop { \hbox{max} }\limits_{m} T_{m}$, then $T^{ - } \le {\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) \le T^{ + }$.

Because the certification process is same with the one of corresponding Boundedness property of the $\hbox{T-SFGMSM}$ operator, we omit it here.

Similarly, we also observe many particular instances about the T-SFWGMSM operator with different parameters. Here, we just introduce a particular instance of this operator.

When $\omega_{m} = \frac{1}{k}$$\left( {m = 1,2, \ldots ,k} \right)$, the proposed ${\hbox{T-SFWGMSM}}$ operator is simplified into the $\hbox{T-SFGMSM}$ operator.

Proof

According to Eq. (18), we should take two cases into account.

• 1. For $1 \le r < k$, we know$\omega_{m} = \frac{1}{k}$$\left( {m = 1,2, \ldots ,k} \right)$, so we can get

$${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = \left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\frac{1}{k}} }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\frac{1}{k}} }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \sum\nolimits_{j = 1}^{r} {\frac{1}{k}} }}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{\frac{{1 - \frac{r}{k}}}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \frac{r}{k}}}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{{1 - \frac{r}{k}}}{{\left( \begin{array}{*{20}l} k - 1 \\ r \\ \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle {\left( {\left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{n \times \lambda_{j} }} } } \right)^{{\frac{1}{{\left( \begin{array}{l} k \\ r \end{array} \right)}}}} } } \right)^{1/n} } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} ,} \right.$$

$$\left. \begin{aligned} \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{1}{{\left( \begin{array}{l} k \\ r \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \prod\limits_{{1 \le m_{1} < \cdots < m_{r} \le k}} {\left( {1 - \prod\limits_{j = 1}^{r} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{1}{{\left( \begin{array}{l} k \\ r \end{array} \right)}}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).$$

• 2. For $r = k$, based on$\omega_{m} = \frac{1}{k}$$\left( {m = 1,2, \ldots ,k} \right)$, we have

$${\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) =$$

$$\left\langle \begin{aligned} &\left( {\prod\limits_{m = 1}^{r} {s_{m}^{{\lambda_{m} + \frac{{1 - k \times \frac{1}{k}}}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} , \hfill \\& \left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - i_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k \times \frac{1}{k}}}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} ,\left( {1 - \left( {\prod\limits_{m = 1}^{r} {(1 - d_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k \times \frac{1}{k}}}{k - 1}}} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{r} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

$$= \left\langle \begin{aligned}& \left( {\prod\limits_{m = 1}^{k} {s_{m}^{{\lambda_{m} }} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{k} }}}} , \hfill \\& \left( {1 - \left( {\prod\limits_{m = 1}^{k} {(1 - i_{m}^{n} )^{{\lambda_{m} }} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{k} }}}} } \right)^{1/n} ,\left( {1 - \left( {\prod\limits_{m = 1}^{k} {(1 - d_{m}^{n} )^{{\lambda_{m} }} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{k} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

By Theorem 1, we can see

$$\hbox{T-SFGMSM}^{{\left( {k, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{k} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{K} } \right) =$$

$$\left\langle \begin{aligned} \left( {\prod\limits_{j = 1}^{k} {s_{{m_{j} }}^{{\lambda_{j} }} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{k} }}}} , \hfill \\ \left( {1 - \left( {\prod\limits_{j = 1}^{k} {(1 - i_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{k} }}}} } \right)^{1/n} ,\left( {1 - \left( {\prod\limits_{j = 1}^{k} {(1 - d_{{m_{j} }}^{n} )^{{\lambda_{j} }} } } \right)^{{\frac{1}{{\lambda_{1} + \lambda_{2} + \cdots + \lambda_{k} }}}} } \right)^{1/n} \hfill \\ \end{aligned} \right\rangle$$

Therefore, when the weights of input arguments are assigned to equal values, the proposed T-SFWGMSM operator is simplified as the T-SFGMSM operator.□