Linguistic single-valued neutrosophic prioritized aggregation operators and their applications to multiple-attribute group decision-making

Abstract

In this paper, we introduce some new linguistic prioritized aggregation operators which simultaneously considers the priority among the attributes and the uncertainty in linguistic terms under the linguistic single-valued neutrosophic set (LSVNS). For this, firstly the operational laws for LSVNSs are introduced along with their properties. Based on these operations, we proposes some prioritized weighted and ordered weighted averaging as well as geometric aggregation operators for a collection of linguistic single-valued neutrosophic numbers. Further, the desirable properties of these operators are studied. Finally, a decision-making approach presents and illustrate with a numerical example.

Appendix

Proof of the Theorem 6

Proof

Firstly, we will prove Eq. (2) by mathematical induction on n. Since for all $i, \beta _{i}=\langle s_{\theta _{i}}, s_{\psi _{i}}, s_{\sigma _{i}}\rangle$ is LSVNN so we have $\theta _{i}+\psi _{i}+\sigma _{i}\le 3t.$ Then the following steps of the mathematical induction have been followed.

• Step 1 For $n=2,$ we have

  $$\begin{aligned} \text {LSVNPWA}(\beta _{1},\beta _{2})= & {} \bigoplus \limits _{i=1}^{2}\frac{T_{i}}{\sum _{i=1}^{2}T_{i}} \beta _{i}\\= & {} \frac{T_{1}}{\sum _{i=1}^{2}T_{i}} \beta _{1} \oplus \frac{T_{2}}{\sum _{i=1}^{2}T_{i}} \beta _{2}\\= & {} \left\langle s_{t\left( 1-\left( 1- \frac{\theta _1}{t}\right) ^{\frac{T_{1}}{\sum \limits _{i=1}^{2}T_{i}}}\right) }, s_{t\left( \left( \frac{\psi _1}{t}\right) ^{\frac{T_{1}}{\sum \limits _{i=1}^{2}T_{i}}}\right) } , s_{t\left( \left( \frac{\sigma _1}{t}\right) ^{\frac{T_{1}}{\sum \limits _{i=1}^{2}T_{i}}}\right) }\right\rangle \\&\oplus \left\langle s_{t\left( 1-\left( 1- \frac{\theta _2}{t}\right) ^{\frac{T_{1}}{\sum \limits _{i=1}^{2}T_{i}}}\right) }, s_{t\left( \left( \frac{\psi _2}{t}\right) ^{\frac{T_{1}}{\sum \limits _{i=1}^{2}T_{i}}}\right) } , s_{t\left( \left( \frac{\sigma _2}{t}\right) ^{\frac{T_{1}}{\sum \limits _{i=1}^{2}T_{i}}}\right) }\right\rangle \\= & {} \left\langle s_{t\bigg (1-\prod \limits _{i=1}^{2}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\bigg )}, s_{t\bigg (\prod \limits _{i=1}^{2}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\bigg )}, \right. \\&\left. \quad s_{t\bigg (\prod \limits _{i=1}^{2}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\bigg )} \right\rangle \end{aligned}$$

  Hence result is true for $n=2.$

• Step 2 Assume the Eq. (2) is true for $n=z,$ then for $n=z+1,$ we have

  $$\begin{aligned}&\bigoplus \limits _{i=1}^{z+1}\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}} \beta _{i}=\bigoplus \limits _{i=1}^{z}\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}} \beta _{i}\oplus \frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}} \beta _{z+1}\\&\quad = \left\langle s_{t\left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{z}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{z}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) } \right\rangle \\&\quad \oplus \left\langle s_{t\left( 1-\left( 1- \frac{\theta _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \left( \frac{\psi _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \left( \frac{\sigma _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }\right\rangle \\&\quad = \left\langle s_{t\left( \left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) + \left( 1-\left( 1- \frac{\theta _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) -\left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) \left( 1-\left( 1- \frac{\theta _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) \right) },\right. \\&\left. \qquad s_{t\left( \prod \limits _{i=1}^{z}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) \left( \left( \frac{\psi _{z+1}}{t} \right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{z}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) \left( \left( \frac{\sigma _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }\right\rangle \\&\quad = \left\langle s_{t\left( 1-\prod \limits _{i=1}^{z+1}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{z+1}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{z+1}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) } \right\rangle \end{aligned}$$

Hence the result given in Eq. (2), is true for all positive integers n.

Further, $0 \le \frac{\theta _i}{t} \le 1 \Rightarrow 0 \le 1- \frac{\theta _i}{t} \le 1 \Rightarrow 0 \le 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le 1 \Rightarrow 0 \le t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le t$. Similarly, we get

$$\begin{aligned} 0 \le t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) , t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le t. \end{aligned}$$

Also,

$$\begin{aligned} t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) +t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) +t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le 3t \end{aligned}$$

Thus, Eq. (2) is still LSVNN, which completes the proof of the theorem. $\square$

Proof of the Theorem 7

Proof

• (P1) Since $\beta _i=\beta$ for all i, then we have

  $$\begin{aligned} \text {LSVNPWA}(\beta _1,\beta _2,\ldots ,\beta _n)= & {} \bigoplus \limits _{i=1}^{n}\frac{T_i}{\sum _{i=1}^n T_i}\beta \\= & {} \sum _{i=1}^{n}\frac{T_i}{\sum _{i=1}^n T_i}\beta \\= & {} \beta \end{aligned}$$

• (P2) Since $s_{\theta _i^{\prime }} \ge s_{\theta _i}$ which implies $\theta _i^{\prime } \ge \theta _i$ and hence, we have

  $$\begin{aligned}&0\le \left( 1-\frac{\theta _i^{\prime }}{t}\right) \le \left( 1-\frac{\theta _i}{t}\right) \le 1 \\&\quad \Rightarrow \prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i^{\prime }}{t}\right) \le \prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t} \right) \\&\quad \Rightarrow \prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}} \le \prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}} \\&\quad \Rightarrow 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}} \ge 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}} \\&\quad \Rightarrow t \left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) \ge t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) \end{aligned}$$

  which implies

  $$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) } \ge s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) } \end{aligned}$$

  (13)

  Similarly, for $s_{\psi _i^{\prime }} \le s_{\psi _i}$ and $s_{\sigma _i^{\prime }} \le s_{\sigma _i}$, we can get

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _{i}^{\prime }}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  (14)

  and

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i^{\prime }}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  (15)

  Thus, from Eqs. (13), (14), (15) and Definition 6, we have

  $$\begin{aligned} \text {LSVNPWA}(\beta _1^{\prime }, \beta _2^{\prime },\ldots , \beta _n^{\prime }) \ge \text {LSVNPWA}(\beta _1,\beta _2,\ldots ,\beta _n) \end{aligned}$$

• (P3) Since $\min \limits _{i}\{s_{\theta _i}\} \le s_{\theta _i} \le \max \limits _{i}\{s_{\theta _i}\}$, $\min \limits _{i}\{s_{\psi _i}\} \le s_{\psi _i} \le \max \limits _{i}\{s_{\psi _i}\}$ and $\min \limits _{i}\{s_{\sigma _i}\} \le s_{\sigma _i} \le \max \limits _{i}\{s_{\sigma _i}\}$, thus from above property, we have

  $$\begin{aligned} \beta ^- \le \text {LSVNPWA}(\beta _1,\beta _2,\ldots ,\beta _n)\le \beta ^+ \end{aligned}$$

$\square$

Proof of the Theorem 10

Proof

We will prove Eq. (5) by mathematical induction on n. Since for all $i, \beta _{i}=\langle s_{\theta _{i}},s_{\psi _{i}}, s_{\sigma _{i}}\rangle$ is LSVNN so we have $0 \le \theta _{i}+\psi _{i}+\sigma _{i}\le 3t.$ Then the following steps of the mathematical induction have been followed.

• Step 1 For $n=2,$ we have

  $$\begin{aligned}&\text {LSVNPWG}(\beta _{1},\beta _{2})=\bigotimes \limits _{i=1}^{2} \beta _{i}^\frac{T_{i}}{\sum _{i=1}^{2}T_{i}}\\&\quad = \beta _{1}^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}} \otimes \beta _{2}^{\frac{T_{2}}{\sum _{i=1}^{2}T_{i}}}\\&\quad = \left\langle s_{t\left( \left( \frac{\theta _1}{t}\right) ^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}}\right) }, s_{t\left( 1-\left( 1- \frac{\psi _1}{t}\right) ^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}}\right) }, s_{t\left( 1-\left( 1- \frac{\sigma _1}{t}\right) ^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}}\right) }\right\rangle \\&\quad \otimes \left\langle s_{t\left( \left( \frac{\theta _2}{t}\right) ^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}}\right) }, s_{t\left( 1-\left( 1- \frac{\psi _2}{t}\right) ^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}}\right) }, s_{t\left( 1-\left( 1- \frac{\sigma _2}{t}\right) ^{\frac{T_{1}}{\sum _{i=1}^{2}T_{i}}}\right) }\right\rangle \\&\quad = \left\langle s_{t\left( \prod \limits _{i=1}^{2}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{2}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{2}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \right\rangle \end{aligned}$$

  Hence result is true for $n=2.$

• Step 2 Assume the Eq. (5) is true for $n=z,$ then for $n=z+1,$ we have

  $$\begin{aligned}&\bigotimes \limits _{i=1}^{z+1} \beta _{i}^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}=\bigotimes \limits _{i=1}^{z} \beta _{i}^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}} }\otimes \beta _{z+1}^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\\&\quad = \left\langle s_{t\left( \prod \limits _{i=1}^{z}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) } \right\rangle \\&\quad \otimes \left\langle s_{t\left( \left( \frac{\theta _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( 1-\left( 1- \frac{\psi _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( 1-\left( 1- \frac{\sigma _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }\right\rangle \\&\quad = \left\langle s_{t\left( \prod \limits _{i=1}^{z}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}} \left( \frac{\theta _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, \right. \\&\quad s_{t\left( \left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) + \left( 1-\left( 1- \frac{\psi _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) - \left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) \left( 1-\left( 1- \frac{\psi _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) \right) }, \\&\left. \quad s_{t\left( \left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) + \left( 1-\left( 1- \frac{\sigma _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) - \left( 1-\prod \limits _{i=1}^{z}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) \left( 1-\left( 1- \frac{\sigma _{z+1}}{t}\right) ^{\frac{T_{z+1}}{\sum _{i=1}^{z+1}T_{i}}}\right) \right) }\right\rangle \\&\quad = \left\langle s_{t\left( \prod \limits _{i=1}^{z+1}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{z+1}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{z+1}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{z+1}T_{i}}}\right) } \right\rangle \end{aligned}$$

Hence the result given in Eq. (5), is true for all positive integers n.

On the other hand, we have, $0 \le \frac{\theta _i}{t} \le 1 \Rightarrow 0 \le \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le 1$ which implies $0 \le t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le t.$ Similarly, we get

$$\begin{aligned}&0 \le t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le t \\ \text {and}&0 \le t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le t \end{aligned}$$

Also, $0 \le t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) + t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) + t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \le 3t.$ Hence, the LSVNPWG operator is LSVNN. $\square$

Proof of the Theorem 11

Proof

• (P1) Since $\beta _i=\beta$ for all i, then we have

  $$\begin{aligned} \text {LSVNPWG}(\beta _1,\beta _2,\ldots ,\beta _n)= & {} \bigotimes \limits _{i=1}^{n}\beta ^{\frac{T_i}{\sum _{i=1}^n T_i}}\\= & {} \beta ^{\frac{\sum _{i=1}^{n}T_i}{\sum _{i=1}^n T_i}} \\= & {} \beta \end{aligned}$$

• (P2) Since $s_{\theta _i^{\prime }} \ge s_{\theta _i}$ for all i which implies

  $$\begin{aligned}&\theta _i^{\prime } \ge \theta _i \\&\quad \Rightarrow \left( \frac{\theta _i^{\prime }}{t} \right) \ge \left( \frac{\theta _i}{t}\right) \\&\quad \Rightarrow \prod \limits _{i=1}^{n}\left( \frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}} \ge \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}} \\&\quad \Rightarrow t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) \ge t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) \end{aligned}$$

  Thus, one has

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i^{\prime }}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \ge s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  (16)

  Similarly, for $s_{\psi _i^{\prime }} \le s_{\psi _i}$ and $s_{\sigma _i^{\prime }} \le s_{\sigma _i}$, we can get

  $$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\psi _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) } \le s_{t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) } \end{aligned}$$

  (17)

  and

  $$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\sigma _i^{\prime }}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) } \le s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_i}{\sum _{i=1}^{n} T_i}}\right) } \end{aligned}$$

  (18)

  Thus, from Eqs. (16), (17), (18) and Definition 6, we have

  $$\begin{aligned} \text {LSVNPWG}(\beta _1^{\prime }, \beta _2^{\prime },\ldots , \beta _n^{\prime }) \ge \text {LSVNPWG}(\beta _1,\beta _2,\ldots ,\beta _n) \end{aligned}$$

• (P3) Since $\min \limits _{i}\{s_{\theta _i}\} \le s_{\theta _i} \le \max \limits _{i}\{s_{\theta _i}\}$, $\min \limits _{i}\{s_{\psi _i}\} \le s_{\psi _i} \le \max \limits _{i}\{s_{\psi _i}\}$ and $\min \limits _{i}\{s_{\sigma _i}\} \le s_{\sigma _i} \le \max \limits _{i}\{s_{\sigma _i}\}$, thus from above property, we have

  $$\begin{aligned} \beta ^- \le \text {LSVNPWG}(\beta _1,\beta _2,\ldots ,\beta _n)\le \beta ^+ \end{aligned}$$

$\square$

Proof of the Theorem 14

Proof

Let $\beta _i= \langle s_{\theta _i}, s_{\psi _i}, s_{\sigma _i} \rangle$, $(i=1,2,\ldots ,n)$ be the collection of LSVNNs and $0 \le 1-(\theta _i/t) \le 1.$ Then by using lemma 1, we have

$$\begin{aligned}&\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le \sum \limits _{i=1}^{n} {\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \left( 1-\frac{\theta _i}{t}\right) \\&\quad \Rightarrow \prod _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le \sum _{i=1}^{n} {\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}- \sum _{i=1}^{n} {\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \left( \frac{\theta _i}{t}\right) \\&\quad \Rightarrow 1-\prod _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \ge \sum \limits _{i=1}^n \frac{T_{i}}{\sum _{i=1}^{n}T_{i}}\left( \frac{\theta _i}{t}\right) \\&\quad \Rightarrow 1-\prod _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \ge \prod _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}. \end{aligned}$$

Thus, we have

$$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) }\ge s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

(19)

Similarly, we get

$$\begin{aligned}&s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \le s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \end{aligned}$$

(20)

$$\begin{aligned} \text {and} \ &s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \le s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \end{aligned}$$

(21)

Thus, from Eqs. (19), (20), (21) and Definition 6, we get

$$\begin{aligned}&\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } , s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \right\rangle \\&\quad \ge \left\langle s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \right\rangle . \end{aligned}$$

Hence,

$$\begin{aligned} \text {LSVNPWA}(\beta _{1},\beta _{2},\ldots ,\beta _{n}) \ge \text {LSVNPWG}(\beta _{1},\beta _{2},\ldots ,\beta _{n}) \end{aligned}$$

$\square$

Proof of the Theorem 15

Proof

We will prove the parts (i) and (iii) and the proofs of remaining are similar.

• (i) Since $\beta _i= \langle s_{\theta _i}, s_{\psi _i}, s_{\sigma _i} \rangle$, $(i=1,2,\ldots ,n)$, $\beta =\langle s_{\theta }, s_{\psi }, s_{\sigma } \rangle$ are LSVNNs and $\lambda>0$, be a real number, then

  $$\begin{aligned}&\text {LSVNPWA}(\beta _{1}\oplus \beta ,\beta _{2}\oplus \beta ,\ldots ,\beta _{n}\oplus \beta )\\&\quad = \left\langle s_{t\left( 1-\prod \limits _{i=1}^{n} \left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t \left( \prod \limits _{i=1}^{n} \left( \frac{\psi _i\psi }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, \right. \\&\left. \qquad \;s_{t \left( \prod \limits _{i=1}^{n} \left( \frac{\sigma _i\sigma }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }\right\rangle \\ \text {and LSVNPWA}&(\beta _{1}\otimes \beta ,\beta _{2}\otimes \beta ,\ldots ,\beta _{n}\otimes \beta )\\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i\theta }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t \left( \prod \limits _{i=1}^{n} \left( 1-\left( 1-\frac{\psi _i}{t}\right) \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) },\right. \\&\left. \qquad \;s_{t \left( \prod \limits _{i=1}^{n} \left( 1-\left( 1-\frac{\sigma _i}{t}\right) \left( 1-\frac{\sigma }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }\right\rangle \\ \end{aligned}$$

  By using Lemma 2, we have

  $$\begin{aligned} {\theta _{i}}/{t} +{\theta }/{t}-({\theta _{i}}/{t})({\theta }/{t})\ge ({\theta _{i}}/{t})({\theta }/{t}) \end{aligned}$$

  which implies

  $$\begin{aligned}&1-\theta _i/t - \theta /t+ \theta _i\theta /t^2 \le 1-({\theta _{i}}/{t})({\theta }/{t}) \\&\quad \Rightarrow \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \le 1-\frac{\theta _i\theta }{t^2} \\&\quad \Rightarrow \prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le \prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i\theta }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \\&\quad \Rightarrow t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \\&\quad \ge t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i\theta }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \end{aligned}$$

  Thus,

  $$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \ge s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i\theta }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \end{aligned}$$

  Also, we have $1-\left( 1-{\psi _{i}}/{t}\right) \left( 1-{\psi }/{t}\right) \ge {\psi _{i}\psi }/{t^2}$, which implies,

  $$\begin{aligned} t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\frac{\psi _i}{t}\right) \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \ge t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i\psi }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \end{aligned}$$

  Thus, we get

  $$\begin{aligned} s_{t\left(\prod \limits _{i=1}^{n}\left( 1- \left( 1-\frac{\psi _i}{t}\right) \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \ge s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi _i\psi }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  Similarly, we get

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left(1- \left( 1-\frac{\sigma _i}{t}\right) \left( 1-\frac{\sigma }{t}\right) \right) ^{ \frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \ge s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma _i\sigma }{t^2}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  Therefore, by using Definition 6, we get

  $$\begin{aligned} \text {LSVNPWA}(\beta _{1}\oplus \beta ,\beta _{2}\oplus \beta ,\ldots ,\beta _{n}\oplus \beta )\ge \text {LSVNPWA}(\beta _{1}\otimes \beta ,\beta _{2}\otimes \beta ,\ldots ,\beta _{n}\otimes \beta ) \end{aligned}$$

  Hence the result.

• (iii) By using Lemma 2, we have

  $$\begin{aligned} \left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) + \left( \frac{\theta }{t}\right) -\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) \right) \left( \frac{\theta }{t}\right) \\ \ge \left( 1-\prod \limits _{i=1}^{n}\left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \left( \frac{\theta }{t}\right) \end{aligned}$$

  Therefore,

  $$\begin{aligned} &1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\ge {} \left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \left( \frac{\theta }{t}\right) \\&\Rightarrow {} t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \\&\ge {} t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \left( \frac{\theta }{t}\right) \end{aligned}$$

  Thus, we have,

  $$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \ge s_{t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\frac{\theta _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \left( \frac{\theta }{t}\right) } \end{aligned}$$

  Also, we have

  $$\begin{aligned}&\prod \limits _{i=1}^{n} \left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}+ \left( \frac{\psi }{t}\right) -\prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) \left( \frac{\psi }{t}\right) \ge \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \left( \frac{\psi }{t}\right) \\&\quad \Rightarrow 1-\left( 1-\prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \left( 1-\frac{\psi }{t}\right) \ge \prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \left( \frac{\psi }{t}\right) \\&\quad \Rightarrow t\left( 1-\left( 1-\prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \left( 1-\frac{\psi }{t}\right) \right) \ge t\left( \prod \limits _{i=1}^{n} \left( \frac{\psi _i\psi }{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \end{aligned}$$

  Hence,

  $$\begin{aligned} s_{t\left( 1-\left( 1-\prod \limits _{i=1}^{n}\left( \frac{\psi _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \left( 1-\frac{\psi }{t}\right) \right) } \ge s_{t\left( \prod \limits _{i=1}^{n} \left( \frac{\psi _i\psi }{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \end{aligned}$$

  Similarly, we can get

  $$\begin{aligned} s_{t\left( 1-\left( 1-\prod \limits _{i=1}^{n}\left( \frac{\sigma _i}{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) \left( 1-\frac{\sigma }{t}\right) \right) } \ge s_{t\left( \prod \limits _{i=1}^{n} \left( \frac{\sigma _i\sigma }{t}\right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) } \end{aligned}$$

  Hence, by Definition 6, we get

  $$\begin{aligned} \text {LSVNPWA}(\beta _{1}\oplus \beta ,\beta _{2}\oplus \beta ,\ldots ,\beta _{n}\oplus \beta )\ge \text {LSVNPWA}(\beta _{1},\beta _{2},\ldots ,\beta _{n})\otimes \beta . \end{aligned}$$

$\square$

Proof of the Theorem 16

Proof

We will prove part (i) and (iii) and the proofs of remaining parts are similar.

• (i) Since $\beta _{i}(i=1,2,\ldots ,n)$ and $\beta$ are the LSVNNs then for $\lambda>0$ we have

  $$\begin{aligned}&\text {LSVNPWA}(\lambda \beta _{1}\oplus \beta ,\lambda \beta _{2}\oplus \beta ,\ldots ,\lambda \beta _{n}\oplus \beta ) \\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) ^{\lambda }\left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \right) }, s_{t \left( \prod \limits _{i=1}^{n}\left( \left( \frac{\psi _i}{t}\right) ^{\lambda } \left( \frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, \right. \\&\left. \qquad \;s_{t \left( \prod \limits _{i=1}^{n}\left( \left( \frac{\sigma _i}{t}\right) ^{\lambda } \left( \frac{\sigma }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \right\rangle \\&\text {LSVNPWA}(\beta _{1}^{\lambda }\otimes \beta ,\beta _{2}^{\lambda }\otimes \beta ,\ldots ,\beta _{n}^{\lambda }\otimes \beta )\\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\left( \frac{\theta _i}{t}\right) ^{\lambda } \left( \frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }, s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\frac{\psi _i}{t}\right) ^{\lambda } \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) },\right. \\&\left. \qquad \; s_{t\left( \prod \limits _{i=1}^{n} \left( 1-\left( 1-\frac{\sigma _i}{t}\right) ^{\lambda } \left( 1-\frac{\sigma }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) }\right\rangle \end{aligned}$$

  By Lemma 3, we have

  $$\begin{aligned}&1-\left( 1-{\theta _{i}}/t\right) ^{\lambda }\left( 1-{\theta }/t\right) \ge \left( {\theta _{i}}/t\right) ^{\lambda }\left( {\theta }/t\right) \\&\quad \Rightarrow \left( 1-{\theta _{i}}/t\right) ^{\lambda }\left( 1-{\theta }/t\right) \le 1-\left( {\theta _{i}}/t\right) ^{\lambda }\left( {\theta }/t\right) \\&\quad \Rightarrow \prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) ^{\lambda }\left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le \prod \limits _{i=1}^{n}\left( 1-\left( \frac{\theta _i}{t}\right) ^{\lambda } \left( \frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \\&\quad \Rightarrow t \left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) ^{\lambda } \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \\&\qquad \ge t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\left( \frac{\theta _i}{t}\right) ^{\lambda } \left( \frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) \end{aligned}$$

  Thus,

  $$\begin{aligned} s_{t \left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) ^{\lambda } \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \ge s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\left( \frac{\theta _i}{t}\right) ^{\lambda } \left( \frac{\theta }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  Also, we have

  $$\begin{aligned}&\left( \frac{\psi _i}{t}\right) ^{\lambda } \left( \frac{\psi }{t}\right) \le 1-\left( 1-\frac{\psi _i}{t}\right) ^{\lambda } \left( 1-\frac{\psi }{t}\right) \\&\quad \Rightarrow t\prod \limits _{i=1}^{n}\left( \left( \frac{\psi _i}{t}\right) ^{\lambda } \left( \frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \le t\prod \limits _{i=1}^{n}\left( 1-\left( 1-\frac{\psi _i}{t}\right) ^{\lambda } \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}} \end{aligned}$$

  Hence,

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \left( \frac{\psi _i}{t}\right) ^{\lambda } \left( \frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\frac{\psi _i}{t}\right) ^{\lambda } \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  Similarly, we have

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \left( \frac{\sigma _i}{t}\right) ^{\lambda } \left( \frac{\sigma }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\frac{\sigma _i}{t}\right) ^{\lambda } \left( 1-\frac{\sigma }{t}\right) \right) ^{\frac{T_{i}}{\sum _{i=1}^{n}T_{i}}}\right) } \end{aligned}$$

  Therefore, by Definition 6, we get

  $$\begin{aligned} \text {LSVNPWA}(\lambda \beta _{1}\oplus \beta ,\lambda \beta _{2}\oplus \beta ,\ldots ,\lambda \beta _{n}\oplus \beta )\ge \text {LSVNPWA}(\beta _{1}^{\lambda }\otimes \beta ,\beta _{2}^{\lambda }\otimes \beta ,\ldots ,\beta _{n}^{\lambda }\otimes \beta ) \end{aligned}$$

  Hence the result.

• (iii) Since $\beta _{i}(i=1,2,\ldots ,n)$ and $\beta$ are the LSVNNs then for $\lambda>0$ we have

  $$\begin{aligned}&\text {LSVNPWA}(\beta _{1}^{\lambda }\oplus \beta ,\beta _{2}^{\lambda }\oplus \beta ,\ldots ,\beta _{n}^{\lambda }\oplus \beta )\\&\quad = \left\langle s_{t\left( 1-\prod \limits _{i=1}^{n} \left( \left( 1-\left( \frac{\theta _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }, s_{t\left( \prod \limits _{i=1}^{n}\left( \left( 1-\left( 1-\frac{\psi _i}{t}\right) ^{\lambda } \right) \left( \frac{\psi }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) },\right. \\&\left. \qquad \;s_{t\left( \prod \limits _{i=1}^{n}\left( \left( 1-\left( 1-\frac{\sigma _i}{t}\right) ^{\lambda }\right) \left( \frac{\sigma }{t}\right) \right) ^{ \frac{T_i}{\sum _{i=1}^n T_i}}\right) } \right\rangle \\&\text {LSVNPWA}(\lambda \beta _{1}\otimes \beta ,\lambda \beta _{2}\otimes \beta ,\ldots ,\lambda \beta _{n}\otimes \beta )\\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( 1-\frac{\theta _i}{t}\right) ^{\lambda }\right) \left( \frac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }, s_{t \left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \frac{\psi _i}{t}\right) ^{\lambda } \right) \left( 1-\frac{\psi }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \right. , \\&\left. \qquad \; s_{t \left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \frac{\sigma _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\sigma }{t}\right) \right) ^{ \frac{T_i}{\sum _{i=1}^n T_i}}\right) } \right\rangle \end{aligned}$$

  From Lemma 3, $\big (1-(\theta _{i}/t)\big )^{\lambda } \big (\theta /t\big )\ge -\big (\theta _{i}/t\big )^{\lambda }\big (1-(\theta /t)\big ),$ which implies, $1-(\theta /t)+\big (1-(\theta _{i}/t)\big )^{\lambda } \big (\theta /t\big )\ge 1-(\theta /t)-\big (\theta _{i}/t\big )^{\lambda }\big (1-(\theta /t)\big )$ $\Rightarrow 1-(\theta /t)\big (1-\big (1-(\theta _{i}/t)\big )^{\lambda }\big ) \ge \big (1-(\theta /t)\big )\big (1-(\theta _{i}/t)^{\lambda }\big ) \Rightarrow$ $\prod \limits _{i=1}^{n}\bigg (1-(\theta /t)\big (1-\big (1-(\theta _{i}/t)\big )^{\lambda }\big )\bigg )^{\frac{T_i}{\sum _{i=1}^n T_i}} \ge \prod \limits _{i=1}^{n}\bigg (\big (1-(\theta /t)\big )\big (1-(\theta _{i}/t)^{\lambda }\big )\bigg )^{\frac{T_i}{\sum _{i=1}^n T_i}} \Rightarrow$ $1-\prod \limits _{i=1}^{n}\bigg (1-(\theta /t)\big (1-\big (1-(\theta _{i}/t)\big )^{\lambda }\big )\bigg )^{\frac{T_i}{\sum _{i=1}^n T_i}} \le 1-\prod \limits _{i=1}^{n}\bigg (\big (1-(\theta /t)\big )\big (1-(\theta _{i}/t)^{\lambda }\big )\bigg )^{\frac{T_i}{\sum _{i=1}^n T_i}}$, which implies

  $$\begin{aligned}&t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta }{t}\right) \left( 1-\left( \frac{\theta _i}{t}\right) ^{ \lambda }\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) \\&\quad \ge t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\left( \frac{\theta }{t}\right) \left( 1-\left( 1-\frac{\theta _i}{t}\right) ^{\lambda } \right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) \end{aligned}$$

  Thus,

  $$\begin{aligned} s_{t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta }{t}\right) \left( 1-\left( \frac{\theta _i}{t}\right) ^{ \lambda }\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \ge s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\left( \frac{\theta }{t}\right) \left( 1-\left( 1-\frac{\theta _i}{t}\right) ^{\lambda } \right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \end{aligned}$$

  Also, we have $\big (1-\psi _{i}/t\big )^{\lambda }\big (\psi /t\big )\ge -\big (\psi _{i}/t\big )^{\lambda }\big (1-(\psi /t)\big ),$ which implies, $\big (\psi /t\big )-\big (1-\psi _{i}/t\big )^{\lambda }\big (\psi /t\big )\le \big (\psi /t\big )+\big (\psi _{i}/t\big )^{\lambda }\big (1-(\psi /t)\big )$ $\Rightarrow$ $\left( \tfrac{\psi }{t}\right) \left( 1-\big (1-\tfrac{\psi }{t}\big )^{\lambda }\right) \le 1-\left( 1-\left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\psi }{t}\right)$ implies,

  $$\begin{aligned}&t\prod \limits _{i=1}^{n}\left( \frac{\psi }{t} \left( 1-\left( 1-\frac{\psi }{t}\right) ^{\lambda }\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}} \\&\quad \le t\prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \frac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\psi }{t} \right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}} \end{aligned}$$

  Thus,

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\psi }{t} \left( 1-\left( 1-\frac{\psi }{t}\right) ^{\lambda }\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \frac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\psi }{t} \right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \end{aligned}$$

  Similarly, we have

  $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( \frac{\sigma }{t} \left( 1-\left( 1-\frac{\sigma }{t}\right) ^{\lambda }\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \frac{\sigma _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\sigma }{t} \right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \end{aligned}$$

  Hence, we get

  $$\begin{aligned} \text {LSVNPWA}(\beta _{1}^{\lambda }\oplus \beta ,\beta _{2}^{\lambda }\oplus \beta ,\ldots ,\beta _{n}^{\lambda }\oplus \beta )\ge \text {LSVNPWA}(\lambda \beta _{1}\otimes \beta ,\lambda \beta _{2}\otimes \beta ,\ldots ,\lambda \beta _{n}\otimes \beta ) \end{aligned}$$

$\square$

Proof of the Theorem 17

Proof

We will prove part (i) and remaining parts are similar.

• (i) For any $\beta _i= \langle s_{\theta _i}, s_{\psi _i}, s_{\sigma _i} \rangle$ and $\beta = \langle s_{\theta }, s_{\psi }, s_{\sigma } \rangle ,$ we have

  $$\begin{aligned}&\text {LSVNPWA}(\lambda \beta _{1}\otimes \beta ,\lambda \beta _{2}\otimes \beta ,\ldots ,\lambda \beta _{n}\otimes \beta )\\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\left( 1-\left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\theta }{t}\right) \right) ^{{\frac{T_i}{\sum _{i=1}^n T_i}}}\right) }, \right. \\&\qquad \;s_{t\left( \prod \limits _{i=1}^{n} \left( 1-\left( 1-\left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\psi }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }, \\&\left. \qquad \;s_{t \left( \prod \limits _{i=1}^{n} \left( 1-\left( 1-\left( \tfrac{\sigma _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\sigma }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \right\rangle ;\\&\text {LSVNPWA}(\lambda \beta _{1},\lambda \beta _{2},\ldots ,\lambda \beta _{n})\\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }, s_{t \left( \prod \limits _{i=1}^{n}\left( \left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }, \right. \\&\left. \qquad \;s_{t \left( \prod \limits _{i=1}^{n}\left( \left( \tfrac{\sigma _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }\right\rangle ;\\ \text {and }&\text {LSVNPWA}(\lambda \beta _{1}\oplus \beta ,\lambda \beta _{2}\oplus \beta ,\ldots ,\lambda \beta _{n}\oplus \beta )\\&\quad =\left\langle s_{t\left( 1-\prod \limits _{i=1}^{n} \left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\left( 1-\tfrac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }, s_{t \left( \prod \limits _{i=1}^{n} \left( \left( \tfrac{\psi _i}{t}\right) ^{\lambda } \left( \tfrac{\psi }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) },\right. \\&\left. \qquad \;s_{t\left( \prod \limits _{i=1}^{n} \left( \left( \tfrac{\sigma _i}{t}\right) ^{\lambda } \left( \tfrac{\sigma }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }\right\rangle \\ \end{aligned}$$

  We shall prove this part in two steps:

  • Step 1 We first prove $\text {LSVNPWA}(\lambda \beta _{1}\otimes \beta ,\lambda \beta _{2}\otimes \beta ,\ldots , \lambda \beta _{n}\otimes \beta )\le \text {LSVNPWA}(\lambda \beta _{1},\lambda \beta _{2},\ldots ,\lambda \beta _{n})$.

    Since, $0 \le 1-\left( 1-{\theta _{i}}/{t}\right) ^{\lambda }\le 1, 0 \le \left( {\theta }/{t}\right) \le 1$, then based on the Lemma 4, we have

    $$\begin{aligned}&1-\big (1-\big (1-{\theta _{i}}/{t}\big )^{\lambda }\big )\big (1-{\theta }/{t}\big ) \ge \big (1-{\theta _{i}}/{t}\big )^{\lambda } \nonumber \\&\quad \Rightarrow \prod \limits _{i=1}^{n}\left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}} \ge \prod \limits _{i=1}^{n}\left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}} \nonumber \\&\quad \Rightarrow t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\left( 1-\left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) \nonumber \\&\qquad \le t\left( 1-\prod \limits _{i=1}^{n} \left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) \nonumber \\&\quad \Rightarrow s_{t\left( 1-\prod \limits _{i=1}^{n} \left( 1-\left( 1-\left( 1-\frac{\theta _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \le s_{t\left( 1-\prod \limits _{i=1}^{n}\left( \left( 1-\frac{\theta _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \end{aligned}$$

    (22)

    Now, we have

    $$\begin{aligned}&1-\left( 1-\left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\psi }{t}\right) \ge \left( \tfrac{\psi _i}{t}\right) ^{\lambda }\nonumber \\&\quad \Rightarrow t\prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\psi }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\ge t \prod \limits _{i=1}^{n} \left( \left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}} \nonumber \\&\quad \Rightarrow s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) \left( 1-\tfrac{\psi }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \ge s_{t\left( \prod \limits _{i=1}^{n} \left( \left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \nonumber \\ \end{aligned}$$

    (23)

    Similarly,

    $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n}\left( 1-\left( 1-\left( \frac{\sigma _i}{t}\right) ^{\lambda }\right) \left( 1-\frac{\sigma }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }\ge s_{t \left( \prod \limits _{i=1}^{n} \left( \left( \frac{\sigma _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \nonumber \\ \end{aligned}$$

    (24)

    Therefore, from Eqs. (22), (23) and (24), we get $\text {LSVNPWA}(\lambda \beta _{1}\otimes \beta ,\lambda \beta _{2}\otimes \beta ,\ldots ,\lambda \beta _{n}\otimes \beta )\le \text {LSVNPWA}(\lambda \beta _{1},\lambda \beta _{2},\ldots ,\lambda \beta _{n})$

  • Step 2 Now, we prove $\text {LSVNPWA}(\lambda \beta _{1},\lambda \beta _{2},\ldots ,\lambda \beta _{n})\le \text {LSVNPWA}(\lambda \beta _{1}\oplus \beta ,\lambda \beta _{2}\oplus \beta ,\ldots ,\lambda \beta _{n}\oplus \beta )$. Since, $0\le \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\le 1$, $0\le \left( 1-\tfrac{\theta }{t}\right) \le 1$, therefore we have

    $$\begin{aligned}&\left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\left( 1-\tfrac{\theta }{t}\right) \le \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\nonumber \\&\quad \Rightarrow \prod \limits _{i=1}^{n}\left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda }\left( 1-\tfrac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\le \prod \limits _{i=1}^{n}\left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda {\frac{T_i}{\sum _{i=1}^n T_i}}}\nonumber \\&\quad \Rightarrow t\left( 1- \prod \limits _{i=1}^{n} \left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda } \left( 1-\tfrac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) \ge t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda {\frac{T_i}{\sum _{i=1}^n T_i}}}\right) \nonumber \\&\quad \Rightarrow s_{t\left( 1- \prod \limits _{i=1}^{n}\left( \left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda } \left( 1-\tfrac{\theta }{t}\right) \right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) }\ge s_{t\left( 1-\prod \limits _{i=1}^{n}\left( 1-\tfrac{\theta _i}{t}\right) ^{\lambda {\frac{T_i}{\sum _{i=1}^n T_i}}}\right) } \end{aligned}$$

    (25)

    Similarly, we get

    $$\begin{aligned}&s_{t\left( \prod \limits _{i=1}^{n}\left( \tfrac{\psi _i}{t}\right) ^{\lambda }\left( \tfrac{\psi }{t}\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \le s_{t\left( \prod \limits _{i=1}^{n} \left( \left( \tfrac{\psi _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \end{aligned}$$

    (26)

    and

    $$\begin{aligned} s_{t\left( \prod \limits _{i=1}^{n} \left( \tfrac{\sigma _i}{t}\right) ^{\lambda } \left( \tfrac{\sigma }{t}\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \le s_{t \left( \prod \limits _{i=1}^{n} \left( \left( \tfrac{\sigma _i}{t}\right) ^{\lambda }\right) ^{\frac{T_i}{\sum _{i=1}^n T_i}}\right) } \end{aligned}$$

    (27)

    Therefore, from Eqs. (25), (26) and (27), we get $\text {LSVNPWA}(\lambda \beta _{1},\lambda \beta _{2},\ldots ,\lambda \beta _{n})\le \text {LSVNPWA}(\lambda \beta _{1}\oplus \beta ,\lambda \beta _{2}\oplus \beta ,\ldots ,\lambda \beta _{n}\oplus \beta )$

  Hence, based on these steps, we get the result.

$\square$