Neutrality operations-based Pythagorean fuzzy aggregation operators and its applications to multiple attribute group decision-making process

Abstract

Pythagorean fuzzy sets accommodate more uncertainties than the intuitionistic fuzzy sets and hence it is one of the most important concepts to describe the fuzzy information in the process of decision making. Under this environment, the main objective of the work is to develop some new operational laws and their corresponding weighted aggregation operators. For it, we define some new neutral addition and scalar multiplication operational laws by incorporating the features of a neutral character towards the membership degrees of the set and the probability sum. Some properties of the proposed laws are investigated. Then, associated with these operational laws, we define some novel Pythagorean fuzzy weighted, ordered weighted and hybrid neutral averaging aggregation operators for Pythagorean fuzzy information, which can neutrally treat the membership and non-membership degrees. The various relations and the characteristics of the proposed operators are discussed. Further, in order to ease with the possible application, we present an algorithm to solve the multiple attribute group decision-making problems under the Pythagorean fuzzy environment. Finally, a practical example is provided to illustrate the approach and show its superiority, advantages by comparing their performance with some several existing approaches.

Appendix

Proof of Proposition 2

Proof

We proof the result by using principle of mathematical induction (PMI) on $\lambda$. For PFN ${\mathcal {A}}=( \zeta _{\mathcal {A}}, \vartheta _{\mathcal {A}})$, the following steps of the induction are executed.

• Step 1: For $\lambda = 2$ and by using Eq. (8), we have

  $$\begin{aligned}&\text {PS}\left( \lambda \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) \\&\quad = \text {PS}\left( \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}, \text {PS}\left( \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) \right) \\&\quad = \text {PS}\left( \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}, \sqrt{1-\pi _{\mathcal {A}}^{2}}\right) \\&\quad = \sqrt{1-(1-\zeta _{\mathcal {A}}^{2}-\vartheta _{\mathcal {A}}^{2})(1-1+\pi _{\mathcal {A}}^{2})} \\&\quad = \sqrt{1-(\pi _{\mathcal {A}}^{2})^2} \end{aligned}$$

  Thus, result is true for $\lambda =2$.

• Step 2: Assume that result holds for $\lambda =n$, then for $\lambda =n+1$, we have

  $$\begin{aligned}&\text {PS}\left( (n+1)\sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) \\&\quad = \text {PS}\left( \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}, \text {PS}\left( n\sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) \right) \\&\quad = \text {PS}\left( \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}, \sqrt{1-(\pi _{\mathcal {A}}^{2})^{n}}\right) \\&\quad = \sqrt{1-(1-\zeta _{\mathcal {A}}^{2}-\vartheta _{\mathcal {A}}^{2})(1-1+(\pi _{\mathcal {A}}^{2})^{n})} \\&\quad = \sqrt{1-(\pi _{\mathcal {A}}^{2})^{n+1}} \end{aligned}$$

  which is true for $\lambda =n+1$. Hence, by PMI, Eq. (9) true for all $\lambda$.

$\square$

Proof of Theorem 2.

Proof

• (i) It can be easily obtained from Eq. (12).

• (ii) For PFNs ${\mathcal {A}}$ and ${\mathcal {B}}$, and by Eq. (7), we get

  $$\begin{aligned}&\lambda {\mathcal {A}} \oplus \lambda {\mathcal {B}} \\&\quad = \left( \begin{aligned} \sqrt{\frac{\text {MCS}^{2}(\lambda {\mathcal {A}}, \lambda {\mathcal {B}})}{\text {MCS}^{2}(\lambda {\mathcal {A}}, \lambda {\mathcal {B}}) + \text {NCS}^{2}(\lambda {\mathcal {A}}, \lambda {\mathcal {B}})} } \cdot \text {PS}\left( \text {PS}\left( \lambda \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) , \text {PS}\left( \lambda \sqrt{\zeta _{\mathcal {B}}^{2}+\vartheta _{\mathcal {B}}^{2}}\right) \right) , \\ \sqrt{\frac{\text {NCS}^{2}(\lambda {\mathcal {A}}, \lambda {\mathcal {B}})}{\text {MCS}^{2}(\lambda {\mathcal {A}}, \lambda {\mathcal {B}}) + \text {NCS}^{2}(\lambda {\mathcal {A}}, \lambda {\mathcal {B}})} } \cdot \text {PS}\left( \text {PS}\left( \lambda \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) , \text {PS}\left( \lambda \sqrt{\zeta _{\mathcal {B}}^{2}+\vartheta _{\mathcal {B}}^{2}}\right) \right) \end{aligned}\right) \\&\quad = \left( \begin{aligned} \sqrt{\frac{\lambda (\zeta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2})}{\lambda (\zeta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2}) + \lambda (\vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2})} } \cdot \text {PS}\left( \sqrt{1-(\pi _{\mathcal {A}}^{2})^\lambda }, \sqrt{1-(\pi _{\mathcal {B}}^{2})^\lambda }\right) , \\ \sqrt{\frac{\lambda (\vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2})}{\lambda (\zeta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2}) + \lambda (\vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2})} } \cdot \text {PS}\left( \sqrt{1-(\pi _{\mathcal {A}}^{2})^{\lambda }}, \sqrt{1-(\pi _{\mathcal {B}}^{2})^{\lambda }}\right) \end{aligned} \right) \nonumber \\&\quad = \left( \begin{aligned} \sqrt{\frac{\zeta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2}}{ \zeta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2} + \vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2}} \left( 1-(\pi _{\mathcal {A}}^{2})^\lambda (\pi _{\mathcal {B}}^{2})^\lambda \right) }, \\ \sqrt{\frac{\vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2}}{\zeta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2} + \vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2}} \left( 1-(\pi _{\mathcal {A}}^{2})^{\lambda }(\pi _{\mathcal {B}}^{2})^{\lambda }\right) } \end{aligned} \right) \\&\quad = \left( \begin{aligned} \sqrt{\frac{\zeta _{\mathcal {A}}^{2} + \zeta _{\mathcal {B}}^{2}}{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2}+\vartheta _{\mathcal {B}}^{2}}}\cdot \text {PS}\left( \lambda \left( \sqrt{1-\pi _{\mathcal {A}}^{2}\pi _{\mathcal {B}}^{2}}\right) \right) , \\ \sqrt{\frac{\vartheta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {B}}^{2}}{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}+\zeta _{\mathcal {B}}^{2}+\vartheta _{\mathcal {B}}^{2}}} \cdot PS\left( \lambda \left( \sqrt{1-\pi _{\mathcal {A}}^{2}\pi _{\mathcal {B}}^{2}}\right) \right) \end{aligned}\right) \\&\quad = \lambda ({\mathcal {A}} \oplus {\mathcal {B}}) \end{aligned}$$

• (iii) For two positive real numbers $\lambda _1$ and $\lambda _2$, we have

  $$\begin{aligned}&\lambda _1 {\mathcal {A}} \oplus \lambda _2 {\mathcal {A}} \\&\quad = \left( \begin{aligned} \sqrt{\frac{\text {MCS}^{2}(\lambda _1 {\mathcal {A}}, \lambda _2 {\mathcal {A}})}{\text {MCS}^{2}(\lambda _1 {\mathcal {A}}, \lambda _2 {\mathcal {A}}) + \text {NCS}^{2}(\lambda _1 {\mathcal {A}}, \lambda _2 {\mathcal {A}})} } \cdot \text {PS}\left( PS\left( \lambda _1 \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) , \text {PS}\left( \lambda _2 \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) \right) , \\ \sqrt{\frac{\text {NCS}^{2}(\lambda _1 {\mathcal {A}}, \lambda _2 {\mathcal {A}})}{\text {MCS}^{2}(\lambda _1 {\mathcal {A}}, \lambda _2 {\mathcal {A}}) + \text {NCS}^{2}(\lambda _1A, \lambda _2 {\mathcal {A}})} } \cdot \text {PS} \left( \text {PS}\left( \lambda _1 \sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) , \text {PS}\left( \lambda _2\sqrt{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}}\right) \right) \end{aligned} \right) \\&\quad = \left( \begin{aligned} \sqrt{\frac{(\lambda _1+\lambda _2) \zeta _{\mathcal {A}}^{2}}{(\lambda _1+\lambda _2) \zeta _{\mathcal {A}}^{2} + (\lambda _1+\lambda _2) \vartheta _{\mathcal {A}}^{2}} } \cdot \text {PS}\left( \sqrt{1-(\pi _{\mathcal {A}}^{2})^{\lambda _1}}, \sqrt{1-(\pi _{\mathcal {A}}^{2})^{\lambda _2}}\right) , \\ \sqrt{\frac{(\lambda _1+\lambda _2)\vartheta _{\mathcal {A}}^{2}}{(\lambda _1+\lambda _2) \zeta _{\mathcal {A}}^{2} + (\lambda _1+\lambda _2)\vartheta _{\mathcal {A}}^{2}} } \cdot \text {PS} \left( \sqrt{1-(\pi _{\mathcal {A}}^{2})^{\lambda _1}}, \sqrt{1-(\pi _{\mathcal {A}}^{2})^{\lambda _2}}\right) \end{aligned} \right) \\&\quad = \left( \sqrt{\frac{\zeta _{\mathcal {A}}^{2}}{ \zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}} \cdot \left( 1-(\pi _{\mathcal {A}}^{2})^{\lambda _1+\lambda _2}\right) }, \sqrt{\frac{\vartheta _{\mathcal {A}}^{2}}{\zeta _{\mathcal {A}}^{2}+\vartheta _{\mathcal {A}}^{2}} \cdot \left( 1-(\pi _{\mathcal {A}}^{2})^{\lambda _1+\lambda _2} \right) } \right) \\&\quad = (\lambda _1+\lambda _2){\mathcal {A}} \end{aligned}$$

$\square$

Proof of Theorem 3

Proof

For PFNs ${\mathcal {A}}_i(i=1,2,\ldots ,n)$ and real numbers $\omega _i>0$, the first result holds immediately from Theorem 1. Now, in order to show Eq. (15) holds, we follows the steps of the PMI on n which are summarized as follows:

• Step 1: For $n=1$, we have ${\mathcal {A}}_i=( \zeta _i, \vartheta _i)$ and $\omega _i=1$. Thus, we can write as

  $$\begin{aligned} \text {PFWNA}({\mathcal {A}}_1)= & {} \omega _1{\mathcal {A}}_1 \\= & {} \big ( \zeta _{1},\vartheta _{1}\big ) \\= & {} \left( \sqrt{\frac{\omega _1 \zeta _{1}^{2}}{ \omega _1(\zeta _{1}^{2}+\vartheta _{1}^{2})} \left( 1-\left( \pi _1^{2}\right) ^{\omega _1}\right) }, \sqrt{\frac{\omega _1 \vartheta _{1}^{2}}{\omega _1(\zeta _{1}^{2}+\vartheta _{1}^{2})}\left( 1-\left( \pi _1^{2}\right) ^{\omega _1}\right) }\right) \end{aligned}$$

  Thus, Eq.(15) holds.

• Step 2: Assume that the Eq.(15) holds for $n=k$, that is

  $$\begin{aligned}&\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_k) \\&\quad = \left( \begin{aligned} \sqrt{\frac{\sum \nolimits _{i=1}^k \omega _i(\zeta _{i}^{2})}{\sum \nolimits _{i=1}^k \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})} \cdot \left( 1-\prod \limits _{i=1}^k \left( \pi _{i}^{2}\right) ^{\omega _i}\right) }, \sqrt{\frac{\sum \nolimits _{i=1}^k \omega _i \vartheta _{i}^{2}}{\sum \nolimits _{i=1}^k \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})} \cdot \left( 1-\prod \limits _{i=1}^k\left( \pi _{i}^{2}\right) ^{\omega _i}\right) } \end{aligned}\right) \end{aligned}$$

  Now, for $n=k+1$, we have

  $$\begin{aligned}&\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_{k+1}) \\&\quad = \text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_k) \oplus (\omega _{k+1}{\mathcal {A}}_{k+1}) \\&\quad = \left( \begin{aligned} \sqrt{\frac{\text {MCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_k),\omega _{k+1}{\mathcal {A}}_{k+1})}{ \left( \begin{aligned} & \text {MCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k),\omega _{k+1}{\mathcal {A}}_{k+1}) \\ & + \text {NCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k),\omega _{k+1}{\mathcal {A}}_{k+1}) \end{aligned} \right) } } \cdot \text {PS}\left( \sqrt{1-\prod \limits _{i=1}^k(\pi _{i}^{2})^{\omega _i}}, \sqrt{1-(\pi _{k+1}^{2})^{\omega _{k+1}}} \right) , \\ \sqrt{\frac{\text {NCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2\ldots ,{\mathcal {A}}_k), \omega _{k+1}{\mathcal {A}}_{k+1})}{ \left( \begin{aligned} \text {MCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k), \omega _{k+1}{\mathcal {A}}_{k+1}) \\ + \text {NCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k), \omega _{k+1}{\mathcal {A}}_{k+1}) \end{aligned} \right) }} \cdot \text {PS}\left( \sqrt{1-\prod \limits _{i=1}^k(\pi _i^{2})^{\omega _i}}, \sqrt{1-(\pi _{k+1}^{2})^{\omega _{k+1}}}\right) \end{aligned} \right) \end{aligned}$$

  By definition of MCS and NCS, we have $\text {MCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k), \omega _{k+1}{\mathcal {A}}_{k+1}) = \text {MCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k)) + \text {MCS}^{2}(\omega _{k+1}{\mathcal {A}}_{k+1})$$=\sum \nolimits _{i=1}^k \omega _i \zeta _i^{2} + \omega _{k+1}\zeta _{k+1}^{2}$$=\sum \limits _{i=1}^{k+1} \omega _i \zeta _i^{2}$. Similarly, we get

  $$\begin{aligned}&\text {NCS}^{2}(\text {PFWNA}({\mathcal {A}}_1,\ldots ,{\mathcal {A}}_k), \omega _{k+1}{\mathcal {A}}_{k+1}) = \sum \limits _{i=1}^{k+1} \omega _i \vartheta _i^{2} \end{aligned}$$

  Also by definition of PS, we have

  $$\begin{aligned}&\text {PS}\left( \sqrt{1-\prod \limits _{i=1}^k (\pi _i^{2})^{\omega _i}}, \sqrt{1-(\pi _{k+1}^{2})^{\omega _{k+1}}} \right) \\&\quad = \sqrt{1-\left( 1-1+\prod \limits _{i=1}^k (\pi _i^{2})^{\omega _i}\right) \left( 1-1+(\pi _{k+1}^{2})^{\omega _{k+1}}\right) } \\&\quad = \sqrt{1-\prod \limits _{i=1}^{k+1} (\pi _i^{2})^{\omega _i}} \end{aligned}$$

  Thus,

  $$\begin{aligned}&\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_{k+1}) \\&\quad = \left( \begin{aligned} \sqrt{\frac{\sum \nolimits _{i=1}^{k+1} \omega _i \zeta _{i}^{2} }{\sum \nolimits _{i=1}^{k+1} \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})} \cdot \left( 1-\prod \limits _{i=1}^{k+1}(\pi _{i}^{2})^{\omega _i}\right) }, \sqrt{\frac{\sum \nolimits _{i=1}^{k+1} \omega _i \vartheta _{i}^{2}}{\sum \nolimits _{i=1}^{k+1} \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})}\cdot \left( 1-\prod \limits _{i=1}^{k+1}(\pi _{i}^{2})^{\omega _i}\right) } \end{aligned} \right) \end{aligned}$$

  i.e., Eq. (15) holds for $n=k+1$. Therefore, by PMI, Eq.(15) holds for all n, which completes the theorem.

$\square$

Proof of Theorem 4

Proof

For a collection of PFNs ${\mathcal {A}}_i=( \zeta _i, \vartheta _i)$ and ${\mathcal {A}}_0=( \zeta _0, \vartheta _0)$ such that ${\mathcal {A}}_i = {\mathcal {A}}_0$ we have $\zeta _i = \zeta _0$ and $\vartheta _i = \vartheta _0$ for all i. Then, by Eq. (15) and by weight vector $\omega _i>0$ with $\sum \nolimits _{i=1}^n \omega _i=1$, we have

$$\begin{aligned}&\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_n) \\&\quad = \left( \begin{aligned} \sqrt{\frac{\sum \nolimits _{i=1}^{n} \omega _i \zeta _{0}^{2} }{\sum \nolimits _{i=1}^{n} \omega _i(\zeta _{0}^{2}+\vartheta _{0}^{2})} \cdot \left( 1-\prod \limits _{i=1}^{n}\left( \pi _{0}^{2}\right) ^{\omega _i}\right) }, \sqrt{\frac{\sum \nolimits _{i=1}^{n} \omega _i \vartheta _{0}^{2}}{\sum \nolimits _{i=1}^{n} \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})}\cdot \left( 1-\prod \limits _{i=1}^{n}\left( \pi _{0}^{2}\right) ^{\omega _i}\right) } \end{aligned} \right) \\&\quad = \left( \sqrt{\frac{\zeta _{0}^{2}}{\zeta _{0}^{2}+\vartheta _{0}^{2}} \cdot \left( 1-(\pi _{0}^{2})^{\sum \nolimits _{i=1}^{n}\omega _i}\right) }, \sqrt{\frac{\vartheta _{0}^{2}}{\zeta _{0}^{2}+\vartheta _{0}^{2}} \cdot \left( 1-(\pi _{0}^{2})^{\sum \nolimits _{i=1}^{n}\omega _i}\right) } \right) \\&\quad = \left( \sqrt{\frac{\zeta _{0}^{2}}{\zeta _{0}^{2}+\vartheta _{0}^{2}} \cdot (\zeta _0^{2}+\vartheta _0^{2})}, \sqrt{\frac{\vartheta _{0}^{2}}{\zeta _{0}^{2}+\vartheta _{0}^{2}} \cdot (\zeta _0^{2}+\vartheta _0^{2})} \right) \\&\quad = {\mathcal {A}}_0 \end{aligned}$$

$\square$

Proof of Theorem 5.

Proof

For a collection of PFNs ${\mathcal {A}}_i=( \zeta _i, \vartheta _i) (i=1,2,\ldots ,n)$, we have

• (i)
  $$\begin{aligned}&\min \limits _i\big \{\zeta _{i}^{2} + \vartheta _{i}^{2} \big \} = 1-\left( 1-\min \limits _i\left\{ \zeta _{i}^{2} + \vartheta _{i}^{2} \right\} \right) ^{\sum \limits _{i=1}^n \omega _i}\\&\quad = 1-\prod \limits _{i=1}^n \left( 1-\min \limits _i\left\{ \zeta _{i}^{2} + \vartheta _{i}^{2} \right\} \right) ^{\omega _i} \le 1 - \prod \limits _{i=1}^n \left( 1-\zeta _{i}^{2} - \vartheta _{i}^{2} \right) ^{\omega _i} \\&\quad \le 1 - \prod \limits _{i=1}^n \left( 1-\max \limits _i\left\{ \zeta _{i}^{2} + \vartheta _{i}^{2} \right\} \right) ^{\omega _i} \\&\quad = 1 - \left( 1-\max \limits _i\left\{ \zeta _{i}^{2} + \vartheta _{i}^{2} \right\} \right) ^{\sum \limits _{i=1}^n \omega _i} = \max \limits _i\left\{ \zeta _{i}^{2} + \vartheta _{i}^{2} \right\} \end{aligned}$$

  Thus, we have

  $$\begin{aligned}&\min \limits _i\big \{\zeta _{i}^{2} + \vartheta _{i}^{2} \big \} \le 1 - \prod \limits _{i=1}^n \left( 1-\zeta _{i}^{2} - \vartheta _{i}^{2} \right) ^{\omega _i} \\&\quad \le \max \limits _i\left\{ \zeta _{i}^{2} + \vartheta _{i}^{2} \right\} \end{aligned}$$

  Now, by Theorem 15, we get

  $$\begin{aligned} \zeta _P= & {} \sqrt{\frac{\sum \nolimits _{i=1}^n \omega _i(\zeta _{i}^{2})}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})} \cdot \left( 1-\prod \limits _{i=1}^n\left( \pi _{i}^{2}\right) ^{\omega _i}\right) } \quad \text { and } \\ \vartheta _P= & {} \sqrt{\frac{\sum \nolimits _{i=1}^n \omega _i \vartheta _{i}^{2}}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{i}^{2}+\vartheta _{i}^{2})} \cdot \left( 1-\prod \limits _{i=1}^n\left( \pi _{i}^{2}\right) ^{\omega _i}\right) } \end{aligned}$$

  Therefore,

  $$\begin{aligned} \zeta _P^{2} + \vartheta _P^{2} = 1 - \prod \limits _{i=1}^n \left( 1-\zeta _{i}^{2} - \vartheta _{i}^{2} \right) ^{\omega _i} \end{aligned}$$

  Hence, we get $\min \limits _i\big \{\zeta _{i}^{2}+\vartheta _{i}^{2}\big \} \le \zeta _{P}^{2}+ \vartheta _{P}^{2} \le \max \limits _i\big \{\zeta _{i}^{2} + \vartheta _{i}^{2} \big \}$.

• (ii) Since, $\zeta _i\ge \min \limits _i \{\zeta _i\}$, so by expression of $\zeta _P$ we have

  $$\begin{aligned} \zeta _P^{2}\ge & {} \frac{\sum \nolimits _{i=1}^n \omega _i(\min \limits _i \{\zeta _{i}^{2}\})}{\sum \nolimits _{i=1}^n \omega _i(\max \limits _i \{\zeta _{i}^{2}+\vartheta _{i}^{2}\})} \left[ 1-\prod \limits _{i=1}^n(1-\min \limits _i \{\zeta _i^{2}+\vartheta _i^{2}\})^{\omega _i} \right] \\= & {} \frac{\min \limits _i \{\zeta _{i}^{2}\}}{\max \limits _i \{\zeta _{i}^{2}+\vartheta _{i}^{2}\}} \left[ 1-(1-\min \limits _i \{\zeta _i^{2}+\vartheta _i^{2}\})^{\sum \nolimits _{i=1}^n \omega _i}\right] \\= & {} \frac{\min \limits _i \{\zeta _i^{2}+\vartheta _i^{2}\} \min \limits _i \{\zeta _{i}^{2}\}}{\max \limits _i \{\zeta _{i}^{2}+\vartheta _{i}^{2}\}} \end{aligned}$$

  Moreover,

  $$\begin{aligned} \zeta _P^{2}\le & {} \frac{\sum \nolimits _{i=1}^n \omega _i(\max \limits _i \{\zeta _{i}^{2}\})}{\sum \nolimits _{i=1}^n \omega _i(\min \limits _i \{\zeta _{i}^{2}+\vartheta _{i}^{2}\})} \left[ 1-\prod \limits _{i=1}^n(1-\max \limits _i \{\zeta _i^{2}+\vartheta _i^{2}\})^{\omega _i} \right] \\= & {} \frac{\max \limits _i \{\zeta _{i}^{2}\}}{\min \limits _i \{\zeta _{i}^{2}+\vartheta _{i}^{2}\}} \left[ 1-(1-\max \limits _i \{\zeta _i^{2}+\vartheta _i^{2}\})^{\sum \nolimits _{i=1}^n \omega _i}\right] \\= & {} \frac{\max \limits _i \{\zeta _i^{2}+\vartheta _i^{2}\} \max \limits _i \{\zeta _{i}^{2}\}}{\min \limits _i \{\zeta _{i}^{2}+\vartheta _{i}^{2}\}} \end{aligned}$$

  Also, by definition of PFN and Theorem 3, we get $\zeta _p^{2}\le 1$. Thus, we have

  $$\begin{aligned}&\frac{\min \limits _i\big \{\zeta _{i}^{2}+\vartheta _{i}^{2}\big \}\cdot \min \limits _i\big \{\zeta _{i}^{2}\big \}}{\max \limits _i\big \{\zeta _{i}^{2}+\vartheta _{i}^{2}\big \}} \le \zeta _P^{2} \\&\quad \le \min \limits _i\Bigg \{\frac{\max \limits _i \big \{\zeta _{i}^{2}+\vartheta _{i}^{2}\big \}\cdot \max \limits _i\big \{\zeta _{i}^{2}\big \}}{\min \limits _i\big \{\zeta _{i}^{2} + \vartheta _{i}^{2}\big \}},1\Bigg \} \end{aligned}$$

• (iii) As similar to part (ii), we can obtain it. So, we omit here.

$\square$

Proof of Theorem 6

Proof

For a collection of PFNs ${\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_n$ and ${\mathcal {B}}_1,{\mathcal {B}}_2,\ldots ,{\mathcal {B}}_n$ and by using Theorem 3, we get $\text {PFWNA}$$({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_n) = ( \zeta _{p_A}, \vartheta _{P_{\mathcal {A}}})$ and $\text {PFWNA}({\mathcal {B}}_1,{\mathcal {B}}_2,\ldots ,{\mathcal {B}}_n) = ( \zeta _{p_B}, \vartheta _{P_{\mathcal {B}}})$ where

$$\begin{aligned} \zeta _{P_{\mathcal {A}}}^{2}= & {} \frac{\sum \nolimits _{i=1}^n \omega _i \zeta _{{\mathcal {A}}_i}^{2}}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {A}}_i}^{2}+\vartheta _{{\mathcal {A}}_i}^{2})}\left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {A}}_i}^{2}-\vartheta _{{\mathcal {A}}_i}^{2})^{\omega _i}\right] ; \\ \vartheta _{P_{\mathcal {A}}}^{2} = & {} \frac{\sum \nolimits _{i=1}^n \omega _i \vartheta _{{\mathcal {A}}_i}^{2}}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {A}}_i}^{2}+\vartheta _{{\mathcal {A}}_i}^{2})} \left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {A}}_i}^{2}-\vartheta _{{\mathcal {A}}_i}^{2})^{\omega _i}\right] ; \\ \zeta _{P_{\mathcal {B}}}^{2}= & {} \frac{\sum \nolimits _{i=1}^n \omega _i \zeta _{{\mathcal {B}}_i}^{2} }{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {B}}_i}^{2}+\vartheta _{{\mathcal {B}}_i}^{2})} \left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {B}}_i}^{2}-\vartheta _{{\mathcal {B}}_i}^{2})^{\omega _i}\right] ; \text {and } \\ \vartheta _{P_{\mathcal {B}}}^{2}= & {} \frac{\sum \nolimits _{i=1}^n \omega _i \vartheta _{{\mathcal {B}}_i}^{2} }{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {B}}_i}^{2}+\vartheta _{{\mathcal {B}}_i}^{2})}\left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {B}}_i}^{2}-\vartheta _{{\mathcal {B}}_i}^{2})^{\omega _i}\right] . \end{aligned}$$

Based on these information, we have

• (i) If $\zeta _{{\mathcal {A}}_i}^{2} + \vartheta _{{\mathcal {A}}_i}^{2} \le \zeta _{{\mathcal {B}}_i}^{2} + \vartheta _{{\mathcal {B}}_i}^{2}$, then we have $\zeta _{P_{\mathcal {A}}}^{2} + \vartheta _{P_{\mathcal {A}}}^{2} \le 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {A}}_i}^{2}-\vartheta _{{\mathcal {A}}_i}^{2})^{\omega _i}$$\le$$1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {B}}_i}^{2}-\vartheta _{{\mathcal {B}}_i}^{2})^{\omega _i}$$=\zeta _{P_{\mathcal {B}}}^{2} + \vartheta _{P_{\mathcal {B}}}^{2}$.

• (ii) If $\zeta _{{\mathcal {A}}_i}^{2} + \vartheta _{{\mathcal {A}}_i}^{2} = \zeta _{{\mathcal {B}}_i}^{2} + \vartheta _{{\mathcal {B}}_i}^{2}$, and $\zeta _{{\mathcal {A}}_i}\le \zeta _{{\mathcal {B}}_i}$, then we have

  $$\begin{aligned} \zeta _{P_{\mathcal {A}}}^{2}= & {} \frac{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {A}}_i}^{2})}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {A}}_i}^{2}+\vartheta _{{\mathcal {A}}_i}^{2})}\left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {A}}_i}^{2}-\vartheta _{{\mathcal {A}}_i}^{2})^{\omega _i}\right] \\\le & {} \frac{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {B}}_i}^{2})}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {B}}_i}^{2}+\vartheta _{{\mathcal {B}}_i}^{2})}\left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {B}}_i}^{2}-\vartheta _{{\mathcal {B}}_i}^{2})^{\omega _i}\right] \\= & {} \zeta _{P_{\mathcal {B}}}^{2} \end{aligned}$$

  and

  $$\begin{aligned} \vartheta _{P_{\mathcal {A}}}^{2}= & {} \frac{\sum \nolimits _{i=1}^n \omega _i(\vartheta _{{\mathcal {A}}_i}^{2})}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {A}}_i}^{2}+\vartheta _{{\mathcal {A}}_i}^{2})}\left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {A}}_i}^{2}-\vartheta _{{\mathcal {A}}_i}^{2})^{\omega _i}\right] \\\ge & {} \frac{\sum \nolimits _{i=1}^n \omega _i(\vartheta _{{\mathcal {B}}_i}^{2})}{\sum \nolimits _{i=1}^n \omega _i(\zeta _{{\mathcal {B}}_i}^{2}+\vartheta _{{\mathcal {B}}_i}^{2})}\left[ 1-\prod \limits _{i=1}^n(1-\zeta _{{\mathcal {B}}_i}^{2}-\vartheta _{{\mathcal {B}}_i}^{2})^{\omega _i}\right] \\= & {} \vartheta _{P_{\mathcal {B}}}^{2} \end{aligned}$$

  Hence, the result.

• (iii) From part (ii), we obtain that $\zeta _{P_{\mathcal {A}}}^{2} \le \zeta _{P_{\mathcal {B}}}^{2}$ and $\vartheta _{P_{\mathcal {A}}}^{2} \ge \vartheta _{P_{\mathcal {B}}}^{2}$. Therefore, by definition of score function given in Eq. (2), we get $\zeta _{P_{\mathcal {A}}}^{2} -\vartheta _{P_{\mathcal {A}}}^{2} \le \zeta _{P_{\mathcal {B}}}^{2} - \vartheta _{P_{\mathcal {B}}}^{2}$. Hence, based on an order relation between PFNs, we have $\text {PFWNA}({\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_n) \le \text {PFWNA}({\mathcal {B}}_1,{\mathcal {B}}_2,\ldots ,{\mathcal {B}}_n)$.

$\square$