Pythagorean Fuzzy Hamy Mean Operators in Multiple Attribute Group Decision Making and Their Application to Supplier Selection

Abstract

In this paper, we extend the Hamy mean (HM) operator and dual Hamy mean (DHM) operator with Pythagorean fuzzy numbers (PFNs) to propose Pythagorean fuzzy Hamy mean (PFHM) operator, weighted Pythagorean fuzzy Hamy mean (WPFHM) operator, Pythagorean fuzzy dual Hamy mean (PFDHM) operator, weighted Pythagorean fuzzy dual Hamy mean (WPFDHM) operator. Then the multiple attribute group decision making (MAGDM) methods are proposed with these operators. In the end, we utilize an applicable example for supplier selection to prove the proposed methods.

1. Introduction

Pythagorean fuzzy set (PFS) [1,2] has been designed with the membership degree and the non-membership degree, whose sum of squares is less than or equal to 1. Zhang and Xu [3] provided TOPSIS for MADM with PFNs. Peng and Yang [4] gave the superiority and inferiority ranking method to analyze the MAGDM with PFNs. Reformat and Yager [5] designed the recommender system with PFNs. Gou et al. [6] studied some characteristic of continuous PFNs. Garg [7] proposed some Einstein operators with PFNs. Zeng et al. [8] defined the hybrid model for MADM with PFNs. Wei [9] defined some novel interaction operators between PFNs. Gao et al. [10] defined some novel interaction operators with PFNs in MADM. Ren et al. [11] expanded TODIM for MADM with PFNs. Wei and Lu [12] extended MSM operator [13] to PFNs. Wu and Wei [14] proposed Hamacher operators with PFNs. Wei and Wei [15] defined some similarity measures between PFNs with cosine function [16,17,18]. Xue et al. [19] developed the LINMAP method with PFNs. Wei and Lu [20] developed the power operators with PFNs. Wan et al. [21] used the mathematical programming to solve MAGDM with PFNs. Baloglu and Demir [22] developed the agent-based methods for demand analysis with PFNs. Liang [23] proposed some Bonferroni mean operators with PFNs based on the traditional Bonferroni mean operators [24,25,26,27,28]. Mandal and Ranadive [29] proposed the decision-theoretic rough sets with PFNs. Chen [30] gave the outranking method under PFNs with a closeness-based assignment model. Garg [31] proposed the linear programming model for MADM with interval-valued Pythagorean fuzzy numbers (IVPFNs). Khan et al. [32] extend TOPSIS model with IVPFNs. Garg [33] defined the exponential operational laws of IVPFNs. Li and Zeng [34] gave some distance measure of PFNs. Gao [35] defined some hamacher prioritized operators in MADM with traditional prioritized aggregation operators [36,37,38,39,40,41]. Wei and Lu [42] developed some Hamacher operators for aggregating dual hesitant PFNs. Lu et al. [43] proposed some hamacher operators with hesitant PFNs and Wei et al. [44] proposed some Pythagorean hesitant fuzzy hamacher operators based on the traditional hamacher operators [45,46,47]. Wei et al. [48], Tang and Wei [49] and Huang and Wei [50] proposed the Pythagorean 2-tuple linguistic operators in MADM with the traditional arithmetic and geometric operators [51,52,53,54,55,56]. Wei et al. [57] proposed some q-Rung Orthopair fuzzy Heronian mean operators in MADM.

And HM operator [58] and DHM operator [59] are famous operators which depict interrelationships among any number of arguments assigned by a variable vector. Therefore, the HM and DHM operators can furnish a robust and flexible mechanism to solve the information fusion in MAGDM problems. Because PFNs can easily capture the fuzzy information and the HM can describe interrelationships among any number of arguments assigned by a variable vector, it is necessary to expand the HM and DHM operators to deal with the PFNs. Thus, how to fuse these PFNs with HM and DHM operators is an interesting topic. In order to accomplish this goal, the remainder of this paper is arranged as follows. In the next section, we introduce some basic concepts related to PFNs. In Section 3, we propose some HM and DHM operators with PFNs. In Section 4, we present some methods for MAGDM problems with PFWHM and PFWDHM operator. In Section 5, we give a numerical example. Finally, a brief conclusion is given in Section 6.

2. Preliminaries

2.1. Pythagorean Fuzzy Sets

Yager [1,2] gave the definition of PFSs.

Definition 1.

Let $X$ be a fixed set. A PFS in X is defined as follows [1,2]

$$\mathrm{P}=\left\{\langle x,\left({\mu }_p(x),{\nu }_p(x)\right)\rangle |x\in X\right\}$$

(1)

where ${\mu }_p(x)\in \left[0,1\right]$ and ${\nu }_p(x\in \left[0,1\right]$ are defined as the degree of membership and non-membership of the element $x\in X$ to $P$, respectively, and satisfying

$${\left({\mu }_p(x)\right)}^2+{\left({\nu }_p(x)\right)}^2\le 1.$$

(2)

Definition 2.

Let $\tilde{a}=(\mu ,\nu )$ be a PFN, then the score function of $\tilde{a}$ is defined [17]

$$S(\tilde{a})=\frac{1}{2}\left(1+{\mu }^2-{\nu }^2\right),S(\tilde{a})\in [0,1].$$

(3)

Definition 3.

Let $\tilde{a}=(\mu ,\nu )$ be a PFN, then the accuracy function of $\tilde{a}$ is defined [11]

$$H(\tilde{a})={\mu }^2+{\nu }^2,H(\tilde{a})\in [0,1].$$

(4)

Definition 4.

Let ${\tilde{a}}_1=({\mu }_1,{\nu }_1)$ and ${\tilde{a}}_2=({\mu }_2,{\nu }_2)$ be two PFNs [17], $S({\tilde{a}}_1)=\frac{1}{2}\left(1+{({\mu }_1)}^2-{({\nu }_2)}^2\right)$ and $S({\tilde{a}}_2)=\frac{1}{2}\left(1+{({\mu }_2)}^2-{({\nu }_2)}^2\right)$ be the scores function of ${\tilde{a}}_1$ and ${\tilde{a}}_2,$ respectively, and let $H({\tilde{a}}_1)={\mu }_1{}^2+{\nu }_1{}^2$ and $H({\tilde{a}}_2)={\mu }_2{}^2+{\nu }_2{}^2$ be the accuracy function of ${\tilde{a}}_1$ and ${\tilde{a}}_2,$ respectively, then if $S({\tilde{a}}_1)<S({\tilde{a}}_2)$, then ${\tilde{a}}_1<{\tilde{a}}_2$ if $S({\tilde{a}}_1)=S({\tilde{a}}_2)$, then

(1) 

If $H({\tilde{a}}_1)=H({\tilde{a}}_2),$ then ${\tilde{a}}_1={\tilde{a}}_2;$

(2) 

If $H({\tilde{a}}_1)<H({\tilde{a}}_2),$ ${\tilde{a}}_1<{\tilde{a}}_2.$

Example 1.

Let ${\tilde{a}}_1=\left(0.5,0.3\right),{\tilde{a}}_2=\left(0.6,0.2\right),{\tilde{a}}_3=\left(0.4,0\right)$, according to Definitions 2–4, we get

$$S\left({\tilde{a}}_1\right)=\frac{1}{2}\left(1+{0.5}^2-{0.3}^2\right)=0.5800,S\left({\tilde{a}}_2\right)=\frac{1}{2}\left(1+{0.6}^2-{0.2}^2\right)=0.6600$$

$$S\left({\tilde{a}}_3\right)=\frac{1}{2}\left(1+{0.4}^2-0^2\right)=0.5800,H\left({\tilde{a}}_1\right)={0.5}^2+{0.3}^2=0.3400$$

$$H\left({\tilde{a}}_2\right)={0.6}^2+{0.2}^2=0.4000,H\left({\tilde{a}}_3\right)={0.4}^2+0^2=0.1600$$

Then we can conclude that ${\tilde{a}}_2>{\tilde{a}}_1>{\tilde{a}}_3$.

Definition 5.

Let ${\tilde{a}}_1=({\mu }_1,{\nu }_1),$ ${\tilde{a}}_2=({\mu }_2,{\nu }_2)$ and $\tilde{a}=(\mu ,\nu )$ be three PFNs, and some basic operations are defined [5]:

(1) 

${\tilde{a}}_1\oplus {\tilde{a}}_2=\left(\sqrt{{({\mu }_1)}^2+{({\mu }_2)}^2-{({\mu }_1)}^2{({\mu }_2)}^2},{\nu }_1{\nu }_2\right);$

(2) 

${\tilde{a}}_1\otimes {\tilde{a}}_2=\left({\mu }_1{\mu }_2,\sqrt{{({\nu }_1)}^2+{({\nu }_2)}^2-{({\nu }_1)}^2{({\nu }_2)}^2}\right);$

(3) 

$\lambda \tilde{a}=\left(\sqrt{1-{(1-\mu )}^{\lambda }},{\nu }^{\lambda }\right),\lambda >0;$

(4) 

${(\tilde{a})}^{\lambda }=\left({\mu }^{\lambda },\sqrt{1-{(1-\nu )}^{\lambda }}\right),\lambda >0;$

(5) 

${\tilde{a}}^c=(\nu ,\mu ).$

Example 2.

Suppose that ${\tilde{a}}_1=\left(0.2,0.3\right),{\tilde{a}}_2=\left(0.6,0.1\right),$ and $\lambda =5$, then we have

(1) 

${\tilde{a}}_1\oplus {\tilde{a}}_2=\left(\sqrt{{0.2}^2+{0.6}^2-{0.2}^2\times {0.6}^2},0.3\times 0.1\right)=\left(0.5456,0.0300\right)$

(2) 

${\tilde{a}}_1\otimes {\tilde{a}}_2=\left(0.2\times 0.6,\sqrt{{0.3}^2+{0.1}^2-{0.3}^2\times {0.1}^2}\right)=\left(0.1200,0.3091\right)$

(3) 

$\lambda {\tilde{a}}_1=\left(\sqrt{1-{(1-0.2)}^{0.5}},{0.3}^{0.5}\right)=\left(0.1056,0.5477\right)$

(4) 

${({\tilde{a}}_1)}^{\lambda }=\left({0.2}^{0.5},\sqrt{1-{(1-0.3)}^{0.5}}\right)=\left(0.4472,0.1633\right)$

(5) 

${\tilde{a}}_1{}^c=(0.3,0.2)$

The following properties are derived based on the Definition 4.

Definition 6.

Let ${\tilde{a}}_1=({\mu }_1,{\nu }_1)and{\tilde{a}}_2=({\mu }_2,{\nu }_2)$ be two PFNs, $\lambda ,{\lambda }_1,{\lambda }_2>0,$ then [5]

(1) 

${\tilde{a}}_1\oplus {\tilde{a}}_2={\tilde{a}}_2\oplus {\tilde{a}}_1$;

(2) 

${\tilde{a}}_1\otimes {\tilde{a}}_2={\tilde{a}}_2\otimes {\tilde{a}}_1$;

(3) 

$\lambda \left({\tilde{a}}_1\oplus {\tilde{a}}_2\right)=\lambda {\tilde{a}}_1\oplus \lambda {\tilde{a}}_2$;

(4) 

${({\tilde{a}}_1\otimes {\tilde{a}}_2)}^{\lambda }={({\tilde{a}}_1)}^{\lambda }\otimes {({\tilde{a}}_2)}^{\lambda };$

(5) 

${\lambda }_1{\tilde{a}}_1\otimes {\lambda }_2{\tilde{a}}_1=({\lambda }_1+{\lambda }_2){\tilde{a}}_1;$

(6) 

${({\tilde{a}}_1)}^{{\lambda }_1}\otimes {({\tilde{a}}_1)}^{{\lambda }_2}={\left({\tilde{a}}_1\right)}^{{\lambda }_1+{\lambda }_2};$

(7) 

${\left({\left({\tilde{a}}_1\right)}^{{\lambda }_1}\right)}^{{\lambda }_2}={\left({\tilde{a}}_1\right)}^{{\lambda }_1{\lambda }_2}$.

2.2. HM Operator

Definition 7.

The HM operator is defined as follows [58]:

$${\mathrm{HM}}^{\left(\mathrm{x}\right)}\left(a_1,a_2,\cdots a_k\right)=\frac{{{\displaystyle \sum _{1\le i_1<\cdots <i_x\le k}\left({\displaystyle \prod _{j=1}^x{a}_{i_j}}\right)}}^{\frac{1}{x}}}{C_k^x}$$

(5)

where $x$ is a parameter and $x=1,2,\cdots ,k,i_1,i_2,\cdots i_x$ are $x$ integer values taken from the set $\{1,2,\cdots ,k\}$ of $k$ integer values, $C_k^x$ denotes the binomial coefficient and $C_k^x=\frac{k!}{x!(k-x)!}$.

The HM operator has three properties.

(i)

when $a_i=a(i=1,2,\cdots ,k),{\mathrm{HM}}^{(x)}\left(a_1,a_2,\cdots a_k\right)=a;$

(ii)

when $a_i\le {\pi }_i(i=1,2,\cdots ,k),{\mathrm{HM}}^{(x)}\left(a_1,a_2,\cdots a_k\right)\le {\mathrm{HM}}^{\left(\mathrm{x}\right)}\left({\pi }_1,{\pi }_2,\cdots {\pi }_k\right);$

(iii)

when $\min \{a_i\}\le {\mathrm{HM}}^{\left(\mathrm{x}\right)}\left(a_1,a_2,\cdots a_k\right)\le \max \{a_i\}.$

Two particular cases of the HM operator are given as follows.

(1)

When $x=1$,${\mathrm{HM}}^{\left(\mathrm{k}\right)}\left(a_1,a_2,\cdots a_k\right)=\frac{1}{k}{\displaystyle \sum _{i=1}^k{a}_i},$ it becomes the arithmetic mean operator.

(2)

When $x=k,{\mathrm{HM}}^{\left(\mathrm{k}\right)}\left(a_1,a_2,\cdots a_k\right)={\left({\displaystyle \prod _{i=1}^k{a}_i}\right)}^{\frac{1}{k}},$ it becomes the geometric mean operator.

3. The HM Operators for PFNs

In this part, we will combine PFNs and HM operator, and propose the PFHM operator and WPFHM operator.

3.1. The PFHM Operator

Definition 8.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a set of PFNs, then we can define PFHM operator as follows:

$${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_k^x}$$

(6)

where $x$ is a parameter and $x=1,2,\cdots ,k,i_1,i_2,\cdots i_x$ are $x$ integer values taken from the set $\{1,2,\cdots ,k\}$ of $k$ integer values, $C_k^x$ denotes the binomial coefficient and $C_k^x=\frac{k!}{x!(k-x)!}$.

Theorem 1.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a set of the PFNs, then the aggregate result of Definition 8 is still a PFNs, and have

$$\begin{array}{l}{\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{1}{C_k^x}\left(\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}},{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)\end{array}$$

(7)

Proof. 

(1)

First of all, we prove (7) is kept.

$$\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}=\left({\displaystyle \prod _{j=1}^x({\mu }_{i_j}}),\sqrt{1-{\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}}\right)$$

(8)

Therefore,

$${\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}=\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}},\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)$$

(9)

Moreover,

$$\begin{array}{l}\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}\\ =\left(\sqrt{1-{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}},{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)\end{array}$$

(10)

Furthermore,

$$\begin{array}{l}{\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{1}{C_k^x}\left(\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}},{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)\end{array}$$

(11)

(2)

Next, we prove (7) is a PFN. Let

$$\begin{array}{l}p=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}},\\ q={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\end{array}$$

Then we prove the following two conditions. (i) $0\le p\le 1,0\le q\le 1;$ (ii) $0\le p^2+q^2\le 1.$

• Since ${\mu }_{i_j}\in [0,1],$ we can get

  $${\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\in [0,1]\Rightarrow {\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\in [0,1]\Rightarrow {\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\in [0,1]\Rightarrow 1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\in [0,1]$$

  $$\Rightarrow {\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\in [0,1]\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}\in [0,1]$$

  $$\Rightarrow \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\in [0,1]\mathrm{i}.\mathrm{e}.,0\le p\le 1.$$
  Similarly, we can get

  $${\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j-1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\in [0,1],\mathrm{i}.\mathrm{e}.,0\le q\le 1.$$
• Obviously, $0\le p^2+q^2\le 1,$ then

  $$\begin{array}{l}{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2\\ \le 1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left({\nu }_{i_j}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_k^x}}+{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_k^x}}=1\end{array}$$

We get $0\le p^2+q^2\le 1.$□

So the aggregated result of Definition 8 is still PFN. Next we will talk about some properties of PFHM operator.

Property 1.

(Idempotency). If ${\tilde{a}}_i(1,2,\cdots ,k)$ and $\tilde{a}$ are PFNs, and ${\tilde{a}}_i=\tilde{a}=({\mu }_i,{\nu }_i)$ for all $i=1,2,\cdots ,k$, then we get

$${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\tilde{a}$$

(12)

Proof. 

Since $\tilde{a}=\left(\mu ,\nu \right)$, based on Theorem 1, we have

$$\begin{array}{l}{\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)\end{array}$$

$$\begin{array}{l}=\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\mu }_i{}^2\right)}\right)}^{\frac{1}{C_k^x}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-\left(1-{({\nu }_i)}^2\right)}}\right)}^{\frac{1}{C_k^x}}\right)\\ =\left(\sqrt{1-\left(1-{\mu }_i{}^2\right),}\sqrt{1-\left(1-{({\nu }_i)}^2\right)}\right)=\left(\sqrt{1-\left(1-{\mu }^2\right),}\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)\right)=\left(\mu ,\nu \right)=\tilde{a}\end{array}$$

□

Property 2.

(Monotonicity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. If ${\mu }_{i_j}\ge {\mu }_{{\theta }_j},{\nu }_{i_j}\le {\nu }_{{\theta }_j}$ for all $j$, then

$${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(13)

Proof. 

Since $x\ge 1,{\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ then

$${\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\ge {\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\Rightarrow 1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\le 1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2$$

$$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}\le {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}$$

$$\Rightarrow \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\ge \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}$$

Similarly, we have

$$1-{({\nu }_{i_j})}^2\ge 1-{({\pi }_{{\theta }_j})}^2\Rightarrow {\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\ge {\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}$$

$$\Rightarrow 1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\le 1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}$$

$$\Rightarrow {\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\le {\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}$$

$$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\le {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}$$

Let $\tilde{a}={\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, $\tilde{\pi }={\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ and $S(\tilde{a}),S(\tilde{\pi })$ be the score values of $a$ and $\pi$ respectively. Based on the score value of PFN in (3) and the above inequality, we can imply that $S(\tilde{a})\ge S(\tilde{\pi }),$ and then we discuss the following cases:

• If $S(\tilde{a})>S(\tilde{\pi }),$ then we can get ${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)>{\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$.
• If $S(\tilde{a})=S(\tilde{\pi }),$ then

  $$\begin{array}{l}\frac{1}{2}\left(1+{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2-{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2\right)\\ =\frac{1}{2}\left(1+{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2-{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2\right)\end{array}$$

Since ${\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ we can deduce that

$$\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}$$

And

$${\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}$$

Therefore, it follows that

$$H(\tilde{a})={\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2$$

$$={\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2=H(\tilde{\pi })$$

The ${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$. □

Property 3.

(Boundedness). Let ${\tilde{a}}_i=({\mu }_{i_j},{\nu }_{i_j}),{\tilde{a}}^{+}=({\mu }_{\max i_j},{\nu }_{\max i_j})(i=1,2,\cdots ,k)$ be a set of PFNs, and ${\tilde{a}}^{-}=({\mu }_{\min i_j},{\nu }_{\min i_j})$ then

$${\tilde{a}}_i{}^{-}<PFH{M}^{(x)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)<{\tilde{a}}_i{}^{+}$$

(14)

Proof. 

Based on Properties 1 and 2, we have

$$\begin{array}{l}{\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}^{-},{\tilde{a}}^{-},\cdots ,{\tilde{a}}^{-}\right)={\tilde{a}}^{-},\\ {\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}^{+},{\tilde{a}}^{+},\cdots ,{\tilde{a}}^{+}\right)={\tilde{a}}^{+}.\end{array}$$

Then we have ${\tilde{a}}^{-}\le {\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\tilde{a}}^{+}.$□

Property 4.

(Commutativity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. Suppose $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, then

$${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(15)

Proof. 

Because $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, then $\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_k^x}=\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{\pi }}_{i_j}\right)}^{\frac{1}{x}}}{C_k^x}$ Thus, ${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right).$ □

Next, we will discuss some particular cases of PFHM operator when $x$ takes different values.

Case 1: When $x=1$, then PFHM operator will become arithmetic average operator of PFNs.

$$\begin{array}{l}{\mathrm{PFHM}}^{\left(1\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\mu }_i{}^2\right)}\right)}^{\frac{1}{k}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-\left(1-{({\nu }_i)}^2\right)}}\right)}^{\frac{1}{k}}\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\mu }_i{}^2\right)}\right)}^{\frac{1}{k}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-\left(1-{({\nu }_i)}^2\right)}}\right)}^{\frac{1}{k}}\right)=\frac{1}{k}\underset{i}{\overset{k}{\oplus }}{\tilde{a}}_i\end{array}$$

(16)

Case 2: When $x=k$, then PFHM operator will become arithmetic average operator of PFNs.

$$\begin{array}{l}{\mathrm{PFHM}}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{1}{C_k^k}\left(\underset{1\le i_1<\cdots <i_x\le k}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_k\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^k{\mu }_{i_j}}\right)}^{\frac{1}{k}}\right)}^2\right)}\right)}^{\frac{1}{C_k^k}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_k\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^k\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{k}}}}\right)}^{\frac{1}{C_k^k}}\right)\\ =\left({\left({\displaystyle \prod _{i=1}^k{\mu }_i}\right)}^{\frac{1}{k}},\sqrt{1-{\left({\displaystyle \prod _{i=1}^k\left(1-{({\nu }_i)}^2\right)}\right)}^{\frac{1}{k}}}\right)=\underset{i=1}{\overset{k}{\otimes }}{\tilde{a}}_i{}^{\frac{1}{k}}\end{array}$$

(17)

Example 3.

Let ${\tilde{a}}_1=\left(0.5,0.2\right),{\tilde{a}}_2=\left(0.6,0.3\right),{\tilde{a}}_3=\left(0.4,0.1\right),{\tilde{a}}_4=\left(0.7,0.3\right)$ be four PFNs. Then we use the proposed PFHM operator to aggregate four PFNs. (suppose $x=2$)

$$\begin{array}{l}\tilde{a}={\mathrm{PFHM}}^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}},{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)\\ =\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}\left(1-{\left({\left({\displaystyle \prod _{j=1}^2{\mu }_{i_j}}\right)}^{\frac{1}{2}}\right)}^2\right)}\right)}^{\frac{1}{C_4^2}}},{\left({\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}\sqrt{1-{\left({\displaystyle \prod _{j=1}^2\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{2}}}}\right)}^{\frac{1}{C_4^2}}\right)\\ =\left(\begin{array}{l}\sqrt{\begin{array}{l}1-\left(1-0.5\times 0.6\right)\times \left(1-0.5\times 0.4\right)\times \left(1-0.5\times 0.7\right)\\ \times \left(1-0.6\times 0.4\right)\times \left(1-0.6\times 0.7\right)\times {\left(1-0.4\times 0.7\right)}^{\frac{1}{6}}\end{array}},\\ {\left(\begin{array}{l}{\left(1-{\left(\left(1-{0.2}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5}\times {\left(1-{\left(\left(1-{0.2}^2\right)\times \left(1-{0.1}^2\right)\right)}^{0.5}\right)}^{0.5}\\ \times {\left(1-{\left(\left(1-{0.2}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5}\times {\left(1-{\left(\left(1-{0.3}^2\right)\times \left(1-{0.1}^2\right)\right)}^{0.5}\right)}^{0.5}\\ \times {\left(1-{\left(\left(1-{0.3}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5}\times {\left(1-{\left(\left(1-{0.1}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5}\end{array}\right)}^{\frac{1}{6}}\end{array}\right)\\ =\left(0.5497,0.2325\right)\end{array}$$

At last, we get ${\mathrm{PFHM}}^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,{\tilde{a}}_3,{\tilde{a}}_4\right)=\left(0.5497,0.2325\right)$.

3.2. The WPFHM Operator

The weights of attributes play an important role in practical decision making, and they can influence the decision result. Therefore, it is necessary to consider attribute weights in aggregating information. It is obvious that the PFHM operator fails to consider the problem of attribute weights. In order to overcome this defect, we propose the WPFHM operator.

Definition 9.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a group of PFNs, $\omega ={\left({\omega }_1,{\omega }_2,\cdots {\omega }_k\right)}^T$ be the weight vector for ${\tilde{a}}_i$$\left(i=1,2,\cdots ,k\right)$, which satisfies ${\omega }_i\in \left[0.1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$, then we can define WPFHM operator as follows:

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\{\begin{array}{cc}\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_{k-1}^x}\hfill & \left(1\le x<k\right)\\ \underset{i=1}{\overset{x}{\otimes }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}\hfill & \left(x=k\right)\end{array}$$

(18)

Theorem 2.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a group of PFNs, and their weight vector meet ${\omega }_i\in \left[0.1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$ then the result from Definition 9 is still a PFN, and have.

$$\begin{array}{l}{\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_{k-1}^x}\\ =\left(\begin{array}{l}\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}},}\\ {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\end{array}\right)\left(1\le x<k\right)\end{array}$$

(19)

Or

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\underset{i=1}{\overset{x}{\otimes }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left({\displaystyle \prod _{i=1}^k{\left({\mu }_i\right)}^{\frac{1-{\omega }_i}{k-1}}},\sqrt{1-{{\displaystyle \prod _{i=1}^k\left(1-{({\nu }_i)}^2\right)}}^{\frac{1-{\omega }_i}{k-1}}}\right)\left(x=k\right)$$

(20)

Proof. 

(1)

First of all, we prove that (19) and (20) are kept. For the first case, when $\left(1\le x<k\right)$, according to the operational laws of PFNs, we get

$${\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}=\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}},\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)$$

(21)

Thereafter,

$$\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}=\left(\sqrt{1-{\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}},{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)$$

(22)

Moreover,

$$\begin{array}{l}\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}\\ =\left(\sqrt{1-{{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)},}{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)\end{array}$$

(23)

Therefore,

$$\begin{array}{l}\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_{k-1}^x}\\ =\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}},{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}\right)\end{array}$$

(24)

For the second case, when $\left(x=k\right)$, we get

$${\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left({\left({\mu }_i\right)}^{\frac{1-{\omega }_i}{k-1}},\sqrt{1-{\left(1-{({\nu }_i)}^2\right)}^{\frac{1-{\omega }_i}{k-1}}}\right)$$

(25)

Then,

$$\underset{i=1}{\overset{k}{\otimes }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left({\displaystyle \prod _{i=1}^k\left({\mu }_i{}^{\frac{1-{\omega }_i}{k-1}}\right)},\sqrt{1-{{\displaystyle \prod _{i=1}^k\left(1-{({\nu }_i)}^2\right)}}^{\frac{1-{\omega }_i}{k-1}}}\right)$$

(26)

(2)

Next, we prove the (19) and (20) are PFNs. For the first case, when $1\le x<k,$

Let

$$\begin{array}{l}p=\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}},\\ q={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}.\end{array}$$

Then we need prove the following two conditions. (i) $0\le p\le 1$, $0\le q\le 1$. (ii) $0\le p^2+q^2\le 1$.

• Since $p\in \left[0,1\right]$, we can get

  $$\begin{array}{l}{\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\in \left[0,1\right]\Rightarrow {\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\in \left[0,1\right]\Rightarrow 1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\in \left[0,1\right]\\ \Rightarrow {{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\in \left[0,1\right]\Rightarrow {\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}\in \left[0,1\right]\\ \Rightarrow \sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\in \left[0,1\right]\end{array}$$
  Therefore, $0\le p\le 1.$
  Similarly, we can get ${\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}\in \left[0,1\right]$
  Therefore, $0\le q\le 1.$
• Since $0\le p^2+q^2\le 1$, we can get the following inequality:

  $$\begin{array}{l}{\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)}^2+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}\right)}^2\\ \le 1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}+{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}=1\end{array}$$

□

For the second case, when $x=k,$ we can easily prove that it is kept. So the aggregation result produced by Definition 8 is still a PFN. Next, we shall deduce some desirable properties of WPFHM operator.

Property 5.

(Idempotency). If ${\tilde{a}}_i\left(i=1,2,\cdots ,k\right)$ are equal, i.e., ${\tilde{a}}_i=\tilde{a}=\left(\mu ,\nu \right),$ and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$ then

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\tilde{a}$$

(27)

Proof. 

Since ${\tilde{a}}_i=\tilde{a}=\left({\mu }_i,{\nu }_i\right)$, based on Theorem 2, we get

(1)

For the first case, when $1\le x<k$.

$$\begin{array}{l}{\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ =\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)\end{array}$$

$$=\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left(\mu \right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)$$

$$=\left(\sqrt{1-{\left({\left(1-{\left(\mu \right)}^2\right)}^{C_k^x-{\displaystyle \sum _{1\le i_1<\cdots <i_x\le k}\left({\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)}^{C_k^x-{\displaystyle \sum _{1\le i_1<\cdots <i_x\le k}\left({\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)$$

$$=\left(\sqrt{1-{\left({\left(1-{\left(\mu \right)}^2\right)}^{C_k^x-{\displaystyle \sum _{k=1}^k{C}_{k-1}^{x-1}}{\omega }_i}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)}^{C_k^x-{\displaystyle \sum _{k=1}^k{C}_{k-1}^{x-1}}{\omega }_i}\right)}^{\frac{1}{C_{k-1}^x}}\right)$$

$$=\left(\sqrt{1-{\left({\left(1-{\left(\mu \right)}^2\right)}^{C_k^x-C_{k-1}^{x-1}{\displaystyle \sum _{k=1}^k{\omega }_i}}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)}^{C_k^x-C_{k-1}^{x-1}{\displaystyle \sum _{k=1}^k{\omega }_i}}\right)}^{\frac{1}{C_{k-1}^x}}\right)$$

Since ${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$ we can get

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left(\sqrt{1-{\left({\left(1-{\left(\mu \right)}^2\right)}^{C_k^x-C_{k-1}^{x-1}}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)}^{C_k^x-C_{k-1}^{x-1}}\right)}^{\frac{1}{C_{k-1}^x}}\right)$$

$$=\left(\sqrt{1-{\left({\left(1-{\left(\mu \right)}^2\right)}^{\frac{(k-1)!}{x!(k-1-x)!}}\right)}^{\frac{x!(k-1-x)!}{(k-1)!}},}{\left({\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)}^{\frac{(k-1)!}{x!(k-1-x)!}}\right)}^{\frac{x!(k-1-x)!}{(k-1)!}}\right)=\left(\mu ,\nu \right)=\tilde{a}$$

(2)

For the second case, when $x=k,$

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left({\displaystyle \prod _{i=1}^k{\left({\mu }_i\right)}^{\frac{1-{\omega }_i}{k-1}}},\sqrt{1-{\left({\displaystyle \prod _{i=1}^k\left(1-{({\nu }_i)}^2\right)}\right)}^{\frac{1-{\omega }_i}{k-1}}}\right)=\left({\left(\mu \right)}^{\frac{k-1}{k-1}},\sqrt{1-{\left(1-{(\nu )}^2\right)}^{\frac{k-1}{k-1}}}\right)$$

Since ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$, we can get ${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left({\left(\mu \right)}^{\frac{k-1}{k-1}},\sqrt{1-{\left(1-{(\nu )}^2\right)}^{\frac{k-1}{k-1}}}\right)=(\mu ,\nu )=\tilde{a},$ which proves the idempotency property of the WPFHM operator. □

Property 6.

(Monotonicity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. If ${\mu }_{i_j}\ge {\mu }_{{\theta }_j},{\nu }_{i_j}\le {\nu }_{{\theta }_j}$ for all $j$, and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$ the $\tilde{a}$ and $\tilde{\pi }$ are equal, then we have

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(28)

Proof. 

Since $x\ge 1,{\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ then

$${\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\ge {\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\Rightarrow {\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\ge {\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\Rightarrow 1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\le 1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2$$

$$\Rightarrow {{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\le {{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}$$

$$\Rightarrow \sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\ge \sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}$$

Similarly, we have

$$1-{({\nu }_{i_j})}^2\ge 1-{({\nu }_{{\theta }_J})}^2\Rightarrow {\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\ge {\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_J})}^2\right)}\right)}^{\frac{1}{x}}$$

$$\Rightarrow {\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\le {\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_J})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}$$

$$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\le {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}$$

Let $a={\mathrm{WPFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, $\pi ={\mathrm{WPFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ and $S(a),S(\pi )$ be the score values of $a$ and $\pi$ respectively. Based on the score value of PFN in (3) and the above inequality, we can imply that $S\left(a\right)\ge S\left(\pi \right),$ and then we discuss the following cases:

(1)

If $S(a)>S(\pi ),$ then we can get ${\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)>{\mathrm{PFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$.

(2)

If $S(a)=S(\pi ),$ then

$$\begin{array}{l}\frac{1}{2}(1+{\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)}^2-{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)}^2)\\ =\frac{1}{2}(1+{\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)}^2-\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right))\end{array}$$

Since ${\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ and based on the Equations (3) and (4), we can deduce that

$$\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}=\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\theta }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}$$

And

$${\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}$$

Therefore, it follows that $\mathrm{H}\left(\tilde{\mathrm{a}}\right)=\mathrm{H}\left(\tilde{\mathsf{\pi }}\right)$, the ${\mathrm{WPFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{WPFHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right).$

When $x=k$, we can prove it in a similar way.

Property 7.

(Boundedness). Let ${\tilde{a}}_i=({\mu }_{i_j},{\nu }_{i_j}),{\tilde{a}}^{+}=({\mu }_{\max i_j},{\nu }_{\max i_j})(i=1,2,\cdots ,k)$ be a set of PFNs, and ${\tilde{a}}^{-}=({\mu }_{\min i_j},{\nu }_{\min i_j})$, and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$ then

$${\tilde{a}}^{-}\le {\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\tilde{a}}^{+}$$

(29)

Proof. 

Based on Properties 5 and 6, we have

$$\begin{array}{l}{\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}^{-},{\tilde{a}}^{-},\cdots ,{\tilde{a}}^{-}\right)={\tilde{a}}^{-},\\ {\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}^{+},{\tilde{a}}^{+},\cdots ,{\tilde{a}}^{+}\right)={\tilde{a}}^{+},\end{array}$$

Then we have ${\tilde{a}}^{-}\le {\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\tilde{a}}^{+}.$□

Property 8.

(Commutativity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. Suppose $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$ the $\tilde{a}$ and $\tilde{\pi }$ are equal, then we have

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(30)

Proof. 

Because $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, then

$$\begin{array}{cc}\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_{k-1}^x}=\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{\pi }}_{i_j}\right)}^{\frac{1}{x}}}{C_{k-1}^x}& \left(1\le x<k\right)\\ \underset{i=1}{\overset{x}{\otimes }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\underset{i=1}{\overset{x}{\otimes }}{\tilde{\pi }}_i{}^{\frac{1-{\omega }_i}{k-1}}& \left(x=k\right)\end{array}$$

Thus, ${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$. Next, we will discuss some particular cases of WPFHM operator for different value $x$.□

Case 1: When $x=1$, the WPFHM will reduce to the following form:

$$\begin{array}{l}{\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{\underset{1\le i_1\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^1{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{1}}}{C_{k-1}^1}\\ =\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1\le k}\left(1-{\left({\mu }_i\right)}^2\right)}}^{\left(1-{\omega }_i\right)}\right)}^{\frac{1}{k-1}}},{\left({\displaystyle \prod _{1\le i_1\le k}{\left({\nu }_i\right)}^{\left(1-{\omega }_i\right)}}\right)}^{{}^{\frac{1}{k-1}}}\right)\end{array}$$

(31)

Case 2: When $x=k$, the proposed WPFHM operator will reduce to the following form:

$${\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\underset{i=1}{\overset{k}{\otimes }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left({\displaystyle \prod _{i=1}^k\left({\mu }_i{}^{\frac{1-{\omega }_i}{k-1}}\right)},\sqrt{1-{{\displaystyle \prod _{i=1}^k\left(1-{({\nu }_i)}^2\right)}}^{\frac{1-{\omega }_i}{k-1}}}\right)$$

(32)

Example 4.

Let ${\tilde{a}}_1=\left(0.6,0.4\right),{\tilde{a}}_2=\left(0.7,0.3\right),{\tilde{a}}_3=\left(0.5,0.1\right),{\tilde{a}}_4=\left(0.4,0.3\right)$ be four PFNs. the weighting vector of attributes be $\omega =\left\{0.1,0.3,0.4,0.2\right\}$, Then we use the proposed WPFHM operator to aggregate four PFNs. (suppose $x=2$)

$$\begin{array}{l}\tilde{a}={\mathrm{WPFHM}}_{\mathsf{\omega }}^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\frac{\underset{1\le i_1<\cdots <i_x\le k}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{\tilde{a}}_{i_j}\right)}^{\frac{1}{x}}}{C_{k-1}^x}\\ =\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)\\ =\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}\left(1-{\left({\left({\displaystyle \prod _{j=1}^2{\mu }_{i_j}}\right)}^{\frac{1}{2}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_3^2}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^2\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{2}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_3^2}}\right)\end{array}$$

$$\begin{array}{l}=\left(\begin{array}{l}\sqrt{1-{\left(\begin{array}{l}{\left(1-0.6\times 0.7\right)}^{1-0.1\times 0.3}\times {\left(1-0.6\times 0.5\right)}^{1-0.1\times 0.4}\times {\left(1-0.6\times 0.4\right)}^{1-0.1\times 0.2}\\ \times {\left(1-0.7\times 0.5\right)}^{1-0.3\times 0.4}\times {\left(1-0.7\times 0.4\right)}^{1-0.3\times 0.2}\times {\left(1-0.5\times 0.4\right)}^{1-0.4\times 0.2}\end{array}\right)}^{\frac{1}{3}}},\\ {\left(\begin{array}{l}{\left(1-{\left(\left(1-{0.4}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5\times (1-0.1\times 0.3)}\times {\left(1-{\left(\left(1-{0.4}^2\right)\times \left(1-{0.1}^2\right)\right)}^{0.5}\right)}^{0.5\times (1-0.1\times 0.4)}\\ \times {\left(1-{\left(\left(1-{0.4}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5\times (1-0.1\times 0.2)}\times {\left(1-{\left(\left(1-{0.3}^2\right)\times \left(1-{0.1}^2\right)\right)}^{0.5}\right)}^{0.5\times (1-0.3\times 0.4)}\\ \times {\left(1-{\left(\left(1-{0.3}^2\right)\times \left(1-{0.1}^2\right)\right)}^{0.5}\right)}^{0.5\times (1-0.3\times 0.2)}\times {\left(1-{\left(\left(1-{0.1}^2\right)\times \left(1-{0.3}^2\right)\right)}^{0.5}\right)}^{0.5\times (1-0.4\times 0.2)}\end{array}\right)}^{\frac{1}{3}}\end{array}\right)\\ =\left(0.7015,0.3114\right)\end{array}$$

At last, we get ${\mathrm{WPFHM}}_{\omega }^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,{\tilde{a}}_3,{\tilde{a}}_4\right)=\left(0.7015,0.3114\right)$.

3.3. The PFDHM Operator

Wu et al. [59] proposed the DHM operator.

Definition 10.

The DHM operator is defined [59]:

$${\mathrm{DHM}}^{(x)}\left({\phi }_1,{\phi }_2,\cdots ,{\phi }_n\right)={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(\frac{{\displaystyle \sum _{j=1}^x{\phi }_{i_j}}}{x}\right)}\right)}^{\frac{1}{C_n^x}}$$

(33)

where $x$ is a parameter and $x=1,2,\dots ,k$, $i_1,i_2,\dots ,i_x$ are $x$ integer values taken from the set $\left\{1,2,\dots ,k\right\}$ of $k$ integer values, $C_n^x$ denotes the binomial coefficient and $C_k^x=\frac{k!}{x!\left(k-x\right)!}$.

In this section, we propose the Pythagorean fuzzy DHM (PFDHM) operator.

Definition 11.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a set of PFNs, then we define PFDHM operator as follows:

$${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}$$

(34)

where $x$ is a parameter and $x=1,2,\dots ,k$, $i_1,i_2,\dots ,i_x$ are $x$ integer values taken from the set $\left\{1,2,\dots ,k\right\}$ of $k$ integer values, $C_n^x$ denotes the binomial coefficient and $C_k^x=\frac{k!}{x!\left(k-x\right)!}$.

Theorem 3.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a collection of the PFNs, then the aggregate result of definition 10 is still a PFNs, and have

$$\begin{array}{l}{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j-1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)\end{array}$$

(35)

Proof. 

(1)

First of all, we prove (35) is kept.

$$\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}=\left(\sqrt{1-{\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}},{\displaystyle \prod _{j=1}^x\left({\nu }_{i_j}\right)}\right)$$

(36)

Then,

$$\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}=\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}},{\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)$$

(37)

Thereafter,

$$\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)=\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}},\sqrt{1-{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}\right)$$

(38)

Furthermore,

$$\begin{array}{l}{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)\end{array}$$

(39)

(2)

Next, we prove (34) is a PFN.

Let

$$p={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}},$$

$$q=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}$$

Then we need to prove that it satisfies the following two conditions. (i) $0\le p\le 1,0\le q\le 1;$ (ii) $0\le p^2+q^2\le 1.$

• Since ${\mu }_{i_j}\in [0,1],$ we can get

  $${({\mu }_{i_j})}^2\in \left[0,1\right]\Rightarrow 1-{({\mu }_{i_j})}^2\in \left[0,1\right]\Rightarrow {\displaystyle \prod _{j-1}^x\left(1-{({\mu }_{i_j})}^2\right)}\in \left[0,1\right]$$

  $${({\mu }_{i_j})}^2\in \left[0,1\right]\Rightarrow 1-{({\mu }_{i_j})}^2\in \left[0,1\right]\Rightarrow {\displaystyle \prod _{j-1}^x\left(1-{({\mu }_{i_j})}^2\right)}\in \left[0,1\right]$$

  $$\Rightarrow 1-{\left({\displaystyle \prod _{j-1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\in \left[0,1\right]\Rightarrow {\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j-1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\in \left[0,1\right]$$

  $$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j-1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\in \left[0,1\right].$$
  Therefore $0\le p\le 1,$ similarly, we can get

  $$\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\in [0,1],\mathrm{therefore},0\le q\le 1.$$
• Obviously, $0\le p^2+q^2\le 1,$ then

  $$\begin{array}{l}{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2+{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2\\ \le {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_k^x}}+1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left({\nu }_{i_j}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_k^x}}=1\end{array}$$

We get $0\le p^2+q^2\le 1.$□

So the aggregated result of Definition 10 is still PFN. Next we will talk about some properties of PFDHM operator.

Property 9.

(Idempotency). If ${\tilde{a}}_i(1,2,\cdots ,k)$ and $\tilde{a}$ are PFNs, and ${\tilde{a}}_i=\tilde{a}=({\mu }_i,{\nu }_i)$ for all $i=1,2,\cdots ,k$, then we get

$${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\tilde{a}$$

(40)

Proof. 

$$\begin{array}{l}{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)\end{array}$$

$$\begin{array}{l}{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)\end{array}$$

$$=\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-\left(1-{({\mu }_i)}^2\right)}}\right)}^{\frac{1}{C_k^x}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\nu }_i{}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)$$

$$=\left(\sqrt{1-\left(1-{({\mu }_i)}^2\right)},\sqrt{1-\left(1-{\nu }_i{}^2\right)}\right)=\left(\sqrt{1-\left(1-{\mu }^2\right),}\left(\sqrt{1-\left(1-{(\nu )}^2\right)}\right)\right)=\left(\mu ,\nu \right)=\tilde{a}$$

□

Property 10.

(Monotonicity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. If ${\mu }_{i_j}\ge {\mu }_{{\theta }_j},{\nu }_{i_j}\le {\nu }_{{\theta }_j}$ for all $j$, then

$${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(41)

Proof. 

Since $x\ge 1,{\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ then

$${\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\le {\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}\Rightarrow 1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\ge 1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}$$

$${\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\le {\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}\Rightarrow 1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\ge 1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}$$

$$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\ge {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}$$

Similarly, we have

$${\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\le {\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\Rightarrow 1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\ge 1-\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)$$

$$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}\ge {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}$$

$$\Rightarrow \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\le \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}$$

Let $\tilde{a}={\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, $\tilde{\pi }={\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ and $S(\tilde{a}),S(\tilde{\pi })$ be the score values of $a$ and $\pi$ respectively. Based on the score value of PFN in (3) and the above inequality, we can imply that $S(\tilde{a})\ge S(\tilde{\pi }),$ and then we discuss the following cases:

• If $S(\tilde{a})>S(\tilde{\pi }),$ then we can get ${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)>{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$
• If $S(\tilde{a})=S(\tilde{\pi }),$ then

  $$\begin{array}{l}\frac{1}{2}\left(1+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2-{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2\right)\\ =\frac{1}{2}\left(1+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2-{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2\right)\end{array}$$

Since ${\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ and based on the Equations (3) and (4), we can deduce that

$${\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}$$

And

$$\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}$$

Therefore, it follows that

$$H(\tilde{a})={\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2+{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2$$

$$={\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}}\right)}^2+{\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)}^2=H(\tilde{\pi })$$

The ${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$

Property 11.

(Boundedness). Let ${\tilde{a}}_i=({\mu }_{i_j},{\nu }_{i_j}),{\tilde{a}}^{+}=({\mu }_{\max i_j},{\nu }_{\max i_j})(i=1,2,\cdots ,k)$ be a set of PFNs, and ${\tilde{a}}^{-}=({\mu }_{\min i_j},{\nu }_{\min i_j})$ then

$${\tilde{a}}_i{}^{-}<{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)<{\tilde{a}}_i{}^{+}$$

(42)

Proof. 

Based on Properties 9 and 10, we have

$$\begin{array}{l}{\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}^{-},{\tilde{a}}^{-},\cdots ,{\tilde{a}}^{-}\right)={\tilde{a}}^{-},\\ {\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}^{+},{\tilde{a}}^{+},\cdots ,{\tilde{a}}^{+}\right)={\tilde{a}}^{+},\end{array}$$

Then we have ${\tilde{a}}^{-}\le {\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\tilde{a}}^{+}.$□

Property 12.

(Commutativity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. Suppose $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, then

$${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(43)

Proof. 

Because $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, then

$${\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{\pi }}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}$$

Thus, ${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right).$

Next, we discuss some particular cases of PFDHM operator.

Case 1: When $x=1$, then PFDHM operator will become arithmetic average operator of PFNs.

$${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left(\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\mu }_i{}^2\right)}\right)}^{\frac{1}{k}},}{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-\left(1-{({\nu }_i)}^2\right)}}\right)}^{\frac{1}{k}}\right)$$

$$=\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-\left(1-{({\mu }_i)}^2\right)}}\right)}^{\frac{1}{k}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\nu }_i{}^2\right)}\right)}^{\frac{1}{k}}}\right)=\frac{1}{k}\underset{i}{\overset{k}{\otimes }}{\tilde{a}}_i$$

Case 2: When $x=k$, then PFDHM operator will become arithmetic average operator of PFNs.

$${\mathrm{PFDHM}}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^k}}$$

$$=\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_k\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^k\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{k}}}}\right)}^{\frac{1}{C_k^k}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_k\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^k{\nu }_{i_j}}\right)}^{\frac{1}{k}}\right)}^2\right)}\right)}^{\frac{1}{C_k^k}}}\right)$$

$$=\left(\sqrt{1-{\left({\displaystyle \prod _{i=1}^k\left(1-{({\mu }_i)}^2\right)}\right)}^{\frac{1}{k}}},{\left({\displaystyle \prod _{i=1}^k{\nu }_i}\right)}^{\frac{1}{k}}\right)=\underset{i=1}{\overset{k}{\oplus }}{\tilde{a}}_i{}^{\frac{1}{k}}$$

Example 5.

Let ${\tilde{a}}_1=\left(0.6,0.2\right),{\tilde{a}}_2=\left(0.5,0.3\right),{\tilde{a}}_3=\left(0.7,0.1\right),{\tilde{a}}_4=\left(0.8,0.2\right)$ be four PFNs. Then we use the PFDHM operator to fuse four PFNs. (suppose $x=2$),

$$\begin{array}{l}\tilde{a}={\mathrm{PFDHM}}^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left(\mu ,\nu \right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_k^x}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}\right)}^{\frac{1}{C_k^x}}}\right)\end{array}$$

$$\begin{array}{l}=\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}\sqrt{1-{\left({\displaystyle \prod _{j=1}^2\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{2}}}}\right)}^{\frac{1}{C_4^2}},\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}\left(1-{\left({\left({\displaystyle \prod _{j=1}^2{\nu }_{i_j}}\right)}^{\frac{1}{2}}\right)}^2\right)}\right)}^{\frac{1}{C_4^2}}}\right)\\ =\left(\begin{array}{l}{\left(\begin{array}{l}{\left(1-{\left(\left(1-{0.6}^2\right)\times \left(1-{0.5}^2\right)\right)}^{0.5}\right)}^{0.5}\times {\left(1-{\left(\left(1-{0.6}^2\right)\times \left(1-{0.7}^2\right)\right)}^{0.5}\right)}^{0.5}\\ \times {\left(1-{\left(\left(1-{0.6}^2\right)\times \left(1-{0.8}^2\right)\right)}^{0.5}\right)}^{0.5}\times {\left(1-{\left(\left(1-{0.5}^2\right)\times \left(1-{0.7}^2\right)\right)}^{0.5}\right)}^{0.5}\\ \times {\left(1-{\left(\left(1-{0.5}^2\right)\times \left(1-{0.8}^2\right)\right)}^{0.5}\right)}^{0.5}\times {\left(1-{\left(\left(1-{0.7}^2\right)\times \left(1-{0.8}^2\right)\right)}^{0.5}\right)}^{0.5}\end{array}\right)}^{\frac{1}{6}},\\ {(1-{\left(\begin{array}{l}\left(1-0.2\times 0.3\right)\times \left(1-0.2\times 0.1\right)\times \left(1-0.2\times 0.2\right)\\ \times \left(1-0.3\times 0.1\right)\times \left(1-0.3\times 0.2\right)\times \left(1-0.1\times 0.2\right)\end{array}\right)}^{\frac{1}{6}})}^{0.5}\end{array}\right)\\ =\left(0.6627,0.1962\right)\end{array}$$

At last, we get ${\mathrm{PFDHM}}^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,{\tilde{a}}_3,{\tilde{a}}_4\right)=\left(0.6627,0.1962\right)$.

3.4. The WPFDHM Operator

The weights of attributes play an important role in practical decision making, and they can influence the decision result. Therefore, it is necessary to consider attribute weights in aggregating information. It is obvious that the PFDHM operator fails to consider the problem of attribute weights. In order to overcome this defect, we propose the WPFDHM operator.

Definition 12.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a group of PFNs, $\omega ={\left({\omega }_1,{\omega }_2,\cdots {\omega }_k\right)}^T$ be the weight vector for ${\tilde{a}}_i$$\left(i=1,2,\cdots ,k\right)$, which satisfies ${\omega }_i\in \left[0.1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$, then we can define WPFHM operator as follows:

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\{\begin{array}{cc}{\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_{k-1}^x}}\hfill & \left(1\le x<k\right)\\ \underset{i=1}{\overset{x}{\oplus }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}\hfill & \left(x=k\right)\end{array}$$

(44)

Theorem 4.

Let ${\tilde{a}}_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,k\right)$ be a group of PFNs, and their weight vector meet ${\omega }_i\in \left[0.1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$ then the result from Definition 11 is still a PFN, and have

$$\begin{array}{l}{\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ ={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_{k-1}^x}}\left(1\le x<k\right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}},\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\end{array}$$

(45)

Or

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\underset{i=1}{\overset{x}{\oplus }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left(\sqrt{1-{{\displaystyle \prod _{i=1}^k\left(1-{({\mu }_i)}^2\right)}}^{\frac{1-{\omega }_i}{k-1}}},{\displaystyle \prod _{i=1}^k{\left({\nu }_i\right)}^{\frac{1-{\omega }_i}{k-1}}}\right)\left(x=k\right)$$

(46)

Proof. 

(1)

First of all, we prove that (45) and (46) are kept.

For the first case, when $\left(1\le x<k\right)$, we get

$$\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}=\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}},{\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)$$

(47)

Then,

$$\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)=\left({\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)},\sqrt{1-{\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)$$

(48)

Thereafter,

$$\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)=\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}},\sqrt{1-{{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)$$

(49)

Furthermore,

$$\begin{array}{l}{\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_{k-1}^x}}\\ =\left(,{\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}},\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\end{array}$$

(50)

For the second case, when $\left(x=k\right)$, we get

$${\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left({\left({\mu }_i\right)}^{\frac{1-{\omega }_i}{k-1}},\sqrt{1-{\left(1-{({\nu }_i)}^2\right)}^{\frac{1-{\omega }_i}{k-1}}}\right),$$

(51)

Then,

$$\underset{i=1}{\overset{x}{\oplus }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left(\sqrt{1-{\left(1-{({\mu }_i)}^2\right)}^{\frac{1-{\omega }_i}{k-1}}},{\left({\nu }_i\right)}^{\frac{1-{\omega }_i}{k-1}}\right)$$

(52)

(2)

Next, we prove the (45) and (46) are PFNs. For the first case, when $1\le x<k$

$$p={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}},q=\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}$$

□

Then we need prove the following two conditions. (i) $0\le p\le 1$,$0\le q\le 1$. (ii) $0\le p^2+q^2\le 1$.

• Since $p\in \left[0,1\right]$, we can get

  $${\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\in \left[0,1\right]\Rightarrow {\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\in \left[0,1\right]\Rightarrow \sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\in \left[0,1\right]$$

  $$\begin{array}{l}\Rightarrow {\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\in \left[0,1\right]\\ \Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\in \left[0,1\right]\end{array}$$
  Therefore, $0\le p\le 1.$ Similarly, we can get

  $$\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\in \left[0,1\right].$$
  Therefore, $0\le q\le 1$.
• Since $0\le p^2+q^2\le 1$, we can get the following inequality:

  $${\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}\right)}^2+{\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)}^2$$

  $$\le {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{{}^{\frac{1}{C_{k-1}^x}}}+1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\nu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}=1$$

For the second case, when $x=k,$ we can easily prove that it is kept. So the aggregation result produced by Definition 8 is still a PFN. Next, we shall deduce some desirable properties of WPFHM operator.

Property 13.

(Idempotency). If ${\tilde{a}}_i\left(i=1,2,\cdots ,k\right)$ are equal, i.e., ${\tilde{a}}_i=\tilde{a}=\left(\mu ,\nu \right)$ and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$ then

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\tilde{a},$$

(53)

Proof. 

Since ${\tilde{a}}_i=\tilde{a}=\left({\mu }_i,{\nu }_i\right)$, based on Theorem 4, we get

(1)

For the first case, when $1\le x<k$.

$$\begin{array}{l}{\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}},\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\end{array}$$

$$=\left({\left({\left(\sqrt{1-\left(1-{(\mu )}^2\right)}\right)}^{C_k^x-{\displaystyle \sum _{1\le i_1<\cdots <i_x\le k}\left({\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}},\sqrt{1-{\left({\left(1-{\left(\nu \right)}^2\right)}^{C_k^x-{\displaystyle \sum _{1\le i_1<\cdots <i_x\le k}\left({\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}}\right)$$

$$\begin{array}{l}=\left({\left({\left(\sqrt{1-\left(1-{(\mu )}^2\right)}\right)}^{C_k^x-{\displaystyle \sum _{k=1}^k{C}_{k-1}^{x-1}}{\omega }_i}\right)}^{\frac{1}{C_{k-1}^x}},\sqrt{1-{\left({\left(1-{\left(\nu \right)}^2\right)}^{C_k^x-{\displaystyle \sum _{k=1}^k{C}_{k-1}^{x-1}}{\omega }_i}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\\ =\left({\left({\left(\sqrt{1-\left(1-{(\mu )}^2\right)}\right)}^{C_k^x-C_{k-1}^{x-1}{\displaystyle \sum _{k=1}^k{\omega }_i}}\right)}^{\frac{1}{C_{k-1}^x}}\sqrt{1-{\left({\left(1-{\left(\nu \right)}^2\right)}^{C_k^x-C_{k-1}^{x-1}{\displaystyle \sum _{k=1}^k{\omega }_i}}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\end{array}$$

Since ${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$ we can get

$$\begin{array}{l}{\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left({\left({\left(\sqrt{1-\left(1-{(\mu )}^2\right)}\right)}^{C_k^x-C_{k-1}^{x-1}}\right)}^{\frac{1}{C_{k-1}^x}},\sqrt{1-{\left({\left(1-{\left(\nu \right)}^2\right)}^{C_k^x-C_{k-1}^{x-1}}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\\ =\left({\left({\left(\sqrt{1-\left(1-{(\mu )}^2\right)}\right)}^{\frac{(k-1)!}{x!(k-1-x)!}}\right)}^{\frac{x!(k-1-x)!}{(k-1)!}},\sqrt{1-{\left({\left(1-{\left(\nu \right)}^2\right)}^{\frac{(k-1)!}{x!(k-1-x)!}}\right)}^{\frac{x!(k-1-x)!}{(k-1)!}}}\right)=\left(\mu ,\nu \right)=\tilde{a}\end{array}$$

(2)

For the second case, when $x=k$

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{x}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left(\sqrt{1-{\left({\displaystyle \prod _{i=1}^k\left(1-{({\mu }_i)}^2\right)}\right)}^{\frac{1-{\omega }_i}{k-1}}},{\displaystyle \prod _{i=1}^k{\left({\nu }_i\right)}^{\frac{1-{\omega }_i}{k-1}}}\right)=\left(\sqrt{1-{\left(1-{(\mu )}^2\right)}^{\frac{k-1}{k-1}}},{\left(\nu \right)}^{\frac{k-1}{k-1}}\right)$$

Since ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$, we can get

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\left(\sqrt{1-{\left(1-{(\nu )}^2\right)}^{\frac{k-1}{k-1}}},{\left(\mu \right)}^{\frac{k-1}{k-1}}\right)=(\mu ,\nu )=\tilde{a},$$

which proves the idempotency property of the WPFDHM operator. □

Property 14.

(Monotonicity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. If ${\mu }_{i_j}\ge {\mu }_{{\theta }_j},{\nu }_{i_j}\le {\nu }_{{\theta }_j}$ for all $j$, and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$ the $\tilde{a}$ and $\tilde{\pi }$ are equal, then we have

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(54)

Proof. 

Since $x\ge 1,{\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ then

$$1-{({\mu }_{i_j})}^2\le 1-{({\mu }_{{\theta }_j})}^2\Rightarrow {\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}\le {\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}$$

$$\Rightarrow {\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\ge {\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}$$

$$\Rightarrow {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\ge {\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}$$

Similarly, we have

$${\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\le {\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\Rightarrow {\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\le {\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\Rightarrow 1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\ge 1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2$$

$$\Rightarrow {{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\le {{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}$$

$$\Rightarrow \sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\le \sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}$$

Let $a={\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, $\pi ={\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ and $S(a),S(\pi )$ be the score values of $a$ and $\pi$ respectively. Based on the score value of PFN in (3) and the above inequality, we can imply that $S\left(a\right)\ge S\left(\pi \right),$ and then we discuss the following cases:

(1)

If $S(a)>S(\pi ),$ then we can get ${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)>{\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$.

(2)

If $S(a)=S(\pi ),$ then

$$\begin{array}{l}\frac{1}{2}(1+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)}^2-{\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\mu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)}^2)\\ =\frac{1}{2}(1+{\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}\right)}^2-{\left(\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)}^2)\end{array}$$

Since ${\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ and based on the Equations (3) and (4), we can deduce that

$${\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}={\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{{\theta }_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}}$$

And

$$\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}=\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{{\theta }_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}$$

Therefore, it follows that $\mathrm{H}\left(\tilde{\mathrm{a}}\right)=\mathrm{H}\left(\tilde{\mathsf{\pi }}\right)$, the ${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$. When $x=k,$ we can prove it in a similar way. □

Property 15.

(Boundedness). Let ${\tilde{a}}_i=({\mu }_{i_j},{\nu }_{i_j}),{\tilde{a}}^{+}=({\mu }_{\max i_j},{\nu }_{\max i_j})(i=1,2,\cdots ,k)$ be a set of PFNs, and ${\tilde{a}}^{-}=({\mu }_{\min i_j},{\nu }_{\min i_j})$, and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1$ then

$${\tilde{a}}^{-}\le {\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\tilde{a}}^{+}$$

(55)

Proof. 

Based on Properties 13 and 14, we have

$$\begin{array}{l}{\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\ge {\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}^{-},{\tilde{a}}^{-},\cdots ,{\tilde{a}}^{-}\right)={\tilde{a}}^{-},\\ {\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}^{+},{\tilde{a}}^{+},\cdots ,{\tilde{a}}^{+}\right)={\tilde{a}}^{+},\end{array}$$

Then we have ${\tilde{a}}^{-}\le {\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\le {\tilde{a}}^{+}.$□

Property 16.

(Commutativity). Let ${\tilde{a}}_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\tilde{\pi }}_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,k)$ be two sets of PFNs. Suppose $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, and weight vector meets ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$ the $\tilde{a}$ and $\tilde{\pi }$ are equal, then we have

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$$

(56)

Proof. 

Because $\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right)$ is any permutation of $\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)$, then

$$\begin{array}{cc}{\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{\pi }}_{i_j}}{x}\right)\right)}^{\frac{1}{C_k^x}}& \left(1\le x<k\right)\\ \underset{i=1}{\overset{x}{\oplus }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\underset{i=1}{\overset{x}{\oplus }}{\tilde{\pi }}_i{}^{\frac{1-{\omega }_i}{k-1}}& \left(x=k\right)\end{array}$$

Thus ${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{\pi }}_1,{\tilde{\pi }}_2,\cdots ,{\tilde{\pi }}_k\right).$ Next, we will discuss some particular cases of WPFDHM operator for different value $x$.

Case 1: When $x=1$, the WPFDHM will reduce to the following form:

$$\begin{array}{l}{\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^1{\omega }_{i_j}}\right)\left(\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}\right)\right)}^{\frac{1}{C_{k-1}^1}}\\ =\left({\left({\displaystyle \prod _{1\le i_1\le k}{\left({\mu }_i\right)}^{\left(1-{\omega }_i\right)}}\right)}^{{}^{\frac{1}{k-1}}},\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1\le k}\left(1-{\left({\nu }_i\right)}^2\right)}}^{\left(1-{\omega }_i\right)}\right)}^{\frac{1}{k-1}}}\right)\end{array}$$

Case 2: When $x=k$, the proposed WPFDHM operator will reduce to the following form:

$${\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(\mathrm{k}\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)=\underset{i=1}{\overset{k}{\oplus }}{\tilde{a}}_i{}^{\frac{1-{\omega }_i}{k-1}}=\left(\sqrt{1-{{\displaystyle \prod _{i=1}^k\left(1-{({\mu }_i)}^2\right)}}^{\frac{1-{\omega }_i}{k-1}}},{\displaystyle \prod _{i=1}^k\left({\nu }_i{}^{\frac{1-{\omega }_i}{k-1}}\right)}\right)$$

Example 6.

Let ${\tilde{a}}_1=\left(0.8,0.2\right),{\tilde{a}}_2=\left(0.6,0.3\right),{\tilde{a}}_3=\left(0.5,0.2\right),{\tilde{a}}_3=\left(0.5,0.4\right)$ be four PFNs. the weighting vector of attributes be $\omega =\left\{0.3,0.2,0.4,0.1\right\}$, Then we use the proposed WPFDHM operator to aggregate four PHNs. (suppose $x=2$)

$$\begin{array}{l}\tilde{a}=\left(\mu ,\nu \right)={\mathrm{WPFDHM}}_{\mathsf{\omega }}^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,\cdots ,{\tilde{a}}_k\right)\\ ={\left(\underset{1\le i_1<\cdots <i_x\le k}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\tilde{a}}_{i_j}}{x}\right)\right)}^{\frac{1}{C_{k-1}^x}}\left(1\le x<k\right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{x}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_{k-1}^x}},\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_x\le k}\left(1-{\left({\left({\displaystyle \prod _{j=1}^x{\nu }_{i_j}}\right)}^{\frac{1}{x}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_{k-1}^x}}}\right)\\ =\left({\left({\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}{\left(\sqrt{1-{\left({\displaystyle \prod _{j=1}^2\left(1-{({\mu }_{i_j})}^2\right)}\right)}^{\frac{1}{2}}}\right)}^{\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)}}\right)}^{\frac{1}{C_3^2}},\sqrt{1-{\left({{\displaystyle \prod _{1\le i_1<\cdots <i_2\le 4}\left(1-{\left({\left({\displaystyle \prod _{j=1}^2{\nu }_{i_j}}\right)}^{\frac{1}{2}}\right)}^2\right)}}^{\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)}\right)}^{\frac{1}{C_3^2}}}\right)\end{array}$$

$$\begin{array}{l}=\left(\begin{array}{l}{\left(\begin{array}{l}{\left({\left(1-{\left(\left(1-{0.8}^2\right)\times \left(1-{0.6}^2\right)\right)}^{0.5}\right)}^{0.5}\right)}^{1-0.3\times 0.2}\times {\left({\left(1-{\left(\left(1-{0.8}^2\right)\times \left(1-{0.5}^2\right)\right)}^{0.5}\right)}^{0.5}\right)}^{1-0.3\times 0.4}\\ \times {\left({\left(1-{\left(\left(1-{0.8}^2\right)\times \left(1-{0.4}^2\right)\right)}^{0.5}\right)}^{0.5}\right)}^{1-0.3\times 0.1}\times {\left({\left(1-{\left(\left(1-{0.6}^2\right)\times \left(1-{0.5}^2\right)\right)}^{0.5}\right)}^{0.5}\right)}^{1-0.4\times 0.2}\\ \times {\left({\left(1-{\left(\left(1-{0.6}^2\right)\times \left(1-{0.5}^2\right)\right)}^{0.5}\right)}^{0.5}\right)}^{1-0.1\times 0.2}\times {\left({\left(1-{\left(\left(1-{0.5}^2\right)\times \left(1-{0.5}^2\right)\right)}^{0.5}\right)}^{0.5}\right)}^{1-0.4\times 0.1}\end{array}\right)}^{\frac{1}{3}},\\ {\left(1-{\left(\begin{array}{l}{\left(1-0.2\times 0.3\right)}^{1-0.3\times 0.2}\times {\left(1-0.2\times 0.2\right)}^{1-0.3\times 0.4}\\ \times {\left(1-0.2\times 0.4\right)}^{1-0.3\times 0.1}\times {\left(1-0.3\times 0.2\right)}^{1-0.2\times 0.4}\\ \times {\left(1-0.3\times 0.4\right)}^{1-0.2\times 0.1}\times {\left(1-0.2\times 0.4\right)}^{1-0.4\times 0.1}\end{array}\right)}^{\frac{1}{3}}\right)}^{0.5}\end{array}\right)\\ =\left(0.6334,0.2636\right)\end{array}$$

At last, we get ${\mathrm{WPFDHM}}_{\omega }^{\left(2\right)}\left({\tilde{a}}_1,{\tilde{a}}_2,{\tilde{a}}_3,{\tilde{a}}_4\right)=\left(0.6334,0.2636\right)$.□

4. A MAGDM Approach Based on the Proposed PFHM Operator

In this part, we apply the WPFHM operator to process the MAGDM problem with the information expressed by PFNs. Let $X=\left\{x_1,x_2,\cdots ,x_m\right\}$ be a set of alternatives, and $C=\left\{c_1,c_2,\cdots ,c_n\right\}$ be a collection attributes, the weighting vector of attributes be $\omega =$ $\left\{{\omega }_1,{\omega }_2\cdots ,{\omega }_n\right\}$, meet ${\omega }_j\in \left[0,1\right],j=1,2,\cdots ,n,{\displaystyle {\sum }_{j=1}^n{\omega }_j}=1$. There are experts $Y=$ $\left\{y_1,y_2,\cdots y_z\right\}$ who are invited to give the evaluation information, and their weighting vector is $w={\left\{w_1,w_2,\cdots w_z\right\}}^T$ with $w_b\in \left[0,1\right],\left(b=1,2,\cdots ,z\right),{\displaystyle {\sum }_{b=1}^z{w}_b}=1$. The expert $y_b$ evaluates each attribute $c_j$ of each alternative $x_i$ by the form of PFN ${\tilde{a}}_{ij}^b=\left({\mu }_{ij}^b,{\nu }_{ij}^b\right)(i=1,2,\cdots ,m,j=$ $1,2,\cdots ,n)$, and then the decision matrix ${\tilde{A}}_b={\left({\tilde{a}}_{ij}^b\right)}_{m\times n}$ $={\left(\left({\mu }_{ij}^b,{\nu }_{ij}^b\right)\right)}_{m\times n}\left(b=1,2,\cdots ,z\right)$ is constructed. The ultimate goal is to give a ranking of all alternatives.

Then, we will give the steps for solving this problem.

Step 1: Based on the WPFHM operator, calculate the collective evaluation value of each attribute for each alternative by ${\tilde{a}}_{ij}^b={\mathrm{WPFHM}}_{\mathrm{w}}\left({\tilde{a}}_{ij}^1,{\tilde{a}}_{ij}^2,\cdots {\tilde{a}}_{ij}^z\right)$

Step 2: Based on the WPFHM operator, calculate the comprehensive decision-making information of each alternative by ${\tilde{a}}_i={\mathrm{WPFHM}}_{\mathsf{\omega }}\left({\tilde{a}}_{i1},{\tilde{a}}_{i2},\cdots {\tilde{a}}_{in}\right)$

Step 3: According to Definition 3, calculate the $S\left(\tilde{a}\right)andH\left(\tilde{a}\right).$

Step 4: Sort all alternatives $\left\{x_1,x_2,\cdots ,x_m\right\}$ and choose the best one.

5. An Illustrate Example

In this section, we will give an example to explain the proposed method. A company wants to select a supplier and now there are four suppliers as candidates $A_i=\left(A_1,A_2,A_3,A_4\right)$. We evaluate each supplier from four aspects $G_i=\left(G_1,G_2,G_3,G_4\right)$, which are “production cost”, “production quality”, “supplier’s service performance”, “risk factor”. The weight vector of attributes is $\omega ={\left(0.1,0.3,0,4,0.2\right)}^T$. There are four experts, and the weight vector of the experts is $w={\left(0.2,0.4,0.1,0.3\right)}^T$. Then the decision matrix ${\tilde{R}}_b={\left({\tilde{a}}_{ij}^b\right)}_{4\times 4}\left(b=1,2,3,4\right)$ are shown in

Table 1. Decision matrix ${\tilde{R}}_1$ .

	${\mathit{G}}_1$	${\mathit{G}}_2$	${\mathit{G}}_3$	${\mathit{G}}_4$
$A_1$	$\left(0.50,0.30\right)$	$\left(0.40,0.20\right)$	$\left(0.50,0.40\right)$	$\left(0.60,0.50\right)$
$A_2$	$\left(0.70,0.20\right)$	$\left(0.60,0.30\right)$	$\left(0.50,0.40\right)$	$\left(0.40,0.30\right)$
$A_3$	$\left(0.80,0.10\right)$	$\left(0.60,0.20\right)$	$\left(0.70,0.20\right)$	$\left(0.90,0.10\right)$
$A_4$	$\left(0.60,0.50\right)$	$\left(0.30,0.40\right)$	$\left(0.50,0.80\right)$	$\left(0.40,0.70\right)$

,

Table 2. Decision matrix ${\tilde{R}}_2$.

	${\mathit{G}}_1$	${\mathit{G}}_2$	${\mathit{G}}_3$	${\mathit{G}}_4$
$A_1$	$\left(0.60,0.50\right)$	$\left(0.70,0.60\right)$	$\left(0.50,0.40\right)$	$\left(0.40,0.20\right)$
$A_2$	$\left(0.80,0.10\right)$	$\left(0.40,0.30\right)$	$\left(0.60,0.10\right)$	$\left(0.80,0.30\right)$
$A_3$	$\left(0.70,0.40\right)$	$\left(0.80,0.20\right)$	$\left(0.70,0.60\right)$	$\left(0.60,0.20\right)$
$A_4$	$\left(0.40,0.60\right)$	$\left(0.50,0.40\right)$	$\left(0.30,0.80\right)$	$\left(0.30,0.60\right)$

,

Table 3. Decision matrix ${\tilde{R}}_3$.

	${\mathit{G}}_1$	${\mathit{G}}_2$	${\mathit{G}}_3$	${\mathit{G}}_4$
$A_1$	$\left(0.40,0.30\right)$	$\left(0.40,0.30\right)$	$\left(0.50,0.70\right)$	$\left(0.40,0.20\right)$
$A_2$	$\left(0.50,0.30\right)$	$\left(0.80,0.60\right)$	$\left(0.50,0.70\right)$	$\left(0.60,0.50\right)$
$A_3$	$\left(0.80,0.20\right)$	$\left(0.40,0.70\right)$	$\left(0.90,0.10\right)$	$\left(0.60,0.30\right)$
$A_4$	$\left(0.50,0.60\right)$	$\left(0.60,0.40\right)$	$\left(0.40,0.60\right)$	$\left(0.50,0.60\right)$

and

Table 4. Decision matrix ${\tilde{R}}_4$.

	${\mathit{G}}_1$	${\mathit{G}}_2$	${\mathit{G}}_3$	${\mathit{G}}_4$
$A_1$	$\left(0.70,0.60\right)$	$\left(0.30,0.40\right)$	$\left(0.40,0.20\right)$	$\left(0.50,0.30\right)$
$A_2$	$\left(0.50,0.80\right)$	$\left(0.70,0.60\right)$	$\left(0.80,0.30\right)$	$\left(0.70,0.20\right)$
$A_3$	$\left(0.70,0.50\right)$	$\left(0.60,0.30\right)$	$\left(0.90,0.10\right)$	$\left(0.80,0.30\right)$
$A_4$	$\left(0.50,0.70\right)$	$\left(0.30,0.50\right)$	$\left(0.40,0.40\right)$	$\left(0.50,0.40\right)$

, and our goal is to rank four suppliers and select the best one.

5.1. Decision-Making Processes

Step 1: Since the four attributes are of the same type, we don’t need to normalize the matrix ${\tilde{R}}_1~{\tilde{R}}_4$.

Step 2: Use WPFHM operator to aggregate four decision matrix ${\tilde{R}}_b={\left({\tilde{a}}_{ij}^b\right)}_{m\times n}$ into a collective matrix $\tilde{R}={\left({\tilde{a}}_{ij}^b\right)}_{m\times n}$ which is listed in

Table 5. The collective decision matrix $\tilde{R}$ .

	${\mathit{G}}_1$	${\mathit{G}}_2$	${\mathit{G}}_3$	${\mathit{G}}_4$
$A_1$	$\left(0.3857,0.1549\right)$	$\left(0.4084,0.2074\right)$	$\left(0.3622,0.1334\right)$	$\left(0.3695,0.1728\right)$
$A_2$	$\left(0.4071,0.1010\right)$	$\left(0.4373,0.0533\right)$	$\left(0.4257,0.3014\right)$	$\left(0.4497,0.2756\right)$
$A_3$	$\left(0.4673,0.0270\right)$	$\left(0.4557,0.1155\right)$	$\left(0.4483,0.1293\right)$	$\left(0.4683,0.1242\right)$
$A_4$	$\left(0.3657,0.3949\right)$	$\left(0.3432,0.3752\right)$	$\left(0.3893,0.3329\right)$	$\left(0.3551,0.2813\right)$

(suppose $x=2$).

Use WPFDHM operator to aggregate four decision matrixes ${\tilde{R}}_b={\left({\tilde{a}}_{ij}^b\right)}_{m\times n}$ into a collective matrix $\tilde{R}={\left({\tilde{a}}_{ij}^b\right)}_{m\times n}$ which is shown in

Table 6. The collective decision matrix $\tilde{R}$ .

	${\mathit{G}}_1$	${\mathit{G}}_2$	${\mathit{G}}_3$	${\mathit{G}}_4$
$A_1$	$\left(0.2751,0.4911\right)$	$\left(0.3346,0.6162\right)$	$\left(0.2014,0.5589\right)$	$\left(0.2609,0.5144\right)$
$A_2$	$\left(0.3357,0.4469\right)$	$\left(0.4803,0.2891\right)$	$\left(0.3954,0.7189\right)$	$\left(0.4951,0.6645\right)$
$A_3$	$\left(0.6078,0.2394\right)$	$\left(0.5182,0.5385\right)$	$\left(0.5584,0.4363\right)$	$\left(0.6114,0.3937\right)$
$A_4$	$\left(0.2323,0.7933\right)$	$\left(0.1602,0.8023\right)$	$\left(0.2737,0.7366\right)$	$\left(0.2046,0.6824\right)$

(suppose $x=2$).

Step 3: Use the WPFHM (WPFDHM) operator to fuse all the attribute values ${\tilde{a}}_{ij},\tilde{a}{\prime }_{ij}(j=1,2,3,4)$ and get the comprehensive evaluation value (suppose $x=2$).

$${\tilde{a}}_1=(0.3015,0.0138),{\tilde{a}}_2=(0.3021,0.0370),{\tilde{a}}_3=(0.3083,0.0443),{\tilde{a}}_4=(0.3003,0.0868).$$

$$\tilde{a}{\prime }_1=(0.0849,0.7369),\tilde{a}{\prime }_2=(0.2034,0.7190),\tilde{a}{\prime }_3=(0.3482,0.5772),\tilde{a}{\prime }_4=(0.0573,0.9129).$$

Step 4: Obtain the score values.

$$S\left({\tilde{a}}_1\right)=0.5382,S\left({\tilde{a}}_2\right)=0.5454,S\left({\tilde{a}}_3\right)=0.5508,S\left({\tilde{a}}_4\right)=0.5272.$$

$$S\left(\tilde{a}{\prime }_1\right)=0.2321,S\left(\tilde{a}{\prime }_2\right)=0.2622,S\left(\tilde{a}{\prime }_3\right)=0.3940,S\left(\tilde{a}{\prime }_4\right)=0.0849.$$

Step 5: Rank all alternatives ${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$, then the best choice is ${\tilde{a}}_3$.

Considering the different parameter values in the WPFHM operator may have an impact on the ordering results, so we calculate the scores produced from the different $x$ and the results are shown in

Table 7. Score and ranking of the alternatives with different parameter values $x$ .

$\mathit{x}$	$\mathbf{Score}\mathbf{of}\mathit{S}\left({\tilde{\mathit{a}}}_{\mathit{i}}\right)$	Ranking
$x=1$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.5693,S\left({\tilde{a}}_2\right)=0.5454,\\ S\left({\tilde{a}}_3\right)=0.7716,S\left({\tilde{a}}_4\right)=0.4473.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
$x=2$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.5382,S\left({\tilde{a}}_2\right)=0.5454,\\ S\left({\tilde{a}}_3\right)=0.5508,S\left({\tilde{a}}_4\right)=0.5272\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
$x=3$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.6529,S\left({\tilde{a}}_2\right)=0.8097,\\ S\left({\tilde{a}}_3\right)=0.9126,S\left({\tilde{a}}_4\right)=0.6028.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
$x=4$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.5251,S\left({\tilde{a}}_2\right)=0.5794,\\ S\left({\tilde{a}}_3\right)=0.7034,S\left({\tilde{a}}_4\right)=0.4030.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$

.

Considering the different parameter values in the WPFDHM operator may have an impact on the ordering results, so we calculate the scores produced from the different $x$ and the results are shown in

Table 8. Score and order of the alternatives with different parameter values $x$ .

$\mathit{x}$	$\mathbf{Score}\mathbf{of}\mathit{S}\left({\tilde{\mathit{a}}}_{\mathit{i}}\right)$	Ranking
$x=1$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.5251,S\left({\tilde{a}}_2\right)=0.5794,\\ S\left({\tilde{a}}_3\right)=0.7034,S\left({\tilde{a}}_4\right)=0.4030.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
$x=2$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.2321,S\left({\tilde{a}}_2\right)=0.2622,\\ S\left({\tilde{a}}_3\right)=0.3940,S\left({\tilde{a}}_4\right)=0.0849.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
$x=3$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.0583,S\left({\tilde{a}}_2\right)=0.0861,\\ S\left({\tilde{a}}_3\right)=0.1941,S\left({\tilde{a}}_4\right)=0.0018.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
$x=4$	$\begin{array}{l}S\left({\tilde{a}}_1\right)=0.8042,S\left({\tilde{a}}_2\right)=0.9026,\\ S\left({\tilde{a}}_3\right)=0.9639,S\left({\tilde{a}}_4\right)=0.6632.\end{array}$	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$

.

From Table 7 and Table 8, we can get following conclusions.

When $x=1$, the sorting of alternatives is ${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4,$ and the best choice is ${\tilde{a}}_3$.

When $x=2,3,4,$ the sorting of alternatives is ${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4,$ and the best choice is ${\tilde{a}}_3$.

5.2. Comparative Analysis

Then, we compare our proposed method with PFWA operator and PFWG operator [60] and the comparative results are listed in

Table 9. Ordering of the green suppliers.

	Ordering
PFWA	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$
PFWG	${\tilde{a}}_3\succ {\tilde{a}}_2\succ {\tilde{a}}_1\succ {\tilde{a}}_4$

.

From Table 9, we can get the same ranking results. However, the PFWA operator and PFWG operator fail to consider the relationship between arguments. Our proposed WPFHM and WPFDHM operators capture the relationship among arguments being aggregated.

6. Conclusions

For this paper, we investigate the MADM problems with PFNs. Then, we utilize the HM operator, DHM operator, WHM operator and WDHM operator to develop the PFHM operator, WPFHM operator, PFDHM operator and WPFDHM operator. The prominent properties of these operators are analyzed. Then, we develop some methods to solve the MADM problems with PFNs. Finally, a practical example for supplier selection is given. In our subsequent works, the extension and application of these operators of PFNs needs to be investigated in other MADM [61,62], risk analysis and uncertain contexts [63,64].