Methods for Multiple Attribute Group Decision Making Based on Intuitionistic Fuzzy Dombi Hamy Mean Operators

Abstract

In this paper, we extended the Hamy mean (HM) operator, the Dombi Hamy mean (DHM) operator, the Dombi dual Hamy mean (DDHM), with the intuitionistic fuzzy numbers (IFNs) to propose the intuitionistic fuzzy Dombi Hamy mean (IFDHM) operator, intuitionistic fuzzy weighted Dombi Hamy mean (IFWDHM) operator, intuitionistic fuzzy Dombi dual Hamy mean (IFDDHM) operator, and intuitionistic fuzzy weighted Dombi dual Hamy mean (IFWDDHM) operator. Following this, the multiple attribute group decision-making (MAGDM) methods are proposed with these operators. To conclude, we utilized an applicable example for the selection of a car supplier to prove the proposed methods.

1. Introduction

Multiple attribute decision making (MADM) is a key branch of decision theory. The definition of intuitionistic fuzzy sets (IFSs) [1,2] has been utilized to deal with uncertainty and imprecision. The introduction of intuitionistic fuzzy entropy by Burillo and Bustince [3] caught the attention of researchers. Xu [4,5] developed a number of aggregation operators with intuitionistic fuzzy numbers (IFNs.) Xu [6] defined the intuitionistic preference relations for multiple attribute group decision making (MAGDM). Li [7] proposed the Linear Programming Technique for Multidimensional Analysis of Preference (LINMAP) models for MADM. Xu [8] developed the Choquet integrals of weighted IFNs. Ye [9] gave some Cosine similarity measures for intuitionistic fuzzy sets (IFSs). Li and Ren [10] considered the amount and reliability of IFNs for MADM. Wei [11] proposed some induced geometric aggregation operators with IFNs. Wei and Zhao [12] gave some induced correlated aggregating operators with IFNs. Wei [13,14] developed the gray relational analysis method for MADM with IFNs. Zhao and Wei [15] defined some Einstein hybrid aggregation operators with IFNs. Garg [16] proposed the generalized interactive geometric interaction operators using Einstein T-norm and T-conorm with IFNs. Chu et al. [17] gave a MAGDM model that considered both the additive consistency and group consensus with IFNs. Wan et al. [18] researched a novel risk attitudinal ranking method for MADM with IFNs. Zhao et al. [19] proposed the VIKOR (VIseKriterijumska Optimizacija I KOmpromisno Resenje) method using IFSs. Liu [20] proposed MADM methods with normal intuitionistic fuzzy interaction operators. Shi [21] developed some constructive methods for intuitionistic fuzzy implication operators. Otay et al. [22] studied the multi-expert performance evaluation of healthcare institutions with intuitionistic fuzzy Analytic Hierarchy Process (AHP) and Data Envelopment Analysis (DEA) methodology. Ai and Xu [23] proposed the multiple definite integrals of intuitionistic fuzzy calculus and isomorphic mappings. Montes et al. [24] defined the entropy measures for IFNs based on divergence. Liu et al. [25] evaluated the commercial bank counterparty credit risk management with IFNs. Some similarity measures and information aggregating operators between intuitionistic fuzzy sets [26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41] and their extension [42,43,44,45,46,47,48,49,50,51,52,53] have been proposed.

Dombi [54] proposed the operations of the Dombi T-norm and T-conorm. Following this, Liu et al. [55] proposed the Dombi operations with IFNs. Chen and Ye [56] proposed the Dombi weighted arithmetic average and geometric average fuse the single-valued neutrosophic numbers (SVNNs). Wei and Wei [57] gave some dombi prioritized weighting aggregation operators with single-valued neutrosophic numbers.

Through existing studies, we can see that the combination Hamy mean (HM) operator [58,59] and Dombi operations are not extended to IFNs so far. In order to develop Hamy mean operators and Dombi operations for IFNs, the main purposes of this study are (1) to develop some Dombi Hamy mean aggregating operators for IFNs and to investigate their properties, and (2) to propose two models to solve the MADM problems based on these operators with IFNs.

To do so, the rest of this paper is organized as follows. In the next section, we introduce some basic concepts of IFSs, Dombi operations and HM operators. In section three we propose some intuition fuzzy Hamy mean operators based on Dombi T-norm and T-conorm. In section four, we have applied these operators to solve the MAGDM problems with IFNs. In section five, a practical example for the selection of a car supplier is given. In section six, we conclude the paper and give some remarks.

2. Preliminaries

In this section, we introduce the concept of IFS, HM operator, and Dombi T-conorm and T-norm.

2.1. Intuitionistic Fuzzy Sets

Definition 1.

Let$X$be a fixed set, with a generic in$X$denoted by$x$. An intuitionistic fuzzy set (IFS)$I$in$X$is following [1,2]:

$$I=\left\{\langle x,{\mu }_i(x),{\nu }_i(x)\rangle x\in X\right\}$$

(1)

where${\mu }_i(x)$is the membership function, and${\nu }_i(x)$is the non-membership function. For each point$x$in$X$, we have${\mu }_i(x),{\nu }_i(x)\in \left[0,1\right]$and$0\le {\mu }_i(x)+{\nu }_i(x)\le 1$.

For each IFS$I$in$X$, let${\pi }_i(x)=1-{\mu }_i(x)-{\nu }_i(x)$,$\forall x\in X$, and we call${\pi }_i(x)$the indeterminacy degree of the element$x$to the set$I$. It can be easily proved that$0\le {\pi }_i(x)\le 1$,$\forall x\in X.$For convenience, we call$k=\left({\mu }_k,{\nu }_k\right)$an IFN, where${\mu }_k\in \left[0,1\right]$,${\nu }_k\in \left[0,1\right]$, and$0\le {\mu }_k+{\nu }_k\le 1$.

Definition 2.

Let$k_1=\left({\mu }_1,{\nu }_1\right)$and$k_2=\left({\mu }_2,{\nu }_2\right)$be two IFNs, then operational laws are defined [4,5].

1.

$k_1\oplus k_2=\left({\mu }_1+{\mu }_2-{\mu }_1{\mu }_2,{\nu }_1{\nu }_2\right)$

2.

$k_1\otimes k_2=\left({\mu }_1{\mu }_2,{\nu }_1+{\nu }_2-{\nu }_1{\nu }_2\right)$

3.

$\lambda k_1=\left(1-{\left(1-{\mu }_1\right)}^{\lambda },{\nu }_1^{\lambda }\right), \lambda >0$

4.

$k_1^{\lambda }=\left({\mu }_1^{\lambda },1-{\left(1-{\nu }_1\right)}^{\lambda }\right), \lambda >0$

Example 1.

Suppose that$k_1=\left(0.4\hspace{1em}0.5\right),k_2=\left(0.6\hspace{1em}0.4\right),$and$\lambda =4$, then we have

1.

$k_1\oplus k_2=\left(0.4+0.6-0.4\times 0.6\hspace{1em}0.5\times 0.4\right)=\left(0.7600\hspace{1em}0.2000\right)$

2.

$k_1\otimes k_2=\left(0.4\times 0.6\hspace{1em}0.5+0.4-0.5\times 0.4\right)=\left(0.2400\hspace{1em}0.7000\right)$

3.

$\lambda k_1=\left(1-{\left(1-0.4\right)}^4,{0.5}^4\right)=\left(0.8704,0.0625\right)$

4.

$k_1^{\lambda }=\left({0.4}^4,1-{\left(1-0.5\right)}^4\right)=\left(0.0256,0.9375\right)$

Definition 3.

Let$k=\left({\mu }_k,{\nu }_k\right)$be an IFN, then a score function is [60]:

$$S\left(k\right)={\mu }_k-{\nu }_k$$

(2)

where$S\left(k\right)\in \left[-1,1\right]$, from (2), we can give the comparison method of IFNs on the basis of the above score function. For the difference${\mu }_k-{\nu }_k$, the larger the$S\left(k\right)$is, the greater the IFN$k$is.

Example 2.

Let$k_1=\left(0.5 0.4\right),k_2=\left(0.6 0.2\right)$be two IFNs, we can get the scores of$k_1$and$k_2$.$S\left(k_1\right)=0.5-0.4=0.1$,$S\left(k_2\right)=0.6-0.2=0.4$, since$S\left(k_2\right)>S\left(k_1\right)$, we get$k_2>k_1$.

Definition 4.

Let$k=\left({\mu }_k,{\nu }_k\right)$be an IFN, then an accuracy function$H$of$k$can be defined as follows [61]:

$$H\left(k\right)={\mu }_k+{\nu }_k$$

(3)

where$H\left(k\right)\in \left[0,1\right]$, for the difference${\mu }_k+{\nu }_k$, the larger the$H\left(k\right)$is, the greater the IFN$k$is.

Xu and Yager [5] develop a comparison method of IFNs.

Definition 5.

Let$k_1=\left({\mu }_1,{\nu }_1\right)$and$k_2=\left({\mu }_2,{\nu }_2\right)$be two IFNs,$S\left(k_1\right)$and$S\left(k_2\right)$are the score function of$k_1$and$k_2$respectively,$H\left(k_1\right)$and$H\left(k_2\right)$are the score function of$k_1$and$k_2$respectively. Then,

(1) 

If$S\left(k_1\right)>S\left(k_2\right)$, then$k_1>k_2$;

(2) 

If$S\left(k_1\right)=S\left(k_2\right)$, then

(3) 

If$H\left(k_1\right)>H\left(k_2\right)$, then$k_1>k_2$;

(4) 

If$H\left(k_1\right)=H\left(k_2\right)$, then$k_1=k_2$.

Example 3.

Let$k_1=\left(0.6,0.3\right),k_2=\left(0.4,0.1\right)$be two IFNs, we can get the scores and the accuracy of$k_1$and$k_2$.$S\left(k_1\right)=0.6-0.3=0.3$,$S\left(k_2\right)=0.4-0.1=0.3$. Since$S\left(k_1\right)=S\left(k_2\right)$, we can’t get the difference of$k_1$and$k_2$, then$H\left(k_1\right)=0.6+0.3=0.9$,$H\left(k_2\right)=0.4+0.1=0.5$, since$H\left(k_1\right)>H\left(k_2\right)$, we can get$k_1>k_2$.

2.2. HM Operator

Definition 6.

The HM operator is defined as follows [58]:

$${\mathrm{HM}}^{(x)}\left(k_1,k_2,\cdots k_n\right)=\frac{{{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left({\displaystyle \prod _{j=1}^x{k}_{i_j}}\right)}}^{\frac{1}{x}}}{C_n^x}$$

(4)

where$x$is a parameter and$x=1,2,\cdots ,n,i_1,i_2,\cdots i_x$are$x$integer values taken from the set$\{1,2,\cdots ,n\}$of$k$integer values,$C_n^x$denotes the binomial coefficient and$C_n^x=\frac{n!}{x!(n-x)!}$. The properties of the operator are shown as follows:

(i) 

When$k_i=k(i=1,2,\cdots ,n),{\mathrm{HM}}^{(x)}\left(k_1,k_2,\cdots k_n\right)=k$;

(ii) 

When$k_i\le {\pi }_i(i=1,2,\cdots ,n),{\mathrm{HM}}^{(x)}\left(k_1,k_2,\cdots k_n\right)\le {\mathrm{HM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots {\pi }_n\right)$;

(iii) 

When$\min \{k_i\}\le {\mathrm{HM}}^{(x)}\left(k_1,k_2,\cdots k_n\right)\le \max \{k_i\}$.

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

(i) 

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

(ii) 

When$x=k$,${\mathrm{HM}}^{(k)}\left(k_1,k_2,\cdots k_k\right)={\left({\displaystyle \prod _{i=1}^n{k}_i}\right)}^{\frac{1}{n}}$, it becomes the geometric mean operator.

2.3. Dombi T-Conorm and T-Norm

Dombi operations involve the Dombi product and Dombi sum, which are special cases of T-norms and T-conorms, respectively.

Definition 7.

Suppose$M=\left\{\langle x,{\mu }_M\left(x\right),{\nu }_M\left(x\right)\rangle \right\}$and$N=\left\{\langle x,{\mu }_N\left(x\right),{\nu }_N\left(x\right)\rangle \right\}$are any two IFNs, then the generalized intersection and generalized union are proposed as follows [54]:

$$M{\cap }_{T,T^{*}}N=\left\{\langle x,T\left({\mu }_M\left(x\right),{\mu }_N\left(x\right)\right),T^{*}\left({\nu }_M\left(x\right),{\nu }_N\left(x\right)\right)\rangle x\in X\right\}$$

(5)

$$M{\cup }_{T,T^{*}}N=\left\{\langle x,T^{*}\left({\mu }_M\left(x\right),{\mu }_N\left(x\right)\right),T\left({\nu }_M\left(x\right),{\nu }_N\left(x\right)\right)\rangle x\in X\right\}$$

(6)

where T denotes a T-norm and$T^{*}$denotes a T-conorm.

Dombi proposed a generator to produce Dombi T-norm and T-conorm which are shown as follows.

$$T_{D,\lambda }\left(x,y\right)=\frac{1}{1+{\left({\left(\frac{1-x}{x}\right)}^{\lambda }+{\left(\frac{1-y}{y}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}$$

(7)

$$T_{D,\lambda }^{*}\left(x,y\right)=1-\frac{1}{1+{\left({\left(\frac{x}{1-x}\right)}^{\lambda }+{\left(\frac{y}{1-y}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}$$

(8)

where $\lambda >0$, $x,y\in \left[0,1\right]$.

Based on the Dombi T-norm and T-conorm, we can give the operational rules of IFNs as follows. Suppose $k_1=\left({\mu }_1,{\nu }_1\right)$ and $k_2=\left({\mu }_2,{\nu }_2\right)$ are any two IFNs, then operational laws of IFNs based on the Dombi T-norm and T-conorm can be defined as follows $\left(\lambda >0\right)$:

• $k_1\oplus k_2=\left(1-\frac{1}{1+{\left({\left(\frac{{\mu }_1}{1-{\mu }_1}\right)}^{\lambda }+{\left(\frac{{\mu }_2}{1-{\mu }_2}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left({\left(\frac{1-{\nu }_1}{{\nu }_1}\right)}^{\lambda }+{\left(\frac{1-{\nu }_2}{{\nu }_2}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$
• $k_1\otimes k_2=\left(\frac{1}{1+{\left({\left(\frac{1-{\mu }_1}{{\mu }_1}\right)}^{\lambda }+{\left(\frac{1-{\mu }_2}{{\mu }_2}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left({\left(\frac{{\nu }_1}{1-{\nu }_1}\right)}^{\lambda }+{\left(\frac{{\nu }_2}{1-{\nu }_2}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$
• $n{k}_1=\left(1-\frac{1}{1+{\left(n{\left(\frac{{\mu }_1}{1-{\mu }_1}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(n{\left(\frac{1-{\nu }_1}{{\nu }_1}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)\hspace{1em}\left(n>0\right)$
• $k_1^n=\left(\frac{1}{1+{\left(n{\left(\frac{1-{\mu }_1}{{\mu }_1}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(n{\left(\frac{{\nu }_1}{1-{\nu }_1}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)\hspace{1em}\left(n>0\right)$

Example 4.

Suppose that$k_1=\left(0.6,0.1\right),k_2=\left(0.7,0.3\right),$and$\lambda =2,n=3$, then we have

(1) 

$k_1\oplus k_2=\left(\begin{array}{l}1-\frac{1}{1+{\left({\left(\frac{0.6}{1-0.6}\right)}^2+{\left(\frac{0.7}{1-0.7}\right)}^2\right)}^{\frac{1}{2}}},\\ \frac{1}{1+{\left({\left(\frac{1-0.1}{0.1}\right)}^2+{\left(\frac{1-0.3}{0.3}\right)}^2\right)}^{\frac{1}{2}}}\end{array}\right)=\left(0.7350,0.0384\right)$

(2) 

$k_1\otimes k_2=\left(\frac{1}{1+{\left({\left(\frac{1-0.6}{0.6}\right)}^2+{\left(\frac{1-0.7}{0.7}\right)}^2\right)}^{\frac{1}{2}}},1-\frac{1}{1+{\left({\left(\frac{0.1}{1-0.1}\right)}^2+{\left(\frac{0.3}{1-0.3}\right)}^2\right)}^{\frac{1}{2}}}\right)=\left(0.5579,0.3069\right)$

(3) 

$n{k}_1=\left(1-\frac{1}{1+{\left(3\times {\left(\frac{0.6}{1-0.6}\right)}^2\right)}^{\frac{1}{2}}},\frac{1}{1+{\left(3\times {\left(\frac{1-0.1}{0.1}\right)}^2\right)}^{\frac{1}{2}}}\right)=\left(0.7221,0.0603\right)$

(4) 

$k_1^n=\left(\frac{1}{1+{\left(3\times {\left(\frac{1-0.6}{0.6}\right)}^2\right)}^{\frac{1}{2}}},1-\frac{1}{1+{\left(3\times {\left(\frac{0.1}{1-0.1}\right)}^2\right)}^{\frac{1}{2}}}\right)=\left(0.4641,0.1613\right)$

3. Intuition Fuzzy Hamy Mean Operators Based on Dombi T-Norm and T-Conorm

In this section, we propose the intuitionistic fuzzy Dombi Hamy mean (IFDHM) operator and intuitionistic fuzzy weighted Dombi Hamy mean (IFWDHM) operator.

3.1. The IFDHM Operator

Definition 8.

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right)\left(i=1,2,\cdots ,n\right)$be a collection of IFNs, then we can define IFDHM operator as follows:

$${\mathrm{IFDHM}}^{\left(x\right)}\left(k_1,k_2,\cdots k_n\right)=\frac{1}{C_n^x}\left(\underset{1\le i_1<\cdots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}\right)$$

(9)

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

Theorem 1.

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right)\left(i=1,2,\cdots ,n\right)$be a collection of IFNs, then the aggregate result of Definition 8 is still an IFN, and have

$$\begin{array}{l}{\mathrm{IFDHM}}^{\left(x\right)}\left(k_1,k_2,\cdots k_n\right)=\frac{1}{C_n^x}\left(\underset{1\le i_1<\cdots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}\right)\\ =\left(1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}, \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\end{array}$$

(10)

Proof. 

• First of all, we prove (10) is kept. According to the operational laws of IFNs, we have

  $$\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}=\left(\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

  (11)

  $${\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}=\left(\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

  (12)
  Moreover,

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

  (13)
  Furthermore,

  $$\begin{array}{l}{\mathrm{IFDHM}}^{\left(x\right)}\left(k_1,k_2,\cdots k_n\right)=\frac{1}{C_n^x}\left(\underset{1\le i_1<\cdots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}\right)\\ =\left(1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\end{array}$$

  (14)
• Next, we prove (10) is an IFN.
  Let

  $$a=1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},b=\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$
  Then we need to prove that the following two conditions which are satisfied,
  (i) $0\le a\le 1,0\le b\le 1$;
  (ii) $0\le a+b\le 1.$
  (i) Since ${\mu }_{i_j}\in \left[0,1\right]$, we can get

  $$\begin{array}{l}1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\ge 1\Rightarrow 1/\left(1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\right)\in \left[0,1\right]\\ \Rightarrow 1-1/\left(1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\right)\in \left[0,1\right]\end{array}$$
  Therefore, $0\le a\le 1$. Similarly, $0\le b\le 1.$
  (ii) Obviously, $0\le a+b\le 1$, then

  $$1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

  $$\le 1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=1$$

We get $0\le a+b\le 1,$ so the aggregated result of Definition 8 is still an IFN. Next we will discuss about some of the properties of the IFDHM operator. □

Property 1 (Idempotency).

If$k_i\left(1,2\cdots ,n\right)$and$k$are IFNs, and$k_i=k=\left({\mu }_i,{\nu }_i\right)$for all$i=1,2\cdots ,n$, then we get

$${\mathrm{IFDHM}}^{\left(x\right)}\left(k_1,k_2,\cdots k_n\right)=k$$

(15)

Proof. 

Since $k=\left(\mu ,\nu \right)$, based on Theorem 1, we have

$$\begin{array}{l}{\mathrm{IFDHM}}^{\left(x\right)}\left(k_1,k_2,\cdots k_n\right)=\frac{1}{C_n^x}\left(\underset{1\le i_1<\cdots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}\right)\\ =\left(1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\end{array}$$

$$\begin{array}{l}=\left(1-\frac{1}{1+{\left(\frac{1}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{1}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\\ =\left(1-\frac{1}{1+{\left(\frac{1}{{\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{1}{{\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)=\left(1-\frac{1}{1+\frac{1}{\frac{1-{\mu }_i}{{\mu }_i}}},\frac{1}{1+\frac{1}{\frac{{\nu }_i}{1-{\nu }_i}}}\right)\\ =\left({\mu }_i,{\nu }_i\right)=\left(\mu ,\nu \right)=k\end{array}$$

 □

Property 2 (Monotonicity).

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

$${\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\ge {\mathrm{IFDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(16)

Proof. 

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

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

$$\Rightarrow \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}\ge \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}$$

$$\begin{array}{l}\Rightarrow 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\ge 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\\ \Rightarrow \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\le \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\\ \Rightarrow 1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\ge 1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}$$

Similarly, we have

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

$$\Rightarrow \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}\ge \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}}^{\lambda }}$$

$$\Rightarrow 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\ge 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}$$

$$\Rightarrow \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\le \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

Let $k={\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)$, $\pi ={\mathrm{IFDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ and $S(k),S(\pi )$ be the score values of $k$ and $\pi$ respectively. Based on the score value of IFN in (2) and the above inequality, we can imply that $S(k)\ge S(\pi ),$ and then we discuss the following cases:

(1)

If $S(k)>S(\pi )$, then we can get ${\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right){>\mathrm{IFDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$.

(2)

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

$$\begin{array}{l}1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\\ =1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}$$

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

$$1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

and

$$\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

Therefore, it follows that

$$\begin{array}{l}H\left(k\right)=1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\\ =1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=H\left(\pi \right)\end{array}$$

$${\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right){=\mathrm{IFDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

 □

Property 3 (Boundedness).

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

$$k^{-}<{\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)<k^{+}$$

(17)

Proof. 

Based on Properties 1 and 2, we have

$$\begin{array}{l}{\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_k\right)\ge {\mathrm{IFDHM}}^{(x)}\left(k^{-},k^{-},\cdots ,k^{-}\right)=k^{-},\\ {\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_k\right)\le {\mathrm{IFDHM}}^{(x)}\left(k^{+},k^{+},\cdots ,k^{+}\right)=k^{+}.\end{array}$$

Then we have $k^{-}<{\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)<k^{+}$. □

Property 4 (Commutativity).

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\pi }_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,n)$be two sets of IFNs. Suppose$\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$is any permutation of$\left(k_1,k_2,\cdots ,k_n\right)$, then

$${\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(18)

Proof. 

Because $\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ is any permutation of $\left(k_1,k_2,\cdots ,k_n\right)$, then $\frac{1}{C_n^x}\left(\underset{1\le i_1<\cdots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}\right)=\frac{1}{C_n^x}\left(\underset{1\le i_1<\cdots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\pi }_{i_j}\right)}^{\frac{1}{x}}\right)$, thus

$${\mathrm{IFDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right).$$

 □

Example 5.

Let$k_1=\left(0.6,0.3\right),k_2=\left(0.5,0.1\right),k_3=\left(0.7,0.2\right),k_4=\left(0.8,0.1\right)$be four IFNs. Then we use the proposed IFDHM operator to aggregate four IFNs (suppose$x=2,\lambda =2$).

Let

$$\begin{array}{l}k={\mathrm{IFDHM}}^{\left(2\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{1}{3}\times \left(\begin{array}{l}\frac{1}{{\left(\frac{1-0.6}{0.6}\right)}^2+{\left(\frac{1-0.5}{0.5}\right)}^2}+\frac{1}{{\left(\frac{1-0.6}{0.6}\right)}^2+{\left(\frac{1-0.7}{0.7}\right)}^2}\\ +\frac{1}{{\left(\frac{1-0.6}{0.6}\right)}^2+{\left(\frac{1-0.8}{0.8}\right)}^2}+\frac{1}{{\left(\frac{1-0.5}{0.5}\right)}^2+{\left(\frac{1-0.7}{0.7}\right)}^2}\\ +\frac{1}{{\left(\frac{1-0.5}{0.5}\right)}^2+{\left(\frac{1-0.8}{0.8}\right)}^2}+\frac{1}{{\left(\frac{1-0.7}{0.7}\right)}^2+{\left(\frac{1-0.8}{0.8}\right)}^2}\end{array}\right)\right)}^{\frac{1}{2}}},\\ \frac{1}{1+{\left(\frac{1}{3}\times \left(\begin{array}{l}\frac{1}{{\left(\frac{0.3}{1-0.3}\right)}^2+{\left(\frac{0.1}{1-0.1}\right)}^2}+\frac{1}{{\left(\frac{0.3}{1-0.3}\right)}^2+{\left(\frac{0.2}{1-0.2}\right)}^2}\\ +\frac{1}{{\left(\frac{0.3}{1-0.3}\right)}^2+{\left(\frac{0.1}{1-0.1}\right)}^2}+\frac{1}{{\left(\frac{0.1}{1-0.1}\right)}^2+{\left(\frac{0.2}{1-0.2}\right)}^2}\\ +\frac{1}{{\left(\frac{0.1}{1-0.1}\right)}^2+{\left(\frac{0.1}{1-0.1}\right)}^2}+\frac{1}{{\left(\frac{0.2}{1-0.2}\right)}^2+{\left(\frac{0.1}{1-0.1}\right)}^2}\end{array}\right)\right)}^{\frac{1}{2}}}\\ =(0.5831,0.3355)\end{array}\right)\end{array}$$

At last, we get${\mathrm{IFDHM}}^{\left(2\right)}\left(k_1,k_2,k_3,k_4\right)=\left(0.5831,0.3355\right)$.

3.2. The IFWDHM 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 IFWDHM operator fails to consider the problem of attribute weights. In order to overcome this defect, we propose the IFWDHM operator.

Definition 9.

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

$${\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\{\begin{array}{cc}\frac{\underset{1\le i_1<\cdots <i_x\le n}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}}{C_{n-1}^x}\hfill & \left(1\le x<n\right)\\ \underset{i=1}{\overset{x}{\otimes }}{k}_i{}^{\frac{1-{\omega }_i}{n-1}}\hfill & \left(x=n\right)\end{array}$$

(19)

Theorem 2.

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

$$\begin{array}{l}{\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\frac{\underset{1\le i_1<\cdots <i_x\le n}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}}{C_{n-1}^x}\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\hspace{1em}\left(1\le x<n\right)\end{array}$$

(20)

or

$$\begin{array}{l}{\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\underset{i=1}{\overset{x}{\otimes }}{k}_i{}^{\frac{1-{\omega }_i}{n-1}}\\ =\left(\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)\hspace{1em}(x=k)\end{array}$$

(21)

Proof. 

(1) First of all, we prove that (20) and (21) are kept. For the first case, when $\left(1\le x<n\right)$, according to the operational laws of IFNs, we get

$$\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}=\left(\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(22)

$${\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}=\left(\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(23)

Thereafter,

$$\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}=\left(\begin{array}{l}1-\frac{1}{1+{\left(\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)$$

(24)

Moreover,

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

(25)

Therefore,

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

(26)

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

$$k_i{}^{\frac{1-{\omega }_i}{n-1}}=\left(\frac{1}{1+{\left(\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(27)

Then,

$$\underset{i=1}{\overset{x}{\otimes }}{k}_i{}^{\frac{1-{\omega }_i}{n-1}}=\left(\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)$$

(28)

(2) Next, we prove the (20) and (21) are IFNs. For the first case, when $1\le x<n$,

Let

$$a=1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}$$

(29)

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

(30)

Then we need prove the following two conditions.

(I) $0\le a\le 1$, $0\le b\le 1$. (II) $0\le a+b\le 1$.

(I) Since $a\in \left[0,1\right]$, we can get

$$1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}\ge 1$$

(31)

$$\Rightarrow \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\in \left[0,1\right]$$

(32)

$$\Rightarrow 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\in \left[0,1\right]$$

(33)

Therefore, $0\le a\le 1$. Similarly, we can get

$$\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\in \left[0,1\right]$$

(34)

Therefore, $0\le b\le 1$.

(II) Since $0\le a+b\le 1$, we can get the following inequality.

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

(35)

For the second case, when $x=n,$ we can easily prove that it is kept. So the aggregation result produced by Definition 9 is still an IFN. Next, we shall deduce some desirable properties of IFWDHM operator. □

Property 5 (Idempotency).

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

$${\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=k$$

(36)

Proof. 

Since $k_i=k=\left(\mu ,\nu \right)$, based on Theorem 2, we get

(1) For the first case, when $1\le x<n$.

$$\begin{array}{l}{\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{1}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left({\displaystyle \sum _{i=1}^x{\omega }_i}\right)}\right)\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left({\displaystyle \sum _{i=1}^x{\omega }_i}\right)}\right)\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\end{array}$$

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

$$\begin{gathered} \begin{array}{l}{\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}\right)\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}\right)\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\end{array} \\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{1}{\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}}\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{1}{\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}}\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right) \\ \begin{array}{l}=\left(1-\frac{1}{1+{\left(\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)\\ =\left(\mu ,\nu \right)\\ =k\end{array} \end{gathered}$$

(2) For the second case, when $x=n$,

$$\begin{array}{l}{\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(\begin{array}{l}\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)=\left(\begin{array}{l}\frac{1}{1+{\left(\frac{n-1}{n-1}{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{n-1}{n-1}{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\end{array}\right)\\ =\left(\mu ,\nu \right)\\ =k\end{array}$$

which proves the idempotency property of the IFWDHM operator. □

Property 6 (Monotonicity).

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\pi }_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,n)$be two sets of IFNs. 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$k$and$\pi$are equal, then we have

$${\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\ge {\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(37)

Proof. 

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

$$\begin{array}{l}{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}\le {\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}\Rightarrow 1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}\ge 1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}\ge \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}\ge \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}\ge \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}}\\ \Rightarrow 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}\ge 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}\\ \Rightarrow \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\le \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\\ \Rightarrow 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\ge 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}$$

Similarly, we have

$$\begin{array}{l}{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}\le {\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}\Rightarrow 1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}\ge 1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}\ge \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}.\\ \Rightarrow \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}\ge \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}}\\ \Rightarrow 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}\ge 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}\\ \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\le \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}$$

Let $a={\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)$, $\pi ={\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ and $S(k),$ $S(\pi )$ be the score values of $a$ and $\pi$ respectively. Based on the score value of IFN in (2) and the above inequality, we can imply that $S\left(k\right)\ge S\left(\pi \right),$ and then we discuss the following cases:

(1) If $S\left(k\right)>S\left(\pi \right),$ then we can get

$${\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)>{\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(2) If $S\left(k\right)=S\left(\pi \right),$ then

$$\begin{gathered} 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}} \\ =1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{{\theta }_j}}{{\mu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{{\theta }_j}}{1-{\nu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}} \end{gathered}$$

Since ${\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ and based on the Equations (2) and (3), we can deduce that

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

and

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

Therefore, it follows that $\mathrm{H}\left(k\right)=\mathrm{H}\left(\mathsf{\pi }\right)$, the ${\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}$ $\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$, When $x=n$, we can prove it in a similar way. □

Property 7 (Boundedness).

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

$$k^{-}\le {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\le k^{+}$$

(38)

Proof. 

Based on Properties 5 and 6, we have

$$\begin{array}{l}{\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\ge {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k^{-},k^{-},\cdots ,k^{-}\right)=k^{-}\\ {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\le {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k^{+},k^{+},\cdots ,k^{+}\right)=k^{+}\end{array}$$

Then we have $k^{-}\le {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\le k^{+}$. □

Property 8 (Commutativity).

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\pi }_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,n)$be two sets of IFNs. Suppose$\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$is any permutation of$\left(k_1,k_2,\cdots ,k_n\right)$, and weight vector meets${\omega }_i\in \left[0,1\right]$and${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$the$k$and$\pi$are equal, then we have

$${\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFWDHM}}_{\omega }^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(39)

Proof. 

Because $\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ is any permutation of $\left(k_1,k_2,\cdots ,k_n\right)$, then

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

Thus, ${\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFWDHM}}_{\omega }^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right).$ □

Example 6.

Let$k_1=\left(0.8,0.2\right),k_2=\left(0.6,0.1\right),k_3=\left(0.7,0.3\right),k_4=\left(0.4,0.2\right)$be four IFNs, the weighting vector of attributes is$\omega =\left\{0.2,0.3,0.4,0.1\right\}$. Then we use the proposed IFWDHM operator to aggregate four IFNs (suppose$x=2,\lambda =2$).

Let

$$\begin{gathered} {\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\frac{\underset{1\le i_1<\cdots <i_x\le n}{\oplus }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\otimes }}{k}_{i_j}\right)}^{\frac{1}{x}}}{C_{n-1}^x} \\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right) \\ =\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{2}{C_{4-1}^2}{\displaystyle \sum _{1\le i_1<\cdots <i_2\le 4}\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^2{\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}^2}}}\right)}^{\frac{1}{2}}},\\ \frac{1}{1+{\left(\frac{2}{C_{4-1}^2}{\displaystyle \sum _{1\le i_1<\cdots <i_2\le 4}\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^2{\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}^2}}}\right)}^{\frac{1}{2}}}\end{array}\right) \\ \begin{array}{l}=\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{2}{3}\left(\begin{array}{l}\frac{1-0.2-0.3}{{\left(\frac{1-0.8}{0.8}\right)}^2+{\left(\frac{1-0.6}{0.6}\right)}^2}+\frac{1-0.2-0.4}{{\left(\frac{1-0.8}{0.8}\right)}^2+{\left(\frac{1-0.7}{0.7}\right)}^2}\\ +\frac{1-0.2-0.1}{{\left(\frac{1-0.8}{0.8}\right)}^2+{\left(\frac{1-0.4}{0.4}\right)}^2}+\frac{1-0.3-0.4}{{\left(\frac{1-0.6}{0.6}\right)}^2+{\left(\frac{1-0.7}{0.7}\right)}^2}\\ +\frac{1-0.3-0.1}{{\left(\frac{1-0.6}{0.6}\right)}^2+{\left(\frac{1-0.4}{0.4}\right)}^2}+\frac{1-0.4-0.1}{{\left(\frac{1-0.7}{0.7}\right)}^2+{\left(\frac{1-0.4}{0.4}\right)}^2}\end{array}\right)\right)}^{\frac{1}{2}}},\\ 1-\frac{1}{1+{\left(\frac{2}{3}\left(\begin{array}{l}\frac{1-0.2-0.3}{{\left(\frac{0.2}{1-0.2}\right)}^2+{\left(\frac{0.1}{1-0.1}\right)}^2}+\frac{1-0.2-0.4}{{\left(\frac{0.2}{1-0.2}\right)}^2+{\left(\frac{0.3}{1-0.3}\right)}^2}\\ +\frac{1-0.2-0.1}{{\left(\frac{0.2}{1-0.2}\right)}^2+{\left(\frac{0.2}{1-0.2}\right)}^2}+\frac{1-0.3-0.4}{{\left(\frac{0.1}{1-0.1}\right)}^2+{\left(\frac{0.3}{1-0.3}\right)}^2}\\ +\frac{1-0.3-0.1}{{\left(\frac{0.1}{1-0.1}\right)}^2+{\left(\frac{0.2}{1-0.2}\right)}^2}+\frac{1-0.4-0.1}{{\left(\frac{0.3}{1-0.3}\right)}^2+{\left(\frac{0.2}{1-0.2}\right)}^2}\end{array}\right)\right)}^{\frac{1}{2}}}\end{array}\right)\\ =\left(0.5302,0.1952\right)\end{array} \end{gathered}$$

At last, we get${\mathrm{IFWDHM}}_{\omega }^{\left(2\right)}\left(k_1,k_2,k_3,k_4\right)=\left(0.5302,0.1952\right)$.

3.3. The IFDDHM Operator

Wu et al. [59] proposed the dual Hamy mean (DHM) operator.

Definition 10.

The DHM operator is defined as follows [59]:

$${\mathrm{DHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(\frac{{\displaystyle \sum _{j=1}^x{k}_{i_j}}}{x}\right)}\right)}^{\frac{1}{C_n^x}}$$

(40)

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

In this section, we will propose the intuitionistic fuzzy Dombi dual Hamy mean DHM (IFDDHM) operator.

Definition 11.

Let$k_i=({\mu }_{i_j},{\nu }_{i_j})\left(i=1,2,\cdots ,n\right)$be a collection of IFNs, then we can define IFDDHM operator as follows:

$${\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\left(\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}$$

(41)

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

Theorem 3.

Let$k_i=({\mu }_i,{\nu }_i)\left(i=1,2,\cdots ,n\right)$be a collection of the IFNs, then the aggregate result of Definition 10 is still an IFNs, and have

$$\begin{array}{l}{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\left(\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\\ =\left(\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\end{array}$$

(42)

Proof. 

(1) First of all, we prove (42) is kept. According to the operational laws of IFNs, we get

$$\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}=\left(1-\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(43)

$$\frac{\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}}{x}=\left(1-\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(44)

Moreover,

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

(45)

Furthermore,

$$\begin{array}{l}{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\left(\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\\ =\left(\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\end{array}$$

(46)

(2) Next, we prove (42) is an IFN.

Let

$$a=\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},b=1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

Then we need to prove that the following two conditions which are satisfied.

(i) $0\le a\le 1,0\le b\le 1$;

(ii) $0\le a+b\le 1$.

(i) Since ${\mu }_{i_j}\in \left[0,1\right]$, we can get

$$1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\ge 1\Rightarrow \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\in \left[0,1\right]$$

Therefore, $0\le a\le 1$. Similarly, we can get $0\le b\le 1$.

(ii) Obviously, $0\le a+b\le 1$, then

$$\begin{array}{l}\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\\ \le \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=1\end{array}$$

We get $0\le a+b\le 1$. so the aggregated result of Definition 10 is still an IFN. Next we will discuss some properties of IFDDHM operator. □

Property 9 (Idempotency).

If$k_i\left(1,2\cdots ,n\right)$and$k$are IFNs, and$k_i=k=\left({\mu }_i,{\nu }_i\right)$for all$i=1,2\cdots ,n$, then we get

$${\mathrm{IFDDHM}}^{\left(x\right)}\left(k_1,k_2,\cdots k_n\right)=k$$

(47)

Proof. 

Since $k=\left(\mu ,\nu \right)$, based on Theorem 3, we have

$$\begin{gathered} \begin{array}{l}{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\left(\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\\ =\left(\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\\ =\left(\frac{1}{1+{\left(\frac{1}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{1}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\end{array} \\ \begin{array}{l}=\left(\frac{1}{1+{\left(\frac{1}{{\left(\frac{{\mu }_i}{1-{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{1}{{\left(\frac{1-{\nu }_i}{{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)=\left(\frac{1}{1+\frac{1}{\frac{{\mu }_i}{1-{\mu }_i}}},1-\frac{1}{1+\frac{1}{\frac{1-{\nu }_i}{{\nu }_i}}}\right)\\ =\left({\mu }_i,{\nu }_i\right)=\left(\mu ,\nu \right)=k\end{array} \end{gathered}$$

 □

Property 10 (Monotonicity).

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

$${\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\ge {\mathrm{IFDDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(48)

Proof. 

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

$$\begin{array}{l}{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }\le {{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }\Rightarrow 1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }\ge 1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }\\ \Rightarrow \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}\ge \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }}\\ \Rightarrow 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}\ge 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}\end{array}$$

Similarly, we have

$$\begin{array}{l}{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }\ge {{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }\Rightarrow 1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }\le 1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }\\ \Rightarrow \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}\le \frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}\\ \Rightarrow 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}\le 1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}\\ \Rightarrow \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\ge \frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\\ \Rightarrow 1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\le 1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}1/{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}$$

Let $k={\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)$, $\pi ={\mathrm{IFDDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ and $S(k),S(\pi )$ be the score values of $k$ and $\pi$ respectively. Based on the score value of IFN in (2) and the above inequality, we can imply that $S(k)\ge S(\pi )$, and then we discuss the following cases:

(1) If $S(k)>S(\pi )$, then we get ${\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right){>\mathrm{IFDDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$.

(2) If $S(k)=S(\pi )$, then

$$\begin{array}{l}\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\\ =\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}$$

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

$$\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

and

$$1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}$$

Therefore, it follows that

$$\begin{array}{l}H\left(k\right)=\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\\ =\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}+1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}=H\left(\pi \right)\end{array}$$

Then ${\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right){=\mathrm{IFDDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right).$ □

Property 11 (Boundedness).

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

$$k^{-}<{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)<k^{+}$$

(49)

Proof. 

Based on Properties 9 and 10, we have

$$\begin{array}{l}{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_k\right)\ge {\mathrm{IFDDHM}}^{(x)}\left(k^{-},k^{-},\cdots ,k^{-}\right)=k^{-},\\ {\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_k\right)\le {\mathrm{IFDDHM}}^{(x)}\left(k^{+},k^{+},\cdots ,k^{+}\right)=k^{+}.\end{array}$$

Then we have $k^{-}<{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)<k^{+}$. □

Property 12 (Commutativity).

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\pi }_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,n)$be two sets of IFNs. Suppose$\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$is any permutation of$\left(k_1,k_2,\cdots ,k_n\right)$, then

$${\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFDDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(50)

Proof. 

Because $\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ is any permutation of $\left(k_1,k_2,\cdots ,k_n\right)$, then

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

thus

$${\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFDDHM}}^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

 □

Example 7.

Let$k_1=\left(0.7,0.3\right),k_2=\left(0.4,0.1\right),k_3=\left(0.5,0.2\right),k_4=\left(0.6,0.2\right)$be four IFNs. Then we use the proposed IFDDHM operator to aggregate four IFNs (suppose$x=2,\lambda =2$).

Let

$$\begin{array}{l}{\mathrm{IFDDHM}}^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\left(\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\\ =\left(\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{x}{C_n^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\frac{1}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\right)\\ =\left(\frac{1}{1+{\left(\frac{2}{C_4^2}{\displaystyle \sum _{1\le i_1<\cdots <i_2\le 4}\frac{1}{{{\displaystyle \sum _{j=1}^4\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^2}}\right)}^{\frac{1}{2}}},1-\frac{1}{1+{\left(\frac{2}{C_4^2}{\displaystyle \sum _{1\le i_1<\cdots <i_4\le 4}\frac{1}{{{\displaystyle \sum _{j=1}^4\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^4}}\right)}^{\frac{1}{2}}}\right)\\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{2}{6}\times \left(\begin{array}{l}\frac{1}{{\left(\frac{0.7}{1-0.7}\right)}^2+{\left(\frac{0.4}{1-0.4}\right)}^2}+\frac{1}{{\left(\frac{0.7}{1-0.7}\right)}^2+{\left(\frac{0.5}{1-0.5}\right)}^2}+\\ \frac{1}{{\left(\frac{0.7}{1-0.7}\right)}^2+{\left(\frac{0.6}{1-0.6}\right)}^2}+\frac{1}{{\left(\frac{0.4}{1-0.4}\right)}^2+{\left(\frac{0.5}{1-0.5}\right)}^2}+\\ \frac{1}{{\left(\frac{0.4}{1-0.4}\right)}^2+{\left(\frac{0.6}{1-0.6}\right)}^2}+\frac{1}{{\left(\frac{0.5}{1-0.5}\right)}^2+{\left(\frac{0.6}{1-0.6}\right)}^2}\end{array}\right)\right)}^{\frac{1}{2}}},\\ 1-\frac{1}{1+{\left(\frac{2}{6}\times \left(\begin{array}{l}\frac{1}{{\left(\frac{1-0.3}{0.3}\right)}^2+{\left(\frac{1-0.1}{0.1}\right)}^2}+\frac{1}{{\left(\frac{1-0.3}{0.3}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}+\\ \frac{1}{{\left(\frac{1-0.3}{0.3}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}+\frac{1}{{\left(\frac{1-0.1}{0.1}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}+\\ \frac{1}{{\left(\frac{1-0.1}{0.1}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}+\frac{1}{{\left(\frac{1-0.2}{0.2}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}\end{array}\right)\right)}^{\frac{1}{2}}}\end{array}\right)\\ =\left(0.5617,0.1596\right)\end{array}$$

At last, we get${\mathrm{IFDDHM}}^{(4)}\left(k_1,k_2,k_3,k_4\right)=\left(0.5617,0.1596\right)$.

3.4. The IFWDDHM 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 IFWDDHM operator fails to consider the problem of attribute weights. In order to overcome this defect, we propose the IFWDDHM operator.

Definition 12.

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

$${\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\{\begin{array}{ll}\frac{\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right){\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}}{C_{n-1}^x}& \left(1\le x<n\right)\\ \underset{i=1}{\overset{x}{\oplus }}{k}_i{}^{\frac{1-{\omega }_i}{n-1}}& \left(x=n\right)\end{array}$$

(51)

Theorem 4.

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right)\left(i=1,2,\cdots ,n\right)$be a group of IFNs, and their weight vector meet${\omega }_i\in \left[0,1\right]$, then the result from Definition 12 is still an IFN, and has

$$\begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\frac{\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right){\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}}{C_{n-1}^x}\\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\hspace{1em}\left(1\le x<n\right)\end{array}$$

(52)

or

$$\begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\underset{i=1}{\overset{x}{\oplus }}{k}_i{}^{\frac{1-{\omega }_i}{n-1}}\\ =\left(1-\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\mu }_i}{1-{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\nu }_i}{{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)(x=n)\end{array}$$

(53)

Proof. 

$$\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}=\left(1-\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left({{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(54)

$${\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}=\left(1-\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left(\frac{1}{x}{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(55)

Thereafter,

$$\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}=\left(\begin{array}{l}\frac{1}{1+{\left(\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\nu }_{i_j}}{1-{\nu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\mu }_{i_j}}{{\mu }_{i_j}}\right)}}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)$$

(56)

Moreover,

$$\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}=\left(\begin{array}{l}\frac{1}{1+{\left({\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left({\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{{\displaystyle \sum _{j=1}^x\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)$$

(57)

Therefore,

$$\begin{array}{l}\frac{1}{C_{n-1}^x}\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right){\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}\\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\end{array}$$

(58)

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

$$k_i{}^{\frac{1-{\omega }_i}{n-1}}=\left(\frac{1}{1+{\left(\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\mu }_i}{{\mu }_i}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\nu }_i}{1-{\nu }_i}\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\right)$$

(59)

Then,

$$\underset{i=1}{\overset{x}{\oplus }}{k}_i{}^{\frac{1-{\omega }_i}{n-1}}=\left(1-\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\mu }_i}{1-{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\nu }_i}{{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)$$

(60)

Next, we prove the (52) and (53) are IFNs. For the first case, when $1\le x<n$,

Let

$$a=\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}$$

$$b=1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}$$

Then we need prove the following two conditions.

(I) $0\le a\le 1$, $0\le b\le 1$;

(II) $0\le a+b\le 1$.

(I) Since $a\in \left[0,1\right]$, we can get

$$\begin{array}{l}1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}>1\\ \Rightarrow \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{x}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\in \left[0,1\right]\end{array}$$

Therefore, $0\le a\le 1$. Similarly, we can get

$$1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\in \left[0,1\right]$$

Therefore, $0\le b\le 1$.

(II) Since $0\le a+b\le 1$, we can get the following inequality.

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

For the second case, when $x=n,$ we can easily prove that it is kept. So the aggregation result produced by Definition 9 is still an IFN. Next, we shall deduce some desirable properties of IFWDDHM operator. □

Property 13 (Idempotency).

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

$${\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=k$$

(61)

Proof. 

Since $k_i=k=\left(\mu ,\nu \right)$, based on Theorem 4, we get

(1) For the first case, when $1\le x<n$.

$$\begin{gathered} \begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\left(\frac{\mu }{1-\mu }\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\left(\frac{1-\nu }{\nu }\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\end{array} \\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\left(\frac{\mu }{1-\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}{\displaystyle \sum _{i=1}^x{\omega }_i}\right)\frac{1}{{\left(\frac{1-\nu }{\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right) \end{gathered}$$

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

$$\begin{gathered} \begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}\right)\frac{1}{{\left(\frac{\nu }{1-\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{1}{C_{n-1}^x}\left(C_n^x-C_{n-1}^{x-1}\right)\frac{1}{{\left(\frac{1-\mu }{\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)\end{array} \\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{1}{\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}}\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}\frac{1}{{\left(\frac{\mu }{1-\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{1}{\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}}\frac{\left(n-1\right)!}{x!\left(k-1-x\right)!}\frac{1}{{\left(\frac{1-\nu }{\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right) \\ \begin{array}{l}=\left(\frac{1}{1+{\left(\frac{1}{{\left(\frac{\mu }{1-\mu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},1-\frac{1}{1+{\left(\frac{1}{{\left(\frac{1-\nu }{\nu }\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\right)\\ =\left(\mu ,\nu \right)\\ =k\end{array} \end{gathered}$$

(2) For the second case, when $x=n$,

$$\begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\\ =\left(\begin{array}{l}1-\frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{{\mu }_i}{1-{\mu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left({\displaystyle \sum _{i=1}^x\left(\frac{1-{\omega }_i}{n-1}\right){\left(\frac{1-{\nu }_i}{{\nu }_i}\right)}^{\lambda }}\right)}^{\frac{1}{\lambda }}}\end{array}\right)=\left(\begin{array}{l}1-\frac{1}{1+{\left(\frac{n-1}{n-1}{\left(\frac{\mu }{1-\mu }\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}},\\ \frac{1}{1+{\left(\frac{n-1}{n-1}{\left(\frac{1-\nu }{\nu }\right)}^{\lambda }\right)}^{\frac{1}{\lambda }}}\end{array}\right)=\left(\mu ,\nu \right)=k\end{array}$$

which proves the idempotency property of the IFWDDHM operator. □

Property 14 (Monotonicity).

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\pi }_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,n)$be two sets of IFNs. 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$k$and$\pi$are equal, then we have

$${\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)\ge {\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(62)

Proof. 

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

$$\begin{array}{l}{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}\ge {\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}\Rightarrow 1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}\le 1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}\le \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}\le \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}}\\ \Rightarrow 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\le 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\\ \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\ge \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}}\end{array}$$

Similarly, we have

$$\begin{array}{l}{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}\ge {\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}\Rightarrow 1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}\le 1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}\le \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}\le \left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}\\ \Rightarrow \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}\le \frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}}\\ \Rightarrow 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\le 1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)1/{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}}\right)}^{\frac{1}{\lambda }}\end{array}$$

Let $a={\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)$, $\pi ={\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ and $S(k)$, $S(\pi )$ be the score values of $a$ and $\pi$ respectively. Based on the score value of IFN in (2) and the above inequality, we can imply that $S\left(k\right)\ge S\left(\pi \right),$ and then we discuss the following cases:

(1) If $S\left(k\right)>S\left(\pi \right),$ then we can get

$${\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)>{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(63)

(2) If $S\left(k\right)=S\left(\pi \right),$ then

$$\begin{gathered} \frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}-\left(1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\right) \\ =\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{{\theta }_j}}{1-{\mu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}-\left(1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{{\theta }_j}}{{\nu }_{{\theta }_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\right) \end{gathered}$$

Since ${\mu }_{i_j}\ge {\mu }_{{\theta }_j}\ge 0,{\nu }_{{\theta }_j}\ge {\nu }_{i_j}\ge 0,$ and based on the Equations (2) and (3), we can deduce

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

and

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

Therefore, it follows that $\mathrm{H}\left(k\right)=\mathrm{H}\left(\mathsf{\pi }\right)$, ${\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}$ $\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$, When $x=n$, we can prove it in a similar way. □

Property 15 (Boundedness).

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

$$k^{-}\le {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\le k^{+}$$

(64)

Proof. 

Based on Properties 13 and 14, we have

$$\begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\ge {\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k^{-},k^{-},\cdots ,k^{-}\right)=k^{-}\\ {\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\le {\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k^{+},k^{+},\cdots ,k^{+}\right)=k^{+}\end{array}$$

Then we have $k^{-}\le {\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)\le k^{+}$. □

Property 16 (Commutativity).

Let$k_i=\left({\mu }_{i_j},{\nu }_{i_j}\right),{\pi }_i=\left({\mu }_{{\theta }_j},{\nu }_{{\theta }_j}\right)(i=1,2,\cdots ,n)$be two sets of IFNs. Suppose$\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$is any permutation of$\left(k_1,k_2,\cdots ,k_n\right)$, and weight vector meets${\omega }_i\in \left[0,1\right]$and${\displaystyle \sum _{i=1}^k{\omega }_i}=1,$the$k$and$\pi$are equal, then we have

$${\mathrm{IFWDDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFWDDHM}}_{\omega }^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$$

(65)

Proof. 

Because $\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$ is any permutation of $\left(k_1,k_2,\cdots ,k_n\right)$, then

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

Thus, ${\mathrm{IFWDHM}}_{\omega }^{(x)}\left(k_1,k_2,\cdots ,k_n\right)={\mathrm{IFWDHM}}_{\omega }^{(x)}\left({\pi }_1,{\pi }_2,\cdots ,{\pi }_n\right)$. □

Example 8.

Let$k_1=\left(0.6,0.1\right),k_2=\left(0.5,0.4\right),k_3=\left(0.8,0.3\right),k_4=\left(0.7,0.2\right)$be four IFNs. Then we use the IFWDDHM operator to fuse four IFNs, the weighting vector of attributes be$\omega =\left\{0.2,0.3,0.4,0.1\right\}$(suppose$x=2,\lambda =2$).

Let

$$\begin{gathered} \begin{array}{l}{\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left(k_1,k_2,\cdots ,k_n\right)=\frac{\underset{1\le i_1<\cdots <i_x\le n}{\otimes }\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right){\left(\underset{j=1}{\overset{x}{\oplus }}{k}_{i_j}\right)}^{\frac{1}{x}}}{C_{n-1}^x}\\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}},\\ 1-\frac{1}{1+{\left(\frac{x}{C_{n-1}^x}{\displaystyle \sum _{1\le i_1<\cdots <i_x\le n}\left(1-{\displaystyle \sum _{j=1}^x{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^x{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^{\lambda }}}}\right)}^{\frac{1}{\lambda }}}\end{array}\right)\end{array} \\ =\left(\begin{array}{l}\frac{1}{1+{\left(\frac{2}{C_3^2}{\displaystyle \sum _{1\le i_1<\cdots <i_2\le 4}\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^2{\left(\frac{{\mu }_{i_j}}{1-{\mu }_{i_j}}\right)}^2}}}\right)}^{\frac{1}{2}}},\\ 1-\frac{1}{1+{\left(\frac{2}{C_3^2}{\displaystyle \sum _{1\le i_1<\cdots <i_2\le 4}\left(1-{\displaystyle \sum _{j=1}^2{\omega }_{i_j}}\right)\frac{1}{{\displaystyle \sum _{j=1}^2{\left(\frac{1-{\nu }_{i_j}}{{\nu }_{i_j}}\right)}^2}}}\right)}^{\frac{1}{2}}}\end{array}\right) \\ \begin{array}{l}=\left(\begin{array}{l}\frac{1}{1+\left(\frac{2}{3}\times +\left(\begin{array}{l}\frac{1-0.2-0.3}{{\left(\frac{0.6}{1-0.6}\right)}^2+{\left(\frac{0.5}{1-0.5}\right)}^2}+\frac{1-0.2-0.4}{{\left(\frac{0.6}{1-0.6}\right)}^2+{\left(\frac{0.8}{1-0.8}\right)}^2}+\\ \frac{1-0.2-0.1}{{\left(\frac{0.6}{1-0.6}\right)}^2+{\left(\frac{0.7}{1-0.7}\right)}^2}+\frac{1-0.3-0.4}{{\left(\frac{0.5}{1-0.5}\right)}^2+{\left(\frac{0.8}{1-0.8}\right)}^2}+\\ \frac{1-0.3-0.1}{{\left(\frac{0.5}{1-0.5}\right)}^2+{\left(\frac{0.7}{1-0.7}\right)}^2}+\frac{1-0.4-0.1}{{\left(\frac{0.8}{1-0.8}\right)}^2+{\left(\frac{0.7}{1-0.7}\right)}^2}\end{array}\right)\right)},\\ 1-\frac{1}{1+\left(\frac{2}{3}\times \left(\begin{array}{l}\frac{1-0.2-0.3}{{\left(\frac{1-0.1}{0.1}\right)}^2+{\left(\frac{1-0.4}{0.4}\right)}^2}+\frac{1-0.2-0.4}{{\left(\frac{1-0.1}{0.1}\right)}^2+{\left(\frac{1-0.3}{0.3}\right)}^2}+\\ \frac{1-0.2-0.1}{{\left(\frac{1-0.1}{0.1}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}+\frac{1-0.3-0.4}{{\left(\frac{1-0.4}{0.4}\right)}^2+{\left(\frac{1-0.3}{0.3}\right)}^2}+\\ \frac{1-0.3-0.1}{{\left(\frac{1-0.4}{0.4}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}+\frac{1-0.4-0.1}{{\left(\frac{1-0.3}{0.3}\right)}^2+{\left(\frac{1-0.2}{0.2}\right)}^2}\end{array}\right)\right)}\end{array}\right)\\ =(0.6592,0.2153)\end{array} \end{gathered}$$

At last, we get${\mathrm{IFWDDHM}}_{\omega }^{\left(2\right)}\left(k_1,k_2,k_3,k_4\right)=\left(0.6592,0.2153\right)$.

4. A MAGDM Approach Based on the Proposed Operators

In this section, we will apply the proposed IFWDHM (IFWDDHM) operator to cope with the MAGDM problem with IFNs. 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 set of 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 $w_t\in \left[0,1\right],\left(t=1,2,\cdots ,z\right),{\displaystyle {\sum }_{t=1}^z{w}_t}=1$. The expert $y_t$ evaluates each attribute $c_j$ of each alternative $x_i$ by the form of IFN $a_{ij}^t=\left({\mu }_{ij}^t,{\nu }_{ij}^t\right)(i=1,2,\cdots ,m,j=1,2,\cdots ,n)$, and then the decision matrix $A_t={\left({\tilde{a}}_{ij}^t\right)}_{m\times n}={\left(\left({\mu }_{ij}^t,{\nu }_{ij}^t\right)\right)}_{m\times n}\left(t=1,2,\cdots ,z\right)$ is constructed. The ultimate goal is to give a ranking of all alternatives.

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

Step1: Calculate the collective evaluation value of each attribute for each alternative by

$${\tilde{a}}_{ij}^t={\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left({\tilde{a}}_{ij}^1,{\tilde{a}}_{ij}^2,\cdots {\tilde{a}}_{ij}^z\right)\mathrm{and}{\tilde{a}}_{ij}^t={\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left({\tilde{a}}_{ij}^1,{\tilde{a}}_{ij}^2,\cdots {\tilde{a}}_{ij}^z\right)$$

Step2: Calculate the overall value of each alternative with the IFWDHM (IFWDDHM) operator

$${\tilde{a}}_i={\mathrm{IFWDHM}}_{\omega }^{\left(x\right)}\left({\tilde{a}}_{i1},{\tilde{a}}_{i2},\cdots {\tilde{a}}_{in}\right),{\tilde{a}}_i={\mathrm{IFWDDHM}}_{\omega }^{\left(x\right)}\left({\tilde{a}}_{i1},{\tilde{a}}_{i2},\cdots {\tilde{a}}_{in}\right)$$

Step3: Calculate the $S\left(\tilde{a}\right)$ and $H\left(\tilde{a}\right).$

Step4: 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 give an example to explain the proposed method. A transportation company wants to pick a car and there are four cars as candidates $M_i=\left(M_1,M_2,M_3,M_4\right)$. We evaluate each supplier from four aspects $E_i=\left(E_1,E_2,E_3,E_4\right)$, which are “production price”, “production quality”, “production’s service performance”, and “risk factor”. The weight vector of attributes is $\omega ={\left(0.1,0.4,0.3,0.2\right)}^T$. There are four experts, and the weight vector of the experts is ${\left(0.3,0.4,0.2,0.1\right)}^T$. Then the decision matrix $R_t={\left(a_{ij}^t\right)}_{4\times 4}\left(t=1,2,3,4\right)$ are shown in

Table 1. Decision matrix $R_1$.

	${\mathit{E}}_{\mathbf{1}}$	${\mathit{E}}_{\mathbf{2}}$	${\mathit{E}}_{\mathbf{3}}$	${\mathit{E}}_{\mathbf{4}}$
$M_1$	(0.5 0.3)	(0.6 0.3)	(0.5 0.2)	(0.6 0.4)
$M_2$	(0.7 0.3)	(0.9 0.1)	(0.8 0.1)	(0.7 0.2)
$M_3$	(0.7 0.2)	(0.5 0.4)	(0.6 0.1)	(0.4 0.2)
$M_4$	(0.5 0.3)	(0.3 0.4)	(0.5 0.4)	(0.4 0.5)

,

Table 2. Decision matrix $R_2$.

	${\mathit{E}}_{\mathbf{1}}$	${\mathit{E}}_{\mathbf{2}}$	${\mathit{E}}_{\mathbf{3}}$	${\mathit{E}}_{\mathbf{4}}$
$M_1$	(0.6 0.2)	(0.7 0.1)	(0.6 0.2)	(0.6 0.3)
$M_2$	(0.9 0.1)	(0.8 0.2)	(0.7 0.1)	(0.6 0.4)
$M_3$	(0.5 0.2)	(0.6 0.3)	(0.7 0.2)	(0.8 0.1)
$M_4$	(0.7 0.2)	(0.4 0.3)	(0.5 0.5)	(0.6 0.3)

,

Table 3. Decision matrix $R_3$.

	${\mathit{E}}_{\mathbf{1}}$	${\mathit{E}}_{\mathbf{2}}$	${\mathit{E}}_{\mathbf{3}}$	${\mathit{E}}_{\mathbf{4}}$
$M_1$	(0.6 0.4)	(0.7 0.2)	(0.6 0.3)	(0.5 0.4)
$M_2$	(0.8 0.6)	(0.7 0.1)	(0.6 0.4)	(0.9 0.1)
$M_3$	(0.5 0.2)	(0.4 0.5)	(0.4 0.3)	(0.5 0.4)
$M_4$	(0.2 0.5)	(0.5 0.4)	(0.7 0.2)	(0.5 0.4)

and

Table 4. Decision matrix $R_4$.

	${\mathit{E}}_{\mathbf{1}}$	${\mathit{E}}_{\mathbf{2}}$	${\mathit{E}}_{\mathbf{3}}$	${\mathit{E}}_{\mathbf{4}}$
$M_1$	(0.6 0.2)	(0.5 0.4)	(0.6 0.4)	(0.4 0.5)
$M_2$	(0.7 0.3)	(0.8 0.1)	(0.6 0.2)	(0.9 0.1)
$M_3$	(0.6 0.4)	(0.3 0.6)	(0.2 0.6)	(0.5 0.3)
$M_4$	(0.3 0.5)	(0.2 0.7)	(0.5 0.4)	(0.3 0.6)

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

5.1. Decision-Making Processes

Step 1: Since the four attributes are of the same type, thus, we don’t need to normalize the matrix $R_1~R_4$.

Step 2: Use IFWDHM operator to fuse four decision matrix $R_t={\left(a_{ij}^t\right)}_{m\times n}$ into a collective matrix $R={\left(a_{ij}^t\right)}_{m\times n}$ which is shown in

Table 5. The collective decision matrix $R$.

	${\mathit{G}}_{\mathbf{1}}$	${\mathit{G}}_{\mathbf{2}}$	${\mathit{G}}_{\mathbf{3}}$	${\mathit{G}}_{\mathbf{4}}$
$A_1$	(0.2976 0.3875)	(0.3504 0.2771)	(0.2156 0.2689)	(0.3996 0.3304)
$A_2$	(0.5818 0.2103)	(0.5000 0.1638)	(0.4172 0.1781)	(0.4554 0.2073)
$A_3$	(0.3282 0.2872)	(0.4554 0.3079)	(0.3095 0.2316)	(0.2857 0.3472)
$A_4$	(0.2411 0.5299)	(0.3671 0.2684)	(0.1813 0.2504)	(0.0371 0.7363)

(suppose $x=2$, $\lambda =2$).

Use IFWDDHM operator to aggregate four decision matrixes $R_t={\left(a_{ij}^t\right)}_{m\times n}$ into a collective matrix $R={\left(a_{ij}^t\right)}_{m\times n}$, which is shown in

Table 6. The collective decision matrix $R$.

	${\mathit{G}}_{\mathbf{1}}$	${\mathit{G}}_{\mathbf{2}}$	${\mathit{G}}_{\mathbf{3}}$	${\mathit{G}}_{\mathbf{4}}$
$A_1$	(0.5372 0.0288)	(0.6100 0.0063)	(0.5618 0.0043)	(0.6535 0.0116)
$A_2$	(0.7000 0.0021)	(0.6667 0.0006)	(0.6422 0.0009)	(0.6646 0.0012)
$A_3$	(0.6300 0.0066)	(0.6344 0.0105)	(0.6066 0.0025)	(0.5969 0.0196)
$A_4$	(0.5827 0.1360)	(0.6729 0.0052)	(0.5818 0.0035)	(0.4172 0.6485)

(suppose $x=2$, $\lambda =2$).

Step 3: Use the IFWDHM (IFWDDHM) operator to aggregate all the attribute values $a_{ij},a{{}^{\prime }}_{ij}(j=1,2,3,4)$ and get the comprehensive evaluation value (suppose $x=2$, $\lambda =2$).

$$\begin{gathered} a_1=(0.0694,0.4051),a_2=(0.5357,0.2264),a_3=(0.1464,0.3736),a_4=(0.0330,0.6366). \\ a{{}^{\prime }}_1=(0.8010,0.0103),a{{}^{\prime }}_2=(0.9380,0.0011),a{{}^{\prime }}_3=(0.8584,0.0087),a{{}^{\prime }}_4=(0.6690,0.0290). \end{gathered}$$

Step 4: Obtain the score values.

$$\begin{gathered} S\left(a_1\right)=-0.3357,S\left(a_2\right)=0.3093,S\left(a_3\right)=-0.2272,S\left(a_4\right)=-\mathrm{0.6036.} \\ S\left(a{{}^{\prime }}_1\right)=0.7907,S\left(a{{}^{\prime }}_2\right)=0.9369,S\left(a{{}^{\prime }}_3\right)=0.8497,S\left(a{{}^{\prime }}_4\right)=\mathrm{0.6399.} \end{gathered}$$

Step 5: Rank all alternatives. $a_2\succ a_3\succ a_1\succ a_4$, then the best choice is $a_2$.

Considering the different parameter values of an IFWDHM operator that may have an impact on the ranking results, we calculated the scores produced from the different $x$ and the results are listed 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
$x=1$	$\begin{array}{l}S\left(a_1\right)=-0.0073,S\left(a_2\right)=0.0352,\\ S\left(a_3\right)=-0.0002,S\left(a_4\right)=-\mathrm{0.0383.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$
$x=2$	$\begin{array}{l}S\left(a_1\right)=-0.3357,S\left(a_2\right)=0.3093,\\ S\left(a_3\right)=-0.2272,S\left(a_4\right)=-0.6036\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$
$x=3$	$\begin{array}{l}S\left(a_1\right)=-0.2988,S\left(a_2\right)=-0.1265,\\ S\left(a_3\right)=-0.2595,S\left(a_4\right)=-\mathrm{0.4853.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$
$x=4$	$\begin{array}{l}S\left(a_1\right)=0.1391,S\left(a_2\right)=0.3007,\\ S\left(a_3\right)=0.1860,S\left({\tilde{a}}_4\right)=-\mathrm{0.1440.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$

.

Considering the different parameter values of an IFWDDHM operator that may have an impact on the ordering results, we calculated the scores with different $x$ and the results are listed in

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

$\mathit{x}$	$\mathbf{Score}\mathbf{of}\mathit{S}\left({\mathit{a}}_{\mathit{i}}\right)$	Ranking
$x=1$	$\begin{array}{l}S\left(a_1\right)=0.0437,S\left(a_2\right)=0.0915,\\ S\left(a_3\right)=0.0548,S\left(a_4\right)=\mathrm{0.0110.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$
$x=2$	$\begin{array}{l}S\left(a_1\right)=0.7907,S\left(a_2\right)=0.9369,\\ S\left(a_3\right)=0.8497,S\left(a_4\right)=\mathrm{0.6399.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$
$x=3$	$\begin{array}{l}S\left(a_1\right)=0.2059,S\left(a_2\right)=0.3597,\\ S\left(a_3\right)=0.2397,S\left(a_4\right)=\mathrm{0.0320.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$
$x=4$	$\begin{array}{l}S\left(a_1\right)=0.2398,S\left(a_2\right)=0.3695,\\ S\left(a_3\right)=0.2477,S\left(a_4\right)=\mathrm{0.0424.}\end{array}$	$a_2\succ a_3\succ a_1\succ a_4$

.

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

When $x=1$, the sorting of alternatives is $a_2\succ a_3\succ a_1\succ a_4$, and the best choice is $a_2$.

When $x=2$, $3$, $4$, the sorting of alternatives is $a_2\succ a_3\succ a_1\succ a_4$, and the best choice is $a_2$.

Although there is the same best selection, the ranking is different. When $x=1,$ the interrelationship between the attributes is not considered, and when $x=2,3,4,$ we can consider the interrelationship for different number of attributes. So these results are reasonable for these two conditions.

5.2. Comparative Analysis

Following this, we compare the proposed method with IFWA operator [4], IFWG operator [5], IFWMM operator [62], and IFDWMM operator [62] and the comparative results are depicted in

Table 9. Ordering of the green suppliers.

	Ordering
IFWA	$a_2\succ a_3\succ a_1\succ a_4$
IFWG	$a_2\succ a_3\succ a_1\succ a_4$
IFWMM	$a_2\succ a_3\succ a_1\succ a_4$
IFDWMM	$a_2\succ a_3\succ a_1\succ a_4$

.

From above analysis, we arrived at the same results. However, the existing operators, such as IFWA operator and IFWG operator do not consider the relationship between arguments, and thus cannot eliminate the corresponding influence of unfair arguments on decision result. The IFWMM operator, IFDWMM operator, IFWDHM and IFWDDHM operators consider the relationship among the arguments.

6. Conclusions

In this paper, we investigated the MADM problems with IFNs. Following this, we utilized the HM operator, DHM operator, DDHM operator, WDHM operator, and WDDHM operator to develop some novel operators with IFNs: Intuitionistic fuzzy DHM (IFDHM) operator, intuitionistic fuzzy WDHM operator, intuitionistic fuzzy DDHM (IFDDHM) operator, and intuitionistic fuzzy WDDHM (IFWDDHM) operator. The prominent characteristic of these proposed operators were studied. Moreover, we have utilized these operators to develop some models to solve the MAGDM problems with IFNs. Finally, a practical example for the selection of a car car supplier was given. In the future, the application of the IFNs needs to be explored in decision-making processes [63,64,65,66,67,68,69,70,71,72], risk analysis [73,74], and other fuzzy environments [75,76,77,78,79,80].