Models for Multiple Attribute Decision Making with Some 2-Tuple Linguistic Pythagorean Fuzzy Hamy Mean Operators

Abstract

The Hamy mean (HM) operator, as a useful aggregation tool, can capture the correlation between multiple integration parameters, and the 2-tuple linguistic Pythagorean fuzzy numbers (2TLPFNs) are a special kind of Pythagorean fuzzy numbers (PFNs), which can easily describe the fuzziness in actual decision making by 2-tuple linguistic terms (2TLTs). In this paper, to consider both Hamy mean (HM) operator and 2TLPFNs, we combine the HM operator, weighted HM (WHM) operator, dual HM (DHM) operator, and dual WHM (DWHM) operator with 2TLPFNs to propose the 2-tuple linguistic Pythagorean fuzzy HM (2TLPFHM) operator, 2-tuple linguistic Pythagorean fuzzy WHM (2TLPFWHM) operator, 2-tuple linguistic Pythagorean fuzzy DHM (2TLPFDHM) operator and 2-tuple linguistic Pythagorean fuzzy DWHM (2TLPFDWHM) operator. Then some multiple attribute decision making (MADM) procedures are developed based on these operators. At last, an applicable example for green supplier selection is given.

1. Introduction

Pythagorean fuzzy set (PFS) [1,2] is the generalization of the intuitionistic fuzzy set (IFS), in which the membership degree (MD) and the non-membership degree (NMD) satisfy the condition that their square sum is equal or less than 1. The PFS has received more and more attention, which has been investigated broadly [3,4,5,6,7,8,9,10,11,12]. Ren et al. [13] developed the Pythagorean fuzzy TODIM (an acronym in Portuguese for Interactive Multi-Criteria Decision Making) approach which consider the DMs’ psychological behaviors. Zhang [14] proposed the hierarchical QUALIFLEX (qualitative flexible multiple criteria method) approach in PFS. TOPSIS (Technique for Order Preference by Similarity to an Ideal Solution) method is used to solve MADM problems in PFS [15,16,17,18,19,20]. Xue et al. [21] proposed the Pythagorean fuzzy LINMAP (Linear Programming Technique for Multidimensional Analysis of Preference) model with entropy theory. Chen [22] developed the Pythagorean fuzzy VIKOR (VIseKriterijumska Optimizacija I KOmpromisno Resenje) models. Peng and Yang [23] proposed the MABAC (multi-attributive border approximation area comparison) method for multiple attribute group decision making (MAGDM) with PFNs. Wan et al. [24] proposed Pythagorean fuzzy mathematical programming method for MAGDM with Pythagorean fuzzy truth degrees. Peng and Dai [25] studied the Pythagorean fuzzy stochastic MADM with prospect theory. Garg [26] proposed linear programming for MADM with interval-valued PFS. Liang et al. [27] gave the projection method for MAGDM with PFNs based on GBM. Some aggregation operators with PFNs are proposed for MADM and decision support [28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44]. Some information measures of PFNs were also investigated [45,46,47,48,49,50,51,52]. The PFS had been generalized to accommodate hesitant fuzzy value [53,54,55], linguistic arguments [56,57,58,59,60,61,62,63,64], interval values [65,66,67], etc.

Although, PFSs theory has been broadly applied to many domains, however, all the above approaches are unsuitable to depict the MD and NMD of an element to a set by 2TLSs. In order to overcome this issue, we develop the definition of 2-tuple linguistic Pythagorean fuzzy sets (2TLPFSs) based on the PFS [1,2] and 2-tuple linguistic terms (2TLTs) [68,69,70,71]. And HM operator [72] and DHM operator [73] are famous aggregation operators which can depict interrelationships among any number of arguments assigned by a variable vector. Therefore, the HM and DHM operators can supply a robust and flexible mechanism to deal with the information fusion in MADM problems. Because 2-tuple linguistic Pythagorean fuzzy numbers (2TLPFNs) can easily describe the fuzzy information, and the HM and DHM operators can capture interrelationships among any number of arguments assigned by a variable vector, it is necessary to extend the HM and DHM operators to deal with the 2TLPFNs.

The aim of this paper is to combine the HM operator, weighted HM (WHM) operator, dual HM (DHM) operator, and dual WHM (DWHM) operator with 2TLPFNs to propose the 2-tuple linguistic Pythagorean fuzzy HM (2TLPFHM) operator, 2-tuple linguistic Pythagorean fuzzy WHM (2TLPFWHM) operator, 2-tuple linguistic Pythagorean fuzzy DHM (2TLPFDHM) operator and 2-tuple linguistic Pythagorean fuzzy DWHM (2TLPFDWHM) operator. Then some multiple attribute decision making (MADM) procedures are developed based on these operators. At last, an applicable example for green supplier selection is given. In order to do so, the rest of this paper is organized as follows. In Section 2, we develop the 2TLPFSs. In Section 3, we develop HM and DHM operators with 2TLPFNs. In Section 4, we present an example for green supplier selection. Conclusions are given in Section 5.

2. Preliminaries

In this section, we briefly introduce some fundamental concepts and theories of the 2TLPFSs [16] based on the PFS [1,2] and 2-tuple linguistic terms (2TLTs) [68,69].

2.1. 2TLTs

Definition 1.

[68,69]. Let$S=\left\{s_i|i=0,1,\dots ,t\right\}$be a linguistic term set with odd cardinality, where$s_i$represents a possible value for a linguistic variable, and it should satisfy the following characteristics:

(1) 

The set is ordered:$s_i>s_j$, if$i>j$;

(2) 

Max operator:$\max \left(s_i,s_j\right)=s_i$, if$s_i\ge s_j$;

(3) 

Min operator:$\min \left(s_i,s_j\right)=s_i$, if$s_i\le s_j$. For example, S can be defined as

$$S=\left\{\begin{array}{l}{s}_0=extremely\hspace{0.17em}poor,s_1=very\hspace{0.17em}poor,s_2=poor,s_3=medium,\\ s_4=good,s_5=very\hspace{0.17em}good,s_6=extremely\hspace{0.17em}good.\end{array}\right\}$$

Herrera and Martinez [68,69] developed the 2-tuple fuzzy linguistic representation model based on the concept of symbolic translation. It is used for representing the linguistic assessment information by means of a 2-tuple $\left(s_i,{\rho }_i\right)$, where $s_i$ is a linguistic label for predefined linguistic term set $S$ and ${\rho }_i$ is the value of symbolic translation, and ${\rho }_i\in \left[-0.5,0.5\right).$

2.2. PFSs

Let $X$ be an ordinary fixed set, denoted by $x$. A Pythagorean fuzzy set (PFS) $A$ in $X$ is defined as following [1,2]:

$$A=\left\{\left(x,u_A\left(x\right),v_A\left(x\right)\right)|x\in X\right\}$$

(1)

where the membership function ${\mathrm{u}}_A\left(x\right)$ and the non-membership function $v_A\left(x\right)$ satisfy $u_A\left(x\right):X\to \left[0,1\right],v_A\left(x\right):X\to \left[0,1\right]$ and ${\left(u_A\left(x\right)\right)}^2+{\left(v_A\left(x\right)\right)}^2\le 1$. Then a simplification of $A$ is denoted by $A=\left\{\langle x,u_A\left(x\right),v_A\left(x\right)\rangle |x\in X\right\}$, which is a PFS.

For a PFS $\left\{\left(x,u_A\left(x\right),v_A\left(x\right)\right)|x\in X\right\}$, the ordered twofold components $\left\{u_A\left(x\right),v_A\left(x\right)\right\}$, are described as a Pythagorean fuzzy number (PFN), and each PFN can be expressed as $A=\left(u_A,v_A\right)$, where $u_A\in \left[0,1\right],v_A\in \left[0,1\right]$ and ${\left(u_A\left(x\right)\right)}^2+{\left(v_A\left(x\right)\right)}^2\le 1$.

2.3. 2TLPFSs

Definition 2.

Assume that$\delta =\left\{{\delta }_0,{\delta }_1,\dots ,{\delta }_t\right\}$is a be a linguistic term set with odd cardinality$t+1$. If$\delta =\left\{\left(s_{\phi },\varphi \right),\left(s_{\vartheta },\theta \right)\right\}$is defined for$s_{\phi },s_{\vartheta }\in \delta \hspace{0.17em}and\hspace{0.17em}\varphi ,\theta \in \left[-0.5,0.5\right),$where$\left(s_{\phi },\varphi \right)and\hspace{0.17em}\left(s_{\vartheta },\theta \right)$express independently the membership degree and non-membership degree by 2-tuple linguistic terms (2TLTs), then 2-tuple linguistic Pythagorean fuzzy sets (2TLPFSs) are defined as follows:

$${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$$

(2)

where$\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)$are2-tuple linguistic terms (2TLTs),$0\le {\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)\le t,0\le {\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)\le t$and$0\le {\left({\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)\right)}^2+{\left({\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)\right)}^2\le t^2.$

Definition 3.

Let${\delta }_1=\left\{\left(s_{{\phi }_1},{\varphi }_1\right),\left(s_{{\vartheta }_1},{\theta }_1\right)\right\}$be a 2TLPFN in$\delta$. Then the score and accuracy functions of${\delta }_1$are given as follows:

$$S\left({\delta }_1\right)=\Delta \left\{t\left({\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_1},{\varphi }_1\right)}{t}\right)}^2-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_1},{\theta }_1\right)}{t}\right)}^2\right)\right\},S\left({\delta }_1\right)\in \left[-t,t\right]$$

(3)

$$H({\delta }_1)=\Delta \left\{t\left({\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_1},{\varphi }_1\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_1},{\theta }_1\right)}{t}\right)}^2\right)\right\},H({\delta }_1)\in \left[0,t^2\right]$$

(4)

Definition 4.

Let${\delta }_1=\left\{\left(s_{{\phi }_1},{\varphi }_1\right),\left(s_{{\vartheta }_1},{\theta }_1\right)\right\}$and${\delta }_2=\left\{\left(s_{{\phi }_2},{\varphi }_2\right),\left(s_{{\vartheta }_2},{\theta }_2\right)\right\}$be two 2TLPFNs, based on the score function S and the accuracy function$H$, in the following, we shall give an order relation between two 2TLPFNs, which is defined as follows:

(1) 

$\mathrm{if}\hspace{0.17em}S\left({\delta }_1\right)<S\left({\delta }_2\right)$,$\mathrm{then}\hspace{0.17em}{\delta }_1<{\delta }_2;$

(2) 

$\mathrm{if}\hspace{0.17em}S\left({\delta }_1\right)>S\left({\delta }_2\right)$,$\mathrm{then}\hspace{0.17em}{\delta }_1>{\delta }_2;$

(3) 

$\mathrm{if}\hspace{0.17em}S\left({\delta }_1\right)=S\left({\delta }_2\right),H\left({\delta }_1\right)<H\left({\delta }_2\right)$,$\mathrm{then}\hspace{0.17em}{\delta }_1<{\delta }_2;$

(4) 

$\mathrm{if}\hspace{0.17em}S\left({\delta }_1\right)=S\left({\delta }_2\right),H\left({\delta }_1\right)>H\left({\delta }_2\right)$,$\mathrm{then}\hspace{0.17em}{\delta }_1>{\delta }_2;$

(5) 

$\mathrm{if}\hspace{0.17em}S\left({\delta }_1\right)=S\left({\delta }_2\right),H\left({\delta }_1\right)=H\left({\delta }_2\right)$,$\mathrm{then}\hspace{0.17em}{\delta }_1={\delta }_2.$

Definition 5.

Let${\delta }_1=\left\{\left(s_{{\phi }_1},{\varphi }_1\right),\left(s_{{\vartheta }_1},{\theta }_1\right)\right\}$and${\delta }_2=\left\{\left(s_{{\phi }_2},{\varphi }_2\right),\left(s_{{\vartheta }_2},{\theta }_2\right)\right\}$be two 2TLPFNs,$\lambda >0$, then some basic operations on them are defined as follows:

(1) 

${\delta }_1\oplus {\delta }_2=\left\{\begin{array}{l}\Delta \left(t\sqrt{1-\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_1},{\varphi }_1\right)}{t}\right)}^2\right)\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_2},{\varphi }_2\right)}{t}\right)}^2\right)}\right),\\ \Delta \left(t\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_1},{\theta }_1\right)}{t}\cdot \frac{{\Delta }^{-1}\left(s_{{\vartheta }_2},{\theta }_2\right)}{t}\right)\right)\end{array}\right\};$

(2) 

${\delta }_1\otimes {\delta }_2=\left\{\begin{array}{l}\Delta \left(t\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_1},{\varphi }_1\right)}{t}\cdot \frac{{\Delta }^{-1}\left(s_{{\phi }_2},{\varphi }_2\right)}{t}\right)\right),\\ \Delta \left(t\sqrt{1-\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_1},{\theta }_1\right)}{t}\right)}^2\right)\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_2},{\theta }_2\right)}{t}\right)}^2\right)}\right)\end{array}\right\};$

(3) 

$\lambda {\delta }_1=\left\{\Delta \left(t\sqrt{1-{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_1},{\varphi }_1\right)}{t}\right)}^2\right)}^{\lambda }}\right),\Delta \left(t{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_1},{\theta }_1\right)}{t}\right)}^{\lambda }\right)\right\};$

(4) 

${\left({\delta }_1\right)}^{\lambda }=\left\{\Delta \left(t{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_1},{\varphi }_1\right)}{t}\right)}^{\lambda }\right),\Delta \left(t\sqrt{1-{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_1},{\theta }_1\right)}{t}\right)}^2\right)}^{\lambda }}\right)\right\};$

2.4. HM Operator

Definition 6.

[72]. The HM operator is given as follows:

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

(5)

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

3. Some 2TLPFHM Operators

3.1. 2TLPFHM Operator

In this section, we expand HM operator with 2TLPFNs and propose the 2-tuple linguistic Pythagorean fuzzy HM (2TLPFHM) operator.

Definition 7.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of2TLPFNs. The 2TLPFHM operator is:

$${2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}}{C_n^x}$$

(6)

Theorem 1.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. The fused result by 2TLPFHM operators is also a 2TLPFN where

$$\begin{array}{c}{2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right),\\ \Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right)\end{array}\right\}\hfill \end{array}$$

(7)

Proof. 

From the basic operations on 2TLPFN which are defined in Definition 5, we can get:

$$\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}=\left\{\Delta \left(t{\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right),\Delta \left(t\sqrt{1-{\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}}\right)\right\}$$

(8)

Thus,

$${\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}=\left\{\Delta \left(t{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{1}{x}}\right),\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}\right)\right\}$$

(9)

Thereafter,

$$\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}=\left\{\begin{array}{l}\Delta \left(t\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}}\right),\\ \Delta \left(t{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)\end{array}\right\}$$

(10)

Therefore,

$$\begin{array}{c}{2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right),\\ \Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right)\end{array}\right\}\hfill \end{array}$$

(11)

Hence, (7) is kept. □

Then we need to give the proving process of that (7) is also a 2TLPFN.

① $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t,0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t.$

② $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2.$

Let

$$\begin{array}{l}\frac{{\Delta }^{-1}\left(s_{\phi },\varphi \right)}{t}=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\\ \frac{{\Delta }^{-1}\left(s_{\vartheta },\theta \right)}{t}={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\end{array}$$

Proof. 

① Since $0\le \frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\le 1,$ we get

$$0\le {\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\le 1\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}0\le 1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\le 1$$

(12)

Then,

$$0\le {\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}\le 1$$

(13)

$$0\le \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\le 1$$

(14)

That means $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t$, similarly, we can have $0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t$. So ① is maintained.

② Since $0\le {\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\le 1$, we can get

$$\begin{array}{c}{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\hfill \\ =\left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)+\left({\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)\hfill \\ \le \left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)+\left({\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)\hfill \\ =1\hfill \end{array}$$

That means $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2,$ so ② is maintained. □

Example 1.

Let$\left\{\left(s_5,0\right),\left(s_2,0\right)\right\},\left\{\left(s_4,0\right),\left(s_3,0\right)\right\},\left\{\left(s_2,0\right),\left(s_5,0\right)\right\}$and$\left\{\left(s_5,0\right),\left(s_1,0\right)\right\}$are four 2TLPFNs, and suppose$x=2$, then according to Equation (7), we have

$$\begin{array}{c}{2\mathrm{TLPFHM}}^{(2)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\Delta \left\{\begin{array}{c}6\times \left(\sqrt{1-{\left(\begin{array}{l}\left(1-{\left(\frac{5}{6}\times \frac{4}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{5}{6}\times \frac{2}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{5}{6}\times \frac{5}{6}\right)}^{\frac{2}{2}}\right)\\ \times \left(1-{\left(\frac{4}{6}\times \frac{2}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{4}{6}\times \frac{5}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{2}{6}\times \frac{5}{6}\right)}^{\frac{2}{2}}\right)\end{array}\right)}^{\frac{1}{C_4^2}}}\right),\hfill \\ 6\times {\left(\begin{array}{c}\sqrt{1-{\left(\left(1-{\left(\frac{2}{6}\right)}^2\right)\times \left(1-{\left(\frac{3}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{2}{6}\right)}^2\right)\times \left(1-{\left(\frac{5}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{2}{6}\right)}^2\right)\times \left(1-{\left(\frac{1}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\hfill \\ \times \sqrt{1-{\left(\left(1-{\left(\frac{3}{6}\right)}^2\right)\times \left(1-{\left(\frac{5}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{3}{6}\right)}^2\right)\times \left(1-{\left(\frac{1}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{5}{6}\right)}^2\right)\times \left(1-{\left(\frac{1}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\hfill \end{array}\right)}^{\frac{1}{C_4^2}}\hfill \end{array}\right\}\hfill \\ =\left\{\left(s_1,0.5955\right),\left(s_2,0.9675\right)\right\}\hfill \end{array}$$

The 2TLPFHM has three properties.

Property 1.

(Idempotency) If${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$are equal, then

$${2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\delta$$

(15)

Proof. 

Since $\delta =\left\{\left(s_{\phi },\varphi \right),\left(s_{\vartheta },\theta \right)\right\}$, then

$$\begin{array}{c}{2\mathrm{TLPFNHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\delta }_{i_j}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t{\left(\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{\phi },\varphi \right)}{t}}\right)}^{\frac{2}{x}}\right)}}\right)}^{\frac{1}{C_n^x}}\right),\\ \Delta \left(t\sqrt{1-{\left(1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{\vartheta },\theta \right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\end{array}\right\}\hfill \end{array}$$

$$\begin{array}{l}=\left\{\begin{array}{l}\Delta \left(t{\left(\sqrt{1-{\left(1-{\left({\left(\frac{{\Delta }^{-1}\left(s_{\phi },\varphi \right)}{t}\right)}^x\right)}^{\frac{2}{x}}\right)}^{\frac{1}{C_n^x}}}\right)}^{\frac{1}{C_n^x}}\right),\\ \Delta \left(t\sqrt{1-{\left(1-{\left(1-{\left({\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{\vartheta },\theta \right)}{t}\right)}^2\right)}^x\right)}^{\frac{1}{x}}\right)}^{\frac{1}{C_n^x}}\right)}^{\frac{1}{C_n^x}}}\right)\end{array}\right\}\\ =\left\{\left(s_{\phi },\varphi \right),\left(s_{\vartheta },\theta \right)\right\}=\delta \end{array}$$

□

Property 2.

(Monotonicity) Let${\delta }_{a_j}=\left\{\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right),\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\right\}(j=1,2,\dots ,n)$and${\delta }_{b_j}=\left\{\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right),\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)\right\}(j=1,2,\dots ,n)$be two sets of 2TLPFNs. If${\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)\le {\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)$$and\hspace{0.17em}{\Delta }^{-1}\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\ge {\Delta }^{-1}\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)$hold for al l$j$, then

$${2\mathrm{TLPFHM}}^{(x)}\left({\delta }_{a_1},{\delta }_{a_2},\cdots ,{\delta }_{a_n}\right)\le {2\mathrm{TLPFHM}}^{(x)}\left({\delta }_{b_1},{\delta }_{b_2},\cdots ,{\delta }_{b_n}\right)$$

(16)

Proof. 

Let ${\delta }_{a_j}=\left\{\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right),\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\right\}$ and ${\delta }_{b_j}=\left\{\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right),\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)\right\}$, suppose that ${\Delta }^{-1}\left(s_{T_{a_j}},{\alpha }_{a_j}\right)\le {\Delta }^{-1}\left(s_{T_{b_j}},{\alpha }_{b_j}\right)$, we can obtain

$${\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)}{t}}\le {\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)}{t}}$$

(17)

$$1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)}{t}}\right)}^{\frac{2}{x}}\ge 1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)}{t}}\right)}^{\frac{2}{x}}$$

(18)

Thereafter,

$${\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}\ge {\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}$$

(19)

Furthermore,

$${\left(\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}}\right)}^{\frac{1}{C_n^x}}\le {\left(\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}}\right)}^{\frac{1}{C_n^x}}$$

(20)

That means ${\Delta }^{-1}\left(s_{{\phi }_a},{\varphi }_a\right)\le {\Delta }^{-1}\left(s_{{\phi }_b},{\varphi }_b\right)$. Similarly, we can get ${\Delta }^{-1}\left(s_{{\vartheta }_a},{\theta }_a\right)\ge {\Delta }^{-1}\left(s_{{\vartheta }_b},{\theta }_b\right)\hspace{0.17em}$.

If ${\Delta }^{-1}\left(s_{{\phi }_a},{\varphi }_a\right)<{\Delta }^{-1}\left(s_{{\phi }_b},{\varphi }_b\right)\hspace{0.17em}and\hspace{0.17em}{\Delta }^{-1}\left(s_{{\vartheta }_a},{\theta }_a\right)\ge {\Delta }^{-1}\left(s_{{\vartheta }_b},{\theta }_b\right)\hspace{0.17em}$

$${2\mathrm{TLPFHM}}^{(x)}\left({\delta }_a,{\delta }_a,\cdots ,{\delta }_a\right)<{2\mathrm{TLPFHM}}^{(x)}\left({\delta }_b,{\delta }_b,\cdots ,{\delta }_b\right)$$

If ${\Delta }^{-1}\left(s_{{\phi }_a},{\varphi }_a\right)={\Delta }^{-1}\left(s_{{\phi }_b},{\varphi }_b\right)\hspace{0.17em}and\hspace{0.17em}{\Delta }^{-1}\left(s_{{\vartheta }_a},{\theta }_a\right)={\Delta }^{-1}\left(s_{{\vartheta }_b},{\theta }_b\right)\hspace{0.17em}$

$$\hspace{0.17em}{2\mathrm{TLPFHM}}^{(x)}\left({\delta }_a,{\delta }_a,\cdots ,{\delta }_a\right){=2\mathrm{TLPFHM}}^{(x)}\left({\delta }_b,{\delta }_b,\cdots ,{\delta }_b\right)$$

So Property 2 is right. □

Property 3.

(Boundedness) Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a set of 2TLPFNs. If${\delta }_i^{+}=\left\{{\max }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\min }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$and${\delta }_i^{-}=\left\{{\min }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\max }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$then

$${\delta }^{-}\le {2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

(21)

From Property 1,

$$\begin{array}{l}{2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1^{-},{\delta }_2^{-},\cdots ,{\delta }_n^{-}\right)={\delta }^{-}\\ {2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1^{+},{\delta }_2^{+},\cdots ,{\delta }_n^{+}\right)={\delta }^{+}\end{array}$$

From Property 2,

$${\delta }^{-}\le {2\mathrm{TLPFHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

3.2. The 2TLPFWHM Operator

In practical MADM, it’s important to consider attribute weights. We shall propose 2-tuple linguistic Pythagorean fuzzy WHM (2TLPFWHM) operator.

Definition 8.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of2TLPFNs with weight vector$w={(w_1,w_2,\dots ,w_n)}^T$, thereby satisfying$w_j\in \left[0,1\right]$and${\displaystyle {\sum }_{j=1}^n{w}_j}=1$. Then we can derive the 2TLPFWHM operator:

$${2\mathrm{TLPFWHM}}_w^{(x)}({\delta }_1,{\delta }_2,\dots ,{\delta }_n)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}}{C_n^x}$$

(22)

Theorem 2.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. The fused result by 2TLPFWHM operator is also a 2TLPFN where

$$\begin{array}{c}{2\mathrm{TLPFWHM}}_w^{(x)}({\delta }_1,{\delta }_2,\dots ,{\delta }_n)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{c}\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right),\hfill \\ \Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right)\hfill \end{array}\right\}\hfill \end{array}$$

(23)

Proof. 

From the basic operations on 2TLPFN which are defined in Definition 5, we can get:

$${\left({\delta }_{i_j}\right)}^{w_{i_j}}=\left\{\Delta \left(t{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}\right),\Delta \left(t\sqrt{1-{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)\right\}$$

(24)

Thus,

$$\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}=\left\{\Delta \left(t{\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right),\Delta \left(t\sqrt{1-{\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}}\right)\right\}$$

(25)

Therefore,

$${\left(\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}=\left\{\begin{array}{l}\Delta \left(t{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right),\\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}\right)\end{array}\right\}$$

(26)

Thereafter,

$$\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}=\left\{\begin{array}{l}\Delta \left(t\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}}\right),\\ \Delta \left(t{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)\end{array}\right\}$$

(27)

Furthermore,

$$\begin{array}{c}{2\mathrm{TLPFWHM}}_w^{(x)}({\delta }_1,{\delta }_2,\dots ,{\delta }_n)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{c}\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right),\hfill \\ \Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right)\hfill \end{array}\right\}\hfill \end{array}$$

(28)

Hence, (23) is kept. □

Then we shall prove that (23) is a 2TLPFN. We shall prove these two conditions.

① $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t,0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t.$

② $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2.$

Let

$$\begin{array}{c}\frac{{\Delta }^{-1}\left(s_{\phi },\varphi \right)}{t}=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\hfill \\ \frac{{\Delta }^{-1}\left(s_{\vartheta },\theta \right)}{t}={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\hfill \end{array}$$

Proof. 

① Since $0\le \frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\le 1,$ we get

$$0\le {\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}\le 1\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}0\le 1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\le 1$$

(29)

Then,

$$0\le {\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\le 1$$

(30)

$$0\le \sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\le 1$$

(31)

That means $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t$, similarly, we can get $0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t$. So ① is maintained.

② Since $0\le {\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\le 1$, we can get

$$\begin{array}{c}{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\hfill \\ =\left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_{i_j}},{\varphi }_{i_j}\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)+{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\hfill \\ \le \left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)+{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{i_j}},{\theta }_{i_j}\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\hfill \\ =1\hfill \end{array}$$

That means $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2,$ so $②$ is maintained. □

Example 2.

Let$\left\{\left(s_5,0\right),\left(s_2,0\right)\right\},\left\{\left(s_4,0\right),\left(s_3,0\right)\right\},\left\{\left(s_2,0\right),\left(s_5,0\right)\right\}$and$\left\{\left(s_5,0\right),\left(s_1,0\right)\right\}$be four 2TLPFNs,$w=\left(0.1,0.3,0.4,0.2\right)$and suppose$x=2$, then according to (23), we have

$$\begin{array}{c}{2\mathrm{TLPFWHM}}_w^{(x)}({\delta }_1,{\delta }_2,\dots ,{\delta }_n)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left({\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\Delta \left\{\begin{array}{c}6\times \left(\sqrt{1-{\left(\begin{array}{c}\left(1-{\left({\left(\frac{5}{6}\right)}^{0.1}\times {\left(\frac{4}{6}\right)}^{0.3}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{5}{6}\right)}^{0.1}\times {\left(\frac{2}{6}\right)}^{0.4}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{5}{6}\right)}^{0.1}\times {\left(\frac{5}{6}\right)}^{0.2}\right)}^{\frac{2}{2}}\right)\hfill \\ \times \left(1-{\left({\left(\frac{4}{6}\right)}^{0.3}\times {\left(\frac{2}{6}\right)}^{0.4}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{4}{6}\right)}^{0.3}\times {\left(\frac{5}{6}\right)}^{0.2}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{2}{6}\right)}^{0.4}\times {\left(\frac{5}{6}\right)}^{0.2}\right)}^{\frac{2}{2}}\right)\hfill \end{array}\right)}^{\frac{1}{C_4^2}}}\right),\hfill \\ 6\times {\left(\begin{array}{c}\sqrt{1-{\left({\left(1-{\left(\frac{2}{6}\right)}^2\right)}^{0.1}\times {\left(1-{\left(\frac{3}{6}\right)}^2\right)}^{0.3}\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left({\left(1-{\left(\frac{2}{6}\right)}^2\right)}^{0.1}\times {\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.4}\right)}^{\frac{1}{2}}}\hfill \\ \times \sqrt{1-{\left({\left(1-{\left(\frac{2}{6}\right)}^2\right)}^{0.1}\times {\left(1-{\left(\frac{1}{6}\right)}^2\right)}^{0.2}\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left({\left(1-{\left(\frac{3}{6}\right)}^2\right)}^{0.3}\times {\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.4}\right)}^{\frac{1}{2}}}\hfill \\ \times \sqrt{1-{\left({\left(1-{\left(\frac{3}{6}\right)}^2\right)}^{0.3}\times {\left(1-{\left(\frac{1}{6}\right)}^2\right)}^{0.2}\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left({\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.4}\times {\left(1-{\left(\frac{1}{6}\right)}^2\right)}^{0.2}\right)}^{\frac{1}{2}}}\hfill \end{array}\right)}^{\frac{1}{C_4^2}}\hfill \end{array}\right\}\hfill \\ =\left\{\left(s_5,0.3725\right),\left(s_1,0.6658\right)\right\}\hfill \end{array}$$

Then we shall discuss two properties of 2TLPFWHM operator.

Property 4.

(Monotonicity) Let${\delta }_{a_j}=\left\{\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right),\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\right\}(j=1,2,\dots ,n)$and${\delta }_{b_j}=\left\{\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right),\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)\right\}(j=1,2,\dots ,n)$be two groups of 2TLPFNs. If${\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)\le {\Delta }^{-1}\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)$$and\hspace{0.17em}{\Delta }^{-1}\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\ge {\Delta }^{-1}\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)$hold for al l$j$, then

$${2\mathrm{TLPFWHM}}^{(x)}\left({\delta }_{a_1},{\delta }_{a_2},\cdots ,{\delta }_{a_n}\right)\le {2\mathrm{TLPFWHM}}^{(x)}\left({\delta }_{b_1},{\delta }_{b_2},\cdots ,{\delta }_{b_n}\right)$$

(32)

The proof is similar to 2TLPFWHM, it’s omitted.

Property 5.

(Boundedness) Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. If${\delta }_i^{+}=\left\{{\max }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\min }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$and${\delta }_i^{-}=\left\{{\min }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\max }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$then

$${\delta }^{-}\le {2\mathrm{TLPFWHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

(33)

From Theorem 2, we get

$$\begin{array}{c}{2\mathrm{TLPFWHM}}_w^{(x)}(\left({\delta }_1^{-},{\delta }_2^{-},\cdots ,{\delta }_n^{-}\right))=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left(\min {\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t{\left(\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\displaystyle \prod _{j=1}^x{\left(\frac{\min {\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^{2w_{i_j}}}\right)}}\right)}^{\frac{1}{C_n^x}}\right),\\ \Delta \left(t\sqrt{1-{\left(1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{\max {\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\end{array}\right\}\hfill \end{array}$$

(34)

$$\begin{array}{c}{2\mathrm{TLPFWHM}}_w^{(x)}\left({\delta }_1^{+},{\delta }_2^{+},\cdots ,{\delta }_n^{+}\right)=\frac{\underset{1\le i_1<\dots <i_x\le n}{\oplus }{\left(\underset{j=1}{\overset{x}{\otimes }}{\left(\max {\delta }_{i_j}\right)}^{w_{i_j}}\right)}^{\frac{1}{x}}}{C_n^x}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t{\left(\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\displaystyle \prod _{j=1}^x{\left(\frac{\max {\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^{2w_{i_j}}}\right)}}\right)}^{\frac{1}{C_n^x}}\right),\\ \Delta \left(t\sqrt{1-{\left(1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{\min {\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\end{array}\right\}\hfill \end{array}$$

(35)

From Property 4, we get

$${\delta }^{-}\le {2\mathrm{TLPFWHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

(36)

It’s obvious that 2TLPFWHM operator doesn’t have the property of idempotency.

3.3. The 2TLPFDHM Operator

Wu et al. [73] proposed the DHM operator.

Definition 9.

[73]. The DHM operator is shown as follows:

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

(37)

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

Then, we shall propose the 2-tuple linguistic Pythagorean fuzzy DHM (2TLPFDHM) operator.

Definition 10.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. The 2TLPFDHM operator is:

$${2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}$$

(38)

Theorem 3.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a set of 2TLPFNs. The fused result by 2TLPFDHM operators is also a 2TLPFN where

$$\begin{array}{c}{2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right),\\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\end{array}\right\}\hfill \end{array}$$

(39)

Proof. 

From the basic operations on 2TLPFN which are defined in Definition 5, we can get:

$$\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}=\left\{\begin{array}{l}\Delta \left(t\sqrt{1-{\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}}\right),\\ \Delta \left(t{\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)\end{array}\right\}$$

(40)

Thus,

$$\frac{\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}}{x}=\left\{\begin{array}{l}\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}\right),\\ \Delta \left(t{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{1}{x}}\right)\end{array}\right\}$$

(41)

Thereafter,

$$\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}}{x}\right)=\left\{\begin{array}{l}\Delta \left(t{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right),\\ \Delta \left(t\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}}\right)\end{array}\right\}$$

(42)

Therefore,

$$\begin{array}{c}{2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\hfill \\ =\left\{\begin{array}{l}\Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right),\\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\end{array}\right\}\hfill \end{array}$$

(43)

Hence, (39) is kept. □

Then we shall prove that (39) is a 2TLPFN. We shall prove these two conditions.

① $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t,0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t.$

② $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2.$

Let

$$\begin{array}{l}\frac{{\Delta }^{-1}\left(s_{\phi },\varphi \right)}{t}={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\\ \frac{{\Delta }^{-1}\left(s_{\vartheta },\theta \right)}{t}=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\end{array}$$

Proof. 

① Since $0\le \frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\le 1,$ we get

$$0\le {\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\le 1\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}0\le 1-\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)\le 1$$

(44)

Then,

$$0\le {\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\le 1$$

(45)

$$0\le {\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\le 1$$

(46)

That means $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t$, similarly, we can derive $0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t$. So ① is maintained.

② Since $0\le {\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\le 1$, we can get

$$\begin{array}{c}{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\hfill \\ ={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_{{}_j}\right)}{t}\right)}^2\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}+\left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)\hfill \\ \le {\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\left(1-\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}\right)}^2\right)\right)}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}+\left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x\frac{{\Delta }^{-1}\left(s_{{\vartheta }_{{}_j}},{\theta }_{{}_j}\right)}{t}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)\hfill \\ =1\hfill \end{array}$$

That means $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2,$ so $②$ is maintained. □

Example 3.

Let$\left\{\left(s_5,0\right),\left(s_2,0\right)\right\},\left\{\left(s_4,0\right),\left(s_3,0\right)\right\},\left\{\left(s_2,0\right),\left(s_5,0\right)\right\}$and$\left\{\left(s_5,0\right),\left(s_1,0\right)\right\}$be four 2TLPFNs, and suppose$x=2$, then according to (39), we get

$$\begin{array}{c}{2\mathrm{TLPFDHM}}^{(2)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\hfill \\ =\Delta \left\{\begin{array}{c}6\times {\left(\begin{array}{c}\sqrt{1-{\left(\left(1-{\left(\frac{5}{6}\right)}^2\right)\times \left(1-{\left(\frac{4}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{5}{6}\right)}^2\right)\times \left(1-{\left(\frac{2}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{5}{6}\right)}^2\right)\times \left(1-{\left(\frac{5}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\hfill \\ \times \sqrt{1-{\left(\left(1-{\left(\frac{4}{6}\right)}^2\right)\times \left(1-{\left(\frac{2}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{4}{6}\right)}^2\right)\times \left(1-{\left(\frac{5}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left(\left(1-{\left(\frac{2}{6}\right)}^2\right)\times \left(1-{\left(\frac{5}{6}\right)}^2\right)\right)}^{\frac{1}{2}}}\hfill \end{array}\right)}^{\frac{1}{C_4^2}},\hfill \\ 6\times \sqrt{1-{\left(\begin{array}{l}\left(1-{\left(\frac{2}{6}\times \frac{3}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{2}{6}\times \frac{5}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{2}{6}\times \frac{1}{6}\right)}^{\frac{2}{2}}\right)\\ \times \left(1-{\left(\frac{3}{6}\times \frac{5}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{3}{6}\times \frac{1}{6}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left(\frac{5}{6}\times \frac{1}{6}\right)}^{\frac{2}{2}}\right)\end{array}\right)}^{\frac{1}{C_4^2}}}\hfill \end{array}\right\}\hfill \\ =\left\{\left(s_4,0.2590\right),\left(s_2,0.6847\right)\right\}\hfill \end{array}$$

Similar to 2TLPFHM operator, we can get three properties.

Property 6.

(Idempotency) If${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$are equal, then

$${2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)=\delta$$

(47)

Property 7.

(Monotonicity) Let${\delta }_{a_j}=\left\{\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right),\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\right\}(j=1,2,\dots ,n)$and${\delta }_{b_j}=\left\{\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right),\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)\right\}(j=1,2,\dots ,n)$be two groups of 2TLPFNs. If${\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)\le {\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)$$and\hspace{0.17em}{\Delta }^{-1}\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\ge {\Delta }^{-1}\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)$hold for al l$j$, then

$${2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_{a_1},{\delta }_{a_2},\cdots ,{\delta }_{a_n}\right)\le {2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_{b_1},{\delta }_{b_2},\cdots ,{\delta }_{b_n}\right)$$

(48)

Property 8.

(Boundedness) Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. If${\delta }_i^{+}=\left\{{\max }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\min }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$and${\delta }_i^{-}=\left\{{\min }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\max }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$then

$${\delta }^{-}\le {2\mathrm{TLPFDHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

(49)

3.4. The 2TLPFDWHM Operator

In practical MADM, it’s very important to consider attribute weights; we shall propose a 2-tuple linguistic Pythagorean fuzzy DWHM (2TLPFDWHM) operator.

Definition 11.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs with weight vector be$w_k={(w_1,w_2,\dots ,w_n)}^T$, thereby satisfying$w_k\in \left[0,1\right]$and${\displaystyle {\sum }_{k=1}^n{w}_k}=1$. If

$${2\mathrm{TLPFWDHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}$$

(50)

Theorem 4.

Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. The fused result by 2TLPFDWHM operators is also a 2TLPFN where

$$\begin{array}{c}{2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\hfill \\ =\left\{\begin{array}{c}\Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right),\hfill \\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\hfill \end{array}\right\}\hfill \end{array}$$

(51)

Proof. 

From the basic operations on 2TLPFN which are defined in Definition 5, we can get:

$$w_{i_j}{\delta }_{i_j}=\left\{\Delta \left(t\sqrt{1-{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right),\Delta \left(t{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}\right)\right\}$$

(52)

Then,

$$\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}=\left\{\Delta \left(t\sqrt{1-{\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}}\right),\Delta \left(t{\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)\right\}$$

(53)

Thus,

$$\frac{\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}}{x}=\left\{\Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}\right),\Delta \left(t{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right)\right\}$$

(54)

Therefore,

$$\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}}{x}\right)=\left\{\begin{array}{l}\Delta \left(t{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right),\\ \Delta \left(t\sqrt{1-{\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}}\right)\end{array}\right\}$$

(55)

Therefore,

$$\begin{array}{c}{2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\hfill \\ =\left\{\begin{array}{c}\Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right),\hfill \\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\hfill \end{array}\right\}\hfill \end{array}$$

(56)

Hence, (51) is kept. □

Then we shall prove that (51) is a 2TLPFN. We shall prove two conditions.

① $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t,0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t.$

② $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2.$

Let

$$\begin{array}{c}\frac{{\Delta }^{-1}\left(s_{\phi },\varphi \right)}{t}={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\hfill \\ \frac{{\Delta }^{-1}\left(s_{\vartheta },\theta \right)}{t}=\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\hfill \end{array}$$

Proof. 

Since $0\le \frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\le 1,$ we get

$$0\le {\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\le 1\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}0\le 1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\le 1$$

(57)

Then,

$$0\le {\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\le 1$$

(58)

$$0\le {\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\le 1$$

(59)

That means $0\le {\Delta }^{-1}\left(s_{\phi },\varphi \right)\le t$, similarly, we can get $0\le {\Delta }^{-1}\left(s_{\vartheta },\theta \right)\le t$. So ① is maintained.

② Since $0\le {\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\le 1$, we can get

$$\begin{array}{c}{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2+{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\hfill \\ ={\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}+\left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)\hfill \\ \le {\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(1-\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^2\right)\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}\right)}\right)}^{\frac{1}{C_n^x}}+\left(1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{{\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}\right)\hfill \\ =1\hfill \end{array}$$

That means $0\le {\left({\Delta }^{-1}\left(s_{\phi },\varphi \right)\right)}^2+{\left({\Delta }^{-1}\left(s_{\vartheta },\theta \right)\right)}^2\le t^2,$ so $②$ is maintained. □

Example 4.

Let$\left\{\left(s_5,0\right),\left(s_2,0\right)\right\},\left\{\left(s_4,0\right),\left(s_3,0\right)\right\},\left\{\left(s_2,0\right),\left(s_5,0\right)\right\}$and$\left\{\left(s_5,0\right),\left(s_1,0\right)\right\}$be four 2TLPFNs,$w=\left(0.1,0.3,0.4,0.2\right)$and suppose$x=2$, then according to (51), we have

$$\begin{array}{c}{2\mathrm{TLPFDWHM}}^{(2)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)={\left(\underset{1\le i_1<\dots <i_x\le n}{\otimes }\left(\frac{\underset{j=1}{\overset{x}{\oplus }}{w}_{i_j}{\delta }_{i_j}}{x}\right)\right)}^{\frac{1}{C_n^x}}\hfill \\ =\Delta \left\{\begin{array}{c}6\times {\left(\begin{array}{c}\sqrt{1-{\left({\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.1}\times {\left(1-{\left(\frac{4}{6}\right)}^2\right)}^{0.3}\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left({\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.1}\times {\left(1-{\left(\frac{2}{6}\right)}^2\right)}^{0.4}\right)}^{\frac{1}{2}}}\hfill \\ \times \sqrt{1-{\left({\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.1}\times {\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.2}\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left({\left(1-{\left(\frac{4}{6}\right)}^2\right)}^{0.3}\times {\left(1-{\left(\frac{2}{6}\right)}^2\right)}^{0.4}\right)}^{\frac{1}{2}}}\hfill \\ \times \sqrt{1-{\left({\left(1-{\left(\frac{4}{6}\right)}^2\right)}^{0.3}\times {\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.2}\right)}^{\frac{1}{2}}}\times \sqrt{1-{\left({\left(1-{\left(\frac{2}{6}\right)}^2\right)}^{0.4}\times {\left(1-{\left(\frac{5}{6}\right)}^2\right)}^{0.2}\right)}^{\frac{1}{2}}}\hfill \end{array}\right)}^{\frac{1}{C_4^2}},\hfill \\ 6\times \sqrt{1-{\left(\begin{array}{l}\left(1-{\left({\left(\frac{2}{6}\right)}^{0.1}\times {\left(\frac{3}{6}\right)}^{0.3}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{2}{6}\right)}^{0.1}\times {\left(\frac{5}{6}\right)}^{0.4}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{2}{6}\right)}^{0.1}\times {\left(\frac{1}{6}\right)}^{0.2}\right)}^{\frac{2}{2}}\right)\\ \times \left(1-{\left({\left(\frac{3}{6}\right)}^{0.3}\times {\left(\frac{5}{6}\right)}^{0.4}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{3}{6}\right)}^{0.3}\times {\left(\frac{1}{6}\right)}^{0.2}\right)}^{\frac{2}{2}}\right)\times \left(1-{\left({\left(\frac{5}{6}\right)}^{0.4}\times {\left(\frac{1}{6}\right)}^{0.2}\right)}^{\frac{2}{2}}\right)\end{array}\right)}^{\frac{1}{C_4^2}}}\hfill \end{array}\right\}\hfill \\ =\left\{\left(s_2,0.1564\right),\left(s_5,0.0455\right)\right\}\hfill \end{array}$$

Then we shall derive some properties of 2TLPFDWHM operator.

Property 9.

(Monotonicity) Let${\delta }_{a_j}=\left\{\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right),\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\right\}(j=1,2,\dots ,n)$and${\delta }_{b_j}=\left\{\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right),\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)\right\}(j=1,2,\dots ,n)$be two groups of 2TLPFNs. If${\Delta }^{-1}\left(s_{{\phi }_{a_j}},{\varphi }_{a_j}\right)\le {\Delta }^{-1}\left(s_{{\phi }_{b_j}},{\varphi }_{b_j}\right)$$and\hspace{0.17em}{\Delta }^{-1}\left(s_{{\vartheta }_{a_j}},{\theta }_{a_j}\right)\ge {\Delta }^{-1}\left(s_{{\vartheta }_{b_j}},{\theta }_{b_j}\right)$hold for all$j$, then

$${2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_{a_1},{\delta }_{a_2},\cdots ,{\delta }_{a_n}\right)\le {2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_{b_1},{\delta }_{b_2},\cdots ,{\delta }_{b_n}\right)$$

(60)

Property 10.

(Boundedness) Let${\delta }_j=\left\{\left(s_{{\phi }_j},{\varphi }_j\right),\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}(j=1,2,\dots ,n)$be a group of 2TLPFNs. If${\delta }_i^{+}=\left\{{\max }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\min }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$and${\delta }_i^{-}=\left\{{\min }_i\left(s_{{\phi }_j},{\varphi }_j\right),{\max }_i\left(s_{{\vartheta }_j},{\theta }_j\right)\right\}$then

$${\delta }^{-}\le {2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

(61)

From Theorem 4,

$$\begin{array}{c}{2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_1^{+},{\delta }_2^{+},\cdots ,{\delta }_n^{+}\right)\hfill \\ =\left\{\begin{array}{c}\Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{\max {\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right),\hfill \\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{\min {\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\hfill \end{array}\right\}\hfill \end{array}$$

(62)

$$\begin{array}{c}{2\mathrm{TLPFWDHM}}^{(x)}\left({\delta }_1^{-},{\delta }_2^{-},\cdots ,{\delta }_n^{-}\right)\hfill \\ =\left\{\begin{array}{c}\Delta \left(t{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\sqrt{1-{\left({\displaystyle \prod _{j=1}^x{\left(1-{\left(\frac{\min {\Delta }^{-1}\left(s_{{\phi }_j},{\varphi }_j\right)}{t}\right)}^2\right)}^{w_{i_j}}}\right)}^{\frac{1}{x}}}}\right)}^{\frac{1}{C_n^x}}\right),\hfill \\ \Delta \left(t\sqrt{1-{\left({\displaystyle \prod _{1\le i_1<\dots <i_x\le n}\left(1-{\left({\displaystyle \prod _{j=1}^x{\left(\frac{\max {\Delta }^{-1}\left(s_{{\vartheta }_j},{\theta }_j\right)}{t}\right)}^{w_{i_j}}}\right)}^{\frac{2}{x}}\right)}\right)}^{\frac{1}{C_n^x}}}\right)\hfill \end{array}\right\}\hfill \end{array}$$

(63)

From Property 9,

$${\delta }^{-}\le {2\mathrm{TLPFDWHM}}^{(x)}\left({\delta }_1,{\delta }_2,\cdots ,{\delta }_n\right)\le {\delta }^{+}$$

(64)

It’s obvious that 2TLPFDWHM operator doesn’t have the property of idempotency.

4. Numerical Example and Comparative Analysis

4.1. Numerical Example

With the damage to the human environment and the earth’s resources increasingly depleted, the traditional supply chain has been gradually not well adapted to the current needs of the times and social demand, thus, the concept of “green supply chain” was introduced. The construction of green supply chain has become the main challenges and trends to provide a green product and towards sustainable development of society for the current time, in which an important part of the core content and implementation of green supply chain is green supplier evaluation and selection, especially sustainable suppliers in line with environmental protection requirements. Since the selection of suppliers plays a decisive role in the green SCM, which directly determines the optimization and core competitiveness of the entire chain of the enterprise, therefore, how to efficiently identify a required supplier from a number of suppliers is a key issue in modern green supply chain management must be solved. Then we propose a example to select green suppliers with 2TLPFNs. There are five possible green suppliers in SCM $A_i\left(i=1,2,3,4,5\right)$ to select. The experts group uses four attribute to assess the five green suppliers: ① G₁ is the price factor; ② G₂ is the delivery factor; ③ G₃ is the environmental factors; ④ G₄ is the product quality factor. The five green suppliers $A_i\left(i=1,2,3,4,5\right)$ are to be assessed with 2TLPFNs (whose weighting vector $\omega =\left(0.26,0.35,0.21,0.18\right)$, expert weighting vector $\omega =\left(0.25,0.35,0.40\right).$), which are listed in

Table 1. 2TLPFN decision matrix $(R_1)$ .

	G1	G2	G3	G4
A1	<(s4,0), (s2,0)>	<(s3,0), (s2,0)>	<(s2,0), (s1,0)>	<(s1,0), (s4,0)>
A2	<(s5,0), (s1,0)>	<(s5,0), (s3,0)>	<(s4,0), (s2,0)>	<(s5,0), (s2,0)>
A3	<(s2,0), (s3,0)>	<(s1,0), (s2,0)>	<(s3,0), (s4,0)>	<(s3,0), (s2,0)>
A4	<(s5,0), (s3,0)>	<(s3,0), (s5,0)>	<(s2,0), (s3,0)>	<(s3,0), (s1,0)>
A5	<(s4,0), (s3,0)>	<(s2,0), (s1,0)>	<(s2,0), (s4,0)>	<(s3,0), (s3,0)>

,

Table 2. 2TLPFN decision matrix $(R_2)$ .

	G1	G2	G3	G4
A1	<(s3,0), (s3,0)>	<(s2,0), (s3,0)>	<(s3,0), (s3,0)>	<(s2,0), (s3,0)>
A2	<(s4,0), (s2,0)>	<(s4,0), (s1,0)>	<(s5,0), (s2,0)>	<(s4,0), (s2,0)>
A3	<(s1,0), (s4,0)>	<(s2,0), (s3,0)>	<(s4,0), (s3,0)>	<(s4,0), (s3,0)>
A4	<(s4,0), (s4,0)>	<(s2,0), (s4,0)>	<(s1,0), (s2,0)>	<(s2,0), (s2,0)>
A5	<(s3,0), (s4,0)>	<(s1,0), (s2,0)>	<(s3,0), (s5,0)>	<(s3,0), (s3,0)>

and

Table 3. 2TLPFN decision matrix $(R_3)$ .

	G1	G2	G3	G4
A1	<(s3,0), (s4,0)>	<(s4,0), (s2,0)>	<(s2,0), (s4,0)>	<(s3,0), (s4,0)>
A2	<(s2,0), (s1,0)>	<(s5,0), (s1,0)>	<(s5,0), (s1,0)>	<(s4,0), (s2,0)>
A3	<(s3,0), (s5,0)>	<(s3,0), (s2,0)>	<(s3,0), (s2,0)>	<(s1,0), (s2,0)>
A4	<(s2,0), (s3,0)>	<(s3,0), (s2,0)>	<(s4,0), (s3,0)>	<(s3,0), (s4,0)>
A5	<(s5,0), (s3,0)>	<(s2,0), (s4,0)>	<(s3,0), (s4,0)>	<(s3,0), (s5,0)>

.

Then, we utilize these operators developed to select best green suppliers.

• Step 1. According to 2TLPFNs $r_{ij}\left(i=1,2,3,4,5,j=1,2,3,4\right)$, we fuse all 2TLPFNs $r_{ij}$ by 2-tuple linguistic Pythagorean fuzzy weighted average (2TLPFWA) operator or 2-tuple linguistic Pythagorean fuzzy weighted geometric (2TLPFWG) operator to get the overall 2TLPFNs of the green suppliers $A_i$. Then the fused results are listed in

  Table 4. The fused results by the 2TLPFNWA operator.

  	${\mathrm{G}}_1$	${\mathrm{G}}_2$
  A1	<(s3,0.3093), (s3,−0.0448)>	<(s3,0.0601), (s2,0.4003)>
  A2	<(s4,0.0210), (s1,0.3660)>	<(s5,−0.3519), (s1,0.3161)>
  A3	<(s2,0.0830), (s4,−0.0199)>	<(s2,0.2139), (s2,0.4003)>
  A4	<(s4,0.0210), (s3,0.4146)>	<(s3,−0.3818), (s3,0.4354)>
  A5	<(s4,0.1069), (s3,0.4146)>	<(s2,−0.3619), (s2,0.0705)>
  	${\mathrm{G}}_3$	${\mathrm{G}}_4$
  A1	<(s3,−0.4790), (s2,0.4850)>	<(s2,0.2193), (s4,−0.4857)>
  A2	<(s5,−0.1806), (s2,−0.3755)>	<(s4,0.3332), (s2,0.0000)>
  A3	<(s4,−0.4771), (s3,−0.1455)>	<(s3,0.2405), (s2,0.4003)>
  A4	<(s3,−0.3420), (s2,0.4997)>	<(s3,−0.3818), (s2,0.0705)>
  A5	<(s3,−0.2021), (s4,0.4225)>	<(s3,0.0000), (s3,0.4968)>

  .

Definition 12.

Let$p_j=\left\{\left(s_{{\varphi }_j},{\phi }_j\right),\left(s_{{\theta }_j},{\vartheta }_j\right)\right\}\left(j=1,2,\dots ,n\right)$be a set of 2TLPFNs with their weight vector be$w_i={(w_1,w_2,\dots ,w_n)}^T$, thereby satisfying$w_i\in \left[0,1\right]$and${\displaystyle {\sum }_{i=1}^n{w}_i}=1$, then we can get

$$\begin{array}{l}2\mathrm{TLPFWA}\left(p_1,p_2,\dots ,p_n\right)={\displaystyle \sum _{j=1}^n{w}_j{p}_j}\\ =\left\{\Delta \left(t\sqrt{1-{\displaystyle \prod _{j=1}^n{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\varphi }_j},{\phi }_j\right)}{t}\right)}^2\right)}^{w_j}}}\right),\Delta \left(t{\displaystyle \prod _{j=1}^n{\left(\frac{{\Delta }^{-1}\left(s_{{\theta }_j},{\vartheta }_j\right)}{t}\right)}^{w_j}}\right)\right\}\end{array}$$

(65)

$$\begin{array}{l}2\mathrm{TLPFWG}\left(p_1,p_2,\dots ,p_n\right)={\displaystyle \prod _{j=1}^n{\left(p_j\right)}^{w_j}}\\ =\left\{\Delta \left(t{\displaystyle \prod _{j=1}^n{\left(\frac{{\Delta }^{-1}\left(s_{{\varphi }_j},{\phi }_j\right)}{t}\right)}^{w_j}}\right),\Delta \left(t\sqrt{1-{\displaystyle \prod _{j=1}^n{\left(1-{\left(\frac{{\Delta }^{-1}\left(s_{{\theta }_j},{\vartheta }_j\right)}{t}\right)}^2\right)}^{w_j}}}\right)\right\}\end{array}$$

(66)

• Step 2. According to Table 4, we fuse all 2TLPFNs $r_{ij}$ by 2TLPFWHM (2TLPFDWHM) operator to derive the overall 2TLPFNs of the green suppliers $A_i$. Let $x=3$, then the fused results are listed in

  Table 5. The fused results of the green suppliers by 2TLPFWHM (2TLPFDWHM) operator.

  	2TLPFWHM	2TLPFDWHM
  A1	<(s5,−0.0266), (s0,0.3603)>	<(s0,0.3817), (s5,−0.0592)>
  A2	<(s6,−0.4283), (s0,0.1014)>	<(s1,0.1114), (s4,0.2563)>
  A3	<(s5,−0.1274), (s0,0.4190)>	<(s0,0.3407), (s5,−0.0111)>
  A4	<(s5,0.0279), (s0,0.4290)>	<(s0,0.4412), (s5,0.0159)>
  A5	<(s5,−0.0943), (s1,−0.4558)>	<(s0,0.4109), (s5,0.0961)>

  .
• Step 3. According to Table 5 and the scores are shown in

  Table 6. The scores of the green suppliers.

  	2TLPFWHM	2TLPFDWHM
  A1	(s5,0.0504)	(s1,−0.0222)
  A2	(s6,−0.4139)	(s2,−0.4067)
  A3	(s5,−0.0361)	(s1,−0.0644)
  A4	(s5,0.0913)	(s1,−0.0804)
  A5	(s5,−0.0192)	(s1,−0.1501)

  .
• Step 4. According to Table 6, the order is shown in

  Table 7. Order of the green suppliers.

  	Order
  2TLPFWHM	A2 > A4 > A1 > A5 > A3
  2TLPFDWHM	A2 > A1 > A3 > A4 > A5

  and the best green supplier is A₂.

4.2. Influence of the Parameter on the Final Result

In order to depict the effects of parameters $x$ in the 2TLPFWHM (2TLPFDWHM) operators, all the results are listed in

Table 8. Order for different parameters of 2TLPFWHM operator.

	s(A1)	s(A2)	s(A3)	s(A4)	s(A5)	Order
$x=1$	(s5,−0.1045)	(s6,−0.4446)	(s5,−0.0918)	(s5,−0.0146)	(s5,−0.0976)	A2 > A4 > A3 > A5 > A1
$x=2$	(s5,−0.1154)	(s6,−0.4589)	(s5,−0.1912)	(s5,−0.0819)	(s5,−0.2093)	A2 > A4 > A1 > A3 > A5
$x=3$	(s5,0.0504)	(s6,−0.4139)	(s5,−0.0361)	(s5,0.0913)	(s5,−0.0192)	A2 > A4 > A1 > A5 > A3
$x=4$	(s5,−0.1208)	(s6,−0.4661)	(s5,−0.2502)	(s5,−0.1206)	(s5,−0.3024)	A2 > A4 > A1 > A3 > A5

and

Table 9. Order for different parameters of 2TLPFDWHM operator.

	s(A1)	s(A2)	s(A3)	s(A4)	s(A5)	Order
$x=1$	(s1,0.0604)	(s2,−0.0648)	(s1,0.0245)	(s1,0.0913)	(s1,−0.1380)	A2 > A4 > A1 > A3 > A5
$x=2$	(s1,0.1320)	(s2,0.0166)	(s1,0.0754)	(s1,0.1140)	(s1,−0.0187)	A2 > A1 > A4 > A3 > A5
$x=3$	(s1,−0.0222)	(s2,−0.4067)	(s1,−0.0644)	(s1,−0.0804)	(s1,−0.1501)	A2 > A1 > A3 > A4 > A5
$x=4$	(s1,0.1687)	(s2,0.0610)	(s1,0.1079)	(s1,0.1296)	(s1,0.0780)	A2 > A1 > A4 > A3 > A5

.

4.3. Comparative Analysis

Then, we compare the proposed methods with LPFWA and LPFWG operator [74]. The comparative order is listed in

Table 10. Order of the green suppliers.

	Order
LPFWAA [74]	A2 > A4 > A1 > A5 > A3
LPFWGA [74]	A2 > A1 > A4 > A3 > A5

.

From the above analysis, it can be seen that two methods have the same best green suppliers A₂ and two methods’ ranking results are slightly different. This verifies that the 2TLPFWHM and 2TLPFDWHM operators we developed are reasonable and valid for MADM problems with 2TLPFNs.

In what follows, the comparisons of proposed approaches and the other methods with regard to some characteristics are shown in Table 8, Table 9 and Table 10. In light of Table 8, Table 9 and Table 10 some conclusions are summarized as follows:

(1)

The methods developed by Garg [74] aggregate the linguistic Pythagorean fuzzy information easily. The drawbacks of Garg’s methods [74] are they assume that the input arguments are not correlated, that is, they fail to consider the relationships between the input arguments. Nevertheless, our developed operators can capture the correlations among all the input arguments, and fuse the 2TLPFNs more flexibly by the parameter vector. Therefore, our developed approaches are more general and flexible comparing with that proposed by Garg’s methods [74].

(2)

Moreover, the methods developed by Garg [74] don’t have the ability that dynamic adjust to the parameter according to the decision maker’s risk attitude, so it is difficult to solve the risk multiple attribute decision making in real practice. Nevertheless, our developed operators have the ability that dynamic adjust to the parameter according to the decision maker’s risk attitude. Thus, our method can overcome the drawbacks of the methods developed by Garg [74], because the 2TLPFWHM and 2TLPFDWHM operators operator can provides more flexible and robust in information fusion and make it more adequate to solve risk multiple attribute decision making in which the attributes are independent.

5. Conclusions

In this paper, we investigate the MADM problems with 2TLPFNs. Then, we utilize the Hamy mean (HM) operator, weighted Hamy mean (WHM) operator, dual Hamy mean (DHM) operator and dual weighted Hamy mean (DWHM) operator to develop some Hamy mean aggregation operators with 2TLPFNs: 2-tuple linguistic Pythagorean fuzzy Hamy mean (2TLPFHM) operator, 2-tuple linguistic Pythagorean fuzzy weighted Hamy mean (2TLPFWHM) operator, 2-tuple linguistic Pythagorean fuzzy dual Hamy mean (2TLPFDHM) operator, 2-tuple linguistic Pythagorean fuzzy dual weighted Hamy mean (2TLPFDWHM) operator. The prominent characteristic of these proposed operators are studied. Then, we have utilized these operators to develop some approaches to solve the MADM problems with 2TLPFNs. Finally, a practical example for green supplier selection is given to verify the developed approach and to demonstrate its practicality and effectiveness. In the future, the application of the proposed aggregating operators of 2TLPFNs needs to be explored in the decision making [75,76,77,78,79,80,81,82,83,84,85,86,87,88,89], risk analysis [90,91] and many other fields under uncertain environments [92,93,94,95,96,97,98,99,100,101,102,103,104,105].