Linguistic Neutrosophic Generalized Partitioned Bonferroni Mean Operators and Their Application to Multi-Attribute Group Decision Making

Abstract

To solve the problems related to inhomogeneous connections among the attributes, we introduce a novel multiple attribute group decision-making (MAGDM) method based on the introduced linguistic neutrosophic generalized weighted partitioned Bonferroni mean operator (LNGWPBM) for linguistic neutrosophic numbers (LNNs). First of all, inspired by the merits of the generalized partitioned Bonferroni mean (GPBM) operator and LNNs, we combine the GPBM operator and LNNs to propose the linguistic neutrosophic GPBM (LNGPBM) operator, which supposes that the relationships are heterogeneous among the attributes in MAGDM. Then, we discuss its desirable properties and some special cases. In addition, aimed at the different importance of each attribute, the weighted form of the LNGPBM operator is investigated, which we call the LNGWPBM operator. Then, we discuss some of its desirable properties and special examples accordingly. In the end, we propose a novel MAGDM method on the basis of the introduced LNGWPBM operator, and illustrate its validity and merit by comparing it with the existing methods.

1. Introduction

The goal of the multiple attribute group decision-making (MAGDM) method is to select the optimal scheme from finite alternatives. First of all, decision makers (DMs) evaluate each alternative under the different attributes. Then, based on the DMs’ evaluation information, the alternatives are ranked in a certain way. As a research hotspot in recent decades, the MAGDM theory and methods have widely been used in all walks of life, such as supplier selection [1,2,3], medical diagnosis, clustering analysis, pattern recognition, and so on [4,5,6,7,8,9,10,11]. When evaluating alternatives, DMs used to evaluate alternatives by crisp numbers, but sometimes it is hard to use precise numbers because the surrounding environment has too much redundant data or interfering information. As a result, DMs have difficulty fully understanding the object of evaluation and exploiting exact information. As an example, when we evaluate people’s morality or vehicle performance, we can easily use linguistic term such as good, fair, or poor, or fuzzy concepts such as slightly, obviously, or mightily, to give evaluation results. For this reason, Zadeh [12] put forward the concept of linguistic variables (LVs) in 1975. Later, Herrera and Herrera-Viedma [5,6] proposed a linguistic assessments consensus model and further developed the steps of linguistic decision analysis. Subsequently, it has become an area of wide concern, and resulted in several in-depth studies, especially in MAGDM [8,11,13,14,15]. In addition, for the reason of fuzziness, Atanassov [16] introduced the intuitionistic fuzzy set (IFS) on the basis of the fuzzy set developed by Zadeh [17]. IFS can embody the degrees of satisfaction and dissatisfaction to judge alternatives, synchronously, and has been studied by large numbers of scholars in many fields [1,2,9,10,18,19,20,21,22,23]. However, intuitionistic fuzzy numbers (IFNs) use the two real numbers of the interval [0,1] to represent membership degree and non-membership degree, which is not adequate or sufficient to quantify DMs’ opinions. Hence, Chen et al. [24] used LVs to express the degrees of satisfaction and dissatisfaction instead of the real numbers of the interval [0,1], and proposed the linguistic intuitionistic fuzzy number (LIFN). LIFNs contain the advantages of both linguistic term sets and IFNs, so that it can address vague or imprecise information more accurately than LVs and IFNs. Since the birth of LIFNs, some scholars have proposed some improved aggregation operators and have applied them to MAGDM problems [10,25,26,27,28].

With the further development of fuzzy theory, Fang and Ye [29] noted while LIFNs can deal with vague or imprecise information more accurately than LVs and IFNs, it can only express incomplete information rather than indeterminate or inconsistent information. Since the indeterminacy of LIFN $I_A\left(x\right)$ is reckoned by $1-T_A\left(x\right)-F_A\left(x\right)$ in default, evaluating the indeterminate or inconsistent information, i.e., $I_A\left(x\right)<1-T_A\left(x\right)-F_A\left(x\right)$ or $I_A\left(x\right)>1-T_A\left(x\right)-F_A\left(x\right)$, is beyond the scope of the LIFN. Hence, a new form of information expression needs to be found. Fortunately, the neutrosophic sets (NSs) developed by Smarandache [30] are able to quantify the indeterminacy clearly, which is independent of truth-membership and false-membership, but NSs are not easy to apply to the MAGDM. So, some stretched form of NS was proposed for solving MAGDM, such as single-valued neutrosophic sets (SVNSs) [31], interval neutrosophic sets (INSs) [32], simplified neutrosophic sets (SNSs) [33], and so on. Meanwhile, they have attracted a lot of research, especially related to MAGDM [34,35,36,37,38,39,40,41]. Due to the characteristic of SNSs that use three crisp numbers of the interval [0,1] to depict truth-membership, indeterminacy-membership, and false-membership, motivated by the narrow scope of the LIFN, Fang and Ye [29] put forward the concept of linguistic neutrosophic numbers (LNNs) by combining linguistic terms and a simplified neutrosophic number (SNN). LNNs use LVs in the predefined linguistic term set to express the truth-membership, indeterminacy-membership, and falsity-membership of SNNs. So, LNNs are more appropriate to depict qualitative information than SNNs, and are also an extension of the LIFNs, obviously. Therefore, in this paper, we tend to study the MAGDM problems with LNNs.

In MAGDM, the key step is how to select the optimal alternative according to the existing information. Usually, we adopt the traditional evaluation methods or the information aggregation operators. The common traditional evaluation methods include TOPSIS [7,9], VIKOR [19], ELECTRE [42], TODIM [20,43], PROMETHE [18], etc., and they can only give the priorities in order regarding alternatives. However, the information aggregation operators first integrate DMs’ evaluation information into a comprehensive value, and then rank the alternatives. In other words, they not only give the prioritization orders of alternatives, they also give each alternative an integrated assessment value, so that the information aggregation operators are more workable than the traditional evaluation approaches in solving MAGDM problems. Hence, our study is concentrated on how to use information aggregation operators to solve the MAGDM problems with LNNs. In addition, in real MAGDM problems, there are often homogeneous connections among the attributes. Using a common example, quality is related to customer satisfaction when picking goods on the Internet. In order to solve this MAGDM problems where the attributes are interrelated, many related results have been achieved as a result, especially information aggregation operators such as the Bonferroni mean (BM) operator [23,44], the Maclaurin symmetric mean (MSM) operator [45], the Hamy mean operator [46], the generalized MSM operator [47], and so forth. However, the heterogeneous connections among the attributes may also exist in real MAGDM problems. For instance, in order to choose a car, we may consider the following attributes: the basic requirements (G₁), the physical property (G₂), the brand influence (G₃), and the user appraisal (G₄), where the attribute G₁ is associated with the attribute G₂, and the attribute G₃ is associated with the attribute G₄, but the attributes G₁ and G₂ are independent of the attributes G₃ and G₄. So, the four attributes can be sorted into two clusters, P₁ and P₂, namely P₁ = {G₁, G₂} and P₂ = {G₃, G₄} meeting the condition where P₁ and P₂ have no relationship. To solve this issue, Dutta and Guha [48] proposed the partition Bonferroni mean (PBM) operator, where all attributes are sorted into several clusters, and the members have an inherent connection in the same clusters, but independence in different clusters. Subsequently, Banerjee et al. [4] extended the PBM operator to the general form that was called the generalized partitioned Bonferroni mean (GPBM) operator, which further clarified the heterogeneous relationship and individually processed the elements that did not belong to any cluster of correlated elements, so the GPBM operator can model the average of the respective satisfaction of the independent and dependent input arguments. Besides, the GPBM operator can be translated into the BM operator, arithmetic mean operator, and PBM operator, so the GPBM operator is a wider range of applications for solving MAGDM problems with related attributes. Therefore, in this paper, we are further focused on how to combine the GPBM operator with LNNs to address the MAGDM problems with heterogeneous relationships among attributes. Inspired by the aforementioned ideas, we aim at:

(1)

establishing a linguistic neutrosophic GPBM (LNGPBM) operator and the weighted form of the LNGPBM operator (the form of shorthand is LNGWPBM).

(2)

discussing their properties and particular cases.

(3)

proposing a novel MAGDM method in light of the proposed LNGWPBM operator to address the MAGDM problems with LNNs and the heterogeneous relationships among its attributes.

(4)

showing the validity and merit of the developed method.

The arrangement of this paper is as follows. In Section 2, we briefly retrospect some elementary knowledge, including the definitions, operational rules, and comparison method of the LNNs. We also review some definitions and characteristics of the PBM operator and GPBM operator. In Section 3, we construct the LNGPBM operator and LNGWPBM operator for LNNs, including their characteristics and some special cases. In Section 4, we propose a novel MAGDM method based on the proposed LNGWPBM operator to address the MAGDM problems where heterogeneous connections exist among the attributes. In Section 5, we give a practical application related to the selection of green suppliers to show the validity and the generality of the MAGDM method, and compare the experimental results of the proposed MAGDM method with the ones of Fang and Ye’s MAGDM method [29] and Liang et al.’s MAGDM method [7]. Section 6 presents the conclusions.

2. Preliminaries

To understand this article much better, this section intends to retrospect some elementary knowledge, including the definitions, operational rules, and comparison method of the LNNs, PBM operator, and generalized PBM operator.

2.1. Linguistic Neutrosophic Set (LNS)

Definition 1 [29].

Let$Z$be the universe of discourse, and$z$be a generic element in$Z$, and let$L=(l_0,l_1,\cdots ,l_s)$be a linguistic term set. A LNS$X$in$Z$is represented by:

$$X=\left\{\left(z,l_{T_X(z)},l_{I_X(z)},l_{F_X(z)}\right)|z\in Z\right\}$$

(1)

where$T_X$,$I_X$, and$F_X$denote the truth-membership function, indeterminacy-membership function, and falsity-membership function of$z$in the set$X$, respectively, and$T_X,I_X,F_X:Z\to \left[0,s\right]$with$s$is an even number.

In [29], Fang and Ye called the pair$\left(l_{\alpha },l_{\beta },l_{\gamma }\right)$an LNN, which meets$\alpha ,\beta ,\gamma :Z\to \left[0,s\right]$, and$s$is an even number.

Definition 2 [29].

Let$z = \left(l_{\alpha },l_{\beta },l_{\gamma }\right)$be an optional LNN in$L$, where the score function$C(z)$of the LNN$z$is defined as shown:

$$C(z)=\frac{2s+\alpha -\beta -\gamma }{3s}$$

(2)

where$\alpha ,\beta ,\gamma \in [0,s]$and$C(z)\in [0,1]$.

Definition 3 [29].

Let$z = \left(l_{\alpha },l_{\beta },l_{\gamma }\right)$be an optional LNN in$L$, where the accuracy function$A(z)$of the LNN$z$is defined as shown:

$$A(z)=\frac{\alpha -\gamma }{s}$$

(3)

where$\alpha ,\beta ,\gamma \in [0,s]$and$A(z)\in [-1,1]$.

Definition 4 [29].

Let$z_1=\left(l_{{\alpha }_1},l_{{\beta }_1},l_{{\gamma }_1}\right)$and$z_2=\left(l_{{\alpha }_2},l_{{\beta }_2},l_{{\gamma }_2}\right)$be two optional LNNs in$L$. Then, the order between$z_1$and$z_2$is given by the following rules:

(1) 

If$C(z_1)>C(z_2)$, then$z_1>z_2$;

(2) 

If$C(z_1)=C(z_2)$, then

If$A(z_1)>A(z_2)$, then$z_1>z_2$;

If$A(z_1)=A(z_2)$, then$z_1=z_2$.

Example 1.

Suppose$L=(l_0,l_1,\cdots ,l_6)$is a linguistic term set, and$z_1=\left(l_6,l_2,l_3\right)$and$z_2=\left(l_4,l_1,l_1\right)$are two LNNs in$L$. Then, we can calculate the values of their score functions and accuracy functions as$C(z_1)=0.7222$,$C(z_2)=0.7778$, $A(z_1)=0.5$, and $A(z_2)=0.5$. According to Definition 4, it is easy to find that$z_1<z_2$.

Definition 5 [29].

Let$L=(l_0,l_1,\cdots ,l_s)$be a linguistic term set, and$z_1=\left(l_{{\alpha }_1},l_{{\beta }_1},l_{{\gamma }_1}\right)$and$z_2=\left(l_{{\alpha }_2},l_{{\beta }_2},l_{{\gamma }_2}\right)$be two haphazard LNNs in$L$. The basic operational laws between the two LNNs are shown as below:

$$z_1\oplus z_2=\left(l_{{\alpha }_1+{\alpha }_2-{\alpha }_1{\alpha }_2/s},l_{{\beta }_1{\beta }_2/s},l_{{\gamma }_1{\gamma }_2/s}\right),$$

(4)

$$z_1\otimes z_2=\left(l_{{\alpha }_1{\alpha }_2/s},l_{{\beta }_1+{\beta }_2-{\beta }_1{\beta }_2/s},l_{{\gamma }_1+{\gamma }_2-{\gamma }_1{\gamma }_2/s}\right),$$

(5)

$$\theta z_1=\left(l_{s-s{\left(1-{\alpha }_1/s\right)}^{\theta }},l_{s{\left({\beta }_1/s\right)}^{\theta }},l_{s{\left({\gamma }_1/s\right)}^{\theta }}\right), where \theta >0,$$

(6)

$$z_1{}^{\theta }=\left(l_{s{\left({\alpha }_1/s\right)}^{\theta }},l_{s-s{\left(1-{\beta }_1/s\right)}^{\theta }},l_{s-s{\left(1-{\gamma }_1/s\right)}^{\theta }}\right), where \theta >0$$

(7)

It is easy to prove the following operational properties of the LNNs, according to Definition 5.

Let $z_1 = \left(l_{{\alpha }_1},l_{{\beta }_1},l_{{\gamma }_1}\right)$ and $z_2 = \left(l_{{\alpha }_2},l_{{\beta }_2},l_{{\gamma }_2}\right)$ be any two LNNs in $L$. Then:

$$z_1\oplus z_2= z_2\oplus z_1,$$

(8)

$$z_1\otimes z_2= z_2\otimes z_1,$$

(9)

$$\theta \left(z_1\oplus z_2\right)=\theta z_1\oplus \theta z_2, where \theta >0,$$

(10)

$${\theta }_1{z}_1\oplus {\theta }_2{z}_1=\left({\theta }_1+{\theta }_2\right)z_1, where {\theta }_1,{\theta }_2>0,$$

(11)

$$z_1{}^{{\theta }_1}\otimes z_1{}^{{\theta }_2}=z_1{}^{{\theta }_1+{\theta }_2}, where {\theta }_1,{\theta }_2>0,$$

(12)

$$z_1^{\theta }\otimes z_2^{\theta }={\left(z_1\otimes z_2\right)}^{\theta }, where \theta >0$$

(13)

2.2. Generalized Partitioned Bonferroni Mean Operators

Definition 6 [48].

Suppose the non-negative real set$A=\left\{a_1,a_2,\cdots ,a_n\right\}$is divided into$t$clusters$P_1,P_2,\cdots ,P_t$, which satisfies$P_x\cap P_y=\varnothing$, x ≠ y and$\underset{r=1}{\overset{t}{\cup }}{P}_r=A$. Then, the partitioned Bonferroni mean (PBM) operator is defined as follows:

$$PB{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)=\frac{1}{t}\left({\displaystyle \sum _{r=1}^t{\left(\frac{1}{h_r}{\displaystyle \sum _{i=1}^{h_r}{a}_i^p\left(\frac{1}{h_r-1}{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{a}_j^q}\right)}\right)}^{\frac{1}{p+q}}}\right)$$

(14)

where$p,q\ge 0$and$p+q>0$,$h_r$indicates the number of elements in partition$P_r$and${\displaystyle \sum _{r=1}^t{h}_r}=n$.

The PBM operator is used to integrate the input arguments of the different clusters, which satisfies that the data has inherent connections in the same clusters, but independence in different clusters. However, sometimes, some of the input arguments have nothing to do with any other argument, that is, it does not exist in any cluster. We can part these arguments and deal with them individually. Hence, we sort the input arguments into two groups: F₁ contains the relevant arguments, and F₂ contains the input arguments that are irrelevant to any argument. These easily derive $F_1\cap F_2=\varnothing$ and $\left|F_1\right|+\left|F_2\right|=n$ where |F₁| and |F₂| denote the numbers of arguments in F₁ and F₂, respectively. According to the upper description, we suppose that the arguments of F₁ are divided into $t$ partitions $P_1,P_2,\cdots ,P_t$ on the basis of the interrelationship pattern [4]. To address this issue, the PBM operator is modified, and the GPBM operator is proposed, as shown in the following.

Definition 7 [4].

Suppose that the non-negative real set$A=\left\{a_1,a_2,\cdots ,a_n\right\}$is sorted into two groups: F₁ and F₂. In F₁, the elements are divided into$t$clusters$P_1,P_2,\cdots ,P_t$, which satisfies$P_x\cap P_y=\varnothing$, x ≠ y and$\underset{r=1}{\overset{t}{\cup }}{P}_r=F_1$; in F₂, the elements are irrelevant to any element. Then, the GPBM operator is defined as follows:

$$GPB{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)={\left(\frac{n-\left|F_2\right|}{n}\left(\frac{1}{t}{\displaystyle \sum _{r=1}^t{\left(\frac{1}{h_r}{\displaystyle \sum _{i=1}^{h_r}{a}_i^p\left(\frac{1}{h_r-1}{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{a}_j^q}\right)}\right)}^{\frac{p}{p+q}}}\right)+\frac{\left|F_2\right|}{n}\left(\frac{1}{\left|F_2\right|}{\displaystyle \sum _{i=1}^{\left|F_2\right|}{a}_i^p}\right)\right)}^{\frac{1}{p}}$$

(15)

where$p,q\ge 0$and$p+q>0$, |F₂| denotes the number of elements in F₂,$h_r$indicates the number of elements in cluster$P_r$and${\displaystyle \sum _{r=1}^t{h}_r}=n-\left|F_2\right|$.

Remark 1.

If |F₂| = 0, we consider the first sum, and if |F₂| = n, we consider the last sum. At the same time, we have made the convention$\frac{0}{0}=0$(we only need to define$\frac{0}{0}$; its conventional real value is not important here).

The interpretation of the GPBM operator is detailed by Banerjee et al. in [4], and the GPBM operator has the following characteristics: idempotency, monotonicity, and boundedness [4].

Based on the characteristics of F₂, there are some special cases of GPBM operator, which are described as follows [4]:

(1)

When |F₂| = 0, all elements belong to the group F₁ and are divided into $t$ clusters.

$$GPB{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)=\frac{1}{t}{\displaystyle \sum _{r=1}^t{\left(\frac{1}{h_r}{\displaystyle \sum _{i=1}^{h_r}{a}_i^p\left(\frac{1}{h_r-1}{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{a}_j^q}\right)}\right)}^{\frac{1}{p+q}}}=PB{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)$$

It is simplified as the PBM operator described in Formula (15).

(2)

When |F₂| = 0 and $t=1$, all elements belong to the same cluster.

$$GPB{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)={\left(\frac{1}{h_r}{\displaystyle \sum _{i=1}^{h_r}{a}_i^p\left(\frac{1}{h_r-1}{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{a}_j^q}\right)}\right)}^{\frac{1}{p+q}}={\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{a}_i^p\left(\frac{1}{n-1}{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^n{a}_j^q}\right)}\right)}^{\frac{1}{p+q}}=B{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)$$

It becomes the BM operator [44].

(3)

When |F₂| = n, all elements are independent.

$$GPB{M}^{p,q}\left(a_1,a_2,\cdots ,a_n\right)={\left(\frac{1}{\left|F_2\right|}{\displaystyle \sum _{i=1}^{\left|F_2\right|}{a}_i^p}\right)}^{\frac{1}{p}}={\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{a}_i^p}\right)}^{\frac{1}{p}}$$

It is simplified as the power root arithmetic mean operator [4].

3. The Linguistic Neutrosophic GPBM Operators

In this section, we will construct the LNGPBM operator from the GPBM operator and LNNs. Moreover, with respect to the different weights of different attributes in real life, we will propose the corresponding weighted operators, and call it the LNGWPBM operator. They are defined as follows.

3.1. The LNGPBM Operator

Definition 8.

Let$z_1,z_2,\cdots$and$z_n$be LNNs, which are sorted into two groups: F₁ and F₂. In F₁, the elements are divided into$t$clusters$P_1,P_2,\cdots ,P_t$, which satisfies$P_x\cap P_y=\varnothing$, x ≠ y and$\underset{r=1}{\overset{t}{\cup }}{P}_r=F_1$; in F₂, the elements are irrelevant to any element. The LNGPBM operator of the LNNs$z_1,z_2,\cdots$and$z_n$is defined as follows:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)={\left(\frac{n-\left|F_2\right|}{n}\left(\frac{1}{t}\underset{r=1}{\overset{t}{\oplus }}{\left(\frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)\right)}^{\frac{p}{p+q}}\right)\oplus \frac{\left|F_2\right|}{n}\left(\frac{1}{\left|F_2\right|}\underset{i=1}{\overset{\left|F_2\right|}{\oplus }}{z}_i^p\right)\right)}^{\frac{1}{p}}$$

(16)

where$z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$and${\alpha }_i,{\beta }_i,{\gamma }_i\in \left[0,s\right]$$(i=1,2,\cdots ,n)$;$p,q\ge 0$and$p+q>0$; |F₂| denotes the number of elements in F₂,$h_r$indicates the number of elements in cluster$P_r$and${\displaystyle \sum _{r=1}^t{h}_r}=n-\left|F_2\right|$.

Theorem 1.

Let$z_1,z_2,\cdots$and$z_n$be LNNs, where$z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$and${\alpha }_i,{\beta }_i,{\gamma }_i\in \left[0,s\right]$$(i=1,2,\cdots ,n)$. The synthesized result of the LNGPBM operator of the LNNs$z_1,z_2,\cdots$and$z_n$is still a LNN, which is shown as follows:

$$\begin{array}{l}LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)=(l_{s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},\\ \hspace{1em}\hspace{1em}{l}_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}})\end{array}$$

(17)

where $H_{\alpha }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}$, $H_{\beta }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\beta }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}$ and $H_{\gamma }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\gamma }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}$.

Proof. 

According to Formula (16), first of all, we can part two steps: the processing of F₁ and F₂, and then combine them to prove.

(i) The processing of F₁:

Based on the operational rules of LNNs, we can get $z_j^q=\left(l_{s{\left({\alpha }_j/s\right)}^q},l_{s-s{\left(1-{\beta }_j/s\right)}^q},l_{s-s{\left(1-{\gamma }_j/s\right)}^q}\right)$ and $\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q=\left(l_{s-s{\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}},l_{s{\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}},l_{s{\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}}\right)$.

Then, we can calculate the average satisfaction of the elements in $P_r$ except $z_i$:

$$\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q=\left(l_{s-s{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}},l_{s{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}},l_{s{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}}\right)$$

and the conjunction of the satisfaction of element $z_i$ with the average satisfaction of the rest of elements in $P_r$:

$$z_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)=\left(l_{s{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)},l_{s-s\left({\left(1-{\beta }_i/s\right)}^p\right)\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)},l_{s-s\left({\left(1-{\gamma }_i/s\right)}^p\right)\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)}\right)$$

Then, the satisfaction of the interrelated elements of $P_r$ is:

$$\begin{array}{cc}\hfill \frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)=& (l_{s-s{\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}},\hfill \\ \hfill & \hspace{1em}{l}_{s{\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\beta }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}},l_{s{\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\gamma }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}})\hfill \end{array}$$

So, the average satisfaction of all of the elements of the $t$ clusters is:

$$A=\frac{1}{t}\underset{r=1}{\overset{t}{\oplus }}{\left(\frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)\right)}^{\frac{p}{p+q}}=(l_{s-s{\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-{\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}},$$

$$l_{s{\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-{\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\beta }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}},l_{s{\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-{\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\gamma }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}})$$

We suppose $H_{\alpha }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}$, $H_{\beta }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\beta }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}$ and $H_{\gamma }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(1-{\gamma }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}$, then the upper formula can be rewritten as:

$$A=\frac{1}{t}\underset{r=1}{\overset{t}{\oplus }}{\left(\frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)\right)}^{\frac{p}{p+q}}=(l_{s-s{\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}},l_{s{\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}},l_{s{\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}})$$

(ii) The processing of F₂:

The average satisfaction of all the elements that are irrelevant to any element is:

$$B=\frac{1}{\left|F_2\right|}\underset{i=1}{\overset{\left|F_2\right|}{\oplus }}{z}_i^p=\left(l_{s-s{\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}},l_{s{\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}},l_{s{\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}}\right)$$

Finally, we can compute the average satisfaction of the elements $z_1,z_2,\cdots$ and $z_n$:

$$\begin{array}{cc}\hfill {\left(\frac{n-\left|F_2\right|}{n}A\oplus \frac{\left|F_2\right|}{n}B\right)}^{\frac{1}{p}}=& (l_{s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}},\hfill \\ \hfill & l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}})\hfill \end{array}$$

That proves that Formula (17) is kept. Then, we prove that the aggregated result of Formula (17) is a LNN. It is easy to prove the following inequalities:

$$0\le s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}\le s,$$

$$0\le s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}\le s,$$

and:

$$0\le s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}\le s.$$

Firstly, we prove $0\le H_{\alpha }\le 1$, $0\le H_{\beta }\le 1$ and $0\le H_{\gamma }\le 1$.

Since ${\alpha }_j,{\beta }_j,{\gamma }_j\in \left[0,s\right]$ and $q\ge 0$, we can get $0\le 1-{\left({\alpha }_j/s\right)}^q\le 1$, $0\le 1-{\left(1-{\beta }_j/s\right)}^q\le 1$ and $0\le 1-{\left(1-{\gamma }_j/s\right)}^q\le 1$. Owing to $h_r>0$, the following inequalities are established:

$$0\le 1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\le 1,\hspace{1em}0\le 1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\beta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\le 1,\hspace{1em}\mathrm{and}\hspace{1em}0\le 1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(1-{\gamma }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\le 1.$$

According to $p\ge 0$, it is easy to obtain the below inequality: $0\le H_{\alpha }\le 1$, $0\le H_{\beta }\le 1$ and $0\le H_{\gamma }\le 1$.

In addition, because $p+q>0$, $t>0$, and $\left|F_2\right|>0$, we can get the following inequalities:

$$0\le {\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\le 1,\hspace{1em}0\le {\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\le 1,\hspace{1em}\mathrm{and}\hspace{1em}0\le {\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\le 1.$$

$$0\le {\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\le 1,\hspace{1em}0\le {\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\le 1,\hspace{1em}\mathrm{and}\hspace{1em}0\le {\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\le 1.$$

Besides, on the basis of the upper inequalities, we can get:

$$0\le {\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}\le 1,$$

$$0\le {\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}\le 1, \mathrm{and}$$

$$0\le {\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}\le 1.$$

which can derive directly:

$$0\le s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}\le s,$$

$$0\le s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}\le s,$$

$$\mathrm{and} 0\le s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}\le s.$$

Therefore, Theorem 1 is kept if some of the partitions only contain one element. ☐

In the following, we will demonstrate the desired properties of the proposed LNGPBM operator:

(1)

Idempotency: If $z_1,z_2,\cdots$ and $z_n$ are LNNs meeting the condition $z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)=z=\left(l_{\alpha },l_{\beta },l_{\gamma }\right)$ $(i=1,2,\cdots ,n)$; then, $LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)=z$.

Proof. 

Since $z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)=z=\left(l_{\alpha },l_{\beta },l_{\gamma }\right)$, we can get:

$$\begin{array}{l}{H}_{\alpha }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(\alpha /s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left(\alpha /s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}\\ \hspace{1em}\hspace{1em}={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(\alpha /s\right)}^p\left(1-{\left({\left(1-{\left(\alpha /s\right)}^q\right)}^{h_r-1}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left(\alpha /s\right)}^p{\left(\alpha /s\right)}^q\right)}\right)}^{\frac{1}{h_r}}=1-{\left(\alpha /s\right)}^{p+q}\end{array}.$$

In the same way, we can obtain $H_{\beta }=1-{\left(1-\beta /s\right)}^{p+q}$ and $H_{\gamma }=1-{\left(1-\gamma /s\right)}^{p+q}$.

According to Theorem 1, we can obtain:

$$\begin{array}{l}LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)=(l_{s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}{l}_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}})\end{array}$$

$$\begin{array}{l}=(l_{s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-\left(1-{\left(\alpha /s\right)}^{p+q}\right)\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(\alpha /s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},\\ \hspace{1em}\hspace{1em}\hspace{1em}{l}_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-\left(1-{\left(1-\beta /s\right)}^{p+q}\right)\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-\beta /s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-\left(1-{\left(1-\gamma /s\right)}^{p+q}\right)\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-\gamma /s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}})\end{array}$$

$$=\left(l_{s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(\alpha /s\right)}^p\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left(1-{\left(\alpha /s\right)}^p\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-\beta /s\right)}^p\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\left({\left(1-{\left(1-\beta /s\right)}^p\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-\gamma /s\right)}^p\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\left({\left(1-{\left(1-\gamma /s\right)}^p\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}}\right)$$

$$=(l_{s{\left(1-\left({\left(1-{\left(\alpha /s\right)}^p\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left(1-{\left(\alpha /s\right)}^p\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-\left({\left(1-{\left(1-\beta /s\right)}^p\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left(1-{\left(1-\beta /s\right)}^p\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-\left({\left(1-{\left(1-\gamma /s\right)}^p\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left(1-{\left(1-\gamma /s\right)}^p\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}})$$

$$\begin{array}{l}=(l_{s{\left(1-\left(1-{\left(\alpha /s\right)}^p\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-\left(1-{\left(1-\beta /s\right)}^p\right)\right)}^{\frac{1}{p}}},l_{s-s{\left(1-\left(1-{\left(1-\gamma /s\right)}^p\right)\right)}^{\frac{1}{p}}})=(l_{s{\left({\left(\alpha /s\right)}^p\right)}^{\frac{1}{p}}},l_{s-s{\left({\left(1-\beta /s\right)}^p\right)}^{\frac{1}{p}}},l_{s-s{\left({\left(1-\gamma /s\right)}^p\right)}^{\frac{1}{p}}})\\ =(l_{\alpha },l_{\beta },l_{\gamma })\end{array}.$$

 ☐

(2)

Monotonicity: If $z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$$\left(i=1,2,\cdots ,n\right)$ and $y_i=\left(l_{{\delta }_i},l_{{\eta }_i},l_{{\sigma }_i}\right)$ $\left(i=1,2,\cdots ,n\right)$ are any two sets of LNNs; they satisfy the condition ${\alpha }_i\ge {\delta }_i$, ${\beta }_i\le {\eta }_i$ and ${\gamma }_i\le {\sigma }_i$, then $LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)\ge LNGPB{M}^{p,q}\left(y_1,y_2,\cdots ,y_n\right)$.

Proof. 

Suppose that $LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)=z=\left(l_{\alpha },l_{\beta },l_{\gamma }\right)$ and $LNGPB{M}^{p,q}\left(y_1,y_2,\cdots ,y_n\right)=y=\left(l_{\delta },l_{\eta },l_{\sigma }\right)$, then:

$$\alpha =s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}},$$

$$\delta =s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\delta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\delta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}},$$

$$\beta =s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}},$$

$$\eta =s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\eta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\eta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}},$$

$$\gamma =s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}},$$

$$\sigma =s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\sigma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left(1-{\sigma }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}.$$

In order to prove this property, we need to compute their score function values $C(z)$ and $C(y)$, and their accuracy values $A(z)$ and $A(y)$ to compare their synthesized result, i.e., $z\ge y$. Firstly, on the basis of the condition ${\alpha }_i\ge {\delta }_i$, ${\beta }_i\le {\eta }_i$, and ${\gamma }_i\le {\sigma }_i$, we can get the compared result of their truth-membership degrees, indeterminacy-membership degrees, and falsity-membership degrees, respectively.

(i)

The comparison of the truth-membership degrees:

Based on ${\alpha }_i\ge {\delta }_i$, we can get:

$$\begin{array}{l}{H}_{\alpha }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\alpha }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\alpha }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}\le {\left({\displaystyle \prod _{i=1}^{h_r}\left(1-{\left({\delta }_i/s\right)}^p\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}\left(1-{\left({\delta }_j/s\right)}^q\right)}\right)}^{\frac{1}{h_r-1}}\right)\right)}\right)}^{\frac{1}{h_r}}=H_{\delta }\\ \Rightarrow {\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\le {\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\delta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\\ \mathrm{and} {\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\le {\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\delta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\end{array}.$$

In accordance with the upper two inequalities, we have:

$$s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\alpha }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}\ge s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-H_{\delta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}\left(1-{\left({\delta }_i/s\right)}^p\right)}\right)}^{\frac{1}{\left|F_2\right|}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}$$

That is, $\alpha \ge \delta$.

(ii)

The comparision of indeterminacy-membership degrees and falsity-membership degrees, respectively:

Based on ${\beta }_i\le {\eta }_i$ and ${\gamma }_i\le {\sigma }_i$, we can also obtain $\beta \le \eta$ and $\gamma \le \sigma$; this process is similar to the process of the truth-membership degrees.

Thus, it can be obtained that $C(z)=\frac{2s+\alpha -\beta -\gamma }{3s}\ge \frac{2s+\delta -\eta -\sigma }{3s}=C(y)$. In the following, we discuss two cases.

(i)

If $C(z)>C(y)$, then $z>y$, according to Definition 2.

(ii)

If $C(z)=C(y)$, then $\left(\alpha -\gamma \right)-\beta =\left(\delta -\sigma \right)-\eta$. Since $\alpha -\gamma \ge \delta -\sigma$ in the light of $\alpha \ge \delta$ and $\gamma \le \sigma$, now we assume $\alpha -\gamma >\delta -\sigma$, then $\beta >\eta$, which is in contradiction with the previous proof $\beta \le \eta$. So, we can conclude that $\alpha -\gamma =\delta -\sigma$. That is, $A(z)=\frac{\alpha -\gamma }{s}=\frac{\delta -\sigma }{s}=A(y)$, which testifies $z=y$.

In conclusion, the synthesized result $z\ge y$, which explains:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)\ge LNGPB{M}^{p,q}\left(y_1,y_2,\cdots ,y_n\right)$$

 ☐

(3)

Boundedness: Let $z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$$\left(i=1,2,\cdots ,n\right)$ be an arbitrary set of LNNs, then:

$$\underset{i}{min}{z}_i\le LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)\le \underset{i}{max}{z}_i$$

Proof. 

Since $z_i\ge \underset{i}{min}{z}_i$, according to the monotonicity and idempotency of the proposed LNGPBM operator, we can obtain the following result:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)\ge LNGPB{M}^{p,q}\left(\underset{i}{min}{z}_i,\underset{i}{min}{z}_i,\cdots ,\underset{i}{min}{z}_i\right)=\underset{i}{min}{z}_i$$

Similarly, we can obtain the corresponding result for $\underset{i}{max}{z}_i$:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)\le LNGPB{M}^{p,q}\left(\underset{i}{max}{z}_i,\underset{i}{max}{z}_i,\cdots ,\underset{i}{max}{z}_i\right)=\underset{i}{max}{z}_i$$

Therefore, $\underset{i}{min}{z}_i\le LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)\le \underset{i}{max}{z}_i$. ☐

Based on the character of F₂, some special cases are discussed about the LNGPBM operator, and shown in the following.

(1)

When |F₂| = 0, all arguments belong to the group F₁, and are divided into $t$ clusters; then, the proposed LNGPBM operator is simplified as the following form:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)=\frac{1}{t}\underset{r=1}{\overset{t}{\oplus }}{\left(\frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)\right)}^{\frac{1}{p+q}}=LNPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)$$

The LNPBM is called the linguistic neutrosophic PBM operator.

(2)

When |F₂| = 0 and $t=1$, all arguments belong to the same cluster, i.e., h_r = n; then, the proposed LNGPBM operator becomes the following form:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)={\left(\frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^p\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^q\right)\right)}^{\frac{1}{p+q}}={\left(\frac{1}{n}\underset{i=1}{\overset{n}{\oplus }}{z}_i^p\otimes \left(\frac{1}{n-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{n}{\oplus }}{z}_j^q\right)\right)}^{\frac{1}{p+q}}=LNB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)$$

The LNBM is called the linguistic neutrosophic BM operator.

(3)

When |F₂| = n, there is no element in group F₁ and all elements are independent; then, the proposed LNGPBM operator reduces to the following form:

$$LNGPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)={\left(\frac{1}{\left|F_2\right|}\underset{i=1}{\overset{\left|F_2\right|}{\oplus }}{z}_i^p\right)}^{\frac{1}{p}}={\left(\frac{1}{n}\underset{i=1}{\overset{n}{\oplus }}{z}_i^p\right)}^{\frac{1}{p}}=LNPRA{M}^p\left(z_1,z_2,\cdots ,z_n\right)$$

The LNPRAM is called the linguistic neutrosophic power root arithmetic mean operator.

Moreover, we can also get some special cases by distributing different values to the parameters p and q.

(1)

When $q\to 0$, the proposed LNGPBM operator becomes the LNPRAM operator, which was described in the previous discussion. Since there is no inner connection in group F₁, all of the elements are independent.

(2)

When $p=1$ and $q\to 0$, the proposed LNGPBM operator reduces to the linguistic neutrosophic arithmetic mean (LNAM) operator, which is shown as follows:

$$LNGPB{M}^{p=1,q\to 0}\left(z_1,z_2,\cdots ,z_n\right)={\left(\frac{1}{n}\underset{i=1}{\overset{n}{\oplus }}{z}_i^p\right)}^{\frac{1}{p}}=\frac{1}{n}\underset{i=1}{\overset{n}{\oplus }}{z}_i=LNAM\left(z_1,z_2,\cdots ,z_n\right)$$

(3)

When $p=2$ and $q\to 0$, the proposed LNGPBM operator is transformed into the linguistic neutrosophic square root arithmetic mean (LNSRAM) operator, which is shown as follows:

$$LNGPB{M}^{p=2,q\to 0}\left(z_1,z_2,\cdots ,z_n\right)={\left(\frac{1}{n}\underset{i=1}{\overset{n}{\oplus }}{z}_i^2\right)}^{\frac{1}{2}}=LNSRAM\left(z_1,z_2,\cdots ,z_n\right).$$

(4)

When $p=q=1$, the proposed LNGPBM operator is simplified as the simplest form of the LNGPBM operator, which is shown as follows:

$$LNGPB{M}^{p=1,q=1}\left(z_1,z_2,\cdots ,z_n\right)=\frac{n-\left|F_2\right|}{n}\left(\frac{1}{t}\underset{r=1}{\overset{t}{\oplus }}{\left(\frac{1}{h_r}\underset{i=1}{\overset{h_r}{\oplus }}{z}_i^{}\otimes \left(\frac{1}{h_r-1}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{z}_j^{}\right)\right)}^{\frac{1}{2}}\right)\oplus \frac{\left|F_2\right|}{n}\left(\frac{1}{\left|F_2\right|}\underset{i=1}{\overset{\left|F_2\right|}{\oplus }}{z}_i^{}\right)$$

It is often used to simplify the calculation in a problem.

3.2. The LNGWPBM Operator

In Definitions 8, we assume that all the input arguments have the same position. However, in many realistic decision-makings, every input argument may have different importance. Accordingly, we give different values to the weights of input arguments, and propose the weighted form of the LNGPBM operator. Let the weight of input argument $z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$ $\left(i=1,2,\cdots ,n\right)$ be ${\omega }_i$, where ${\omega }_i\in \left[0,1\right]$ and ${\displaystyle \sum _{i=1}^n{\omega }_i}=1$. The weighted form of the LNGPBM operator is shown in the following.

Definition 9.

Let$z_1,z_2,\cdots$and$z_n$be LNNs that are sorted into two groups: F₁ and F₂. In F₁, the elements are divided into$t$clusters$P_1,P_2,\cdots ,P_t$, which satisfy$P_x\cap P_y=\varnothing$, x ≠ y and$\underset{r=1}{\overset{t}{\cup }}{P}_r=F_1$; in F₂, the elements are irrelevant to any element. The weighted form of the LNGPBM operator of the LNNs$z_1,z_2,\cdots$and$z_n$is defined as follows:

$$LNGWPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)={\left(\frac{n-\left|F_2\right|}{n}\left(\frac{1}{t}\underset{r=1}{\overset{t}{\oplus }}{\left(\frac{1}{{\displaystyle \sum _{i=1}^{h_r}{\omega }_i}}\underset{i=1}{\overset{h_r}{\oplus }}{\omega }_i{z}_i^p\otimes \left(\frac{1}{{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{\omega }_j}}\underset{\begin{array}{l}j=1\\ j\ne i\end{array}}{\overset{h_r}{\oplus }}{\omega }_j{z}_j^q\right)\right)}^{\frac{p}{p+q}}\right)\oplus \frac{\left|F_2\right|}{n}\left(\frac{1}{{\displaystyle \sum _{i=1}^{\left|F_2\right|}{\omega }_i}}\underset{i=1}{\overset{\left|F_2\right|}{\oplus }}{\omega }_i{z}_i^p\right)\right)}^{\frac{1}{p}}$$

(18)

where$z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$and${\alpha }_i,{\beta }_i,{\gamma }_i\in \left[0,s\right]$$(i=1,2,\cdots ,n)$;${\omega }_i$is the weight of input argument$z_i$meeting${\omega }_i\in \left[0,1\right]$and${\displaystyle \sum _{i=1}^n{\omega }_i}=1$;$p,q\ge 0$and$p+q>0$; |F₂| denotes the number of elements in F₂;$h_r$indicates the number of elements in partition$P_r$; and${\displaystyle \sum _{r=1}^t{h}_r}=n-\left|F_2\right|$. Then, we call it a linguistic neutrosophic generalized weighted PBM (LNGWPBM) operator.

Theorem 2.

Let$z_1,z_2,\cdots$and$z_n$be LNNs, where$z_i=\left(l_{{\alpha }_i},l_{{\beta }_i},l_{{\gamma }_i}\right)$and${\alpha }_i,{\beta }_i,{\gamma }_i\in \left[0,s\right]$$(i=1,2,\cdots ,n)$, and let the weight of input argument$z_i$be${\omega }_i$, where${\omega }_i\in \left[0,1\right]$and${\displaystyle \sum _{i=1}^n{\omega }_i}=1$. Then, the synthesized result of the LNGWPBM operator of the LNNs$z_1,z_2,\cdots$and$z_n$is still a LNN, which is shown as follows:

$$\begin{array}{l}LNGWPB{M}^{p,q}\left(z_1,z_2,\cdots ,z_n\right)=(l_{s{\left(1-\left({\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-K_{\alpha }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}\right)\left({\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}{\left(1-{\left({\alpha }_i/s\right)}^p\right)}^{{\omega }_i}}\right)}^{\frac{1}{{\displaystyle \sum _{i=1}^{\left|F_2\right|}{\omega }_i}}}\right)}^{\frac{\left|F_2\right|}{n}}\right)\right)}^{\frac{1}{p}}},\\ \hspace{1em}\hspace{1em}\hspace{1em}{l}_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-K_{\beta }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}{\left(1-{\left(1-{\beta }_i/s\right)}^p\right)}^{{\omega }_i}}\right)}^{\frac{1}{{\displaystyle \sum _{i=1}^{\left|F_2\right|}{\omega }_i}}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}},l_{s-s{\left(1-{\left({\left({\displaystyle \prod _{r=1}^t\left(1-{\left(1-K_{\gamma }\right)}^{\frac{p}{p+q}}\right)}\right)}^{\frac{1}{t}}\right)}^{\frac{n-\left|F_2\right|}{n}}{\left({\left({\displaystyle \prod _{i=1}^{\left|F_2\right|}{\left(1-{\left(1-{\gamma }_i/s\right)}^p\right)}^{{\omega }_i}}\right)}^{\frac{1}{{\displaystyle \sum _{i=1}^{\left|F_2\right|}{\omega }_i}}}\right)}^{\frac{\left|F_2\right|}{n}}\right)}^{\frac{1}{p}}})\end{array}$$

(19)

where $K_{\alpha }={\left({\displaystyle \prod _{i=1}^{h_r}\left(1-\left(1-{\left(1-{\left({\alpha }_i/s\right)}^p\right)}^{{\omega }_i}\right)\left(1-{\left({\displaystyle \prod _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{\left(1-{\left({\alpha }_j/s\right)}^q\right)}^{{\omega }_j}}\right)}^{\frac{1}{{\displaystyle \sum _{\begin{array}{l}j=1\\ j\ne i\end{array}}^{h_r}{\omega }_j}}}\right)\right)}\right)}^{\frac{1}{{\displaystyle \sum _{i=1}^{h_r}}}}$