Some Operations and Properties of the Cubic Intuitionistic Set with Application in Multi-Criteria Decision-Making

Abstract

This paper proposes some operations on the cubic intuitionistic set along with useful properties. We propose the internal cubic intuitionistic set (ICIS), the external cubic intuitionistic set (ECIS), P-order, R-order order (P-(R-) order), P-union, R-union (P-(R-) union), P-intersection, and R-intersection (P-(R-) intersection). We further investigate several properties of the P-(R-) union and P-(R-) intersection of ICISs and ECISs, and present some examples in this context. Some important theorems related to ICISs and ECISs are also presented with proof. Finally, an application example is given to measure the effectiveness and significance of the proposed operations by solving a multi-criteria decision-making (MCDM) problem.

1. Introduction

Zadeh [1] proposed the idea of fuzzy sets in 1965 and further extended this idea to an interval-valued fuzzy set (IVFS) [2]. Some complex decision-making problems in the economy, engineering, social science, environmental science, etc., exist that cannot be completely modeled by methods of classical mathematics because of the presence of various types of uncertainties. Others, on the other hand, use certain data processed by methods that are hybrid approaches, such as the INVAR method [3] or the CODAS-COMET method [4]. However, to handle the vagueness and uncertainty occurring in such decision-making problems, some well-known mathematical theories have been introduced, such as fuzzy set theory [1], intuitionistic fuzzy set (IFS) theory [5], interval-valued intuitionistic fuzzy set (IVIFS) theory [6,7], hesitant fuzzy set theory [8], hesitant fuzzy linguistic set theory [9], soft set theory [10], fuzzy soft set theory [11], etc. An example of this could be the use of triangular fuzzy numbers in a fuzzy extension of a simplified best–worst method [12].

At times, uncertainty research uses generalized approaches to better cope with the decision-making process via approaches related to the Dempster–Shafer evidence theory (DSET) [13], or quantum evidence theory (QET) [14]. Other ways are to use methods based on either entropy [15] or distance measures [16]. Most of the researchers studied IVFS [12]. For example, Zhang et al. [17] investigated the entropy of IVFSs based on distance measures. Zeng and Guo [18] discussed the similarity measure, inclusion of the measure, and entropy of IVFSs, while Grzegorzewski [19] proposed IVFSs based on the Hausdorff metric. Furthermore, IVFSs have been widely used and applied in real-life applications. For example, Sambuc [20] and Kohout [21] used the concept of IVFSs in medical diagnoses in thyroid pathology and medicine in a CLINAID system, respectively. Gorzalczany [22] used the idea of IVFSs in approximate reasoning. Turksen [23,24] further used the same idea of IVFSs in interval-valued logic in preference modeling [25].

Jun et al. [26] proposed the idea of a cubic set and presented its two important types, called the internal cubic set and the external cubic set by using the idea of the fuzzy set and IVFS. They further introduced some operations of union and intersection regarding the cubic sets, such as the P-(R-) union and P-(R-) intersection, and studied important related properties. Jun [27] further extended the idea of the cubic set, introduced the notion of the cubic intuitionistic set, and discussed its useful applications in BCK/BCI-algebras. Recently, studies on the cubic set theory have rapidly grown. For example, Jun et al. [28] proposed the concept of cubic IVIFS and discussed its important applications in BCK/BCI-algebra. With the help of using a cubic set and a neutrosophic set, Ali et al. [29] presented the notion of a neutrosophic cubic set and studied some useful properties. Kang and Kim [30] investigated the images and inverse images of almost-stable cubic sets and discussed the complement, the P-union, and the P-intersection of inverse images of almost-stable cubic sets. Chinnadurai et al. [31] investigated several properties of the P-(R-) union and P-(R-) intersection of cubic sets and studied some properties of cubic ideals of near rings. Jun et al. [32] proposed the ideas of cubic $\alpha$-ideals and cubic p-ideals and studied several useful properties.

Cubic sets are widely studied and are important in many areas, as discussed in the literature by various researchers. Motivated by the advantages of cubic sets, this paper proposes the notion of CIS based on IVFSs and intuitionistic fuzzy sets. Although Jun [27] previously introduced the idea of CIS as cubic intuitionistic sets and discussed their applications in BCK/BCI-algebras, this paper presents a completely different research work under the framework of CIS. We first propose two important types of CIS, named ICIS and ECIS. We then investigate the complement of CIS, the P-(R-) cubic intuitionistic subsets, and the P-(R-) union and the intersection of CISs. Furthermore, we prove various important theorems and results related to the proposed union and intersection operations. Finally, we present an application example to demonstrate the validity of the proposed operations by solving a MCDM problem.

The remainder of the paper can be summarized briefly as follows. Some basic concepts related to the work are presented in Section 2. The notions of CIS, ICIS, and ECIS are introduced in Section 3. We further investigate P-(R-)order, P-(R-)union, P-(R-)intersection, and related important properties with proof in the same section. A MCDM approach using CISs is presented in Section 4 along with an application example. We conclude the paper with some concluding remarks in Section 5.

2. Preliminary

This section introduces necessary notions and presents a few auxiliary results that we need in the rest of the paper. Throughout this paper, we let $\left[I\right]$, $I^X$, and ${\left[I\right]}^X$ stand for the set of all closed subintervals of $[0,1]$, the collection of all fuzzy sets in a set X, and IVFSs in X, respectively.

Definition 1.

Let X be a non-empty set. A fuzzy set in set X is defined as function $f:X\to [0,1]$. the relation ≤, join $(\vee )$, meet $(\wedge )$, and complement of $I^X$for all$x\in X$ can be defined, respectively, as follows:

$$f_1\le f_2\iff f_1(x)\le f_2(x) forall f_1,f_2\in I^X,$$

$$(f_1\vee f_2)(x)=f_1(x)\vee f_2(x)=max\{f_1(x),f_2(x)\},$$

$$(f_1\wedge f_2)(x)=f_1(x)\wedge f_2(x)=min\{f_1(x),f_2(x)\},$$

$$f_1^c(x)=1-f_1(x),$$

where $f_1^c$ represents the complement of $f_1$.

Definition 2.

By an interval number, we mean a closed sub-interval $a=[a^{-},a^{+}]$ of I where $0\le a^{-}\le a^{+}\le 1$. The complement $a^c$ of $a\in \left[I\right]$ is defined as follows:

$$a^c=[1-a^{+},1-a^{-}].$$

The refined minimum and refined maximum (briefly, rmin and rmax) and the symbols ⪰, ⪯, = of the elements $a_1=[a_1^{-},a_1^{+}]$ and $a_2=[a_2^{-},a_2^{+}]$ of $\left[I\right]$ is defined as follows:

$$rmin\left\{a_1,a_2\right\}=[min\left\{a_1^{-},a_2^{-}\right\},min\left\{a_1^{+},a_2^{+}\right\}],$$

$$rmax\left\{a_1,a_2\right\}=[max\left\{a_1^{-},a_2^{-}\right\},max\left\{a_1^{+},a_2^{+}\right\}],$$

$$a_1⪰a_2 \mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if} a_1^{-}\ge a_2^{-} and a_1^{+}\ge a_2^{+}.$$

Similarly, we can define $a_1⪯a_2$ and $a_1=a_2$.

Definition 3.

For a non-empty set X, a function $A:X\to \left[I\right]$ is called an IVFS in X. The element $A=[A^{-}(x),A^{+}(x)]$ for every $A\in {\left[I\right]}^X$ and $x\in X$, is called the membership degree of an element x to the set A. The IVFS is simply denoted as $A=[A^{-},A^{+}]$. The complement $A^c$ of A can be defined as $A^c=[1-A^{+},1-A^{-}].$

 For every $A_1,A_2\in {\left[I\right]}^X$, the following are true:

$$A_1\subseteq A_2\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}{A}_1⪯A_2,$$

$$A_1=A_2\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}{A}_1=A_2.$$

Definition 4

([5]). Let E be a crisp set. An IFS $\tilde{A}$ can be defined as

$$\tilde{A}=\{\langle x,{\mu }_{\tilde{A}}(x),{\nu }_{\tilde{A}}(x)\rangle :x\in E\}.$$

where ${\mu }_{\tilde{A}}:E\to [0,1]$ and ${\nu }_{\tilde{A}}:E\to [0,1]$ indicate, respectively, the membership and non-membership degrees of $x\in E$ with the condition $0\le {\mu }_{\tilde{A}}(x)+{\nu }_{\tilde{A}}(x)\le 1$ for every $x\in E$.

Definition 5

([6]). An expression of the form given by

$$B=\{\langle x,M_B(x),N_B(x)\rangle :x\in X\}$$

is called the IVIFS in X, where $M_B:X\to \left[I\right]$ and $N_B:X\to \left[I\right]$ are IVFSs with the condition that

$$0\le M_B^{+}(x)+N_B^{+}(x)\le 1 for all x\in X.$$

The intervals $M_B$ and $N_B$ denote, respectively, the membership and non-membership degrees of $x\in X$.

Definition 6

([26]). A mathematical structure of the form

$$A=\{\langle x,A(x),\lambda (x)\rangle :x\in X\},$$

is called the cubic set in X, where A and λ are, respectively, the IVFS and a fuzzy set in X. Jun [27] introduced the notion of the cubic intuitionistic set as follows:

Definition 7

([27]). A mathematical structure of the form

$$A=\{\langle x,A(x),\lambda (x)\rangle :x\in X\},$$

is called the cubic intuitionistic set where A is an IVIFS in X and λ is an IFS in X.

3. Some Operations on the Cubic Intuitionistic Set

This section introduces the concept of CIS with some modifications as proposed by Jun in [27] as follows:

Definition 8.

By CIS in a non-empty set X, we mean a mathematical structure of the form

$$\mathbf{A}=\left\{\langle x,M_A(x)/{\alpha }_A(x),N_A(x)/{\beta }_A(x)\rangle \right|x\in X\}$$

where $M_A:X\to \left[I\right]$ and $N_A:X\to \left[I\right]$ are IVFSs of the form $M_A(x)=[M^{-}(x),M^{+}(x)]$, $N_A(x)=[N^{-}(x),N^{+}(x)]$ with the conditions that

$$0\le M_A^{+}(x)+N_A^{+}(x)\le 1 and 0\le {\alpha }_A(x)+{\beta }_A(x)\le 1 for all x\in X.$$

$M_A(x)$ and $N_A(x)$ denote, respectively, the membership and non-membership degrees of x and ${\alpha }_A:X\to [0,1]$, ${\beta }_A:X\to [0,1]$ are fuzzy sets in X. For simplicity, we denote $CIS(X)$ as the collection of all CISs $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ in X. In the rest of the paper, we will use the same notations with symbols for CIS as presented in the above definition.

Remark 1.

For any non-empty set X, let $1(x)=1$ and $0(x)=0$ for all $x\in X.$ Then, $\mathbf{A}=\langle M_A/1,N_A/0(x)\rangle$, $\mathbf{B}=\langle M_B/0,N_B/1\rangle$ and $\mathbf{C}=\langle M_C/\frac{M_C^{-}+M_C^{+}}{2},N_C/\frac{N_C^{-}+N_C^{+}}{2}\rangle$ are all CISs in X.

Definition 9.

For $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle \in CIS(X)$, the score value of $\mathbf{A}$ is defined as

$$Sc(\mathbf{A})=\frac{1}{3}\left[\left(M_A^{-}+M_A^{+}+{\alpha }_A\right)-\left(N^{-}(x)+N^{+}(x)+{\beta }_A\right)\right]$$

where $Sc(\mathbf{A})\in \left[-1,1\right]$.

Definition 10.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle \in CIS(X)$, then

(i) 

$\mathbf{A}=\mathbf{B}\iff M_A=M_B,{\alpha }_A={\alpha }_B;N_A=N_B,{\beta }_A={\beta }_B$ (Equality)

(ii) 

$\mathbf{A}{\subseteq }_P\mathbf{B}\iff M_A\subseteq M_B,{\alpha }_A\le {\alpha }_B;N_A\supseteq N_B,{\beta }_A\ge {\beta }_B$ (P-order)

(iii) 

$\mathbf{A}{\subseteq }_R\mathbf{B}\iff M_A\subseteq M_B,{\alpha }_A\ge {\alpha }_B;N_A\supseteq N_B,{\beta }_A\le {\beta }_B$ (R-order)

Definition 11.

Let $\mathbf{0}=[0,0] and \mathbf{1}=[1,1]$. Then, a CIS $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ in which $M_A=\mathbf{0}$, ${\alpha }_A=1$, $N_A=\mathbf{1}$ and ${\beta }_A=0$ (respectively, $M_A=\mathbf{1}$, ${\alpha }_A=0$, $N_A=\mathbf{0}$ and ${\beta }_A=1)$ is denoted by $\ddot{0}$ (respectively $\ddot{1}$).

A CIS $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ in which $M_B=\mathbf{0}$, ${\alpha }_B=0$, $N_B=\mathbf{1}$, ${\beta }_B=1$ (respectively $M_B=\mathbf{1}$, ${\alpha }_B=1$, $N_B=\mathbf{0}$ and ${\beta }_A=0)$ is denoted by $\widehat{0}$ (respectively, $\widehat{1}$).

We can see that the score values of $\ddot{0},\ddot{1},\widehat{0}$ and $\widehat{1}$ can be computed, respectively, as $Sc(\ddot{0})=-0.33$, $Sc(\ddot{1})=0.33$, $Sc(\widehat{0})=-1$ and $Sc(\widehat{1})=1$.

Definition 12.

Consider the family of CISs ${\mathbf{A}}_i=\langle M_i/{\alpha }_i,N_i/{\beta }_i\rangle$, $i\in \mho$ in X, we define

(a) 

P-union

$$\underset{i\in \mho }{{\cup }_P}{\mathbf{A}}_i=\langle \underset{i\in \mho }{\cup }{M}_i/\underset{i\in \mho }{\vee }{\alpha }_i,\underset{i\in \mho }{\cap }{N}_i/\underset{i\in \mho }{\wedge }{\beta }_i\rangle$$

(b) 

P-intersection

$$\underset{i\in \mho }{{\cap }_P}{\mathbf{A}}_i=\langle \underset{i\in \mho }{\cap }{M}_i/\underset{i\in \mho }{\wedge }{\alpha }_i,\underset{i\in \mho }{\cup }{N}_i/\underset{i\in \mho }{\vee }{\beta }_i\rangle$$

(c) 

R-union

$$\underset{i\in \mho }{{\cup }_R}{\mathbf{A}}_i=\langle \underset{i\in \mho }{\cup }{M}_i/(\underset{i\in \mho }{\wedge }{\alpha }_i,\underset{i\in \mho }{\cap }{N}_i/\underset{i\in \mho }{\vee }{\beta }_i\rangle$$

(d) 

R-intersection

$$\underset{i\in \mho }{{\cap }_R}{\mathbf{A}}_i=\langle \underset{i\in \mho }{\cap }{M}_i/\underset{i\in \mho }{\vee }{\alpha }_i),\underset{i\in \mho }{\cup }{N}_i/(\underset{i\in \mho }{\wedge }{\beta }_i\rangle$$

Remark 2.

The complement of $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ is defined as

$${\mathbf{A}}^c=\langle M_A^c/1-{\alpha }_A,N_A^c/1-{\beta }_A\rangle .$$

Obviously, ${({\mathbf{A}}^c)}^c=\mathbf{A}$, ${\ddot{0}}^c=\ddot{1}$, ${\ddot{1}}^c=\ddot{0}$, ${\widehat{0}}^c=\widehat{1}$, ${\widehat{1}}^c=\widehat{0}$.

Remark 3.

For the family of CISs ${\mathbf{A}}_i=\langle M_i/{\alpha }_i,N_i/{\beta }_i\rangle$, $i\in \mho$ in X, we have ${(\underset{i\in \mho }{{\cup }_P}{\mathbf{A}}_i)}^c=\underset{i\in \mho }{{\cap }_P}{({\mathbf{A}}_i)}^c$, ${(\underset{i\in \mho }{{\cap }_P}{\mathbf{A}}_i)}^c=\underset{i\in \mho }{{\cup }_P}{({\mathbf{A}}_i)}^c,$${(\underset{i\in \mho }{{\cup }_R}{\mathbf{A}}_i)}^c=\underset{i\in \mho }{{\cap }_R}{({\mathbf{A}}_i)}^c$ and ${(\underset{i\in \mho }{{\cap }_R}{\mathbf{A}}_i)}^c=\underset{i\in \mho }{{\cup }_R}{({\mathbf{A}}_i)}^c.$

Definition 13.

Let X be a non-empty set.

1 

A CIS $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ is said to be ICIS if $M_A^{-}\le {\alpha }_A\le M_A^{+}$ and

$N_A^{-}\le {\beta }_A\le N_A^{+}$.

2 

A CIS $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ in X is said to be ECIS if ${\alpha }_B\notin (M_B^{-},M_B^{+})$ and ${\beta }_B\notin (N_B^{-},N_B^{+}).$

Example 1.

For a non-empty set X,

1 

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ be a CIS with $M_A=[0.1,0.3]$, ${\alpha }_A=0.2$, $N_A=[0.4,0.6]$ and ${\beta }_A=0.5$, then $\mathbf{A}$ is ICIS.

2 

Let $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be a CIS with $M_B=[0.2,0.4]$, ${\alpha }_B=0.1$, $N_B=[0.5,0.6]$ and ${\beta }_B=0.7$, then $\mathbf{B}$ is ECIS.

Remark 4.

Every CIS in X can be considered a Zadeh fuzzy set, IFS, IVFS, IVIFS, and cubic set according to $(M=N=\mathbf{0},\beta =0)$, $(M=N=\mathbf{0})$, $(N=\mathbf{0},\beta =0)$, $(\beta =\alpha =0)$ and $(N=\mathbf{0},\beta =0)$, respectively.

Theorem 1.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ be A CIS which is not an ECIS in X. Then there exist $x\in X$ such that ${\alpha }_A(x)\in (M_A^{-}(x),M_A^{+}(x))$ and ${\beta }_A(x)\in (N_A^{-}(x),N_A^{+}(x)).$

Proof. 

Straightforward. □

Theorem 2.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ be A CIS in X. If $\mathbf{A}$ is both ICIS and ECIS, then $\alpha (x)\in U(M)\cup L(M)$ and $\beta (x)\in U(N)\cup L(N)$ $for all x\in X$ where $U(M)=\left\{M^{+}(x)\right|x\in X\}$, $L(M)=\left\{M^{-}(x)\right|x\in X\}$, $U(N)=\left\{N^{+}(x)\right|x\in X\}$ and $L(N)=\left\{N^{-}(x)\right|x\in X\}.$

Proof. 

Assume that $\mathbf{A}$ is both ICIS and ECIS. Then, using Definition 13, we have $M^{-}(x)\le \alpha (x)\le M^{+}(x),N^{-}(x)\le \beta (x)\le N^{+}(x)$ and $\alpha (x)\notin (M^{-}(x),M^{+}(x)),\beta (x)\notin (N^{-}(x),$ $N^{+}(x))$ for all $x\in X.$ Thus $\alpha (x)=M^{-}(x) \mathrm{or} \alpha (x)=M^{+}(x) \mathrm{and} \beta (x)=N^{-}(x) \mathrm{or} \beta (x)=N^{+}(x).$ Hence $\alpha (x)\in U(M)\cup L(M) \mathrm{and} \beta (x)\in U(N)\cup L(N)$ $\mathrm{for} \mathrm{all} x\in X.$  □

Theorem 3.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ be A CIS in X. If $\mathbf{A}$ is ICIS (respectively, ECIS), then $A^c$ is ICIS (respectively ECIS).

Proof. 

Since $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ is ICIS in X, we have

$$M_A^{-}\le {\alpha }_A\le M_A^{+} and N_A^{-}\le {\beta }_A\le N_A^{+}$$

$$\left(\mathrm{respectively},{\alpha }_A\notin (M_A^{-},M_A^{+}) and {\beta }_A\notin (N_A^{-},N_A^{+})\right).$$

This implies that

$$1-M_A^{+}\le 1-{\alpha }_A\le 1-M_A^{-} and 1-N_A^{+}\le 1-{\beta }_A\le 1-N_A^{-}$$

$$\left(\mathrm{respectively},1-{\alpha }_A\notin (1-M_A^{+},1-M_A^{-}) and 1-{\beta }_A\notin (1-N_A^{+},1-N_A^{-})\right).$$

Hence ${\mathbf{A}}^c=\langle M_A^c/1-{\alpha }_A,N_A^c/1-{\beta }_A\rangle$ is ICIS (respectively, ECIS) □

We will show (through the following example) that the P-union and P-intersections of ECISs are not necessarily ECISs.

Example 2.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X. Let $M_A=[0.1,0.3]$, ${\alpha }_A=0.5$, $N_A=[0.4,0.6]$, ${\beta }_A=0.2$, $M_B=[0.4,0.6]$, ${\alpha }_B=0.2$, $N_B=[0.1,0.3]$ and ${\beta }_B=0.5$ for all $x\in X$. Then $\mathbf{A}{\cup }_p\mathbf{B}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$ and $\mathbf{A}{\cap }_p\mathbf{B}=\langle M_A/{\alpha }_B,N_A/{\beta }_B\rangle$. Hence, $\mathbf{A}{\cup }_p\mathbf{B}$ and $\mathbf{A}{\cap }_p\mathbf{B}$ are not ECISs.

From the following example, it can be easily seen that the R-union and R-intersection of ICIS need not be ICISs.

Example 3.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ICISs in X. Let $M_A=[0.1,0.3]$, ${\alpha }_A=0.2$, $N_A=[0.5,0.7]$, ${\beta }_A=0.6$, $M_B=[0.5,0.7]$, ${\alpha }_B=0.6$, $N_B=[0.1,0.3]$ and ${\beta }_B=0.2$ for all $x\in X$. Then $\mathbf{A}{\cup }_R\mathbf{B}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$ and $\mathbf{A}{\cap }_R\mathbf{B}=\langle M_A/{\alpha }_B,N_A/{\beta }_B\rangle$. Hence, $\mathbf{A}{\cup }_R\mathbf{B}$ and $\mathbf{A}{\cap }_p\mathbf{B}$ are not ICISs.

In the following examples, we will show that the R-union and R-intersection of ECIS may not be ECIS.

Example 4.

1 

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X. Let $M_A=[0.1,0.3]$, ${\alpha }_A=0.5$, $N_A=[0.35,0.4]$, ${\beta }_A=0.2$, $M_B=[0.4,0.6]$, ${\alpha }_B=0.7$, $N_B=[0.2,0.3]$ and ${\beta }_B=0.1$ for all $x\in X$. Then $\mathbf{A}{\cup }_R\mathbf{B}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$ and note that ${\alpha }_A\in (M_B^{-},M_B^{+})$; therefore, $\mathbf{A}{\cup }_R\mathbf{B}$ is not ECIS.

2 

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X. Let $M_A=[0.2,0.4]$, ${\alpha }_A=0.1$, $N_A=[0.4,0.6]$, ${\beta }_A=0.5$, $M_B=[0.5,0.7]$, ${\alpha }_B=0.3$, $N_B=[0.1,0.3]$ and ${\beta }_B=0.6$ for all $x\in X$. Then $\mathbf{A}{\cap }_R\mathbf{B}=\langle M_A/{\alpha }_B,N_A/{\beta }_A\rangle$ and, hence, $\mathbf{A}{\cap }_R\mathbf{B}$ is not ECIS.

Theorem 4.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ICISs in X, such that $max\{M_A^{-},M_B^{-}\}\le ({\alpha }_A\wedge {\alpha }_B)$ and $min\{N_A^{+},N_B^{+}\}\ge ({\beta }_A\vee {\beta }_B)$. Then the R-union and R-intersection of $\mathbf{A}$ and $\mathbf{B}$ are ICISs.

Proof. 

$\mathbf{A}$ and $\mathbf{B}$ are ICISs; therefore,

$$M_A^{-}\le {\alpha }_A\le M_A^{+},N_A^{-}\le {\beta }_A\le N_A^{+}$$

$$M_B^{-}\le {\alpha }_B\le M_B^{+} and N_B^{-}\le {\beta }_B\le N_B^{+}$$

which implies that

$$({\alpha }_A\wedge {\alpha }_B)\le {(M_A\cup M_B)}^{+} and ({\beta }_A\vee {\beta }_B)\ge {(N_A\cap N_B)}^{-}.$$

It follows that

$${(M_A\cup M_B)}^{-}=max\{M_A^{-},M_B^{-}\}\le ({\alpha }_A\wedge {\alpha }_B)\le {(M_A\cup M_B)}^{+}$$

and

$${(N_A\cap N_B)}^{-}\le ({\beta }_A\vee {\beta }_B)\le min\{N_A^{+},N_B^{+}\}={(N_A\cap N_B)}^{+}.$$

Hence, $\mathbf{A}{\cup }_R\mathbf{B}$ is ICIS. Similar arguments work in the case of $\mathbf{A}{\cap }_R\mathbf{B}$. □

Given two CISs $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ in X. If we exchange ${\alpha }_A$ for ${\alpha }_B$ and ${\beta }_A$ for ${\beta }_B$, we denote these CISs by ${\mathbf{A}}^{*}=\langle M_A/{\alpha }_B,N_A/{\beta }_B\rangle$ and ${\mathbf{B}}^{*}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$, respectively.

The next example shows that, for any two ECISs in X, ${\mathbf{A}}^{*}$ and ${\mathbf{B}}^{*}$ need not be ICISs in X.

Example 5.

1 

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be ICIS in X. Let $M_A=[0.1,0.3]$, ${\alpha }_A=0.7$, $N_A=[0.5,0.7]$, ${\beta }_A=0.15$, $M_B=[0.4,0.6]$, ${\alpha }_B=0.35$, $N_B=[0.2,0.3]$ and ${\beta }_B=0.1$ for all $x\in X$. Then it is easy to see that ${\mathbf{A}}^{*}=\langle M_A/{\alpha }_B,N_A/{\beta }_B\rangle$ and ${\mathbf{B}}^{*}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$ are not ICISs in X.

2 

Let $X=\{a,b\}$. Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X defined in

Table 1. CISs $\mathbf{A}$ and $\mathbf{B}$.

X	${\mathit{M}}_{\mathit{A}}/{\mathit{\alpha }}_{\mathit{A}}$	${\mathit{N}}_{\mathit{A}}/{\mathit{\beta }}_{\mathit{A}}$	${\mathit{M}}_{\mathit{B}}/{\mathit{\alpha }}_{\mathit{B}}$	${\mathit{N}}_{\mathit{B}}/{\mathit{\beta }}_{\mathit{B}}$
a	$[0.4,0.6]/0.65$	$[0.1,0.2]/0.35$	$[0.1,0.3]/0.35$	$[0.4,0.5]/0.25$
b	$[0.2,0.4]/0.1$	$[0.4,0.6]/0.7$	$[0.4,0.5]/0.15$	$[0.1,0.3]/0.35$

. Moreover, ${\mathbf{A}}^{*}=\langle M_A/{\alpha }_B,N_A/{\beta }_B\rangle$ and ${\mathbf{B}}^{*}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$ are not ICISs in X because ${\alpha }_B(a)=0.35\notin [0.4,0.6]=M_A(a)$, ${\beta }_B(a)=0.25\notin [0.1,0.2]=N_A(a)$. Moreover, ${\alpha }_B(b)=0.15\notin [0.2,0.4]=M_A(b)$ and ${\beta }_B(b)=0.35\notin [0.4,0.6]=N_A(b)$.

We will show through the following example that the P-union of two ECISs in X may not be an ICIS in X.

Example 6.

Consider again two ECISs, $\mathbf{A}$ and $\mathbf{B}$, as shown in Table 1. In this case, $\mathbf{A}{\cup }_P\mathbf{B}$ is not ICIS in X because $({\alpha }_A\vee {\alpha }_B)(a)=0.65\notin [0.4,0.6]=M_A\cup M_B,({\beta }_A\wedge {\beta }_B)(a)=0.25\notin [0.1,0.2]=N_A\cap N_B.$

 In the following result, we will find a condition for the P-union of two ECISs to be an ICIS.

Theorem 5.

For two ECISs $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ in X. If ${\mathbf{A}}^{*}=\langle M_A/{\alpha }_B,N_A/{\beta }_B\rangle$ and ${\mathbf{B}}^{*}=\langle M_B/{\alpha }_A,N_B/{\beta }_A\rangle$ are ICISs in X. Then $\mathbf{A}{\cup }_P\mathbf{B}$ and $\mathbf{A}{\cap }_P\mathbf{B}$ are ICISs in X.

Proof. 

Since $\mathbf{A}$ and $\mathbf{B}$ are ECISs in X, then

$${\alpha }_A\notin (M_A^{-},M_A^{+}),{\beta }_A\notin (N_A^{-},N_A^{+}),$$

$${\alpha }_B\notin (M_B^{-},M_B^{+}) and {\beta }_B\notin (N_B^{-},N_B^{+}).$$

For all $x\in X.$ Since ${\mathbf{A}}^{*}$ and ${\mathbf{B}}^{*}$ are ICISs in X, then

$$M_A^{-}\le {\alpha }_B\le M_A^{+},N_A^{-}\le {\beta }_B\le N_A^{+}$$

$$M_B^{-}\le {\alpha }_A\le M_B^{+} and N_B^{-}\le {\beta }_A\le N_B^{+}$$

for all $x\in X$. Thus, we can consider the following cases for any $x\in X$.

Case 1

$${\alpha }_A\le M_A^{-}\le {\alpha }_B\le M_A^{+},{\beta }_A\le N_A^{-}\le {\beta }_B\le N_A^{+},$$

$${\alpha }_B\le M_B^{-}\le {\alpha }_A\le M_B^{+} and {\beta }_B\le N_B^{-}\le {\beta }_A\le N_B^{+}.$$

Case 2

$$M_A^{-}\le {\alpha }_B\le M_A^{+}\le {\alpha }_A,N_A^{-}\le {\beta }_B\le N_A^{+}\le {\beta }_A,$$

$$M_B^{-}\le {\alpha }_A\le M_B^{+}\le {\alpha }_B and N_B^{-}\le {\beta }_A\le N_B^{+}\le {\beta }_B.$$

Case 3

$${\alpha }_A\le M_A^{-}\le {\alpha }_B\le M_A^{+},{\beta }_A\le N_A^{-}\le {\beta }_B\le N_A^{+},$$

$$M_B^{-}\le {\alpha }_A\le M_B^{+}\le {\alpha }_B and N_B^{-}\le {\beta }_A\le N_B^{+}\le {\beta }_B.$$

Case 4

$$M_A^{-}\le {\alpha }_B\le M_A^{+}\le {\alpha }_A,N_A^{-}\le {\beta }_B\le N_A^{+}\le {\beta }_A,$$

$${\alpha }_B\le M_B^{-}\le {\alpha }_A\le M_B^{+} and {\beta }_B\le N_B^{-}\le {\beta }_A\le N_B^{+}.$$

The arguments in all cases are similar; therefore, we consider the first case.

We have ${\alpha }_A=M_A^{-}=M_B^{-}={\alpha }_B \mathrm{and} {\beta }_A=N_A^{-}=N_B^{-}={\beta }_B.$

Since ${\mathbf{A}}^{*}$ and ${\mathbf{B}}^{*}$ are ICISs in X, then

$${\alpha }_B\le M_A^{+},{\alpha }_A\le M_B^{+},{\beta }_B\le N_A^{+} and {\beta }_A\le N_B^{+}.$$

It follows that

$${(M_A\cup M_B)}^{-}=max\{M_A^{-},M_B^{-}\}=({\alpha }_A\vee {\alpha }_B)$$

$$\le max\{M_A^{+},M_B^{+}\}={(M_A\cup M_B)}^{+} and$$

$${(N_A\cap N_B)}^{-}=min\{N_A^{-},N_B^{-}\}=({\beta }_A\wedge {\beta }_B)$$

$$\le min\{N_A^{+},N_B^{+}\}={(N_A\cup N_B)}^{+}.$$

Hence, $\mathbf{A}{\cup }_P\mathbf{B}$ is ICIS. Similar steps can be used for $\mathbf{A}{\cap }_P\mathbf{B}$. □

From Example 2, it can be easily seen that the P-union and P-intersections of ECISs are not necessarily the ECISs in X. In the next result, we will show when the P-union and P-intersection of two ECISs are ECISs in X.

Theorem 6.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X, such that

$$min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\}\ge ({\alpha }_A\wedge {\alpha }_B)$$

$$>max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\} and$$

$$min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\}>({\beta }_A\vee {\beta }_B)$$

$$\ge max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\},$$

then $\mathbf{A}{\cap }_P\mathbf{B}$ is ECIS in X.

Proof. 

Take

$${\alpha }_x=min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\},$$

$${\beta }_x=max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\},$$

$${\alpha }_x^{*}=min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\} and$$

$${\beta }_x^{*}=max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\}$$

then ${\alpha }_x$ is one of $M_A^{-},M_B^{-},M_A^{+},M_B^{+}$ and ${\alpha }_x^{*}$ is one of $N_A^{-},N_B^{-},N_A^{+},N_B^{+}$. We will consider the case when ${\alpha }_x=M_A^{-}$ and ${\alpha }_x^{*}=N_A^{-}$ or ${\alpha }_x=M_A^{+}$ and ${\alpha }_x^{*}=N_A^{+}$. Similar arguments will work for all remaining cases.

If ${\alpha }_x=M_A^{-}$ and ${\alpha }_x^{*}=N_A^{-}$, then

$$M_B^{-}\le M_B^{+}\le M_A^{-}\le M_A^{+}$$

$$N_B^{-}\le N_B^{+}\le N_A^{-}\le N_A^{+}$$

and so ${\beta }_x=M_B^{+}$ and ${\beta }_x^{*}=N_B^{+}$. Thus,

$$M_B^{-}={(M_A\cap M_B)}^{-}\le {(M_A\cap M_B)}^{+}=M_B^{+}={\beta }_x<({\alpha }_A\wedge {\alpha }_B),$$

$$N_A^{-}={(N_A\cup N_B)}^{-}={\alpha }_x>({\beta }_A\vee {\beta }_B)$$

and, hence,

$$({\alpha }_A\wedge {\alpha }_B)\notin ({(M_A\cap M_B)}^{-},{(M_A\cap M_B)}^{+}) and$$

$$({\beta }_A\vee {\beta }_B)\notin ({(N_A\cup N_B)}^{-},{(N_A\cup N_B)}^{+}).$$

If ${\alpha }_x=M_A^{+}$ and ${\alpha }_x^{*}=N_A^{+}$, then

$$M_B^{-}\le M_A^{+}\le M_B^{+} and N_B^{-}\le N_A^{+}\le N_B^{+}$$

so

$${\beta }_x=max\{M_A^{-},M_B^{-}\} and {\beta }_x^{*}=max\{N_A^{-},N_B^{-}\}.$$

Assume that ${\beta }_x=M_A^{-}$ and ${\beta }_x^{*}=N_A^{-}$, then

$$M_B^{-}\le M_A^{-}<({\alpha }_A\wedge {\alpha }_B)\le M_A^{+}\le M_B^{+} and$$

$$N_B^{-}\le N_A^{-}\le ({\beta }_A\vee {\beta }_B)<N_A^{+}\le N_B^{+}.$$

From the above inequality, we have the following cases

Case-1

$$M_B^{-}\le M_A^{-}<({\alpha }_A\wedge {\alpha }_B)<M_A^{+}\le M_B^{+} and$$

$$N_B^{-}\le N_A^{-}<({\beta }_A\vee {\beta }_B)<N_A^{+}\le N_B^{+}$$

Case-2

$$M_B^{-}\le M_A^{-}<({\alpha }_A\wedge {\alpha }_B)=M_A^{+}\le M_B^{+} and$$

$$N_B^{-}\le N_A^{-}=({\beta }_A\vee {\beta }_B)\le N_A^{+}\le N_B^{+}.$$

Case-1 contradicts the fact that CISs $\mathbf{A}$ and $\mathbf{B}$ are ECISs. From Case-2, it implies that

$$({\alpha }_A\wedge {\alpha }_B)\notin ({(M_A\cap M_B)}^{-},{(M_A\cap M_B)}^{+}) and$$

$$({\beta }_A\vee {\beta }_B)\notin ({(N_A\cup N_B)}^{-},{(N_A\cup N_B)}^{+})$$

since

$$({\alpha }_A\wedge {\alpha }_B)=M_A^{+}={(M_A\cap M_B)}^{+} and$$

$$({\beta }_A\vee {\beta }_B)=N_A^{-}={(N_A\cup N_B)}^{-}.$$

Assume that ${\beta }_x=M_B^{-}$ and ${\beta }_x^{*}=N_B^{-}$, then

$$M_A^{-}\le M_B^{-}<({\alpha }_A\wedge {\alpha }_B)\le M_A^{+}\le M_B^{+} and$$

$$N_A^{-}\le N_B^{-}\le ({\beta }_A\vee {\beta }_B)\le N_A^{+}\le N_B^{+}.$$

We now have two cases.

Case-1

$$M_A^{-}\le M_B^{-}<({\alpha }_A\wedge {\alpha }_B)<M_A^{+}\le M_B^{+} and$$

$$N_A^{-}\le N_B^{-}<({\beta }_A\vee {\beta }_B)<N_A^{+}\le N_B^{+}.$$

Case-2

$$M_A^{-}\le M_B^{-}<({\alpha }_A\wedge {\alpha }_B)=M_A^{+}\le M_B^{+} and$$

$$N_A^{-}\le N_B^{-}=({\beta }_A\vee {\beta }_B)<N_A^{+}\le N_B^{+}.$$

Case-1 contradicts that $\mathbf{A}$ and $\mathbf{B}$ are ECISs. From Case-2, it implies that

$$({\alpha }_A\wedge {\alpha }_B)\notin ({(M_A\cap M_B)}^{-},{(M_A\cap M_B)}^{+}) and$$

$$({\beta }_A\vee {\beta }_B)\notin ({(N_A\cup N_B)}^{-},{(N_A\cup N_B)}^{+})$$

since

$$({\alpha }_A\wedge {\alpha }_B)=M_A^{+}={(M_A\cap M_B)}^{+} and$$

$$({\beta }_A\vee {\beta }_B)=N_B^{-}={(N_A\cup N_B)}^{-}.$$

Similar results can be obtained if we assume

$${\beta }_x=M_B^{-} and {\beta }_x^{*}=N_A^{-}or{\beta }_x=M_A^{-} and {\beta }_x^{*}=N_B^{-}$$

Hence, the P-intersection of $\mathbf{A}$ and $\mathbf{B}$ is ECIS in X. □

Theorem 7.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X, such that

$$min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\}>({\alpha }_A\vee {\alpha }_B)$$

$$\ge max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\} and$$

$$min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\}\ge ({\beta }_A\wedge {\beta }_B)$$

$$>max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\},$$

then $\mathbf{A}{\cup }_P\mathbf{B}$ is ECIS in X.

Proof. 

The proof is similar to Theorem 6; therefore, we omit the details. □

Example 7.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in $X=\{a,b,c\}$ as shown in

Table 2. CISs $\mathbf{A}$ and $\mathbf{B}$.

X	${\mathit{M}}_{\mathit{A}}/{\mathit{\alpha }}_{\mathit{A}}$	${\mathit{N}}_{\mathit{A}}/{\mathit{\beta }}_{\mathit{A}}$	${\mathit{M}}_{\mathit{B}}/{\mathit{\alpha }}_{\mathit{B}}$	${\mathit{N}}_{\mathit{B}}/{\mathit{\beta }}_{\mathit{B}}$
a	$[0.1,0.2]/0.2$	$[0.45,0.6]/0.45$	$[0.05,0.3]/0.03$	$[0.4,0.5]/0.6$
b	$[0.1,0.4]/0.05$	$[0.5,0.6]/0.7$	$[0.2,0.3]/0.3$	$[0.55,0.65]/0.55$
c	$[0.6,0.7]/0.7$	$[0.1,0.15]/0.1$	$[0.5,0.8]/0.4$	$[0.05,0.2]/0.3$

. Then, $\mathbf{A}$ and $\mathbf{B}$ always satisfy the following conditions.

$$min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\}=({\alpha }_A\vee {\alpha }_B)$$

$$>max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\} and$$

$$min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\}>({\beta }_A\wedge {\beta }_B)$$

$$=max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\}.$$

However, the P-union of $\mathbf{A}$ and $\mathbf{B}$ is not ECIS because $({\alpha }_A\vee {\alpha }_B)(a)=0.2\in [0.1,0.3]=[{(M_A\cup M_B)}^{-}(a),{(M_A\cup M_B)}^{+}(a)]$ and $({\beta }_A\wedge {\beta }_B)(a)=0.45\in [0.4,0.5]=[{(N_A\cap N_B)}^{-}(a),{(N_A\cap N_B)}^{+}(a)].$

From Example 4, it can be easily observed that the R-union and R-intersection of ECISs may not be ECISs in X. In the next result, we will show that the R-union and R-intersection of two ECISs are ECISs in X.

Theorem 8.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X, such that

$$min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\}>({\alpha }_A\wedge {\alpha }_B)$$

$$\ge max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\} and$$

$$min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\}\ge ({\beta }_A\vee {\beta }_B)$$

$$>max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\},$$

then $\mathbf{A}{\cup }_R\mathbf{B}$ is ECIS in X.

Proof. 

Take

$${\alpha }_x=min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\},$$

$${\beta }_x=max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\},$$

$${\alpha }_x^{*}=min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\} and$$

$${\beta }_x^{*}=max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\}$$

then ${\alpha }_x$ is one of $M_A^{-},M_B^{-},M_A^{+},M_B^{+}$ and ${\alpha }_x^{*}$ is one of $N_A^{-},N_B^{-},N_A^{+},N_B^{+}$. We will consider the case when ${\alpha }_x=M_B^{-}$ and ${\alpha }_x^{*}=N_B^{-}$ or ${\alpha }_x=M_B^{+}$ and ${\alpha }_x^{*}=N_B^{+}$. Similar arguments will work for all remaining cases.

If ${\alpha }_x=M_B^{-}$ and ${\alpha }_x^{*}=N_B^{-}$, then

$$M_A^{-}\le M_A^{+}\le M_B^{-}\le M_B^{+} and$$

$$N_A^{-}\le N_A^{+}\le N_B^{-}\le N_B^{+}$$

so ${\beta }_x=M_A^{+}$ and ${\beta }_x^{*}=N_A^{+}$. Thus,

$$M_B^{-}={(M_A\cup M_B)}^{-}={\alpha }_x>({\alpha }_A\wedge {\alpha }_B) and$$

$$N_A^{+}={(N_A\cap N_B)}^{+}={\beta }_x^{*}<({\beta }_A\wedge {\beta }_B)$$

and, hence,

$$({\alpha }_A\wedge {\alpha }_B)\notin ({(M_A\cup M_B)}^{-},{(M_A\cup M_B)}^{+}) and$$

$$({\beta }_A\vee {\beta }_B)\notin ({(N_A\cap N_B)}^{-},{(N_A\cap N_B)}^{+}).$$

If ${\alpha }_x=M_B^{+}$ and ${\alpha }_x^{*}=N_B^{+}$, then

$$M_A^{-}\le M_B^{+}\le M_A^{+} and N_A^{-}\le N_B^{+}\le N_A^{+}$$

and so

$${\beta }_x=max\{M_A^{-},M_B^{-}\} and {\beta }_x^{*}=max\{N_A^{-},N_B^{-}\}.$$

Assume that ${\beta }_x=M_A^{-}$ and ${\beta }_x^{*}=N_A^{-}$, then

$$M_B^{-}\le M_A^{+}<({\alpha }_A\wedge {\alpha }_B)<M_B^{+}\le M_A^{+} and$$

$$N_B^{-}\le N_A^{-}<({\beta }_A\vee {\beta }_B)\le N_B^{+}\le N_A^{+}.$$

We have two cases

Case-1

$$M_B^{-}\le M_A^{-}<({\alpha }_A\wedge {\alpha }_B)<M_B^{+}\le M_A^{+} and$$

$$N_B^{-}\le N_A^{-}<({\beta }_A\vee {\beta }_B)<N_B^{+}\le N_A^{+}.$$

Case-2

$$M_B^{-}\le M_A^{-}=({\alpha }_A\wedge {\alpha }_B)\le M_B^{+}\le M_A^{+} and$$

$$N_B^{-}\le N_A^{-}<({\beta }_A\vee {\beta }_B)=N_B^{+}\le N_A^{+}.$$

Case-1 contradicts the fact that CISs $\mathbf{A}$ and $\mathbf{B}$ are ECISs. From Case-2, it implies that

$$({\alpha }_A\wedge {\alpha }_B)\notin ({(M_A\cup M_B)}^{-},{(M_A\cup M_B)}^{+}) and$$

$$({\beta }_A\vee {\beta }_B)\notin ({(N_A\cap N_B)}^{-},{(N_A\cap N_B)}^{+})$$

since

$$({\alpha }_A\wedge {\alpha }_B)=M_A^{-}={(M_A\cup M_B)}^{+} and$$

$$({\beta }_A\vee {\beta }_B)=N_B^{+}={(N_A\cap N_B)}^{+}.$$

Assume that ${\beta }_x=M_B^{-}$ and ${\beta }_x^{*}=N_B^{-}$, then

$$M_A^{-}\le M_B^{-}\le ({\alpha }_A\wedge {\alpha }_B)\le M_B^{+}\le M_A^{+} and$$

$$N_A^{-}\le N_B^{-}<({\beta }_A\vee {\beta }_B)\le N_B^{+}\le N_A^{+}.$$

We have two cases

Case-1

$$M_A^{-}\le M_B^{-}<({\alpha }_A\wedge {\alpha }_B)<M_B^{+}\le M_A^{+} and$$

$$N_A^{-}\le N_B^{-}<({\beta }_A\vee {\beta }_B)<N_B^{+}\le N_A^{+}$$

Case-2

$$M_A^{-}\le M_B^{-}=({\alpha }_A\wedge {\alpha }_B)<M_B^{+}\le M_A^{+} and$$

$$N_A^{-}\le N_B^{-}<({\beta }_A\vee {\beta }_B)=N_B^{+}\le N_A^{+}.$$

Case-1 contradicts the fact that CISs $\mathbf{A}$ and $\mathbf{B}$ are ECISs. From Case-2, it implies that

$$({\alpha }_A\wedge {\alpha }_B)\notin ({(M_A\cup M_B)}^{-},{(M_A\cup M_B)}^{+}) and$$

$$({\beta }_A\vee {\beta }_B)\notin ({(N_A\cap N_B)}^{-},{(N_A\cap N_B)}^{+})$$

since

$$({\alpha }_A\wedge {\alpha }_B)=M_B^{-}={(M_A\cup M_B)}^{-} and$$

$$({\beta }_A\vee {\beta }_B)=N_B^{+}={(N_A\cap N_B)}^{+}.$$

Similar results can be obtained if we assume

$${\beta }_x=M_B^{-} and {\beta }_x^{*}=N_A^{-}or{\beta }_x=M_A^{-} and {\beta }_x^{*}=N_B^{-}$$

Hence $\mathbf{A}{\cup }_R\mathbf{B}$ is ECIS in X. □

Example 8.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in a set $X=\{a,b,c\}$ as shown in

Table 3. CISs $\mathbf{A}$ and $\mathbf{B}$.

X	${\mathit{M}}_{\mathit{A}}/{\mathit{\alpha }}_{\mathit{A}}$	${\mathit{N}}_{\mathit{A}}/{\mathit{\beta }}_{\mathit{A}}$	${\mathit{M}}_{\mathit{B}}/{\mathit{\alpha }}_{\mathit{B}}$	${\mathit{N}}_{\mathit{B}}/{\mathit{\beta }}_{\mathit{B}}$
a	$[0.6,0.7]/0.7$	$[0.1,0.15]/0.1$	$[0.5,0.8]/0.9$	$[0.05,0.2]/0.03$
b	$[0.1,0.4]/0.5$	$[0.5,0.6]/0.5$	$[0.2,0.3]/0.3$	$[0.55,0.65]/0.55$
c	$[0.1,0.2]/0.2$	$[0.45,0.6]/0.45$	$[0.05,0.3]/0.4$	$[0.4,0.5]/0.3$

. Then it is easy to see that $\mathbf{A}$ and $\mathbf{B}$ satisfy the conditions

$$min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\}=({\alpha }_A\wedge {\alpha }_B)$$

$$>max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\} and$$

$$min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\}>({\beta }_A\vee {\beta }_B)$$

$$=max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\}.$$

However, $\mathbf{A}{\cup }_R\mathbf{B}$ is not ECIS because$({\alpha }_A\wedge {\alpha }_B)(a)=0.7\in [0.6,0.8]=[{(M_A\cup M_B)}^{-}(a),{(M_A\cup M_B)}^{+}(a)]$ and $({\beta }_A\vee {\beta }_B)(a)=0.1\in [0.05,0.15]=[{(N_A\cap N_B)}^{-}(a),{(N_A\cap N_B)}^{+}(a)].$

The following theorems can be easily verified and proved; therefore, we omit the details.

Theorem 9.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ECISs in X, such that

$$min\{max\{M_A^{+},M_B^{-}\},max\{M_A^{-},M_B^{+}\}\}\ge ({\alpha }_A\vee {\alpha }_B)$$

$$>max\{min\{M_A^{+},M_B^{-}\},min\{M_A^{-},M_B^{+}\}\} and$$

$$min\{max\{N_A^{+},N_B^{-}\},max\{N_A^{-},N_B^{+}\}\}>({\beta }_A\wedge {\beta }_B)$$

$$\ge max\{min\{N_A^{+},N_B^{-}\},min\{N_A^{-},N_B^{+}\}\},$$

then $\mathbf{A}{\cap }_R\mathbf{B}$ is also an ECIS in X.

Theorem 10.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ICISs in $X.$ If

$$({\alpha }_A\wedge {\alpha }_B)\le max\{M_A^{-},M_B^{-}\}$$

$$({\beta }_A\vee {\beta }_B)\ge min\{N_A^{+},N_B^{+}\},$$

then $\mathbf{A}{\cup }_R\mathbf{B}$ is an ECIS in X.

Theorem 11.

Let $\mathbf{A}=\langle M_A/{\alpha }_A,N_A/{\beta }_A\rangle$ and $\mathbf{B}=\langle M_B/{\alpha }_B,N_B/{\beta }_B\rangle$ be two ICISs in $X.$ If

$$({\alpha }_A\vee {\alpha }_B)\ge min\{M_A^{+},M_B^{+}\}$$

$$({\beta }_A\wedge {\beta }_B)\le max\{N_A^{-},N_B^{-}\},$$

then $\mathbf{A}{\cap }_R\mathbf{B}$ is ECIS in X.

4. MCDM Method Based on Cubic Intuitionistic Sets

In this section, we will apply the proposed operations to deal with the MCDM problems using CISs.

Let $A=\{A_1,A_2,\dots ,A_m\}$ be a set of alternatives, $C=\{C_1,C_2,\dots ,C_n\}$ be a set of criteria, and $E=\{e_1,e_2,\dots ,e_K\}$ be a set of experts. Suppose each alternative $A_i(i=1,2,\dots ,m)$ is assessed by the expert $e_k(k=1,2,\dots ,K)$ with respect to the criteria $C_j(j=1,2,\dots ,n)$ using CISs. The proposed MCDM method is based on the following steps.

Step 1

Construct the decision matrices $R^k={(r_{ij}^k)}_{m\times n}$ based on the assessed values of expert $e_k(k=1,2,\dots ,K)$ in the form of CISs $r_{ij}^k$.

Step 2

Calculate the aggregated decision matrix $R={(r_{ij})}_{m\times n}$ by using the proposed operations as discussed in Definition 12 where $r_{ij}=\underset{k=1,2,\dots K}{{\cup }_P}{r}_{ij}^k$ or $r_{ij}=\underset{k=1,2,\dots K}{{\cup }_R}{r}_{ij}^k$.

Step 3

Calculate the score value of each $r_{ij}$ of the aggregated decision matrix R by using Definition 9.

Step 4

Calculate the preference values of each alternative $A_i(i=1,2,\dots ,m)$ where $P(A_i)={\sum }_{i=1}^m{\sum }_{j=1}^n{r}_{ij}$.

Step 5

Generate the ranking order of alternatives according to the non-increasing order of the preference values.

An Application Example

Let us suppose that a technical committee composed of three technicians/experts $E=\{e_1,e_2,e_3\}$ wishes to select the best available washing machine on the market. Suppose, there are four types of washing machines $A=\{A_1,A_2,A_3,A_4\}$ available in the market and the experts are requested to select the best one amongst the four with respect to the criteria set $C=\{C_1=\mathrm{eco}-\mathrm{friendly},C_2=\mathrm{capacity},C_3=\mathrm{price}\}$. Suppose the expert $e_k(k=1,2,3)$ assessed each alternative $A_i(i=1,2,\dots ,4)$ under the criteria $C_j(j=1,2,3)$ by using the CISs. We will now proceed with the following steps.

Step 1

According to the expert’s opinion, the individual decision matrices $R^1,R^2,R^3$ are constructed, which can be seen in

Table 4. Decision matrix $R^1$ provided by expert $e_1$.

Alt.	${\mathit{C}}_1$	${\mathit{C}}_2$	${\mathit{C}}_3$
$A_1$	$\langle [0.7,0.8]/0.35,[0.1,0.2]/0.6\rangle$	$\langle [0.5,0.6]/0.5,[0.2,0.3]/0.2\rangle$	$\langle [0.6,0.7]/0.4,[0.1,0.2]/0.7\rangle$
$A_2$	$\langle [0.2,0.3]/0.25,[0.3,0.4]/0.7\rangle$	$\langle [0.3,0.4]/0.65,[0.5,0.6]/0.2\rangle$	$\langle [0.2,0.3]/0.7,[0.4,0.5]/0.2\rangle$
$A_3$	$\langle [0.8,0.9]/0.7,[0.05,0.1]/0.3\rangle$	$\langle [0.7,0.8]/0.2,[0.1,0.2]/0.4\rangle$	$\langle [0.6,0.7]/0.3,[0.2,0.3]/0.6\rangle$
$A_4$	$\langle [0.5,0.6]/0.5,[0.1,0.2]/0.3\rangle$	$\langle [0.6,0.7]/0.3,[0.2,0.3]/0.4\rangle$	$\langle [0.4,0.5]/0.6,[0.2,0.3]/0.2\rangle$

,

Table 5. Decision matrix $R^2$ provided by expert $e_2$.

Alt.	${\mathit{C}}_1$	${\mathit{C}}_2$	${\mathit{C}}_3$
$A_1$	$\langle [0.6,0.7]/0.3,[0.1,0.2]/0.5\rangle$	$\langle [0.45,0.5]/0.6,[0.25,0.35]/0.3\rangle$	$\langle [0.5,0.6]/0.7,[0.2,0.3]/0.2\rangle$
$A_2$	$\langle [0.25,0.4]/0.5,[0.4,0.5]/0.4\rangle$	$\langle [0.4,0.5]/0.6,[0.3,0.4]/0.3\rangle$	$\langle [0.3,0.4]/0.4,[0.5,0.6]/0.6\rangle$
$A_3$	$\langle [0.7,0.8]/0.8,[0.1,0.2]/0.1\rangle$	$\langle [0.8,0.9]/0.7,[0,0.1]/0.3\rangle$	$\langle [0.5,0.6]/0.8,[0.1,0.2]/0.2\rangle$
$A_4$	$\langle [0.4,0.5]/0.4,[0.1,0.2]/0.5\rangle$	$\langle [0.5,0.6]/0.4,[0.2,0.3]/0.6\rangle$	$\langle [0.5,0.6]/0.5,[0.2,0.3]/0.4\rangle$

and

Table 6. Decision matrix $R^3$ provided by expert $e_3$.

Alt.	${\mathit{C}}_1$	${\mathit{C}}_2$	${\mathit{C}}_3$
$A_1$	$\langle [0.6,0.7]/0.7,[0.2,0.3]/0.2\rangle$	$\langle [0.55,0.6]/0.8,[0.2,0.3]/0.1\rangle$	$\langle [0.65,0.7]/0.6,[0.2,0.3]/0.3\rangle$
$A_2$	$\langle [0.2,0.3]/0.5,[0.4,0.5]/0.4\rangle$	$\langle [0.3,0.4]/0.6,[0.4,0.5]/0.1\rangle$	$\langle [0.25,0.3]/0.4,[0.5,0.6]/0.5\rangle$
$A_3$	$\langle [0.7,0.85]/0.6,[0.1,0.15]/0.2\rangle$	$\langle [0.75,0.8]/0.6,[0.1,0.2]/0.3\rangle$	$\langle [0.6,0.7]/0.8,[0.1,0.2]/0.2\rangle$
$A_4$	$\langle [0.5,0.6]/0.7,[0.2,0.3]/0.2\rangle$	$\langle [0.5,0.6]/0.5,[0.2,0.3]/0.4\rangle$	$\langle [0.4,0.5]/0.7,[0.1,0.2]/0.1\rangle$

.

Step 2

The aggregated decision matrix $R={(r_{ij})}_{4\times 3}$ is calculated with the help of the proposed operation (P-union) as introduced in Definition 12 where $r_{ij}=\underset{k=1,2,3}{{\cup }_P}{r}_{ij}^k$. The aggregated decision matrix R is shown in

Table 7. Aggregated decision matrix R by applying the P-union operation.

Alt.	${\mathit{C}}_1$	${\mathit{C}}_2$	${\mathit{C}}_3$
$A_1$	$\langle [0.7,0.8]/0.7,[0.1,0.2]/0.2\rangle$	$\langle [0.55,0.6]/0.8,[0.2,0.3]/0.1\rangle$	$\langle [0.65,0.7]/0.7,[0.1,0.2]/0.2\rangle$
$A_2$	$\langle [0.25,0.4]/0.5,[0.3,0.4]/0.4\rangle$	$\langle [0.4,0.5]/0.65,[0.3,0.4]/0.1\rangle$	$\langle [0.3,0.4]/0.7,[0.4,0.5]/0.2\rangle$
$A_3$	$\langle [0.8,0.9]/0.8,[0.05,0.1]/0.1\rangle$	$\langle [0.8,0.9]/0.7,[0,0.1]/0.3\rangle$	$\langle [0.6,0.7]/0.8,[0.1,0.2]/0.2\rangle$
$A_4$	$\langle [0.5,0.6]/0.7,[0.1,0.2]/0.2\rangle$	$\langle [0.6,0.7]/0.5,[0.2,0.3]/0.4\rangle$	$\langle [0.5,0.6]/0.7,[0.1,0.2]/0.1\rangle$

.

Step 3

By using Definition 9, we will calculate the score value of each $r_{ij}$ of the aggregated decision matrix R. The matrix of the score values of the elements of R is shown in

Table 8. Score values of the aggregated decision matrix.

Alt.	${\mathit{C}}_1$	${\mathit{C}}_2$	${\mathit{C}}_3$
$A_1$	$0.5667$	$0.4500$	$0.5167$
$A_2$	$0.0167$	$0.2500$	$0.1000$
$A_3$	$0.7500$	$0.6667$	$0.5333$
$A_4$	$0.4333$	$0.3000$	$0.4667$

.

Step 4,5

Finally, the preference value $P(A_i),i=1,2,...,4$ of each alternative is calculated where $P(A_i)={\sum }_{i=1}^4{\sum }_{j=1}^3{r}_{ij}$. The preference values of alternatives by using the P-union operation are given below:

$$P(A_1)=0.5111,P(A_2)=0.1222,P(A_3)=0.6500,P(A_4)=0.4000.$$

We can see that the ranking order of alternatives according to the non-increasing order of their preference values is $A_3⪰A_1⪰A_4⪰A_2$. Similarly, the preference value of each alternative by using the R-union operation is calculated and given as follows:

$$P(A_1)=0.2778,P(A_2)=-0.0444,P(A_3)=0.4389,P(A_4)=0.2333$$

In this case, the ranking order of alternatives is $A_3⪰A_1⪰A_4⪰A_2$.

We can observe that the ranking order of alternatives by using the R-union operation is exactly the same as that obtained with the help of the P-union operation, which shows the robustness of the proposed approach. We can easily see that by using the P-intersection and R-intersection operations as discussed in Definition 12, the ranking order of alternatives will lead to the reverse order of the raking orders obtained in the P-union and R-union operations, respectively.

5. Conclusions

In this research work, we introduced a new modified form of CIS and discussed some of its related properties. We further introduced two types of CISs, i.e., ICIS and ECIS. The P-(R-) order, P-(R-) union, P-(R-) intersection of CISs, and some useful properties were also discussed with necessary examples. As a supplement, we proved that the P-union and P-intersection of ICISs are also ICISs. Some conditions for the P-(R-) union and P-(R-) intersection of two ECISs to be ICISs were also provided in this paper. We also provided a few conditions for the P-(R-) union and P-(R-) intersection of two ECISs to be ECISs. To check the effectiveness and validity of the proposed operations, we provided an application example at the end by solving a MCDM problem.

In future work, more research can be conducted regarding the intuitionistic cubic soft set and its application in information science and knowledge systems. We intend to apply the intuitionistic cubic soft sets to algebraic structures.