A novel type of soft rough covering and its application to multicriteria group decision making

Abstract

In this paper, we contribute to a recent and successful modelization of uncertainty, which the practitioner often encounters in the formulation of multicriteria group decision making problems. To be precise, in order to approach the uncertainty issue we introduce a novel type of soft rough covering by means of soft neighborhoods, and then we use it to improve decision making in a multicriteria group environment. Our research method is as follows. Firstly we introduce the soft covering upper and lower approximation operators of soft rough coverings. Then its relationships with well-established types of soft rough coverings are analyzed. Secondly, we define and investigate the measure degree of our novel soft rough covering. With this tool we produce a new class of soft rough sets. Finally, we propose an application of such soft rough covering model to multicriteria group decision making by means of an algorithmic solution. A fully developed example supports the implementability of this decision making method.

Appendices

Appendix I: Examples

Example I.1

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS, where $U=\{u_1, \ldots ,u_4\}, A=\{e_1,e_2,e_3\}$ and ${\mathbf {G}}=(F,A)$ is characterized by $F(e_1)=\{u_1,u_2\}, F(e_2)=\{u_1,u_2,u_3\}$ and $F(e_3)=U$.

• (1) Some simple calculations show $N_S(u_1)=\{u_1,u_2\}=N_S(u_2), N_S(u_3)=\{u_1,u_2,u_3\}$ and $N_S(u_4)=U$.

• (2) $Md_S(u_1)=\{F(e_1)\}, Md_S(u_2)=\{F(e_1)\}, Md_S(u_3)=\{F(e_2)\}$ and $Md_S(u_4)=\{F(e_3)\}$. Thus ${\mathbf {G}}$ is soft unary.

• (3) Clearly, $u_1\in F(e_3), Md_S(u_1)=\{F(e_1)\},$ but $F(e_3)\nsubseteq \bigcup Md_S(u_1)=F(e_1)$. Hence ${\mathbf {G}}$ is not SPC.

Example I.2

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS, where $A=\{e_1, \ldots , e_4\}$, $U=\{u_1,u_2,u_3\}$ and ${\mathbf {G}}=(F,A)$ is characterized by $F(e_1)=\{u_1,u_2\}, F(e_2)=\{u_1,u_3\}, F(e_3)=\{u_2,u_3\}$ and $F(e_4)=U$. Now ${\mathbf {G}}$ is SPC because $Md_S(u_1)=\{F(e_1), F(e_2)\}, Md_S(u_2)=\{F(e_1), F(e_3)\}$, $Md_S(u_3)=\{F(e_2), F(e_3)\}$.

Example I.3

Define the SCAS $S=(U,C_{{\mathbf {G}}})$, where $U=\{u_1, \ldots ,u_5\}, A=\{e_1, \ldots ,e_4\}$ and ${\mathbf {G}}=(F,A)$ is characterized by the data in Table 1.

Table 1 Tabular representation of ${\mathbf {G}}=(F,A)$ in Example I.3

Let $X_1=\{u_1,u_2,u_3\}$. Some simple calculations show $ZS^*(X_1)=\{u_1,u_2,u_3,u_4\}$ and $ZS_*(X_1)=\{u_1,u_2,u_3\}$. Thus, $X_1$ is a Z-SRC.

Let $X_1=\{u_5\}$. Some simple calculations show $ZS^*(X_2)=ZS_*(X_2)=\{u_5\}$. Thus $X_2$ is Z-SC-definable.

Example I.4

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS, where $U=\{u_1,\ldots ,u_6\}, A=\{e_1,\ldots ,e_4\}$ and ${\mathbf {G}}=(F,A)$ is characterized by the data in Table 2.

Table 2 The tabular representation of ${\mathbf {G}}=(F,A)$ in Example I.4

Some simple calculations ensure

$$\begin{aligned} N_S(u_1)= & {} F(e_1)\cap F(e_2)=\{u_1,u_2\},\\ N_S(u_2)= & {} F(e_1)\cap F(e_2)\cap F(e_3)=\{u_2\},\\ N_S(u_3)= & {} F(e_1)\cap F(e_3)\cap F(e_4)=\{u_3\},\\ N_S(u_4)= & {} F(e_2)\cap F(e_3)\cap F(e_4)=\{u_4, u_5\},\\ N_S(u_5)= & {} F(e_2)\cap F(e_3)\cap F(e_4)=\{u_4,u_5\} \end{aligned}$$

and $N_S(u_6)=F(e_6)=\{u_3,u_4,u_5,u_6\}$.

Let $X=\{u_1,u_4,u_5\}$ and $Y=\{u_2,u_3,u_4,u_5\}$, and so, $X\cup Y=\{u_1,u_2,u_3,u_4,u_5\}$.

By calculations, $ZS_*(X)=\{u_4,u_5\}$ and $ZS_*(Y)=\{u_2,u_3,u_4,u_5\}$, and so $ZS_*(X)\cup ZS_*(Y)=\{u_2,u_3,u_4,u_5\}$. But, $ZS_*(X\cup Y)=\{u_1,u_2,u_3,u_4,u_5\}$. In conclusion, $ZS_*(X\cup Y)\supsetneq ZS_*(X)\cup ZS_*(Y).$

Example I.5

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS, where $U=\{u_1, \ldots ,u_8\}, A=\{e_1,\ldots ,e_6\}$ and ${\mathbf {G}}=(F,A)$ is characterized by the data in Table 3.

Table 3 The tabular representation of ${\mathbf {G}}=(F,A)$ in Example I.5

Some simple calculations ensure

$$\begin{aligned} N_S(u_1)= & {} F(e_1)\cap F(e_4)=\{u_1,u_4\},\\ N_S(u_2)= & {} F(e_1)\cap F(e_2)\cap F(e_3)\cap F(e_5)=\{u_2,u_4\},\\ N_S(u_3)= & {} F(e_1)\cap F(e_2)\cap F(e_3)\cap F(e_6)=\{u_3,u_4\},\\ N_S(u_4)= & {} F(e_1)\cap \cdots \cap F(e_6)=\{u_4\},\\ N_S(u_5)= & {} F(e_2)\cap F(e_4)\cap F(e_5)\cap F(e_6)=\{u_4,u_5,u_6\},\\ N_S(u_6)= & {} F(e_2)\cap F(e_4)\cap F(e_5)\cap F(e_6)=\{u_4,u_5,u_6\},\\ N_S(u_7)= & {} F(e_3)\cap \cdots \cap F(e_6)=\{u_4,u_7,u_8\}, \end{aligned}$$

and $N_S(u_8)=F(e_3)\cap \cdots \cap F(e_6)=\{u_4,u_7,u_8\}$.

Let $X=\{u_1,u_5,u_7\}$ and $Y=\{u_3,u_4,u_5,u_7,u_8\}$, hence $X\cap Y=\{u_5,u_7\}$.

By direct calculation one obtains $ZS^*(X)=\{u_1,u_5,u_6,u_7,u_8\}$ and $ZS_*(Y)=U$, and so $ZS_*(X)\cap ZS_*(Y)=\{u_1,u_5,u_6,u_7,u_8\}$. But $ZS_*(X\cap Y)=\{u_5,u_6,u_7,u_8\}$. This means that $ZS_*(X\cap Y)\subsetneq ZS_*(X)\cap ZS_*(Y).$

Example I.6

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS, where ${\mathbf {G}}=(F,A)$ is defined by Table 4 ($A=\{e_1,e_2,e_3\}, U=\{u_1,u_2,u_3,u_4\}$).

Table 4 Tabular representation of ${\mathbf {G}}=(F,A)$ in Example I.6

Let $X=\{u_2\}$. By simple calculations one obtains ${ FS}_*(X)=\emptyset$ and $ZS_*(X)=\{u_2\}$. This means that ${ FS}_*\subsetneq ZS_*$.

Example I.7

Consider $S=(U,C_{{\mathbf {G}}})$ as in Example I.3.

Let $X=\{u_3,u_4\}$. By calculations, $FS^*(X)=U$, $SS^*(X)=\{u_3,u_4\}$, $TS^*(X)=\{u_1,u_2,u_3,u_4\}$ and $ZS^*(X)=\{u_3,u_4\}$. This means that $SS^*\subsetneq TS^*$, $TS^*\subsetneq FS^*$ and $ZS^*\subsetneq FS^*$.

Example I.8

• (1) Consider $S=(U,C_{{\mathbf {G}}})$ as in Example I.6. Let $X=\{u_2\}$. By calculations, then $ZS^*(X)=\{u_1,u_2\}$ and $TS^*(X)=\{u_1,u_2,u_3,u_4\}.$ This means that $ZS^*\subsetneq TS^*$.

• (2) Let us define $U=\{u_1,u_2,u_3,u_4\}$ and let $S=(U,C_{{\mathbf {G}}})$ be a SCAS with $A=\{e_1,e_2\}$, where ${\mathbf {G}}=(F,A)$ is defined by $F(e_1)=\{u_1\}$ and $F(e_2)=U$. Then $ZS^*(X)=U$ and $TS^*(X)=\{u_1\}$. This means that $TS^*\subsetneq ZS^*$.

Example I.9

• (1) Consider $S=(U,C_{{\mathbf {G}}})$ as in Example I.6. Let $X=\{u_2\}$. By calculations, then $ZS^*(X)=\{u_1,u_2\}$ and $SS^*(X)=\{u_1,u_2,u_3,u_4\}.$ This means that $ZS^*\subsetneq SS^*$.

• (2) Consider $S=(U,C_{{\mathbf {G}}})$ as in Example I.8 (2).

Then $ZS^*(X)=\{u_1,u_2,u_3,u_4\}$ and $SS^*(X)=\{u_1\}$. This means that $SS^*\subsetneq ZS^*$.

Example I.10

Let $U=\{u_1, \ldots , u_6\}$ and let $S=(U,C_{{\mathbf {G}}})$ be a SCAS with $A=\{e_1, \ldots , e_5\}$. We define ${\mathbf {G}}=(F,A)$ according to Table 5 below.

Table 5 Tabular representation of the soft set ${\mathbf {G}}=(F,A)$ in Example I.10

By simple calculations we compute $N_S(u_1)=\{u_1\}$, $N_S(u_2)=\{u_2\}$, $N_S(u_3)=\{u_1,u_2,u_3\}$, $N_S(u_4)=\{u_4\}$, $N_S(u_5)=\{u_5\}$ and $N_S(u_6)=\{u_6\}$.

Let $X=\{u_2,u_3,u_5\}$. Then by calculations, $D_{{\mathbf {G}}}(u_1,X)=0$, $D_{{\mathbf {G}}}(u_2,X)=1$, $D_{{\mathbf {G}}}(u_3,X)=\frac{1}{3}$, $D_{{\mathbf {G}}}(u_4,X)=0$, $D_{{\mathbf {G}}}(u_5,X)=1$ and $D_{{\mathbf {G}}}(u_6,X)=0$.

Example I.11

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS, where $U=\{u_1, \ldots ,u_8\}, A=\{e_1, \ldots ,e_6\}$ and ${\mathbf {G}}=(F,A)$ is defined by Table 6.

Table 6 The tabular representation of the soft set ${\mathbf {G}}=(F,A)$ in Example I.11

By direct calculations we have

$N_S(u_1)=\{u_1,u_2\}$, $N_S(u_2)=\{u_2\}$, $N_S(u_3)=\{u_3\}$, $N_S(u_4)=\{u_4\}$, $N_S(u_5)=\{u_5\}$, $N_S(u_6)=\{u_2,u_5,u_6\}$, $N_S(u_7)=\{u_5,u_7\}$ and $N_S(u_8)=\{u_5,u_7,u_8\}$.

• (1) Let $X=\{u_1\}$ and $Y=\{u_2\}$, and so, $X\cup Y=\{u_1,u_2\}$. By calculations,

  $D_{X}^{{\mathbf {G}}}(u_1)=\frac{1}{2}$ and $D_{Y}^{{\mathbf {G}}}(u_1)=\frac{1}{2}$, and so, $(D_{X}^{{\mathbf {G}}}\cup D_{Y}^{{\mathbf {G}}}) (u_1)=\frac{1}{2}$, but $D_{X\cup Y}^{{\mathbf {G}}}(u_1)=1$. This shows that $D_{X\cup Y}^{{\mathbf {G}}}\supsetneq D_{X}^{{\mathbf {G}}}\cup D_{Y}^{{\mathbf {G}}}$.

• (2) $X\cap Y=\emptyset$. By calculations, we have

  $(D_{X}^{{\mathbf {G}}}\cap D_{Y}^{{\mathbf {G}}}) (u_2)=\frac{1}{2}$, but $D_{X\cap Y}^{{\mathbf {G}}}(u_2)=0$. This shows that $D_{X\cap Y}^{{\mathbf {G}}}\subsetneq D_{X}^{{\mathbf {G}}}\cap D_{Y}^{{\mathbf {G}}}$.

Example I.12

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS with $U=\{u_1, \ldots ,u_5\}, A=\{e_1,\ldots , e_4\}$ and ${\mathbf {G}}=(F,A)$ is characterized in Table 7.

Table 7 The tabular representation of ${\mathbf {G}}=(F,A)$ in Example I.12

By calculations, $N_S(u_1)=\{u_1\}$, $N_S(u_2)=\{u_2,u_3\}$, $N_S(u_3)=\{u_3\}$, $N_S(u_4)=\{u_4\}$ and $N_S(u_5)=\{u_4,u_5\}$.

Let $X=\{u_1,u_2\}$. Then by calculations, $D_{X}^{{\mathbf {G}}}(u_1)=1$, $D_{X}^{{\mathbf {G}}}(u_2)=\frac{1}{2}$, $D_{X}^{{\mathbf {G}}}(u_3)=0$, $D_{X}^{{\mathbf {G}}}(u_4)=0$ and $D_{X}^{{\mathbf {G}}}(u_5)=0$. Then $S_*(X)=\{u_1\}$ and $S^*(X)=\{u_1,u_2\}$.

Example I.13

Let $S=(U,C_{{\mathbf {G}}})$ be a SCAS as in Example I.12. By calculations, $N_S(u_1)=\{u_1\}$, $N_S(u_2)=\{u_2,u_3\}$, $N_S(u_3)=\{u_3\}$, $N_S(u_4)=\{u_4\}$ and $N_S(u_5)=\{u_4,u_5\}$.

Let $X=\{u_1,u_2\}$ and $Y=\{u_3\}$. Then by calculations, $D_{X}^{{\mathbf {G}}}(u_1)=1$, $D_{X}^{{\mathbf {G}}}(u_2)=\frac{1}{2}$, $D_{X}^{{\mathbf {G}}}(u_3)=D_{X}^{{\mathbf {G}}}(u_4)=D_{X}^{{\mathbf {G}}}(u_5)=0$; $D_{Y}^{{\mathbf {G}}}(u_1)=D_{Y}^{{\mathbf {G}}}(u_4)=D_{Y}^{{\mathbf {G}}}(u_5)=0$, $D_{Y}^{{\mathbf {G}}}(u_2)=\frac{1}{2}$ and $D_{X}^{{\mathbf {G}}}(u_3)=1$.

Hence, $S_*(X)=\{u_1\}$, $S^*(X)=\{u_1,u_2\}$, $S_*(Y)=\{u_3\}$ and $S^*(Y)=\{u_2,u_3\}$.

• (1) Since $X\cup Y=\{u_1,u_2,u_3\}$. Then $D_{X\cup Y}^{{\mathbf {G}}}(u_1)=D_{X\cup Y}^{{\mathbf {G}}}(u_2)=D_{X\cup Y}^{{\mathbf {G}}}(u_3)=1$ and $D_{X\cup Y}^{{\mathbf {G}}}(u_4)=D_{X\cup Y}^{{\mathbf {G}}}(u_5)=0$. Hence $S_*(X\cup Y)=\{u_1,u_2,u_3\}$. This means that $S_*(X)\cup S_*(Y)\subsetneq S_*(X\cup Y)$.

• (2) $S^*(X\cap Y)=\emptyset$ because X and Y are disjoint subsets. Hence, $S^*(X\cap Y)\subsetneq S^*(X)\cap S^*(Y)$.

Appendix II: Proofs

In this section we prove the results that we have stated.

Proof of Theorem 4.3

It is clear that (1L), (1H), (2L) and (2H) hold true.

• (3LH) From the definition, it is immediate to deduce ${ FS}_*(X)\subseteq X$. Let $y\in X$, then $\exists a\in A$ with the property $y\in F(a)$ since ${\mathbf {G}}$ is a SC over U, therefore $X\cap F(a)\ne \emptyset$. This means that $y\in FS^*(X)$ which proves $X\subseteq FS^*(X)$.

• (4L) Let $X\subseteq Y$. If $x\in { FS}_*(X)$, there must be $e\in A$ ensuring $u\in F(e)\subseteq X\subseteq Y$ thus $u\in { FS}_*(Y)$. This ensures ${ FS}_*(X)\subseteq FS^*(Y)$.

• (4H) This property is similar to (4L).

• (5L) (1) Since $Y\cap X\subseteq X$, then by (4L), ${ FS}_*(X\cap Y)\subseteq { FS}_*(X)$. Similarly, ${ FS}_*(X\cap Y)\subseteq { FS}_*(Y)$. Hence ${ FS}_*(X\cap Y)\subseteq { FS}_*(X)\cap { FS}_*(Y)$.

• (2) Since $X\subseteq X\cup Y$, then by (4L), ${ FS}_*(X)\subseteq { FS}_*(X\cup Y)$. Similarly, ${ FS}_*(Y)\subseteq { FS}_*(X\cup Y)$. Hence ${ FS}_*(X)\cup { FS}_*(Y)\subseteq { FS}_*(X\cup Y)$.

• (5H) (1) This property can be proven by arguments as in the proof of (5L)(1).

• (2) Let $u\in FS^*(X\cup Y)$, there exists $a\in A$ with the properties $u\in F(a)$ and $F(a)\cap (X\cup Y)\ne \emptyset$. The latter statement implies either $F(a)\cap X\ne \emptyset$ or $F(a)\cap Y\ne \emptyset$, that is, $u\in FS^*(X)$ or $u\in FS^*(Y)$. We deduce that $FS^*(X\cup Y)\subseteq FS^*(X)\cup FS^*(Y)$.

  By application of (4H) to $X\subseteq X\cup Y$, one obtains $FS^*(X)\subseteq FS^*(X\cup Y)$, and similarly, $FS^*(Y)\subseteq FS^*(X\cup Y)$. Hence $FS^*(X)\cup FS^*(Y)\subseteq FS^*(X\cup Y)$.

  By coupling these conclusions we deduce $FS^*(X)\cup FS^*(Y)=FS^*(X\cup Y)$.

• (6) (1) The case $\thicksim { FS}_*(X)=\emptyset$ obviously entails $\thicksim { FS}_*(X)\subseteq FS^*(\thicksim X)$. Otherwise, pick any $x\in \thicksim { FS}_*(X)$. Since ${\mathbf {G}}$ is a SC over U, there exists $e\in A$ for which $x\in F(e)$. Note that $\thicksim { FS}_*(X)=\{u\in U: \text{ for } \text{ each } a\in A, u\in F(a) \text{ implies } \thicksim X \cap F(a)\ne \emptyset \}.$ It follows that $F(e)\cap (\thicksim X)\ne \emptyset$, therefore $x\in FS^*(\thicksim X)$. Thus $\thicksim { FS}_*(X)\subseteq FS^*(\thicksim X).$

• (2) The case $\thicksim FS^*(X)=\emptyset$ obviously entails $\thicksim FS^*(X)\subseteq { FS}_*(\thicksim X)$. Otherwise, pick any $x\in \thicksim FS^*(X)$. Since ${\mathbf {G}}$ is a SC over U, there exists $e\in A$ for which $x\in F(e)$. Note that $\thicksim FS^*(X)=\{x\in U\vert \forall a\in A, x\in F(a) \text{ implies } F(a)\subseteq \thicksim X\}.$ We deduce that $F(e)\subseteq \thicksim X$, therefore $x\in FS^*(\thicksim X)$. Thus $\thicksim FS^*(X)\subseteq { FS}_*(\thicksim X).$

• (7L) We prove this claim by double inclusion.

  A routine application of (3LH) produces ${ FS}_*({ FS}_*(X))\subseteq { FS}_*(X)$.

  For each $x\in { FS}_*(X)$ there exists $e\in A$ with the property $x\in F(e)\subseteq X$. It is clear that $x\in F(e)$ and $F(e)\subseteq { FS}_*(X)$, therefore $x\in { FS}_*({ FS}_*(X))$. Hence ${ FS}_*(X)\subseteq { FS}_*({ FS}_*(X))$.

• (7H) Let $x\in FS^*(X)$, there exists $e\in A$ with the properties $x\in F(e)$ and $F(e)\cap X\ne \emptyset$. It is clear that $x\in F(e)$ and $F(e)\cap FS^*(X)\ne \emptyset$, therefore $x\in FS^*(FS^*(X))$. Hence, $FS^*(X)\subseteq FS^*(FS^*(X))$.

• (8L) For any $a\in A$, from the definition of ${ FS}_*$ we can easily derive ${ FS}_*(F(a))=F(a)$.$\square$

Proof of Theorem 5.2

It is clear that (1L), (1H), (2L) and (2H) hold.

• (3LH) We prove the two inclusions separately.

  When $x\in ZS_*(X)$, one necessarily has the inclusion $N_S(x)\subseteq X$. Since $x\in N_S(x)$, then $x\in X$. Therefore we obtain the inclusion $ZS_*(X)\subseteq X$.

  Furthermore, select an element x from X, then $X\cap N_S(x)\ne \emptyset$ thus $x\in ZS^*(X)$. This proves the inclusion $X\subseteq ZS^*(X)$.

• (4L) Fix subsets X, Y such that $X\subseteq Y$. If $x\in ZS_*(X)$, then $N_S(x)\subseteq X\subseteq Y$, and so, $x\in ZS_*(Y)$. This means that the inclusion $ZS_*(X)\subseteq ZS_*(Y)$ holds.

• (4H) It is similar to (4L).

• (5L) (1)

  $$\begin{aligned} ZS_*(X\cap Y)= & {} \left\{ y\in U\vert N_S(y)\subseteq X\cap Y\right\} \\= & {} \left\{ y\in U\vert N_S(y)\subseteq X\wedge N_S(y)\subseteq Y\right\} \\= & {} \left\{ y\in U\vert N_S(y)\subseteq X\right\} \cap \left\{ y\in U\vert N_S(y)\subseteq Y\right\} \\= & {} ZS_*(X)\cap ZS_*(Y). \end{aligned}$$

• (2) Since $X\subseteq X\cup Y$, then by (4L), $ZS_*(X)\subseteq ZS_*(X\cup Y)$. Similarly, one obtains the inclusion $ZS_*(Y)\subseteq ZS_*(X\cup Y)$. Hence $ZS_*(X)\cup ZS_*(Y)\subseteq ZS_*(X\cup Y)$.

• (5H) (1) It is analogous to the proof of (5L)(2).

• (2) We demonstrate this set equality by double inclusion.

  Fix $x\in ZS^*(X\cup Y)$, then $N_S(x)\cap (X\cup Y)\ne \emptyset$, which implies, $N_S(x)\cap X\ne \emptyset$ or $N_S(x)\cap Y\ne \emptyset$, that is, $x\in ZS^*(X)$ or $x\in ZS^*(Y)$. This means that $ZS^*(X\cup Y)\subseteq ZS^*(X)\cup ZS^*(Y)$.

  Furthermore, by application of (4H) to $X\subseteq X\cup Y$ we know $ZS^*(X)\subseteq ZS^*(X\cup Y)$, and similarly $ZS^*(Y)\subseteq ZS^*(X\cup Y)$. Hence $ZS^*(X)\cup ZS^*(Y)\subseteq ZS^*(X\cup Y)$.

• (6) (1) Pick $x\in ZS_*(X)$, then one has the chain of equivalences $x\in ZS_*(X)\Leftrightarrow N_S(x)\subseteq X\Leftrightarrow N_S(\thicksim X)=\emptyset \Leftrightarrow x\notin ZS^*(\thicksim X)\Leftrightarrow \thicksim ZS_*(\thicksim X).$ This means that $ZS_*(\thicksim X)=\thicksim ZS^*(X).$

• (2) It is similar to case (1).

• (7L) We prove the two inclusions separately.

  Let $x\in ZS_*(X)$, then $N_S(x)\subseteq X$. By (4L), $ZS_*(N_S(X))\subseteq ZS_*(X)$. Let $y\in N_S(x)$, then $N_S(y)\subseteq N_S(x)$, that is, $y\in ZS_*(N_S(x))$. Thus, $N_S(x)\subseteq ZS_*(N_S(x))\subseteq ZS_*(X)$, that is, $x\in ZS_*(ZS_*(X))$. Hence, $ZS_*(X)\subseteq ZS_*(ZS_*(X))$.

  Since $ZS_*(X)\subseteq X$, by (3L), we have $ZS_*(ZS_*(X))\subseteq ZS_*(X)$.

• (7H) It is similar to (7L).

• (8L) By (3L), $ZS_*(F(a))\subseteq F(a)$, for all $a\in A$. Furthermore, if $y\in F(a)$ then $y\in N_S(y)\subseteq F(a)$, we have $y\in ZS_*(F(a))$ therefore $F(a)\subseteq ZS_*(F(a)))$. Thus $F(a)= ZS_*(F(a)))$. $\square$

Proof of Theorem 6.1

• (1) Let $X\subseteq U$. For any $x\in F(a)$ and $x\in A$, $N_S(x)\subseteq F(a)$.

  If $x\in { FS}_*(X)$, then $F(a)\subseteq X$, and so, $N_S(x)\subseteq F(a)\subseteq X$. Thus, $x\in ZS_*(X)$. This means that ${ FS}_*(X)\subseteq ZS_*(X)$.

• (2) Since $SS^*(X)={ FS}_*(X)\bigcup \{Md_S(x)\vert x\in X-{ FS}_*(X)\}$ and $TS^*(X)=\bigcup \{Md_S(x)\vert x\in X\}=\bigcup \{Md_S(x)\vert x\in { FS}_*(X)\}\bigcup \{Md_S(x)\vert x\in X-{ FS}_*(X)\}$, which implies, $SS^*(X)\subseteq TS^*(X)$.

  Clearly, $TS^*(X)\subseteq FS^*(X)$. This shows that (2) holds.

• (3) Pick $x\in ZS^*(X)$. From $N_S(x)\cap X\ne \emptyset$ we deduce that F(a) intersects X for all $a\in A$. Thus, $x\in FS^*(X)$, and so, $ZS^*(X)\subseteq FS^*(X)$.$\square$

Proof of Theorem 6.4

“$\Rightarrow$” An inspection of the definition of ${ FS}_*$ ensures this implication.

“$\Leftarrow$” Assume $X\subseteq U, ZS_*(X)$ is a union of F(a) for all $a\in A$, we let $ZS_*(X)=F(a_1)\cup F(a_2)\cup \cdots \cup F(a_k)$ for all $a_i\in A, i=1,2, \ldots ,k$, and so, $F(a_i)\subseteq X$ for all $1\le i\le k$. The definition of ${ FS}_*$ yields $F(a_j)\subseteq { FS}_*(X)$ for every $1\le j\le k$. This implies that $ZS_*(X)=\bigcup \{ F(a_i)\}\subseteq { FS}_*(X)$. By Theorem 6.1, ${ FS}_*(X)\subseteq ZS_*(X)$. Thus, ${ FS}_*(X)=ZS_*(X)$. $\square$

Proof of Theorem 6.5

“$\Leftarrow$” For any $X\subseteq U$, $SS^*(X)={ FS}_*(X)\bigcup \{Md_S(x)\vert x\in X-{ FS}_*(X)\}$ and $TS^*(X)=\bigcup \{Md_S(x)\vert x\in X\}=\bigcup \{Md_S(x)\in { FS}_*(X)\}\bigcup \{Md_S(x)\vert x\in X-{ FS}_*(X)\}$. This shows that it suffices to prove the claim $\bigcup \{Md_S(x)\in { FS}_*(X)\}\subseteq { FS}_*(X)$.

In fact, $\forall x\in { FS}_*(X)$, there exists $a\in A$ with $x\in F(a)\subseteq X$. Let F(b) be the only element of $Md_S(x)$ for some $b\in A$. Let us prove that $F(b)\subseteq F(a)$. Assume by way of contradiction that $F(b)\subseteq F(a)$ is false. Then there must be $y\in F(b)$ for which $y\notin F(a)$. The construction of $Md_S(x)$ ensures the existence of $c\in A$ for which $F(c)\subseteq F(a)$ and $F(c)\in Md_S(x)$. The facts $y\notin F(a)$ and $y\notin F(c)$ imply that $Md_S(x)$ has at least two elements. This contradicts the assumption that ${\mathbf {G}}$ is soft unary.

Now $x\in F(b)\subseteq X$, therefore $F(b)\subseteq { FS}_*(X)$. This implies that our claim holds. Thus $SS^*(X)=TS^*(X)$.

“$\Rightarrow$” Suppose that ${\mathbf {G}}$ is not soft unary. Then there is $x\in U$ for which $Md_S(x)$ has at least two elements. In other words, there are $a,b\in A$ with the property $F(a) = F(b)$ is false.

Using the definition of $SS^*$, by Theorem 4.6 (8H) we have $SS^*(F(a))=F(a)$. Also $TS^*(F(a))=\bigcup \{Md_S(x)\vert x\in F(a)\}$, therefore $F(b)\subseteq TS^*(F(a))$. By hypothesis, $SS^*(X)=TS^*(X)$ thus we have $F(b)\subseteq F(a)$. We can proceed symmetrically with b in order to deduce $F(a)\subseteq F(b)$. Both statements contradict the fact that $F(a) \ne F(b)$. Hence ${\mathbf {G}}$ is soft unary. $\square$

Proof of Theorem 6.6

“$\Rightarrow$” Let $TS^*(X)=FS^*(X)$. For any $a\in A$ and $y\in U$ for which $y\in F(a)$, it must be the case that $y\in F(a)\subseteq FS^*(\{y\})=TS^*(\{y\})=\bigcup Md_S(y)$. This proves that ${\mathbf {G}}$ is SPC.

“$\Leftarrow$” Assume that ${\mathbf {G}}$ is SPC. Let $X\subseteq U$, $FS^*(X)=\bigcup _{a\in A}\{F(a)\vert F(a)\cap X \ne \emptyset \}$. For any $a\in A$, let $x\in F(a)\cap X$, then $x\in F(a)$ and $x\in X$. By the definition of SPC one has $F(a)\subseteq \bigcup Md_S(x)=TS^*(X)$, which proves that $FS^*(X)\subseteq TS^*(X)$.

In addition, by Theorem 6.1 (2) we have $TS^*(X)\subseteq FS^*(X)$.

The latter two statements prove $TS^*(X)= FS^*(X)$. $\square$

Proof of Lemma 6.8

“$\Leftarrow$” For any $a,b\in A$. Since ${\mathbf {G}}$ is partition, then $F(a)\cap F(b)=\emptyset$, and so, ${\mathbf {G}}$ is soft unary. Then by Proposition 6.7, we know that $F(a)\subsetneq F(b)$ and $F(b)\subsetneq F(a)$ do not hold. Let $x\in F(a)$, then $F(a)\subseteq \bigcup Md_S(x)$ holds. This means that ${\mathbf {G}}$ is SPC.

“$\Rightarrow$” Assume that ${\mathbf {G}}$ is soft unary and SPC. Let us prove that $F(a)\cap F(b)=\emptyset$ when $a,b\in A$. Otherwise, by Proposition 6.7, $F(a)\subsetneq F(b)$ or $F(b)\subsetneq F(a)$ holds. No generality is lost if we only study the case $x\in F(a)\subsetneq F(b)$. It is clear that $Md_S(x)=F(a)$, but $F(b)\nsubseteq Md_S(x)$, which contradicts the definition of SPC. Because $F(a)\cap F(b)=\emptyset$ throughout, ${\mathbf {G}}$ is partition. $\square$

Proof of Theorem 6.9

“$\Leftarrow$” If ${\mathbf {G}}$ is partition, then by Lemma 6.8, ${\mathbf {G}}$ is both soft unary and SPC. By Theorems 6.5 and 6.6, we have $SS^*(X)=TS^*(X)$ and $TS^*(X)=FS^*(X)$, and so, $SS^*(X)=FS^*(X)$.

“$\Rightarrow$” Let $SS^*(X)=FS^*(X)$. By Theorem 6.1(2), $SS^*\subseteq TS^*\subseteq FS^*$, we can obtain $SS^*= TS^*=FS^*$. By Theorems 6.5 and 6.6, we know that ${\mathbf {G}}$ is both soft unary and SPC. It follows from Lemma 6.8 that ${\mathbf {G}}$ is partition. $\square$

Proof of Theorem 6.11

“$\Leftarrow$” If ${\mathbf {G}}$ is partition, then for any $a,b\in A$, $F(a)\cap F(b)=\emptyset$. This means that for any $x\in F(a)$, $N_S(x)=F(a)$. Let $X\subseteq U$, then $ZS^*(X)=\{x\in U\vert N_S(X)\cap X\ne \emptyset \}=\bigcup _{a\in A}\{F(a)\vert F(a)\cap X\ne \emptyset \} =FS^*(X).$

“$\Rightarrow$” For any $a,b\in A$ and $F(a)\ne F(b)$. Assume that $F(a)\cap F(b)\ne \emptyset$. From the definition of ${ RS}^*$, it is trivial ${ RS}^*(F(a))=F(a), \forall a\in A$. Since $ZS^*(X)=FS^*(X)$, by Proposition 6.10, ${ RS}^*=FS^*$, and so, $FS^*(F(a))=F(a)$. By the definition of $FS^*$, $F(b)\subseteq FS^*(F(a))$. By hypothesis, $F(a)\ne F(b)$, then $F(b)\subsetneq F(a)$. Again, ${ RS}^*(F(b))=F(b)$. By the definition of $FS^*$, $F(a)\subseteq FS^*(F(b))$.

Hence ${ RS}^*(F(b))=F(b)\subsetneq F(a)\subseteq FS^*(F(b))$, a contradiction. This shows that $F(a)\cap F(b)=\emptyset$, which implies, ${\mathbf {G}}$ is partition. $\square$

Proof of Theorem 7.4

• (1) Let $X\subseteq Y$ and $u\in U$. Then $|N_S(u)\cap X|\le |N_S(u)\cap Y|$, and so, $\frac{|N_S(u)\cap X|}{|N_S(u)|}\le \frac{|N_S(u)\cap Y|}{|N_S(u)|}$, that is, $D_{{\mathbf {G}}}(u,X)\le D_{{\mathbf {G}}}(u,Y).$

• (2) Let $X,Y\subseteq U$, we have

  $$\begin{aligned} D_{X\cup Y}^{{\mathbf {G}}}\left( u\right)= & {} \frac{|N_S\left( u\right) \cap \left( X\cup Y\right) |}{|N_S\left( u\right) |} \\= & {} \frac{|\left( N_S\left( u\right) \cap X\right) \cup \left( N_S\left( u\right) \cap Y\right) |}{|N_S\left( u\right) |} \\\ge & {} \frac{\max \left\{ |N_S\left( u\right) \cap X|, |N_S\left( u\right) \cap Y|\right\} }{|N_S\left( u\right) |} \\= & {} \max \left\{ \frac{|N_S\left( u\right) \cap X|}{|N_S\left( u\right) |}, \frac{|N_S\left( u\right) \cap Y|}{|N_S\left( u\right) |} \right\} \\= & {} \max \left\{ D_{X}^{{\mathbf {G}}}\left( u\right) , D_{Y}^{{\mathbf {G}}}\left( u\right) \right\} \\= & {} \left( D_{X}^{{\mathbf {G}}}\cup D_{X}^{{\mathbf {G}}}\right) \left( u\right) , \end{aligned}$$

  which implies, $D_{X\cup Y}^{{\mathbf {G}}} \supseteq D_{X}^{{\mathbf {G}}}\cup D_{X}^{{\mathbf {G}}}$.

• (3) By (2), we can easily prove (3) holds.

• (4) It is similar to (2).

• (5) By (4), we can easily prove (5) holds.

• (6) $D_{{\mathbf {G}}}(u,X)+D_{{\mathbf {G}}}(u,\thicksim X)=\frac{|N_S(u)\cap X|+|N_S(u)\cap (\thicksim X)|}{|N_S(u)|}= \frac{|N_S(u)|}{|N_S(u)|}=1$.$\square$

Proof of Theorem 7.6

• (1L) For any $u\in U$, $D_U^{{\mathbf {G}}}(u)=\frac{|N_S(u)\cap U|}{|N_S(u)|}=1$, which implies $S_*(U)=U$.

• (1H) By (1L), for any $u\in U$, $D_U^{{\mathbf {G}}}(u)=1>0$, which implies $S^*(U)=U$.

• (2L) For any $u\in U$, $D_{\emptyset }^{{\mathbf {G}}}(u)=\frac{|N_S(u)\cap \emptyset |}{|N_S(u)|}=0$, which implies $S_*(\emptyset )=\emptyset$.

• (2H) By (2L), for any $u\in U$, $D_{\emptyset }^{{\mathbf {G}}}(u)=0$, which implies $S^*(\emptyset )=\emptyset$.

• (3LH) We prove the two inclusions separately.

  For any $u\in S_*(X)$, $D_X^{{\mathbf {G}}}(u)=1$. By Theorem 7.2(1), $u\in ZS_*(X)$, which implies, $N_S(u)\subseteq X$. Since $u\in N_S(u)$, then $u\in X$, which implies $S_*(X)\subseteq X$.

  Furthermore, let $u\in X$. Now $X\cap N_S(u)\ne \emptyset$ thus $u\in ZS^*(X)$. By Theorem 7.2(2) and (3), $D_X^{{\mathbf {G}}}(u)>0$, therefore $u\in S^*(X)$, which implies $X\subseteq S^*(X)$. This ensures $X\subseteq S^*(X)$.

• (4L) Let $X\subseteq Y$. If $u\in S_*(X)$, then $D_X^{{\mathbf {G}}}(u)=1$. By Theorem 7.4(1), $D_X^{{\mathbf {G}}}(u)\le D_Y^{{\mathbf {G}}}(u)$ since $X\subseteq Y$, and so, $D_Y^{{\mathbf {G}}}(u)\ge 1$, that is, $D_Y^{{\mathbf {G}}}(u)=1$, therefore $u\in S_*(Y)$. This means that $S_*(X)\subseteq S_*(Y)$.

• (4H) Let $X\subseteq Y$. If $u\in S^*(X)$, then $D_X^{{\mathbf {G}}}(u)>0$. By Theorem 7.4(1), $D_X^{{\mathbf {G}}}(u)\le D_Y^{{\mathbf {G}}}(u)$ since $X\subseteq Y$, and so, $D_Y^{{\mathbf {G}}}(u)\ge 0$, and so, $u\in S^*(Y)$. This means $S^*(X)\subseteq S^*(Y)$.

• (5L) (1) We prove this set-theoretic equality by double inclusion. By (4L), $S_*(X\cap Y)\subseteq S_*(X)$ since $X\cap Y\subseteq X$. Similarly, $S_*(X\cap Y)\subseteq S_*(Y)$, and so, $S_*(X\cap Y)\subseteq S_*(X)\cap S_*(Y)$.

Furthermore, let $u\in S_*(X)\cap S_*(Y)$, then $u\in S_*(X)$ and $u\in S_*(Y)$, and so, $D_X^{{\mathbf {G}}}(u)=1$ and $D_Y^{{\mathbf {G}}}(u)=1$. By Theorem 7.2(1), $u\in ZS_*(X)$ and $u\in ZS_*(Y)$, therefore $N_S(u)\subseteq X$ and $N_S(u)\subseteq Y$. Hence, $N_S(u)\subseteq X\cap Y$, that is, $u\in ZS_*(X\cap Y)$. By Theorem 7.2(1), $D_{X\cap Y}^{{\mathbf {G}}}(u)=1$, therefore $u\in S_*(X\cap Y)$, which implies, $S_*(X)\cap S_*(Y)\subseteq S_*(X\cap Y)$.

• (2) Since $X\subseteq X\cup Y$, then by (4L), $S_*(X)\subseteq S_*(X\cup Y)$. Similarly, $S_*(Y)\subseteq S_*(X\cup Y)$. Hence $S_*(X)\cup S_*(Y)\subseteq S_*(X\cup Y)$.

• (5H) (1) Since $X\cap Y\subseteq X$, then by (4H), $S^*(X\cap Y)\subseteq S^*(X)$. Similarly, $S^*(X\cap Y)\subseteq S^*(Y)$. Hence $S^*(X\cap Y)\subseteq S^*(X)\cap S^*(Y)$.

• (2) We prove this set-theoretic equality by double inclusion.

Let $u\in S^*(X\cup Y)$, then $D_{X\cup Y}^{{\mathbf {G}}}(u)>0$. By Theorem 7.2 (1) and (3), $u\in ZS^*(X\cup Y)$, which implies, $N_S(u)\cap (X\cup Y)\ne \emptyset$, and so, $N_S(u)\cap X\ne \emptyset$ or $N_S(u)\cap Y\ne \emptyset$, that is, $u\in ZS^*(X)$ or $u\in ZS^(Y)$. By Theorem 7.2 (3), $D_{X}^{{\mathbf {G}}}(u)>0$ or $D_{Y}^{{\mathbf {G}}}(u)>0$, that is, $u\in S^*(X)$ or $u\in S^*(Y)$, and so, $u\in S^*(X)\cup S^*(Y)$. This means that $S^*(X\cup Y)\subseteq S^*(X)\cup S^*(Y)$.

Furthermore, $X\subseteq X\cup Y$, then by (4H), $S^*(X)\subseteq S^*(X\cup Y)$. Similarly, $S^*(Y)\subseteq S^*(X\cup Y)$. Hence $S^*(X)\cup S^*(Y)\subseteq S^*(X\cup Y)$.

• (6) For any $X\subseteq U$, we have

  $$\begin{aligned} \thicksim S_*\left( \thicksim X\right)= & {} \thicksim \left\{ u\in U\vert D_{\thicksim X}^{{\mathbf {G}}}\left( u\right) >0\right\} \\= & {} \thicksim \left\{ u\in U\vert D_{X}^{{\mathbf {G}}}\left( u\right) <1\right\} \left( \mathrm {By\ Theorem}\,7.4\left( 6\right) \right) \\= & {} \left\{ u\in U\vert D_{X}^{{\mathbf {G}}}\left( u\right) \ge 1\right\} \\= & {} \left\{ u\in U\vert D_{X}^{{\mathbf {G}}}\left( u\right) = 1\right\} \\= & {} S_*\left( X\right) . \end{aligned}$$

  Similarly, $\thicksim S_*(\thicksim X)=\ \thicksim S^*(X).$

• (7L) We prove this set-theoretic equality by double inclusion.

  Let $u\in S_*(X)$, then $D_{X}^{{\mathbf {G}}}(u)=1$. By Theorem 7.2(1), $u\in ZS_*(X)$. By Theorem 5.2(7L), $ZS_*(ZS_*(X))=ZS_*(X)$, that is, $u\in ZS_*(ZS_*(X))$. Hence, by Theorem 7.2(1) again, $D_{ZS_*(X)}^{{\mathbf {G}}}(u)=1$, that is, $u\in S_*(S_*(X))$, which implies, $S_*(X)\subseteq S_*(S_*(X))$.

  Since $S_*(X)\subseteq X$, by (3L), we have $S_*(S_*(X))\subseteq S_*(X)$.

• (7H) It is similar to (7L).

• (8L) We prove this set-theoretic equality by double inclusion. Whenever $a\in A$, by (3L), $S_*(F(a))\subseteq F(a)$. Furthermore, let $u\in F(a)$, since $N_S(u)\subseteq F(a)$, then $D_{F(a)}^{{\mathbf {G}}}(u)=1$, and so, $u\in S_*(F(a))$. Hence $F(a)\subseteq S_*(F(a))$. $\square$