Comparison of Overlap and Grouping Functions

Abstract

This paper investigates the pointwise comparability of overlap and grouping functions which obtained by Bustince et al.’s and Bedregal et al.’s generator pairs, respectively. Some necessary and sufficient conditions for the comparison of these functions are proved. We also introduce some compositions of these functions and study the order preservation of these compositions.

1. Introduction

Overlap function introduced by Bustince et al. [1] is a particular type of aggregation function [2]. Its dual concept is the grouping function [3]. In recent years, those two concepts have attracted a wide range of interests. For applications, they have been successfully applied to many domains, such as image processing [1,4], classification [5,6] and decision making [7,8]. For theoretical research, general overlap and grouping functions [9,10], N-dimensional overlap functions [11], Archimedean overlap functions [12], general interval-valued overlap functions [13], complex-valued overlap and grouping functions [14,15], quasi-homogeneous overlap functions [16], pseudo-homogeneous overlap and grouping functions [17], overlap functions on bounded lattices [18], overlap and grouping functions on complete lattices [19] have been introduced. Many fuzzy concepts derived from overlap and grouping functions, such as generalized interval-valued OWA operators [20], residual implications [21,22], (G,N)-implications [23], binary relations [24], (IO, O)-fuzzy rough sets [25] and so on.

In the study of overlap and grouping functions, the study of their properties accounts for a large proportion and play an important role. Bustince et al. [1] gave an alternative characterization of overlap functions by their generator pairs. Bedregal et al. [26] gave an alternative characterization of grouping functions in a similar way. Dimuro et al. [27] introduced the additive generators of overlap and grouping functions. Qiao and Hu [28] studied the interval additive generators of interval overlap and grouping functions. They [29] also introduced the multiplicative generators of overlap and grouping functions.

We have already known that there is a partial order between two t-norms $T_1$ and $T_2$, i.e., $T_1\le T_2$ if $T_1(a,b)\le T_2(a,b)$ for all $(a,b)\in {[0,1]}^2$ (see [30], Chapter 6). Klement et al. [31] presented a necessary and sufficient condition for the comparability of continuous Archimedean t-norms. There also exist some pointwise comparison results of fuzzy implications (see [32], Chapter 1). However, comparatively little investigation has been made on the comparability of overlap/grouping functions. Bustince et al. [1] defined the pointwise order of two overlap functions $O_1$ and $O_2$, i.e., $O_1\le O_2$ if $O_1(a,b)\le O_2(a,b)$ for all $(a,b)\in {[0,1]}^2$. Bedregal et al. [26] defined the pointwise order of two grouping functions in a similar way. Dai et al. [33] showed that the meet operation, join operation, convex combination, and ⊛-composition of overlap and grouping functions are order preserving. But the research on the pointwise comparability of overlap and grouping functions have not been studied in details. Therefore, in this paper, we study the pointwise comparability of overlap and grouping functions involving Bustince et al. [1] and Bedregal et al. [26] generators. We present some necessary and sufficient conditions for their comparability. We also investigate order preservation of some compositions of overlap and grouping functions.

The paper is organized as follows: In Section 2, we recall the concepts of overlap/grouping functions and their order relationship. In Section 3, we study the pointwise comparability of overlap functions involving Bustince et al. [1] generators. In Section 4, we study the pointwise comparability of grouping functions involving Bedregal et al. [26] generators. In Section 5, we introduce some compositions of overlap/grouping functions and study properties preservation of these compositions. In Section 6, our researches are concluded.

2. Preliminaries

2.1. Overlap and Grouping Functions

In this section, we recall the basic theory of overlap and grouping functions. More details can be found in [1,11,26,28].

Definition 1

([1]). A bivariate function $O:{[0,1]}^2\to [0,1]$ is a overlap function if, for any $a,b\in [0,1]$, it has the following properties:

(O1)

O is commutative;

(O2)

$O(a,b)=0$ if and only if $ab=0$;

(O3)

$O(a,b)=1$ if and only if $ab=1$;

(O4)

O is non-decreasing;

(O5)

O is continuous.

Definition 2

([3]). A bivariate function $G:{[0,1]}^2\to [0,1]$ is a grouping function if, for any $a,b\in [0,1]$, it has the following properties:

(G1)

G is commutative;

(G2)

$G(a,b)=0$ if and only if $a=b=0$;

(G3)

$G(a,b)=1$ if and only if $a=1$ or $b=1$.

(G4)

G is non-decreasing;

(G5)

G is continuous.

Denote by $\mathcal{O}$ the set of all overlap functions, and $\mathcal{G}$ the set of all grouping functions.

Let O be an overlap function, the dual grouping function of O is defined as $G_O(a,b)=1-O(1-a,1-b)$.

Example 1

([1,26]). The following are typical examples of overlap and grouping functions, where $p>0$,

•

$O_{nm}(a,b)=min(a,b)max(a^2,b^2)$;

•

$O_p(a,b)=a^p{b}^p$;

•

$O_{mp}(a,b)=min(a^p,b^p)$;

•

$O_{Mp}(a,b)=1-max({(1-a)}^p,{(1-b)}^p)$;

•

$O_{DB}(a,b)=\left\{\begin{array}{cc}\frac{2ab}{a+b},\hfill & ifa+b\ne 0,\hfill \\ 0,\hfill & ifa+b=0.\hfill \end{array}\right.$

•

$G_{nm}(a,b)=1-min(1-a,1-b)max({(1-a)}^2,{(1-b)}^2)$;

•

$G_p(a,b)=1-{(1-a)}^p{(1-b)}^p$;

•

$G_{mp}(a,b)=1-min({(1-a)}^p,{(1-b)}^p)$;

•

$G_{Mp}(a,b)=max(a^p,b^p)$;

•

$G_{DB}(a,b)=\left\{\begin{array}{cc}\frac{a+b-2ab}{2-a-b},\hfill & ifa\ne 1orb\ne 1,\hfill \\ 1,\hfill & ifa=b=1.\hfill \end{array}\right.$

2.2. Orders of Overlap and Grouping Functions

Bustince et al. [1] and Bedregal et al. [26] introduced the following partial order for overlap and grouping functions, respectively.

Definition 3

([1,26]). Let $f_1,f_2\in \mathcal{O}$ (or both $f_1,f_2\in \mathcal{G}$),

(i)

we say that $f_1$ is weaker than $f_2$, denote $f_1⪯f_2$, if $f_1(a,b)\le f_2(a,b)$ holds for all $(a,b)\in {[0,1]}^2$.

(ii)

we write $f_1\prec f_2$ if $f_1⪯f_2$ and $f_1\ne f_2$.

Proposition 1.

Let $O_1$ and $O_2$ be two overlap functions, if $O_1⪯O_2$, then $G_{O_2}⪯G_{O_1}$, where $G_{O_1}$ and $G_{O_2}$ are the dual grouping functions of $O_1$ and $O_2$, respectively.

Proof. 

First $O_1⪯O_2$ means $O_1(a,b)\le O_2(a,b)$ holds for all $(a,b)\in {[0,1]}^2$. Then $O_1(1-a,1-b)\le O_2(1-a,1-b)$ holds for all $(a,b)\in {[0,1]}^2$.

Afterwards we have $1-O_1(1-a,1-b)\ge 1-O_2(1-a,1-b)$ holds for all $(a,b)\in {[0,1]}^2$. Thus $G_{O_2}⪯G_{O_1}$, i.e., $G_{O_2}(a,b)\le G_{O_1}(a,b)$ holds for all $(a,b)\in {[0,1]}^2$. □

Example 2.

Consider the overlap and grouping functions in Example 1, we have

•

$O_{nm}⪯O_{mp}$, where $0<p\le 1$;

•

$O_{mp}⪯O_{nm}$, where $p\ge 3$;

•

$O_p⪯O_{mp}$;

•

$O_p⪯O_{DB}$, where $p\ge 1$;

•

$G_{mp}⪯G_{nm}$, where $0<p\le 1$;

•

$G_{nm}⪯G_{mp}$, where $p\ge 3$;

•

$G_{mp}⪯G_p$;

•

$G_{DB}⪯G_p$, where $p\ge 1$.

Remark 1.

$⪯$ is a partial order, but not a linear order. For example, consider the $O_{mp}$ with $p=2$ and $O_{nm}$, $O_{mp}(a,b)=min(a^2,b^2)$ and $O_{nm}$ are incomparable since $O_{mp}(1,\frac{1}{2})=\frac{1}{4}<O_{nm}(1,\frac{1}{2})=\frac{1}{2}$ and $O_{mp}(\frac{1}{2},\frac{1}{2})=\frac{1}{4}>O_{nm}(\frac{1}{2},\frac{1}{2})=\frac{1}{8}$.

3. Comparison of Overlap Functions

Bustince et al. [1] gave an alternative characterization of overlap functions.

Theorem 1

([1]). The bivariate function $O_{fg}:{[0,1]}^2\to [0,1]$ is an overlap function if and only if

$$\begin{array}{c}\hfill O_{fg}(a,b)=\frac{f(a,b)}{f(a,b)+g(a,b)}\end{array}$$

(1)

for some $f,g:{[0,1]}^2\to [0,1]$ satisfying the following conditions

(F1)

f and g are symmetric;

(F2)

f is non decreasing and g is non increasing;

(F3)

$f(a,b)=0$ if and only if $ab=0$;

(F4)

$g(a,b)=0$ if and only if $ab=1$;

(F5)

f and g are continuous functions.

For any overlap function $O_{fg}$ characterized by Equation (1), $(f,g)$ is said to be the generator pair of $O_{fg}$.

We give the following necessary and sufficient condition for the comparison of overlap functions characterized by different generator pairs.

Theorem 2.

Let $O_{f_1{g}_1}$ and $O_{f_2{g}_2}$ be two overlap functions with generator pair $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$, respectively. Then $O_{f_1{g}_1}⪯O_{f_2{g}_2}$ if and only if $f_1{g}_2\le f_2{g}_1$, i.e., for all $a,b\in [0,1]$,

$$\begin{array}{c}\hfill f_1(a,b)g_2(a,b)\le f_2(a,b)g_1(a,b).\end{array}$$

(2)

Proof. 

(⇒) From the definition of the generator pair in Equation (1), if $O_{f_1{g}_1}⪯O_{f_2{g}_2}$, then for all $a,b\in [0,1]$, by Definition 3,

$$\frac{f_1(a,b)}{f_1(a,b)+g_1(a,b)}\le \frac{f_2(a,b)}{f_2(a,b)+g_2(a,b)}.$$

From Theorem 1 (F3) and (F4), we have $f_1(a,b)+g_1(a,b)>0$ and $f_2(a,b)+g_2(a,b)>0$ for all $a,b\in [0,1]$, and we have

$$f_1(a,b)[f_2(a,b)+g_2(a,b)]\le f_2(a,b)[f_1(a,b)+g_1(a,b)].$$

Then for all $a,b\in [0,1]$, it holds that

$$f_1(a,b)g_2(a,b)\le f_2(a,b)g_1(a,b)$$

Thus $f_1{g}_2\le f_2{g}_1$.

(⇐) If $f_1{g}_2\le f_2{g}_1$, i.e., for all $a,b\in [0,1]$, it holds that

$$f_1(a,b)g_2(a,b)\le f_2(a,b)g_1(a,b).$$

By adding $f_1(a,b)f_2(a,b)$ in both sides of this inequality, we obtain

$$f_1(a,b)f_2(a,b)+f_1(a,b)g_2(a,b)\le f_1(a,b)f_2(a,b)+f_2(a,b)g_1(a,b),$$

i.e.,

$$f_1(a,b)[f_2(a,b)+g_2(a,b)]\le f_2(a,b)[f_1(a,b)+g_1(a,b)].$$

From $f_1(a,b)+g_1(a,b)>0$ and $f_2(a,b)+g_2(a,b)>0$ for all $a,b\in [0,1]$, one has that $[f_2(a,b)+g_2(a,b)][f_1(a,b)+g_1(a,b)]>0$ for all $a,b\in [0,1]$.

Then by dividing both sides of the equation by $[f_2(a,b)+g_2(a,b)][f_1(a,b)+g_1(a,b)]$, we get for all $a,b\in [0,1]$

$$\frac{f_1(a,b)}{f_1(a,b)+g_1(a,b)}\le \frac{f_2(a,b)}{f_2(a,b)+g_2(a,b)}.$$

Thus by Definition 3, $O_{f_1{g}_1}⪯O_{f_2{g}_2}$. □

Corollary 1.

Let $O_{f_1{g}_1}$ and $O_{f_2{g}_2}$ be two overlap functions with generator pair $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$, respectively. Then $O_{f_1{g}_1}⪯O_{f_2{g}_2}$ if and only if $\frac{f_1}{f_2}\le \frac{g_1}{g_2}$, i.e., for all $(a,b)\in {(0,1]}^2\backslash \left\{(1,1)\right\}$,

$$\begin{array}{c}\hfill \frac{f_1(a,b)}{f_2(a,b)}\le \frac{g_1(a,b)}{g_2(a,b)}.\end{array}$$

(3)

Corollary 2.

Let $O_{f_{1}g}$ and $O_{f_{2}g}$ be two overlap functions with generator pair $(f_1,g)$ and $(f_2,g)$, respectively. If $f_1\le f_2$, i.e., $f_1(a,b)\le f_2(a,b)$ for all $a,b\in [0,1]$. Then $O_{f_{1}g}⪯O_{f_{2}g}$.

Corollary 3.

Let $O_{f{g}_1}$ and $O_{f{g}_2}$ be two overlap functions with generator pair $(f,g_1)$ and $(f,g_2)$, respectively. If $g_1\le g_2$, i.e., $g_1(a,b)\le g_2(a,b)$ for all $a,b\in [0,1]$. Then $O_{f{g}_2}⪯O_{f{g}_1}$.

Example 3.

Consider the following functions $f_1,g_1,f_2,g_2:{[0,1]}^2\to [0,1]$, defined by

$$\begin{array}{cc}\hfill f_1(a,b)& =\sqrt{ab},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill f_2(a,b)& =a^2{b}^2,\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill g_1(a,b)& =1-ab,\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill g_2(a,b)& =max(1-a,1-b).\hfill \end{array}$$

(7)

Obviously, they satify the conditions of Theorem 1. We also have $f_2\le f_1$ and $g_2\le g_1$. Then it holds that $f_2{g}_2\le f_1{g}_1$.

From Theorem 2, we obtain $O_{f_2{g}_1}⪯O_{f_1{g}_2}$, i.e.,

$$\frac{a^2{b}^2}{a^2{b}^2+1-ab}\le \frac{\sqrt{ab}}{\sqrt{ab}+max(1-a,1-b)}$$

for all $a,b\in [0,1]$.

Moreover,

$$\frac{f_1(a,b)}{f_2(a,b)}=\frac{\sqrt{ab}}{a^2{b}^2}=\frac{1}{a^{3/2}{b}^{3/2}}$$

and

$$\frac{g_1(a,b)}{g_2(a,b)}=\frac{1-ab}{max(1-a,1-b)}$$

are incomparable since

$$\frac{f_1(0.9,0.9)}{f_2(0.9,0.9)}=\frac{1}{0.9^3}\approx 1.372<\frac{g_1(0.9,0.9)}{g_2(0.9,0.9)}=1.9$$

and

$$\frac{f_1(0.1,0.1)}{f_2(0.1,0.1)}=\frac{1}{0.1^3}=1000>\frac{g_1(0.1,0.1)}{g_2(0.1,0.1)}=1.1.$$

Then $O_{f_1{g}_1}(a,b)=\frac{\sqrt{ab}}{\sqrt{ab}+1-ab}$ and $O_{f_2{g}_2}(a,b)=\frac{a^2{b}^2}{a^2{b}^2+max(1-a,1-b)}$ are incomparable because of Corollary 1.

4. Comparison of Grouping Functions

Bedregal et al. [26] gave an alternative characterization of grouping functions.

Theorem 3

([26]). The bivariate function $G_{fg}:{[0,1]}^2\to [0,1]$ is a grouping function if and only if

$$\begin{array}{c}\hfill G_{fg}(a,b)=1-\frac{f(a,b)}{f(a,b)+g(a,b)}\end{array}$$

(8)

for some $f,g:{[0,1]}^2\to [0,1]$ satisfying the following conditions

(T1)

f and g are symmetric;

(T2)

f is non increasing and g is non decreasing;

(T3)

$f(a,b)=0$ if and only if $a=1$ or $b=1$;

(T4)

$g(a,b)=0$ if and only if $a=b=0$;

(T5)

f and g are continuous functions.

For any grouping function $G_{fg}$ characterized by Equation (8), $(f,g)$ is said to be the generator pair of $G_{fg}$.

We give the following necessary and sufficient condition for the comparison of grouping functions characterized by different generator pairs.

Theorem 4.

Let $G_{f_1{g}_1}$ and $G_{f_2{g}_2}$ be two grouping functions with generator pair $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$, respectively. Then $G_{f_1{g}_1}⪯G_{f_2{g}_2}$ if and only if $f_2{g}_1\le f_1{g}_2$, i.e., for all $a,b\in [0,1]$,

$$\begin{array}{c}\hfill f_2(a,b)g_1(a,b)\le f_1(a,b)g_2(a,b).\end{array}$$

(9)

Proof. 

(⇒) From the definition of the generator pair in Equation (8), if $G_{f_1{g}_1}⪯G_{f_2{g}_2}$, then for all $a,b\in [0,1]$, by Definition 3,

$$1-\frac{f_1(a,b)}{f_1(a,b)+g_1(a,b)}\le 1-\frac{f_2(a,b)}{f_2(a,b)+g_2(a,b)}.$$

This is

$$\frac{f_2(a,b)}{f_2(a,b)+g_2(a,b)}\le \frac{f_1(a,b)}{f_1(a,b)+g_1(a,b)}.$$

From Theorem 3 (T3) and (T4), we have $f_1(a,b)+g_1(a,b)>0$ and $f_2(a,b)+g_2(a,b)>0$ for all $a,b\in [0,1]$, and we have

$$f_2(a,b)[f_1(a,b)+g_1(a,b)]\le f_1(a,b)[f_2(a,b)+g_2(a,b)].$$

Then for all $a,b\in [0,1]$, it holds that

$$f_2(a,b)g_1(a,b)\le f_1(a,b)g_2(a,b).$$

Thus $f_2{g}_1\le f_1{g}_2$.

(⇐) If $f_2{g}_1\le f_1{g}_2$, i.e., for all $a,b\in [0,1]$, it holds that

$$f_2(a,b)g_1(a,b)\le f_1(a,b)g_2(a,b).$$

By adding $f_1(a,b)f_2(a,b)$ in both sides of this inequality, we obtain

$$f_1(a,b)f_2(a,b)+f_2(a,b)g_1(a,b)\le f_1(a,b)f_2(a,b)+f_1(a,b)g_2(a,b),$$

i.e.,

$$f_2(a,b)[f_1(a,b)+g_1(a,b)]\le f_1(a,b)[f_2(a,b)+g_2(a,b)].$$

From $f_1(a,b)+g_1(a,b)>0$ and $f_2(a,b)+g_2(a,b)>0$ for all $a,b\in [0,1]$, one has that $[f_2(a,b)+g_2(a,b)][f_1(a,b)+g_1(a,b)]>0$ for all $a,b\in [0,1]$.

Then by dividing both sides of the equation by $[f_2(a,b)+g_2(a,b)][f_1(a,b)+g_1(a,b)]$, we get for all $a,b\in [0,1]$

$$\frac{f_2(a,b)}{f_2(a,b)+g_2(a,b)}\le \frac{f_1(a,b)}{f_1(a,b)+g_1(a,b)}.$$

So we have, for all $a,b\in [0,1]$

$$1-\frac{f_1(a,b)}{f_1(a,b)+g_1(a,b)}\le 1-\frac{f_2(a,b)}{f_2(a,b)+g_2(a,b)}.$$

Thus by Definition 3, $G_{f_1{g}_1}⪯G_{f_2{g}_2}$. □

Corollary 4.

Let $G_{f_1{g}_1}$ and $G_{f_2{g}_2}$ be two grouping functions with generator pair $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$, respectively. Then $G_{f_1{g}_1}⪯G_{f_2{g}_2}$ if and only if $\frac{g_1}{g_2}\le \frac{f_1}{f_2}$, i.e., for all $(a,b)\in {[0,1)}^2\left\{(0,0)\right\}$,

$$\begin{array}{c}\hfill \frac{g_1(a,b)}{g_2(a,b)}\le \frac{f_1(a,b)}{f_2(a,b)}.\end{array}$$

(10)

Corollary 5.

Let $G_{f_{1}g}$ and $G_{f_{2}g}$ be two grouping functions with generator pair $(f_1,g)$ and $(f_2,g)$, respectively. If $f_1\le f_2$, then $G_{f_{2}g}⪯G_{f_{1}g}$.

Corollary 6.

Let $G_{f{g}_1}$ and $G_{f{g}_2}$ be two grouping functions with generator pair $(f,g_1)$ and $(f,g_2)$, respectively. If $g_1\le g_2$, then $G_{f{g}_1}⪯G_{f{g}_2}$.

Example 4.

Consider the following functions $f_1,g_1,f_2,g_2:{[0,1]}^2\to [0,1]$, defined by

$$\begin{array}{cc}\hfill f_1(a,b)& =(1-a)(1-b),\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill f_2(a,b)& =(1-a^2)(1-b^2),\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill g_1(a,b)& =\frac{a+b}{2},\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill g_2(a,b)& =min(a,b).\hfill \end{array}$$

(14)

Obviously, they satify the conditions of Theorem 3. We also have $f_1\le f_2$ and $g_2\le g_1$. Then it holds that $f_1{g}_2\le f_2{g}_1$.

From Theorem 4, we obtain $G_{f_2{g}_2}⪯G_{f_1{g}_1}$, i.e.,

$$1-\frac{(1-a^2)(1-b^2)}{(1-a^2)(1-b^2)+min(a,b)}\le 1-\frac{(1-a)(1-b)}{(1-a)(1-b)+\frac{a+b}{2}}$$

for all $a,b\in [0,1]$.

Moreover, for all $(a,b)\in {[0,1)}^2\backslash \left\{(0,0)\right\}$

$$\frac{f_1(a,b)}{f_2(a,b)}=\frac{(1-a)(1-b)}{(1-a^2)(1-b^2)}=\frac{1}{(1+a)(1+b)}$$

and

$$\frac{g_2(a,b)}{g_1(a,b)}=\frac{min(a,b)}{\frac{a+b}{2}}=\frac{2min(a,b)}{a+b}$$

are incomparable since

$$\frac{f_1(0.25,0.25)}{f_2(0.25,0.25)}=\frac{1}{1.{25}^2}=0.64<\frac{g_2(0.25,0.25)}{g_1(0.25,0.25)}=1$$

and

$$\frac{f_1(0.1,0.9)}{f_2(0.1,0.9)}=\frac{1}{1.1\ast 1.9}\approx 0.4785>\frac{g_1(0.1,0.9)}{g_2(0.1,0.9)}=0.2.$$

Then $G_{f_1{g}_2}(a,b)=1-\frac{(1-a)(1-b)}{(1-a)(1-b)+min(a,b)}$ and $G_{f_2{g}_1}(a,b)=1-\frac{(1-a^2)(1-b^2)}{(1-a^2)(1-b^2)+\frac{a+b}{2}}$ are incomparable because of Corollary 4.

5. Order Preservation of Some Compositions of Overlap and Grouping Functions

In this section, we consider the following problem.

Problem 1.

Whether we have

$$\begin{array}{c}\hfill H_{f_1{g}_1}⪯H_{f_2{g}_2},H_{f_3{g}_3}⪯H_{f_4{g}_4}\Rightarrow H_{\left(f_1{\circ }_1{f}_3\right)\left(g_1{\circ }_1{g}_3\right)}⪯H_{\left(f_2{\circ }_2{f}_4\right)\left(g_2{\circ }_2{g}_4\right)}\end{array}$$

(15)

for some operations ${\circ }_1$ and ${\circ }_2$ of bivariate functions, where $H_{f_i{g}_i}$, with $i=1,...,4$ are all overlap functions or all grouping functions?

Let $h_1$ and $h_2$ be two bivariate functions, their meet, join and product operations are defined as

$$\begin{array}{cc}\hfill (h_1\vee h_2)(a,b)& =max\left(h_1(a,b),h_2(a,b)\right),\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill (h_1\wedge h_2)(a,b)& =min\left(h_1(a,b),h_2(a,b)\right),\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill (h_1\times h_2)(a,b)& =h_1(a,b)h_2(a,b),\hfill \end{array}$$

(18)

for all $(a,b)\in {[0,1]}^2$.

First, we prove the closures of the proposed compositions of overlap (or grouping) functions $H_{(f_1\circ f_2)(g_1\circ g_2)}$, where $\circ \in \{\vee ,\wedge ,\times \}$.

Lemma 1.

If $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$ satisfy the following conditions (F1)–(F5) of Theorem 1. Then $(f_1\vee f_2,g_1\vee g_2)$, $(f_1\wedge f_2,g_1\wedge g_2)$ and $(f_1\times f_2,g_1\times g_2)$ also satisfy these conditions.

Proof. 

The cases for (F1), (F2), and (F5) are straightforward.

(F3) (⇒) If $(f_1\vee f_2)(a,b)=max\left(f_1(a,b),f_2(a,b)\right)=0$. Then, $f_1(a,b)=f_2(a,b)=0$, thus $ab=0$.

If $(f_1\wedge f_2)(a,b)=min\left(f_1(a,b),f_2(a,b)\right)=0$. Case I, $f_1(a,b)=0$ then $ab=0$. Case II, $f_2(a,b)=0$ then $ab=0$.

If $(f_1\times f_2)(a,b)=f_1(a,b)f_2(a,b)=0$. Case I, $f_1(a,b)=0$ then $ab=0$. Case II, $f_2(a,b)=0$ then $ab=0$.

(⇐) is straightforward.

(F4) (⇒) If $(g_1\vee g_2)(a,b)=max\left(g_1(a,b),g_2(a,b)\right)=0$. Then $g_1(a,b)=g_2(a,b)=0$, thus $ab=1$.

If $(g_1\wedge g_2)(a,b)=min\left(g_1(a,b),g_2(a,b)\right)=0$. Case I, $g_1(a,b)=0$ then $ab=1$. Case II, $g_2(a,b)=0$ then $ab=1$.

If $(g_1\times g_2)(a,b)=g_1(a,b)g_2(a,b)=0$. Case I, $g_1(a,b)=0$ then $ab=1$. Case II, $g_2(a,b)=0$ then $ab=1$.

(⇐) is straightforward. □

Corollary 7.

If $O_{f_1{g}_1}$ and $O_{f_2{g}_2}$ be two overlap function functions with generator pair $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$, respectively. Then $O_{(f_1\circ f_2)(g_1\circ g_2)}$ is an overlap function, where $\circ \in \{\vee ,\wedge ,\times \}$

Lemma 2.

If $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$ satisfy the following conditions (T1)-(T5) of Theorem 3. Then $(f_1\vee f_2,g_1\vee g_2)$, $(f_1\wedge f_2,g_1\wedge g_2)$ and $(f_1\times f_2,g_1\times g_2)$ also satisfy these conditions.

Proof. 

The cases for (T1), (T2), and (T5) are straightforward.

(T3) (⇒) If $(f_1\vee f_2)(a,b)=max\left(f_1(a,b),f_2(a,b)\right)=0$. Then, $f_1(a,b)=f_2(a,b)=0$, thus $a=1$ or $b=1$.

If $(f_1\wedge f_2)(a,b)=min\left(f_1(a,b),f_2(a,b)\right)=0$. Case I, $f_1(a,b)=0$ then $a=1$ or $b=1$. Case II, $f_2(a,b)=0$ then $a=1$ or $b=1$.

If $(f_1\times f_2)(a,b)=f_1(a,b)f_2(a,b)=0$. Case I, $f_1(a,b)=0$ then $a=1$ or $b=1$. Case II, $f_2(a,b)=0$ then $a=1$ or $b=1$.

(⇐) is straightforward.

(T4) (⇒) If $(g_1\vee g_2)(a,b)=max\left(g_1(a,b),g_2(a,b)\right)=0$. Then $g_1(a,b)=g_2(a,b)=0$, thus $a=b=0$.

If $(g_1\wedge g_2)(a,b)=min\left(g_1(a,b),g_2(a,b)\right)=0$. Case I, $g_1(a,b)=0$ then $a=b=0$. Case II, $g_2(a,b)=0$ then $a=b=0$.

If $(g_1\times g_2)(a,b)=g_1(a,b)g_2(a,b)=0$. Case I, $g_1(a,b)=0$ then $a=b=0$. Case II, $g_2(a,b)=0$ then $a=b=0$.

(⇐) is straightforward. □

Corollary 8.

If $G_{f_1{g}_1}$ and $G_{f_2{g}_2}$ be two grouping functions with generator pair $f_1,g_1:{[0,1]}^2\to [0,1]$ and $f_2,g_2:{[0,1]}^2\to [0,1]$, respectively. Then $G_{(f_1\circ f_2)(g_1\circ g_2)}$ is a grouping function, where $\circ \in \{\vee ,\wedge ,\times \}$

Lemma 3.

Let $a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4\in [0,1]$. If $a_1{b}_2\le a_2{b}_1$ and $a_3{b}_4\le a_4{b}_3$. Then

(1)

$\left(a_1{a}_3\right)\left(b_2{b}_4\right)\le \left(a_2{a}_4\right)\left(b_1{b}_3\right)$.

(2)

$(a_1\wedge a_3)(b_2\wedge b_4)\le (a_2\vee a_4)(b_1\vee b_3)$.

Theorem 5.

Let $O_{f_1{g}_1}$, $O_{f_2{g}_2}$, $O_{f_3{g}_3}$ and $O_{f_4{g}_4}$ be four overlap functions with generator pair $(f_1,g_1)$, $(f_2,g_2)$, $(f_3,g_3)$ and $(f_4,g_4)$, respectively. If $O_{f_1{g}_1}⪯O_{f_2{g}_2}$ and $O_{f_3{g}_3}⪯O_{f_4{g}_4}$, then

(1)

$O_{(f_1\times f_3)(g_1\times g_3)}⪯O_{(f_2\times f_4)(g_2\times g_4)}$;

(2)

$O_{(f_1\wedge f_3)(g_1\wedge g_3)}⪯O_{(f_2\vee f_4)(g_2\vee g_4)}$.

Proof. 

From $O_{f_1{g}_1}⪯O_{f_2{g}_2}$ and $O_{f_3{g}_3}⪯O_{f_4{g}_4}$, by Theorem 1, it hold that $f_1{g}_2\le f_2{g}_1$ and $f_3{g}_4\le f_4{g}_3$. By Lemma 3(1) and (2), we have $(f_1\times f_3)(g_2\times g_4)\le (f_2\times f_4)(g_1\times g_3)$ and $(f_1\wedge f_3)(g_2\wedge g_4)\le (f_2\vee f_4)(g_1\vee g_3)$, respectively. Thus we obtain $O_{(f_1\times f_3)(g_1\times g_3)}⪯O_{(f_2\times f_4)(g_2\times g_4)}$ and $O_{(f_1\wedge f_3)(g_1\wedge g_3)}⪯O_{(f_2\vee f_4)(g_2\vee g_4)}$. □

Theorem 6.

Let $G_{f_1{g}_1}$, $G_{f_2{g}_2}$, $G_{f_3{g}_3}$ and $G_{f_4{g}_4}$ be four grouping functions with generator pair $(f_1,g_1)$, $(f_2,g_2)$, $(f_3,g_3)$ and $(f_4,g_4)$, respectively. If $G_{f_1{g}_1}⪯G_{f_2{g}_2}$ and $G_{f_3{g}_3}⪯G_{f_4{g}_4}$, then

(1)

$G_{(f_1\times f_3)(g_1\times g_3)}⪯G_{(f_2\times f_4)(g_4\times g_4)}$;

(2)

$G_{(f_1\vee f_3)(g_1\wedge g_3)}⪯G_{(f_2\wedge f_4)(g_2\vee g_4)}$.

Proof. 

From $G_{f_1{g}_1}⪯G_{f_2{g}_2}$ and $G_{f_3{g}_3}⪯G_{f_4{g}_4}$, by Theorem 4, it hold that $f_2{g}_1\le f_1{g}_2$ and $f_4{g}_3\le f_3{g}_4$. By Lemma 3(1) and (2), we have $(f_2\times f_4)(g_1\times g_3)\le (f_1\times f_3)(g_2\times g_4)$ and $(f_2\wedge f_4)(g_1\wedge g_3)\le (f_1\vee f_3)(g_2\vee g_4)$, respectively. Thus we obtain $G_{(f_1\times f_3)(g_1\times g_3)}⪯G_{(f_2\times f_4)(g_2\times g_4)}$ and $G_{(f_1\vee f_3)(g_1\wedge g_3)}⪯G_{(f_2\wedge f_4)(g_2\vee g_4)}$. □

6. Conclusions

This paper studies the pointwise comparability of overlap and grouping functions, respectively. We give some necessary and sufficient conditions for the comparison of overlap functions characterized by Bustince et al. generator pairs [1] and grouping functions characterized by Bedregal et al. generator pairs [26]. We present some more general results on order preservation with respect to some compositions of overlap and grouping functions.

In this paper, we only focus on overlap and grouping functions characterized by Bustince et al. [1] and Bedregal et al. [26] generators, respectively. Naturally, a more detailed discussion of other generators of overlap and grouping functions, such as additive generators proposed by Dimuro et al. [27] and multiplicative generators proposed by Qiao and Hu [29], will be both necessary and interesting.

It was observed that there are various compositions of overlap and grouping functions. Obviously, the results presented in this paper are particular cases of order preservation with respect to some compositions of overlap and grouping functions. Therefore, it will be meaningful to further discuss order preservation of other compositions.