On Equivalence Operators Derived from Overlap and Grouping Functions

Abstract

This paper introduces the concept of equivalence operators based on overlap and grouping functions where the associativity property is not strongly required. Overlap functions and grouping functions are weaker than positive and continuous t-norms and t-conorms, respectively. Therefore, these equivalence operators do not necessarily satisfy certain properties, such as associativity and the neutrality principle. In this paper, two models of fuzzy equivalence operators are obtained by the composition of overlap functions, grouping functions and fuzzy negations. Their main properties are also studied.

1. Introduction

In classical logic, the equivalence operator can be represented as

$$\begin{array}{c}\hfill s\equiv t\stackrel{def}{=}(s\to t)\wedge (t\to s)=(s\wedge t)\vee (\neg s\wedge \neg t).\end{array}$$

(1)

It is also interpreted as the bi-implication logical connective denoted by using a double-headed arrow $s\leftrightarrow t$.

Different versions of fuzzy equivalence operator have been investigated in the literature. Hájek [1] considered the equivalence connective as a derived connective which is defined by

$$\begin{array}{c}\hfill s\equiv t\stackrel{def}{=}(s\to t)\&(t\to s)\end{array}$$

(2)

in several fuzzy logics, where → is a fuzzy implication and & is a t-norm. Novák et al. [2,3] developed the EQ-logic and EQ-algebras in which fuzzy equivalence operator is a basic connective. Dombi et al. [4,5] studied several different types of equivalence operators in Pliant systems and nilpotent systems. Hu et al. [6] studied the asymmetric equivalences in fuzzy logics. Fuzzy equivalence operators are highly applied in fuzzy theories and fuzzy methods. Based on the Formula (2) wherein & represents the minimum t-norm, i.e., $\&=\wedge$, Pan et al. [7] explored the properties of robustness of the lattice-valued similarity, while Jin et al. [8] and Dai et al. [9] investigated the robustness of fuzzy reasoning, Georgescu [10] explored the similarity of fuzzy choice functions, Wang et al. [11] and Duan et al. [12] investigated fuzzy metric spaces.

Notably, Fodor and Keresztfalvi [13] and Bustince et al. [14,15] mentioned applications where the associativity properties of the t-norm and t-conorm are not rrquired. Consequently, Bustince et al. [14,15] introduced the overlap and grouping functions as two types of non-necessarily associative bivariate operators. By replacing t-norms and t-conorms with overlap and grouping functions respectively, researchers devoloped many new concepts, such as symmetric differences [16,17], $R_O$-implications [18,19], (G,N)-implications [20], binary relations ${\le }_O$[21], (IO, O)-fuzzy rough sets [22], and $(O,G)$-fuzzy rough sets [23] where G is a grouping function and O is an overlap function.

In this paper, we aim to define equivalence operators by considering (G,N)-implications, overlap functions and grouping functions in the previously mentioned formulas. This approach is particularly significant given that overlap and grouping functions are weaker than the positive and continuous t-norms and t-conorms. Therefore, the new equivalence operators may not necessarily satisfy certain properties, such as associativity and the neutrality principle.

The paper is organized as follows. Section 2 presents the the concept of overlap and grouping functions, fuzzy implication and (G, N)-implication. Section 3 focuses on the model $G(O_1(s,t),O_2(N_1\left(s\right),N_2\left(t\right)))$ of equivalence operator. Section 4 addresses the model $O(G_1(N_1\left(s\right),t),G_2(N_2\left(t\right),s))$ of the equivalence operator. Section 5 constructes fuzzy symmetric differences from fuzzy equivalences. Lastly, Section 6 provides a comparative study, and the conclusion is presented in Section 7.

2. Preliminaries

Definition 1

([14]). A binary function $O:{[0,1]}^2\to [0,1]$ is said to be an overlap function if the following conditions hold: $\forall s,t\in [0,1]$,

(O1)

$O(s,t)=O(t,s)$;

(O2)

$O(s,t)=0⟺st=0$;

(O3)

$O(s,t)=1⟺st=1$;

(O4)

O is increasing;

(O5)

O is continuous.

Definition 2

([15]). A binary function $O:{[0,1]}^2\to [0,1]$ is said to be a grouping function if the following conditions hold: $\forall s,t\in [0,1]$,

(G1)

$G(s,t)=G(t,s)$;

(G2)

$G(s,t)=0⟺s=t=0$;

(G3)

$G(s,t)=1⟺s=1\mathrm{or}t=1$;

(G4)

G is increasing;

(G5)

G is continuous.

Definition 3

([24]). A binary function $I:{[0,1]}^2\to [0,1]$ is said to be a fuzzy implication if the following conditions hold: $\forall s,t,u\in [0,1]$,

(F1)

$s\le t⟹I(t,u)\le I(s,u)$;

(F2)

$s\le t⟹I(u,s)\le I(u,t)$;

(F3)

$I(0,0)=1$;

(F4)

$I(1,1)=1$;

(F5)

$I(1,0)=0$.

Definition 4

([24,25]). A function $N:[0,1]\to [0,1]$ is said to be a fuzzy negation if it is decreasing and satisfies $N\left(0\right)=1$ and $N\left(1\right)=0$.

Moreover, we say N is strong if $N\left(N\right(s\left)\right)=s,\forall s\in [0,1].$ The standard negation is $N\left(s\right)=1-s,\forall s\in [0,1].$

The overlap function O given by

$$\begin{array}{c}\hfill O(s,t)=N\left(G\right(N\left(s\right),N\left(t\right)\left)\right),\forall s,t\in [0,1].\end{array}$$

(3)

is the dual overlap function of G for N and, analogously, the grouping function G given by

$$\begin{array}{c}\hfill G(s,t)=N\left(O\left(N\left(s\right),N\left(t\right)\right)\right),\forall s,t\in [0,1].\end{array}$$

(4)

is the dual grouping function of O for N.

Definition 5

([20]). Let G be a grouping function and N be a fuzzy negation, we say that the function $I_{G,N}:{[0,1]}^2\to [0,1]$ defined by

$$\begin{array}{cc}\hfill I_{G,N}(s,t)& =G\left(N\left(s\right),t\right)\hfill \end{array}$$

(5)

is a (G,N)-implication.

Example 1.

Some examples of overlap and grouping functions are given in [15,26]

•

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

•

$O_p(s,t)=s^p{t}^p$, $G_p(s,t)=1-{(1-s)}^p{(1-t)}^p$, where $p>0$;

•

$O_{mp}(s,t)=min(s^p,t^p)$, $G_{mp}(s,t)=1-min\left({(1-s)}^p,{(1-t)}^p\right)$, where $p>0$;

•

$O_{Mp}(s,t)=1-max\left({(1-s)}^p,{(1-t)}^p\right)$, $G_{Mp}(s,t)=max(s^p,t^p)$, where $p>0$.

3. The Model $\mathit{G}\left({\mathit{O}}_{\mathbf{1}}(\mathit{s},\mathit{t}),{\mathit{O}}_{\mathbf{2}}\left({\mathit{N}}_{\mathbf{1}}\left(\mathit{s}\right),{\mathit{N}}_{\mathbf{2}}\left(\mathit{t}\right)\right)\right)$

In this section, we present our first model of the equivalence operator. We then illustrate its properties and conclude with two examples.

The analogue of the formula $a\equiv b\stackrel{def}{=}(a\wedge b)\vee (\neg a\wedge \neg b)$ in fuzzy set theory is

$$\begin{array}{c}\hfill s{\equiv }_{1}t\stackrel{def}{=}G(O_1(s,t),O_2(N_1\left(s\right),N_2\left(t\right)))\end{array}$$

(6)

where, G is a grouping function, $O_1$ and $O_2$ are overlap functions, and $N_1$ and $N_2$ are fuzzy negations.

Theorem 1.

The function ${\equiv }_1:{[0,1]}^2\to [0,1]$ given by Equation (6) satisfies: $\forall s,t\in [0,1]$

(i)

If $N_1=N_2$, then $s{\equiv }_{1}t=t{\equiv }_{1}s$;

(ii)

$1{\equiv }_{1}0=0{\equiv }_{1}1=0$;

(iii)

$1{\equiv }_{1}1=0{\equiv }_{1}0=1$;

(iv)

If both $N_1$ and $N_2$ are continuous, then ${\equiv }_1$ is continuous;

(v)

If $O_1$ has 1 as neutral element, G has 0 as neutral element, then $1{\equiv }_{1}s=s{\equiv }_{1}1=s$;

(vi)

If $O_2$ has 1 as neutral element, G has 0 as neutral element, then $0{\equiv }_{1}s=N_2\left(s\right)$ and $s{\equiv }_{1}0=N_1\left(s\right)$;

(vii)

If both $N_1$ and $N_2$ are defined as

$$\begin{array}{c}\hfill N_{\top }\left(s\right)=\left\{\begin{array}{cc}0,\hfill & ifs=1,\hfill \\ 1,\hfill & ifs<1.\hfill \end{array}\right.\end{array}$$

(7)

then $s{\equiv }_{1}s=1$;

(viii)

If $O_1=O_2$ and $N_1$ and $N_2$ are strong, then ${\equiv }_1$ is invariant with the pair of negations ($N_1,N_2$), i.e., $s{\equiv }_{1}t=N_1\left(s\right){\equiv }_1{N}_2\left(t\right)$.

Proof. (i) If $N_1=N_2$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}t\hfill \\ \hfill & =G\left(O_1(s,t),O_2\left(N_1\left(s\right),N_1\left(t\right)\right)\right),byassumptionandEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(t,s),O_2\left(N_1\left(t\right),N_1\left(s\right)\right)\right),by\left(O1\right)\hfill \\ \hfill & =t{\equiv }_{1}s,byEquation\left(6\right).\hfill \end{array}\end{array}$$

(ii) Taking $s=1$ and $t=0$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 1{\equiv }_{1}0\hfill \\ \hfill & =G\left(O_1(1,0),O_2\left(N_1\left(1\right),N_2\left(0\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(1,0),O_2(0,1)\right)\hfill \\ \hfill & =G\left(0,0\right),by\left(O2\right)\hfill \\ \hfill & =0,by\left(G2\right).\hfill \end{array}\end{array}$$

Similarly, we have $0{\equiv }_{1}1=0$.

(iii) Taking $s=0$ and $t=0$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 0{\equiv }_{1}0\hfill \\ \hfill & =G\left(O_1(0,0),O_2\left(N_1\left(0\right),N_2\left(0\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(0,0),O_2(1,1)\right)\hfill \\ \hfill & =G\left(0,1\right),by\left(O2\right)and\left(O3\right)\hfill \\ \hfill & =1,by\left(G3\right).\hfill \end{array}\end{array}$$

Taking $s=t=1$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 1{\equiv }_{1}1\hfill \\ \hfill & =G\left(O_1(1,1),O_2\left(N_1\left(1\right),N_2\left(1\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(1,1),O_2(0,0)\right)\hfill \\ \hfill & =G\left(1,0\right),by\left(O2\right)and\left(O3\right)\hfill \\ \hfill & =1,by\left(G3\right).\hfill \end{array}\end{array}$$

(iv) It is easy to be obtained from the continuity of $O_1$, $O_2$, G, $N_1$ and $N_2$.

(v) Since $O_1$ has 1 as neutral element, G has 0 as neutral element, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 1{\equiv }_{1}s\hfill \\ \hfill & =G\left(O_1(1,s),O_2\left(N_1\left(1\right),N_2\left(s\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(1,s),O_2\left(0,N_2\left(s\right)\right)\right)\hfill \\ \hfill & =G\left(s,0\right),byassumptionand\left(O2\right)\hfill \\ \hfill & =s,byassumption.\hfill \end{array}\end{array}$$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}1\hfill \\ \hfill & =G\left(O_1(s,1),O_2\left(N_1\left(s\right),N_2\left(1\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(s,1),O_2(N_1\left(s\right),0)\right)\hfill \\ \hfill & =G\left(s,0\right),byassumptionand\left(O2\right)\hfill \\ \hfill & =s,byassumption.\hfill \end{array}\end{array}$$

(vi) Since $O_2$ has 1 as neutral element, G has 0 as neutral element, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 0{\equiv }_{1}s\hfill \\ \hfill & =G\left(O_1(0,s),O_2\left(N_1\left(0\right),N_2\left(s\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(0,s),O_2(1,N_2\left(s\right))\right)\hfill \\ \hfill & =G\left(0,N_2\left(s\right)\right),byassumptionand\left(O2\right)\hfill \\ \hfill & =N_2\left(s\right),byassumption.\hfill \end{array}\end{array}$$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}0\hfill \\ \hfill & =G\left(O_1\left(s,0),O_2(N_1\left(s\right),N_2\left(0\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(s,0),O_2(N_1\left(s\right),1)\right)\hfill \\ \hfill & =G\left(0,N_1\left(s\right)\right),byassumptionand\left(O2\right)\hfill \\ \hfill & =N_1\left(s\right),byassumption.\hfill \end{array}\end{array}$$

(vii) If $N_1=N_2=N_{\mathbf{T}}$, case 1, if $s=0$ or $s=1$, then $s{\equiv }_{1}s=1$. Case 2, if $s\in (0,1)$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}s\hfill \\ \hfill & =G\left(O_1(s,s),O_2\left(N_{\top }\left(s\right),N_{\top }\left(s\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(s,s),O_2(1,1)\right)\hfill \\ \hfill & =G\left(O_1(s,s),1\right),by\left(03\right)\hfill \\ \hfill & =1.\hfill \end{array}\end{array}$$

(viii) If $O_1=O_2$ and $N_1$ and $N_2$ are strong, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}t\hfill \\ \hfill & =G\left(O_1(s,t),O_1\left(N_1\left(s\right),N_2\left(t\right)\right)\right),byEquation\left(6\right)\hfill \\ \hfill & =G\left(O_1(N_1\left(s\right),N_2\left(t\right)),O_1(s,t)\right),by\left(G1\right)\hfill \\ \hfill & =G\left(O_1\left(N_1\left(s\right),N_2\left(t\right)\right),O_1\left(N_1\left(N_1\left(s\right)\right),N_2\left(N_2\left(t\right)\right)\right)\right)\left(N_{1}and{N}_{2}arestrong\right)\hfill \\ \hfill & =N_1\left(s\right){\equiv }_1{N}_2\left(t\right),byEquation\left(6\right).\hfill \end{array}\end{array}$$

□

Example 2.

Consider the $O_1(s,t)=O_2(s,t)=O_{m0.5}(s,t)=min\{\sqrt{s},\sqrt{t}\}$, $N_1$ and $N_2$ are standard negation, and $G(s,t)=G_{M2}(s,t)=max\{s^2,t^2\}$. Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}t\hfill \\ \hfill & =G\left(O_1(s,t),O_2\left(N_1\left(s\right),N_2\left(t\right)\right)\right)\hfill \\ \hfill & =max\left({\left(min(\sqrt{s},\sqrt{t})\right)}^2,{\left(min(\sqrt{1-s},\sqrt{1-t})\right)}^2\right)\hfill \\ \hfill & =max\left(min(s,t),min\left(\right(1-s),(1-t\left)\right)\right).\hfill \end{array}\end{array}$$

(8)

The characteristics of this equivalence operator are shown in Figure 1. The lines are turning progressively yellower, indicating that the values are on the rise.

Example 3.

Consider the $O_1(s,t)=O_{m0.5}(s,t)=min\{\sqrt{s},\sqrt{t}\}$, $O_2(s,t)=O_{p=3}(s,t)=s^3{t}^3$,$N_1\left(s\right)=N_2\left(s\right)=N\left(s\right)=1-s^2$, and $G(s,t)=G_{M2}(s,t)=max\{s^2,t^2\}$. Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{1}t\hfill \\ \hfill & =G(O_1(s,t),O_2(N_1\left(s\right),N_2\left(t\right)))\hfill \\ \hfill & =max\left({\left(min(\sqrt{s},\sqrt{t})\right)}^2,{\left({(1-s^2)}^3{(1-t^2)}^3\right)}^2\right)\hfill \\ \hfill & =max\left(min(s,t),{(1-s^2)}^6{(1-t^2)}^6\right).\hfill \end{array}\end{array}$$

(9)

The characteristics of this equivalence operator are shown in Figure 2.

4. The Model $\mathit{O}\left({\mathit{G}}_{\mathbf{1}}({\mathit{N}}_{\mathbf{1}}\left(\mathit{s}\right),\mathit{t}),{\mathit{G}}_{\mathbf{2}}({\mathit{N}}_{\mathbf{2}}\left(\mathit{t}\right),\mathit{s})\right)$

In this section, we present our second model of the equivalence operator. We then illustrate its properties and conclude with two examples.

The analogue of the formula $a\equiv b\stackrel{def}{=}(a\to b)\wedge (b\to a)$ in fuzzy set theory is

$$\begin{array}{c}\hfill s{\equiv }_{2}t\stackrel{def}{=}O\left(\left(s{\to }_{1}t\right),\left(t{\to }_{2}s\right)\right)\end{array}$$

(10)

where O is a overlap function and ${\to }_1$ and ${\to }_2$ are fuzzy implications. Here we consider the (G,N)-implication, i.e., $s\to t=G\left(N\right(s),t)$. Then we obtain the following model for fuzzy equivalence operator:

$$\begin{array}{c}\hfill s{\equiv }_{2}t\stackrel{def}{=}O\left(G_1(N_1\left(s\right),t),G_2(N_2\left(t\right),s)\right)\end{array}$$

(11)

where, $G_1$ and $G_2$ are grouping functions, O is a overlap function, and $N_1$ and $N_2$ are fuzzy negations.

Theorem 2.

The function ${\equiv }_2$ given by formula (11) satisfies: $\forall s,t\in [0,1]$

(i)

If $G_1=G_2$ and $N_1=N_2$, then $s{\equiv }_{2}t=t{\equiv }_{2}s$;

(ii)

$1{\equiv }_{2}0=0{\equiv }_{2}1=0$;

(iii)

$1{\equiv }_{2}1=0{\equiv }_{2}0=1$;

(iv)

If $N_1$ and $N_2$ are continuous, then ${\equiv }_2$ is continuous;

(v)

If O has 1 as neutral element, $G_1$ has 0 as neutral element, then $1{\equiv }_{2}s=s$ and $s{\equiv }_{2}0=N_1\left(s\right)$;

(vi)

If O has 1 as neutral element, $G_2$ has 0 as neutral element, then $s{\equiv }_{2}1=s$ and $0{\equiv }_{2}s=N_2\left(s\right)$;

(vii)

If $N_1=N_2=N_{\top }$, then $s{\equiv }_{2}s=1$;

(viii)

If $G_1=G_2$, $N_1=N_2$, and $N_1$ is strong, then ${\equiv }_2$ is invariant with the negation $N_1$, i.e., $s{\equiv }_{2}t=N_1\left(s\right){\equiv }_2{N}_1\left(t\right)$.

Proof. (i) If $G_1=G_2$ and $N_1=N_2$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}t\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),t),G_1(N_1\left(t\right),s)\right),byassumptionandEquation\left(6\right)\hfill \\ \hfill & =O\left(G_1(N_1\left(t\right),s),G_1(N_1\left(s\right),t)\right),by\left(O1\right)\hfill \\ \hfill & =t{\equiv }_{2}s,byassumptionandEquation\left(6\right).\hfill \end{array}\end{array}$$

(ii) Taking $s=1$ and $t=0$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 1{\equiv }_{2}0\hfill \\ \hfill & =O\left(G_1(N_1\left(1\right),0),G_2(N_2\left(0\right),1)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(0,0),G_2(1,1)\right)\hfill \\ \hfill & =O\left(0,1\right),by\left(G3\right)and\left(O2\right)\hfill \\ \hfill & =0,by\left(O2\right).\hfill \end{array}\end{array}$$

Similarly, we can obtain $0{\equiv }_{2}1=0$.

(iii) Taking $s=0$ and $t=0$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 0{\equiv }_{2}0\hfill \\ \hfill & =O\left(G_1(N_1\left(0\right),0),G_2(N_2\left(0\right),0)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(1,0),G_2(1,0)\right)\hfill \\ \hfill & =O\left(1,1\right),by\left(G3\right)\hfill \\ \hfill & =1,by\left(O3\right).\hfill \end{array}\end{array}$$

Taking $s=t=1$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 1{\equiv }_{2}1\hfill \\ \hfill & =O\left(G_1(N_1\left(1\right),1),G_2(N_2\left(1\right),1)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(0,1),G_2(0,1)\right)\hfill \\ \hfill & =O\left(1,1\right),by\left(G3\right)\hfill \\ \hfill & =1,by\left(O3\right).\hfill \end{array}\end{array}$$

(iv) It is easy to be obtained from the continuity of $G_1$, $G_2$, O, $N_1$ and $N_2$.

(v) Taking $t=0$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}0\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),0),G_2(N_2\left(0\right),s)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),0),G_2(1,s)\right)\hfill \\ \hfill & =O\left(N_1\left(s\right),1\right),byassumptionand\left(G3\right)\hfill \\ \hfill & =N_1\left(s\right),byassumption.\hfill \end{array}\end{array}$$

Similarly, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 1{\equiv }_{2}s\hfill \\ \hfill & =O\left(G_1(N_1\left(1\right),s),G_2(N_2\left(s\right),1)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(0,s),G_2(N_2\left(s\right),1)\right)\hfill \\ \hfill & =O\left(s,1\right),byassumptionand\left(G3\right)\hfill \\ \hfill & =s,byassumption.\hfill \end{array}\end{array}$$

(vi) Taking $t=1$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}1\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),1),G_2(N_2\left(1\right),s)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),1),G_2(0,s)\right)\hfill \\ \hfill & =O\left(1,s\right),byassumptionand\left(G3\right)\hfill \\ \hfill & =s,byassumption.\hfill \end{array}\end{array}$$

Similarly, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & 0{\equiv }_{2}s\hfill \\ \hfill & =O\left(G_1(N_1\left(0\right),s),G_2(N_2\left(s\right),0)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(1,s),G_2(N_2\left(s\right),0)\right)\hfill \\ \hfill & =O\left(1,N_2\left(s\right)\right),byassumptionand\left(G3\right)\hfill \\ \hfill & =N_2\left(s\right),byassumption.\hfill \end{array}\end{array}$$

(vii) If $N_1=N_2=N_{\top },$ there are two cases. In case 1, if $s=0$ or $s=1$, then $s{\equiv }_{2}s=1$. In case 2, if $s\in (0,1)$, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}s\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),s),G_2(N_2\left(s\right),s)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(1,s),G_2(1,s)\right)\hfill \\ \hfill & =O(1,1),by\left(G3\right)\hfill \\ \hfill & =1,by\left(O3\right).\hfill \end{array}\end{array}$$

(viii) If $G_1=G_2$, $N_1=N_2$, and $N_1$ is strong, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}t\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),t),G_1(N_1\left(t\right),s)\right),byEquation\left(11\right)\hfill \\ \hfill & =O\left(G_1(s,N_1\left(t\right)),G_1\left(t,N_1\left(s\right)\right)\right),by\left(G1\right)and\left(O1\right)\hfill \\ \hfill & =O\left(G_1\left(N_1\left(N_1\left(s\right)\right),N_1\left(t\right)\right),G_1\left(N_1\left(N_1\left(t\right)\right),N_1\left(s\right)\right)\right)\left(N_{1}and{N}_{2}arestrong\right)\hfill \\ \hfill & =N_1\left(s\right){\equiv }_1{N}_1\left(t\right),byEquation\left(6\right).\hfill \end{array}\end{array}$$

This completes the proof. □

Example 4.

Consider the $G_1(s,t)=G_2(s,t)=max\{s^2,t^2\}$, $O(s,t)=O_{m2}(s,t)=min\{s^2,t^2\}$, $N_1$ and $N_2$ are the standard negation. Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}t\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),t),G_2(N_2\left(t\right),s)\right)\hfill \\ \hfill & =min\left({\left(max({(1-s)}^2,t^2)\right)}^2,{\left(max({(1-t)}^2,s^2)\right)}^2\right)\hfill \\ \hfill & =min\left(max\left({(1-s)}^4,t^4\right),max\left({(1-t)}^4,s^4\right)\right).\hfill \end{array}\end{array}$$

(12)

The characteristics of this equivalence operator are shown in Figure 3.

Example 5.

Consider the $G_1(s,t)=max\{s^2,t^2\}$, $G_2(s,t)=max\{s^3,t^3\}$, $O(s,t)=O_{p=2}(s,t)=s^2{t}^2$, $N_1$ and $N_2$ are the standard negation. Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}t\hfill \\ \hfill & =O\left(G_1(N_1\left(s\right),t),G_2(N_2\left(t\right),s)\right)\hfill \\ \hfill & ={\left(max({(1-s)}^2,t^2)\right)}^2·{\left(max({(1-t)}^3,s^3)\right)}^2\hfill \\ \hfill & =max\left({(1-s)}^4,t^4\right)·max\left({(1-t)}^6,s^6\right).\hfill \end{array}\end{array}$$

(13)

The characteristics of this equivalence operator are shown in Figure 4.

5. Comparative Study

In this section, we show a short comparison of the proposed equivalence operators with some existing equivalence operators.

In [27], Li et al. defined the fuzzy equivalence operator as a binary function $EN:{[0,1]}^2\to [0,1]$ satisfying

($EN1$)

$E(s,t)=E(t,s)$;

($EN2$)

$E(s,1)=s$;

($EN3$)

$E(0,0)=0$.

Li et al. [27] gave the following two models,

$$\begin{array}{c}\hfill E{N}_1(s,t)=S\left(T(s,t),T\left(N\left(s\right),N\left(t\right)\right)\right);\end{array}$$

(14)

$$\begin{array}{c}\hfill E{N}_2(s,t)=T\left(S\left(N\right(s),t),S\left(N\right(t),s)\right).\end{array}$$

(15)

In [4,5], Dombi, Csiszár listed some important properties for equivalence operators: $\forall s,t,u\in [0,1]$

(E1)

Symmetry, $s\equiv t=t\equiv s$;

(E2)

Compatibility, $0\equiv 1=1\equiv 0=0$ and $0\equiv 0=1\equiv 1=1$;

(E3)

Reflexivity, $s\equiv s=1$;

(E4)

Associativity, $s\equiv (t\equiv u)=(s\equiv t)\equiv u$;

(E5)

Neutrality principle, $1\equiv s=s$.

The comparison is demonstrated in the following:

Remark 1.

${\equiv }_1$ and ${\equiv }_2$ are two developed models of $E{N}_1$ and $E{N}_2$ by replacing t-norms and t-conorms with overlap and grouping functions respectively.

Remark 2.

${\equiv }_1$, ${\equiv }_2$ and fuzzy equivalences in [6] drop symmetry (E1). Fuzzy equivalences in [4,5,27] satisfy symmetry.

Remark 3.

${\equiv }_1$, ${\equiv }_2$ and fuzzy equivalences in [4,5,6,27] satisfy the compatibility (E2), so they are generalizations of classical equivalence.

Remark 4.

Both ${\equiv }_1$ and ${\equiv }_2$ drop reflexivity (E3). Fuzzy equivalences in [4,5,6,27] satisfy reflexivity.

Remark 5.

${\equiv }_1$, ${\equiv }_2$ and fuzzy equivalences in [5] drop associativity (E4). Fuzzy equivalences in [4,27] satisfy associativity.

Remark 6.

Both ${\equiv }_1$ and ${\equiv }_2$ drop the neutrality principle (E5). Fuzzy equivalences in [4,5,6,27] satisfy the eutrality principle.

Table 1. Comparison of fuzzy equivalences.

Property	${\equiv }_1$	${\equiv }_2$	Fuzzy Equivalences in [6]	Fuzzy Equivalences in [5]	Fuzzy Equivalences in [4,27]
${\mathit{E}}_{\mathbf{1}}$	×	×	×	√	√
${\mathit{E}}_{\mathbf{2}}$	√	√	√	√	√
${\mathit{E}}_{\mathbf{3}}$	×	×	√	√	√
${\mathit{E}}_{\mathbf{4}}$	×	×	√	×	√
${\mathit{E}}_{\mathbf{5}}$	×	×	√	√	√

provides a comprehensive comparison of various fuzzy equivalences. As can be seen from the table, fuzzy equivalences ${\equiv }_1$ and ${\equiv }_2$ introduced in this paper have few restrictions, thus providing greater flexibility and functionality.

6. Fuzzy Symmetric Differences from Fuzzy Equivalences

Fuzzy symmetric difference is a dual concept of fuzzy equivalence. We can define two kinds od fuzzy symmetric differences as

$$\begin{array}{cc}\hfill s{\Delta }_{1}t\stackrel{def}{=}& N\left(s{\equiv }_{1}t\right)=N\left(G\left(O_1(s,t),O_2\left(N_1\left(s\right),N_2\left(t\right)\right)\right)\right);\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill s{\Delta }_{2}t\stackrel{def}{=}& N\left(s{\equiv }_{2}t\right)=N\left(O\left(G_1(N_1\left(s\right),t),G_2(N_2\left(t\right),s)\right)\right).\hfill \end{array}$$

(17)

Fuzzy symmetric differences ${\Delta }_1$ and ${\Delta }_2$ are generalizations of classical symmetric difference since $1{\Delta }_{1}1=0{\Delta }_{1}0=1{\Delta }_{2}1=0{\Delta }_{2}0=0$ and $0{\Delta }_{1}1=1{\Delta }_{1}0=0{\Delta }_{2}1=1{\Delta }_{2}0=1$.

Hu et al. [16] introduced the following two models of fuzzy symmetric differences

$$\begin{array}{cc}\hfill s{▴}_{1}t\stackrel{def}{=}& G\left(O_1\left(s,N_1\left(t\right)\right),O_2\left(N_2\left(s\right),t\right)\right);\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill s{▴}_{2}t\stackrel{def}{=}& O\left(G(s,t),N\left(O(s,t)\right)\right).\hfill \end{array}$$

(19)

We show that they are connected as follows.

Theorem 3.

If $N=N_1=N_2$ is a strong negation, let $O=O_1=O_2$ and G is the dual grouping functions of O for N. Then $s{\Delta }_{1}t=s{▴}_{2}t$.

Proof. 

From the assumptions, it follows that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill s{\equiv }_{1}t& =N\left(G\right(O(s,t),O\left(N\right(s),N(t\left)\right)\left)\right)\hfill \\ \hfill & =O\left(N\left(O(s,t)\right),N\left(O\left(N\left(s\right),N\left(t\right)\right)\right)\right)\hfill \\ \hfill & =O\left(N\left(O(s,t)\right),G\left(N\left(N\right(s\left)\right),N\left(N\right(t\left)\right)\right)\right)\hfill \\ \hfill & =O\left(N\left(O(s,t)\right),G(s,t)\right)\hfill \\ \hfill & =O\left(G(s,t),N\left(O(s,t)\right)\right)\hfill \\ \hfill & =s{▴}_{2}t.\hfill \end{array}\end{array}$$

(20)

This completes the proof. □

Theorem 4.

If $N=N_1=N_2$ is a strong negation, let G is the dual grouping function of O for N and $O_1$ and $O_2$ are the dual overlap functions of $G_1$ and $G_2$ for N respectively. Then $s{\Delta }_{2}t=G\left(O_1\left(s,N\left(t\right)\right),O_2\left(t,N\left(s\right)\right)\right)$.

Proof. 

From the assumptions, it follows that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & s{\equiv }_{2}t\hfill \\ \hfill & =N\left(O\left(G_1(N\left(s\right),t),G_2(N\left(t\right),s)\right)\right)\hfill \\ \hfill & =G\left(N\left(G_1(N\left(s\right),t)\right),N\left(G_2(N\left(t\right),s)\right)\right)\hfill \\ \hfill & =G\left(N\left(G_1(N\left(s\right),t)\right),N\left(G_2(N\left(t\right),s)\right)\right)\hfill \\ \hfill & =G\left(O_1\left(N\left(N\right(s\left)\right),N\left(t\right)\right),O_2\left(N\left(N\right(t\left)\right),N\left(s\right)\right)\right)\hfill \\ \hfill & =G\left(O_1\left(s,N\left(t\right)\right),O_2\left(t,N\left(s\right)\right)\right).\hfill \end{array}\end{array}$$

(21)

This completes the proof. □

Corollary 1.

If $N=N_1=N_2$ is a strong negation, let $O=O_1=O_2$, $G=G_1=G_2$ and G is the dual grouping function of O for N. Then $s{\Delta }_{2}t=s{▴}_{1}t$.

7. Conclusions

This paper introduces fuzzy equivalences based on overlap functions and grouping functions instead of t-norms and t-conorms, aiming to provide more flexible fuzzy equivalences that do not necessarily conform to certain properties such as associativity and the neutrality principle. The fuzzy equivalences proposed in this paper are more flexible due to their limited restrictions, compared to those based on t-norms and t-conorms (refer to Table 1). The study of two models of equivalence operators has revealed the potential for further investigation.

For further work, some possible topics are given.

(1)

To expand on this research, it would be valuable to explore different types of overlap and grouping functions, including interval-valued, complex-valued, and lattice-valued functions. This exploration could lead to the development of new interval-valued, complex-valued, and lattice-valued operators based on these functions.

(2)

Equivalence operators play a critical role in constructing similarity measures that are essential for various aspects of fuzzy theory and applications. They contribute to the robustness of fuzzy reasoning and the stabilization of fuzzy control systems. The equivalence operators presented in this paper offer promising prospects for enhancing fuzzy theory and applications.

(3)

The additive and multiplicative generators of overlap and grouping functions are given in [28,29], suggesting that further exploration into the use of these generators to obtain equivalence operators would be a worthwhile pursuit.