Developing Constrained Interval Operators for Fuzzy Logic with Interval Values

Abstract

A well-known problem in the interval analysis literature is the overestimation and loss of information. In this article, we define new interval operators, called constrained interval operators, that preserve information and mitigate overestimation. These operators are investigated in terms of correction, algebraic properties, and orders. It is shown that a large part of the properties studied is preserved by this operator, while others remain preserved with the condition of continuity, as is the case of the exchange principle. In addition, a comparative study is carried out between this operator $\ddot{g}$ and the best interval representation: $\widehat{g}$. Although $\ddot{g}\subseteq \widehat{g}$ and $\ddot{g}$ do not preserve the Moore correction, we do not have a loss of relevant information since everything that is lost is irrelevant, mitigating the overestimation.

1. Introduction

Fuzzy logic is very successful in dealing with uncertainties related to fuzzy rules; however, this success does not extend to when referring to input–outputs, since they are assumed to be exact, i.e., values in $[0,1]$. In cases in which the fuzzification of inputs derives from choices made by experts and they are unsure about the value to use in to system, fuzzy logic is not always suitable. Another case lies in the situation in which the input data have impreciseness. In order to overcome this structure, interval-valued fuzzy logic has been suggested. This interval logic is a particular case of interval type-2 fuzzy logic [1,2,3,4,5,6].

Interval-valued fuzzy logic has been applied in a wide variety of domains; for example, Melin and Castillo [7] used it in the context of plant control; Figueroa et al. [8] used it for non-autonomous robots in the context of a robot football game; Lynch et al. [9] built an interval control system for large marine diesel engines; Chourasia et al. [10] developed a new method for assessing fetal health status based on interval type-2 fuzzy logic through fetal phonocardiography; Nguyen et al. [11] used the wavelet feature in an interval type-2 fuzzy logic system (IT2FLS) to reduce the computation burden and time of IT2FLS; Leow et al. [12] developed a hybrid of generalized adaptive resonance theory (GART) and an interval type-2 fuzzy logic system algorithm; among many others [5,6].

On the other hand, another field of great importance for this work, interval analysis, was developed in order to limit rounding errors and deal with inaccuracies. This field was first presented by Warmus [13], Sunaga [14], and Moore [15]. However, it can be said that interval mathematics and analysis began with the appearance of R. E. Moore’s book, Interval Analysis, in 1966 [16]. Moore’s arithmetic is accepted as the standard approach and is called standard interval arithmetic (SIA). Interval analysis has been applied in several areas, like electrical power systems [17], mechanical engineering [18], chemical engineering [19], artificial intelligence [20], multi-agent systems [21], and geophysics [22].

In interval analysis, there are two most important criteria, namely correctness (accuracy) and optimality [23,24]. The first criterion establishes that the result of an interval computation must always contain the value of the respective real function (see [24] (Theorem 3.1)). Although correctness is a desirable property, not every interval method is correct. Santiago et al. [25,26] and Bedregal and Santiago [27] investigate the notion of correctness for interval functions and its impact on some interval topological viewpoints. They call correctness interval representations since interval entities (algorithms and intervals) are seen as entities that represent exact entities (functions and numbers). An interval function, F, is said to represent a real function, f, whenever it satisfies the following property: $\forall x\in [a,b]\Rightarrow f\left(x\right)\in F\left(\right[a,b\left]\right)$ and $F\left(\right[a,a\left]\right)=f\left(a\right)$. The second criterion establishes that the resulting interval of a computation should not be greater than necessary, which is captured by the notion of canonical interval representation.

In order to solve the algebraic incompatibility between the real arithmetic and Moore arithmetic, Lodwick [28,29] defined a new interpretation for intervals called constraint intervals. With this new approach, Lodwick proposed an alternative to Moore arithmetic on intervals in order to have $X-X=[0,0]$, $X÷X=[1,1]$ if $0\notin X$ and the distributive property. This new arithmetic, called constrained interval arithmetic (CIA), is an extension of Moore’s interval arithmetic, in the sense that they coincide in the case where there is no variable dependence and are distinct when there are dependencies. In this case, the CIA presents a smaller width interval, thus improving the overestimation of Moore’s arithmetic [30]. Actually, CIA has been used to solve several mathematical problems, such as in [30,31,32].

In order to solve, or mitigate, the problems related to overestimation and loss of information when working with intervals, in this article, we present a new approach to fuzzy logic with interval values, in which interval operators preserve some of the main algebraic properties, the overestimation problem is mitigated, and there is no loss of information. These operators were defined by using the constrained intervals introduced by Lodwick [28]. In what follows, it is described how this study was performed and also the contributions that were studied or compared with this new approach.

The new operators, called constrained interval operators, in the proposed new approach to interval-valued fuzzy logic are very close to Moore’s correctness. In fact, they satisfy a new correction, which was called here constraint interval correctness, which is as efficient as Moore’s correction, in the sense of not losing information; however, they are less demanding and, in turn, work with less irrelevant information. The main properties of this operator have been verified about two orders, namely Kulisch–Miranker and Moore order. It has been found that this new approach guarantees the extension of many algebraic properties and that its fuzzy operators are extended to their respective constrained interval operators when considering the order of Kulisch–Miranker as in [27,33].

This paper is organized as follows: Section 2 recalls some definitions and concepts used throughout the text in order to provide a self-contained document. Section 3 proposes a new way of making interval-valued fuzzy logic in which the operators are called constrained interval operators. The last chapters includes the conclusions and bibliography.

2. Preliminaries

In this section, we present the concepts and results that we will use in this work.

2.1. Fuzzy Connectives

Classical conjunctions, negations, and implications are generalized to fuzzy logic by the operators t-norms, fuzzy negations, and fuzzy implications, respectively, and these are among the most important operators in fuzzy logic. The definitions are provided below:

Definition 1 

([34]). A binary function T: ${[0,1]}^2\to [0,1]$ is said to be a triangular norm (t-norm for short) if it is increasing, symmetric, associative, and satisfies the boundary condition $T(x,1)=x$ for all $x\in [0,1]$.

In the following, we recall the notion of fuzzy negation.

Definition 2. 

A unary function N: $[0,1]\to [0,1]$ is called fuzzy negation if the following conditions are satisfied:

(N1)

$N\left(0\right)=1$ and $N\left(1\right)=0$;

(N2)

$N\left(x\right)\le N\left(y\right)$ whenever $y\le x$.

It is called strong if it satisfies $N\left(N\right(x\left)\right)=x$ for all $x\in [0,1]$.

Example 1. 

Here are two special examples of fuzzy negation:

$$\begin{array}{c}\hfill N_{\perp }\left(x\right)=\left\{\begin{array}{c}1,ifx=0\hfill \\ 0,ifx>0\hfill \end{array}\right.\end{array}$$

and

$$\begin{array}{c}\hfill N_{\top }\left(x\right)=\left\{\begin{array}{c}0,ifx=1\hfill \\ 1,ifx<1\hfill \end{array}.\right.\end{array}$$

The first is known as less fuzzy negation, $N_{\perp }$, since $N_{\perp }\le N$, and the second is known as greater fuzzy negation, $N_{\top }$, since $N\le N_{\top }$, for every fuzzy negation N.

Fuzzy implication generalizes the usual material implication. In what follows, we present the definition and some of its properties.

Definition 3. 

A binary function $I:{[0,1]}^2\to [0,1]$ is called fuzzy implication if it satisfies the following conditions for all $x,y,z\in [0,1]$:

(I1)

If $x\le y$, then $I(y,z)\le I(x,z)$ for all $z\in [0,1]$;

(I2)

If $y\le z$, then $I(x,y)\le I(x,z)$ for all $x\in [0,1]$;

(I3)

$I(0,y)=1$ for all $y\in [0,1]$;

(I4)

$I(x,1)=1$ for all $x\in [0,1]$;

(I5)

$I(1,0)=0$.

We will denote the set of all fuzzy implications by $\mathcal{FI}$.

We present below some important properties of some fuzzy implications that will be used in this work (see [35,36,37]). To perform this, consider a fuzzy implication I, $\forall x,y,z\in [0,1]$:

(IP)

The identity property: $I(x,x)=1;$

(NP)

The left neutrality property: $I(1,y)=y;$

(OP)

The order property: $I(x,y)=1$ iff $x\le y;$

(EP)

The exchange principle: $I(x,I(y,z\left)\right)=I(y,I(x,z\left)\right);$

(CP)

Contraposition law with respect to N: $I(x,y)=I\left(N\right(y),N(x\left)\right);$

(LCP)

Left contraposition law with respect to N: $I\left(N\right(x),y)=I\left(N\right(y),x);$

(RCP)

Right contraposition law with respect to N: $I(x,N(y\left)\right)=I(y,N(x\left)\right).$

We will use the notation $CP\left(N\right)$ to represent that I satisfies the contraposition property in relation to fuzzy negation N. Similarly, we will use $LCP\left(N\right)$ and $RCP\left(N\right)$ for the left and right contraposition properties, respectively.

2.2. Interval Operators and Moore’s Correctness

We will denote by $U\left(\right[0,1\left]\right)$ the set of all intervals closed in $[0,1]$; that is, $U\left(\right[0,1\left]\right)=\left\{\right[x,y]\mid 0\le x\le y\le 1\}$. We will further denote $X=[\underline{x},\overline{x}]$ for all $X\in U\left(\right[0,1\left]\right)$. Let $\langle U\left(\right[0,1\left]\right),⪯\rangle$ be a bounded poset $[0,0]$ as the bottom element and $[1,1]$ as the top element.

Definition 4 

([38]). A fuzzy negation $\mathbb{N}:U\left(\right[0,1\left]\right)\to U\left(\right[0,1\left]\right)$ on $\langle U\left(\right[0,1\left]\right),⪯\rangle$, called interval fuzzy negation, is decreasing and satisfies $\mathbb{N}\left(\right[0,0\left]\right)=[1,1]$ and $\mathbb{N}\left(\right[1,1\left]\right)=[0,0]$. If $\mathbb{N}\left(\mathbb{N}\right(X\left)\right)=X$, $\forall X\in U\left(\right[0,1\left]\right)$; then, $\mathbb{N}$ is called strong interval fuzzy negation.

Definition 5 

([38]). A function $\mathbb{T}:U{\left([0,1]\right)}^2\to U\left([0,1]\right)$ is an interval triangular norm (it-norm) on $\langle U\left(\right[0,1\left]\right),⪯\rangle$ if it is commutative, associative, increasing, and satisfies $\mathbb{T}(X,[1,1\left]\right)=X$ for all $X\in U\left(\right[0,1\left]\right)$.

Definition 6. 

A fuzzy implication $\mathbb{I}:U{\left([0,1]\right)}^2\to U\left([0,1]\right)$ on $\langle U\left(\right[0,1\left]\right),⪯\rangle$, called interval fuzzy implication, satisfies the following conditions, $\forall A_1,A_2,A_3\in U\left([0,1]\right)$:

(II1)

If $A_1⪯A_2$ then $\mathbb{I}(A_2,A_3)⪯\mathbb{I}(A_1,A_3)$;

(II2)

If $A_2⪯A_3$ then $\mathbb{I}(A_1,A_2)⪯\mathbb{I}(A_1,A_3)$;

(II3)

$\mathbb{I}([0,0],A_2)=[1,1]$;

(II4)

$\mathbb{I}(A_1,[1,1])=[1,1]$;

(II5)

$\mathbb{I}\left(\right[1,1],[0,0\left]\right)=[0,0]$.

We present below some important properties of some interval fuzzy implications that will be verified in this work.

Definition 7. 

An interval fuzzy implication $\mathbb{I}$ on $\langle U\left(\right[0,1\left]\right),⪯\rangle$ is said to satisfy, $\forall A_1,A_2,A_3\in U\left([0,1]\right)$:

(i) 

The interval identity property if

$\mathbb{I}(A_1,A_1)=1$;   (I-IP)

(ii) 

The interval left neutrality property if

$\mathbb{I}(1,A_2)=A_2$;   (I-NP)

(iii) 

The interval order property if

$\mathbb{I}(A_1,A_2)=1$ iff $A_1⪯A_2$;   (I-OP)

(iv) 

The interval exchange principle if

$\mathbb{I}(A_1,\mathbb{I}(A_2,A_3))=\mathbb{I}(A_2,\mathbb{I}(A_1,A_3)).$   (I-EP)

Definition 8 

([36]). Let $\mathbb{I}$ be an interval fuzzy implication and $\mathbb{N}$ be an interval fuzzy negation on $\langle U\left(\right[0,1\left]\right),⪯\rangle$. $\mathbb{I}$ is said to satisfy, $\forall A_1,A_2\in U\left([0,1]\right)$:

(i) 

Interval contraposition law with respect to $\mathbb{N}$ if

$\mathbb{I}(A_1,A_2)=\mathbb{I}(\mathbb{N}\left(A_2\right),\mathbb{N}\left(A_1\right))$;   (I-CP)

(ii) 

Interval left contraposition law with respect to $\mathbb{N}$ if

$\mathbb{I}(\mathbb{N}\left(A_1\right),A_2)=\mathbb{I}(\mathbb{N}\left(A_2\right),A_1)$;   (I-LCP)

(iii) 

Interval right contraposition law with respect to $\mathbb{N}$ if

$\mathbb{I}(A_1,\mathbb{N}\left(A_2\right))=\mathbb{I}(A_2,\mathbb{N}\left(A_1\right))$.   (I-RCP)

In interval mathematics, there are two important criteria called correctness (accuracy) and optimality [23,24]. They were formalized by Santiago et al. in [26] and called interval representation and best interval representation:

Definition 9 

([39]). An interval operation $F:U{\left([0,1]\right)}^n\to U\left([0,1]\right)$ is Moore-correct or interval representation with respect to a function $f:{[0,1]}^n\to [0,1]$ whenever, for all $(A_1,\dots ,A_n)\in U{\left([0,1]\right)}^n$ and $a_i\in A_i$, $f(a_1,\dots ,a_n)\in F(A_1,\dots ,A_n)$. In addition, F is the best interval representation with respect to function f, denoted by $\widehat{f}$, if $F(A_1,\dots ,A_n)$ is the last interval containing the set $\{f(a_1,\dots ,a_n)\mid a_i\in A_i\}$ for all $A_i\in U\left([0,1]\right)$ and $i\in \{1,\dots ,n\}$, i.e.,

$$\begin{array}{c}\hfill \widehat{f}(A_1,\dots ,A_n)=\left[inf\{f(a_1,\dots ,a_n)\mid a_i\in A_i\},sup\{f(a_1,\dots ,a_n)\mid a_i\in A_i\}\right].\end{array}$$

(1)

2.3. Constrained Interval Arithmetic

One of the problems with Moore’s arithmetic is that the equalities $A-A=[0,0]$ and $A÷A=[1,1]$ are not always valid. Then, in 1999, Lodwick [28] proposed a new way of interpreting intervals, called constraint intervals, in which the previously mentioned problems were solved for all $A\in U\left(\right[0,1\left]\right)$ such that $0\in A$. In what follows, we will provide a slight change in the definition and notation given by Lodwick:

Definition 10. 

Given an interval $A=[\underline{a},\overline{a}]\in U\left([0,1]\right)$ a constrained interval associated to A is the function $f_A:[0,1]\to [0,1]$ such that, for $0\le {\lambda }_a\le 1$,

$$\begin{array}{c}\hfill f_A\left({\lambda }_a\right)=(1-{\lambda }_a)\underline{a}+{\lambda }_a\overline{a}=\underline{a}+{\lambda }_a{\omega }_a,\end{array}$$

where ${\omega }_a=\overline{a}-\underline{a}$ (the width of A).

Remark 1. 

Given $A\in U\left(\right[0,1\left]\right)$, for each $a\in A$ there exists ${\lambda }_a\in [0,1]$ such that $a=f_A\left({\lambda }_a\right)$; that is, each $a\in A$ is the image of ${\lambda }_a$ by the increasing convex constrained function $f_A$.

The resulting arithmetic, called constrained interval arithmetic (CIA), is defined as follows:

$$\begin{array}{c}\hfill f_{A_1}\left({\lambda }_{a_1}\right)\ast f_{A_2}\left({\lambda }_{a_2}\right),\end{array}$$

(2)

for ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$ and $\ast \in \{+,-,\times ,÷\}$, where the resulting interval is:

$$\begin{array}{c}\hfill A_1★A_2=\left[\underset{0\le {\lambda }_{a_1},{\lambda }_{a_2}\le 1}{min}\{f_{A_1}\left({\lambda }_{a_1}\right)\ast f_{A_2}\left({\lambda }_{a_2}\right)\},\underset{0\le {\lambda }_{a_1},{\lambda }_{a_2}\le 1}{max}\{f_{A_1}\left({\lambda }_{a_1}\right)\ast f_{A_2}\left({\lambda }_{a_2}\right)\}\right].\end{array}$$

For this arithmetic, the desirable properties apply; that is, $A-A=[0,0]$, $A÷A=[1,1]$ and the distributive property $A_1\times (A_2+A_3)=(A_1\times A_2)+(A_1\times A_3)$. Note that Moore’s interval arithmetic is extended by this new arithmetic since, when there is no dependence, they coincide, and when there is dependence, they are distinct. We can conclude that CIA arithmetic improves the overestimation of Moore’s arithmetic since it presents a smaller width interval.

Example 2. 

Given the intervals $A_1=[1,2]$ and $A_2=[-1,1]$, we will calculate the expression $A_1+A_2-A_1$ using the two arithmetics—Moore and CIA—and see the improvement in interval overestimation in CIA arithmetic. First, using Moore’s arithmetic, we obtain $A_1+A_2-A_1=([1,2]+[-1,1])-[1,2]=[0,3]-[1,2]=[-2,2]$. Now, using CIA arithmetic, $f_{A_1}\left({\lambda }_{a_1}\right)=1+{\lambda }_{a_1}$ and $f_{A_2}\left({\lambda }_{a_2}\right)=-1+2{\lambda }_{a_2}$; then, $(f_{A_1}\left({\lambda }_{a_1}\right)+f_{A_2}\left({\lambda }_{a_2}\right))-f_{A_1}\left({\lambda }_{a_1}\right)=f_{A_2}\left({\lambda }_{a_2}\right)$ and therefore $A_1+A_2-A_1=A_2=[-1,1]$.

3. Constraint Interval Fuzzy Operators

In this section, following the same approach as Lodwick [28], we introduce the constrained interval operators, in which the two lambdas, including ${\lambda }_1$ and ${\lambda }_2$, are considered when there is no dependence. The constrained interval operators were investigated in terms of correction, algebraic properties, and orders. In addition, a comparative study was carried out between these operators and an operator well known in the literature.

Definition 11. 

Let $A=[\underline{a},\overline{a}]\in U\left([0,1]\right)$ in any interval. Then,

(i) 

A function $F_A:[0,1]\to [0,1]$ such that

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_a\le 1}{inf}{F}_A\left({\lambda }_a\right)=\underline{a}\end{array}$$

and

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_a\le 1}{sup}{F}_A\left({\lambda }_a\right)=\overline{a},\end{array}$$

will be called a constraint function associated to A.

(ii) 

Associated to the interval A, we define the increasing convex constraint function $f_A$: $[0,1]\to [0,1]$ by means

$$\begin{array}{c}\hfill f_A\left({\lambda }_a\right)=(1-{\lambda }_a)\underline{a}+{\lambda }_a\overline{a},0\le {\lambda }_a\le 1,\end{array}$$

or, equivalently,

$$\begin{array}{c}\hfill f_A\left({\lambda }_a\right)=\underline{a}+{\lambda }_a(\overline{a}-\underline{a}),0\le {\lambda }_a\le 1.\end{array}$$

(iii) 

Associated to the interval A, we define the decreasing convex constraint function $f_A^{{}^{\prime }}$: $[0,1]\to [0,1]$ by means

$$\begin{array}{c}\hfill f_A^{{}^{\prime }}\left({\lambda }_a\right)={\lambda }_a\underline{a}+(1-{\lambda }_a)\overline{a},0\le {\lambda }_a\le 1,\end{array}$$

or, equivalently,

$$\begin{array}{c}\hfill f_A^{{}^{\prime }}\left({\lambda }_a\right)=\overline{a}+{\lambda }_a(\underline{a}-\overline{a}),0\le {\lambda }_a\le 1.\end{array}$$

Observe that the same interval can be determined by different constraint functions. However, each interval is associated with a single increasing convex constraint function and two different intervals determine two different increasing convex constraint functions (i.e., there is a bijection between the space of intervals and the space of increasing (decreasing) convex constrained functions).

Remark 2. 

Given $A\in U\left(\right[0,1\left]\right)$, for each $a\in A$, there exists ${\lambda }_a\in [0,1]$ such that $a=f_A\left({\lambda }_a\right)$; that is, each $a\in A$ is the image of ${\lambda }_a$ by the function $f_A$.

An analogous development can be carried out using the decreasing convex constraint functions. However, in this article, we will use increasing convex constraint functions.

Definition 12. 

The operator $\ddot{g}$: $U{\left([0,1]\right)}^n\to U\left([0,1]\right)$, called the constraint interval operator, associated with the function g: ${[0,1]}^n\to [0,1]$ is defined by, for all intervals $A_1,\dots ,A_n$,

1.

If $A_j\ne A_k$ for any $j\ne k$, then $\ddot{g}(A_1,\dots ,A_n)$ is

$$\begin{array}{c}\hfill \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\}\right];\end{array}$$

2.

If $A_{j_1}=\dots =A_{j_k}=A$ for $k\le n$ and for some $j_i\in \{1,\dots ,n\}$ with $j_r\ne j_s$ if $r\ne s$, then $\ddot{g}(A_1,\dots ,A,\dots ,A,\dots ,A_n)$ is

$$\begin{array}{ccc}\hfill [\underset{0\le {\lambda }_{a_i},{\lambda }_a\le 1}{inf}\{g& (& f_{A_1}\left({\lambda }_{a_1}\right),\dots ,f_{A_{j_1}}\left({\lambda }_a\right),\dots ,f_{A_{j_k}}\left({\lambda }_a\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right)\left)\right\},\hfill \\ & & \underset{0\le {\lambda }_{a_i},{\lambda }_a\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),\dots ,f_{A_{j_1}}\left({\lambda }_a\right),\dots ,f_{A_{j_k}}\left({\lambda }_a\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\}].\hfill \end{array}$$

We denote $\underline{\ddot{g}(A_1,\dots ,A_n)}$ and $\overline{\ddot{g}(A_1,\dots ,A_n)}$ as the endpoints

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\}\end{array}$$

and

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\},\end{array}$$

respectively.

It can be easily noticed that when there is no dependence on the input intervals, we have a different parameter for each interval; that is, different intervals are traversed at different levels (${\lambda }_{a_i}$, for $i\in \{1,2,\cdots ,n\}$). In the case of dependency, we have a single parameter for equal intervals, so they are traversed at the same level (${\lambda }_a$). Therefore, the resulting interval has less irrelevant information, and pieces of information are not lost; see Remark 6.

Example 3. 

1.

Let N: $[0,1]\to [0,1]$ be the standard fuzzy negation; that is, $N\left(x\right)=1-x$, so, $\forall A\in U\left(\right[0,1\left]\right)$,

$$\begin{array}{c}\hfill \ddot{N}\left(A\right)=\left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{N\left(f_A\left({\lambda }_a\right)\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{N\left(f_A\left({\lambda }_a\right)\right)\right\}\right]=[1-\overline{a},1-\underline{a}].\end{array}$$

2.

Let T: ${[0,1]}^2\to [0,1]$ be the minimum t-norm; that is, $T(x,y)=min(x,y)$, so, $\forall A_1,A_2\in U\left([0,1]\right)$,

$$\begin{array}{ccc}\hfill \ddot{T}(A_1,A_2)& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\{min(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\{min(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\}\right]\hfill \\ & =& [min(\underline{a_1},\underline{a_2}),min(\overline{a_1},\overline{a_2})],\hfill \end{array}$$

since $f_{A_1}$ and $f_{A_2}$ are increasing.

3.

Consider the aggregation function g: ${[0,1]}^n\to [0,1]$ given by $g(x_1,\dots ,x_n)=\frac{x_1+\dots +x_n}{n}$. For all $A_1,\dots ,A_n\in U\left([0,1]\right)$, since $\underline{a_i}=f_{A_i}\left(0\right)\le f_{A_i}\left({\lambda }_{a_i}\right)$ for all ${\lambda }_{a_i}\in [0,1]$, we have

$$\begin{array}{c}\hfill \frac{\underline{a_1}+\dots +\underline{a_n}}{n}\le \frac{f_{A_1}\left({\lambda }_{a_1}\right)+\dots +f_{A_n}\left({\lambda }_{a_n}\right)}{n}\end{array}$$

for all ${\lambda }_{a_i}\in [0,1]$ and $i\in \{1,\dots ,n\}$. So,

$$\begin{array}{c}\hfill \frac{\underline{a_1}+\dots +\underline{a_n}}{n}=\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{\frac{f_{A_1}\left({\lambda }_{a_1}\right)+\dots +f_{A_n}\left({\lambda }_{a_n}\right)}{n}\right\}.\end{array}$$

and, analogously,

$$\begin{array}{c}\hfill \frac{\overline{a_1}+\dots +\overline{a_n}}{n}=\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{\frac{f_{A_1}\left({\lambda }_{a_1}\right)+\dots +f_{A_n}\left({\lambda }_{a_n}\right)}{n}\right\}.\end{array}$$

Therefore,

$$\begin{array}{c}\hfill \ddot{g}(A_1,\dots ,A_n)=\left[\frac{\underline{a_1}+\dots +\underline{a_n}}{n},\frac{\overline{a_1}+\dots +\overline{a_n}}{n}\right].\end{array}$$

The following proposition ensures that a constrained interval operator $\ddot{g}$ coincides with its best representation $\widehat{g}$ whenever g is increasing.

Proposition 1. 

Let $g:{[0,1]}^n\to [0,1]$ be an increasing function. Then, $\forall A_1,A_2,\dots ,$$A_n\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \ddot{g}(A_1,A_2,\dots ,A_n)=\left[g(\underline{a_1},\underline{a_2},\dots ,\underline{a_n}),g(\overline{a_1},\overline{a_2},\dots ,\overline{a_n})\right].\end{array}$$

Proof. 

Given any $A_1,A_2,\dots ,A_n\in U\left([0,1]\right)$, since the function $f_{A_i}$ is increasing for all $i\in \{1,2\dots ,n\}$, we have $\underline{a_i}=f_{A_i}\left(0\right)\le f_{A_i}\left({\lambda }_{a_i}\right),\forall {\lambda }_{a_i}\in [0,1]$. So, $g(\underline{a_1},\underline{a_2},\dots ,\underline{a_n})\le g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))$ for all ${\lambda }_{a_i}\in [0,1]$ with $i\in \{1,2,\dots ,n\}$. Therefore,

$$\begin{array}{c}\hfill g(\underline{a_1},\underline{a_2},\dots ,\underline{a_n})=\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\}.\end{array}$$

Analogously,

$$\begin{array}{c}\hfill g(\overline{a_1},\overline{a_2},\dots ,\overline{a_n})=\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right),\dots ,f_{A_n}\left({\lambda }_{a_n}\right))\right\}.\end{array}$$

Thus, $\ddot{g}(A_1,A_2\dots ,A_n)=\left[g(\underline{a_1},\underline{a_2},\dots ,\underline{a_n}),g(\overline{a_1},\overline{a_2},\dots ,\overline{a_n})\right]$. □

Remark 3. 

An analogous result can be obtained if g is decreasing. In this case, we would have $\ddot{g}(A_1,A_2,\dots ,A_n)=\left[g(\overline{a_1},\overline{a_2},\dots ,\overline{a_n}),g(\underline{a_1},\underline{a_2},\dots ,\underline{a_n})\right]$ for all $A_1,A_2,\dots ,A_n\in U\left([0,1]\right)$.

Remark 4. 

We can conclude from the previous results that, if g is monotone, $\ddot{g}$ is correct (Moore’s correctness) and coincides with the best interval representation of g; that is, $\ddot{g}=\widehat{g}$ (see Equation (1)).

Therefore, given an increasing function g, we have that $\ddot{g}(A_1,A_2)=\widehat{g}(A_1,A_2),\forall A_1,A_2\in U\left([0,1]\right)$. However, some fuzzy logical operators like implications are neither increasing nor decreasing, in which case we have situations like:

Example 4. 

Consider the Łukasiewicz implication $I_{{}_{LK}}(x,y)=min(1,1-x+y)$ and $A\in U\left(\right[0,1\left]\right)$ as non-degenerate. Then, since $I_{{}_{LK}}\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)=min(1,1-f_A\left({\lambda }_a\right)+f_A\left({\lambda }_a\right))$ = 1,

$$\begin{array}{c}\hfill {\ddot{I}}_{{}_{LK}}(A,A)=\left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{I_{{}_{LK}}\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{I_{{}_{LK}}\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\}\right]=[1,1]\end{array}$$

and $I_{{}_{LK}}(\overline{a},\underline{a})=min(1,1-\overline{a}+\underline{a})\stackrel{\underline{a}<\overline{a}}{=}1-\overline{a}+\underline{a}<1$, in which case we have that ${\ddot{I}}_{{}_{LK}}$ is not correct, since $I_{{}_{LK}}(\overline{a},\underline{a})\notin {\ddot{I}}_{{}_{LK}}(A,A)$. However, the fuzzy implications satisfy the property presented in the following proposition:

Proposition 2. 

Let I be a fuzzy implication. Then, $\ddot{I}(A_1,A_2)\subseteq \widehat{I}(A_1,A_2)$ for all $A_1,A_2\in U\left([0,1]\right)$.

Proof. 

By Definition 12, for all $A\in U\left(\right[0,1\left]\right)$,

$$\begin{array}{c}\hfill \ddot{I}(A,A)=\left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{I\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{I\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\}\right].\end{array}$$

So, since

$$\begin{array}{c}\hfill \{I(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right))\mid {\lambda }_a\in [0,1]\}\subseteq \{I(f_A\left({\lambda }_a\right),f_A\left({\lambda }_{\tilde{a}}\right))\mid {\lambda }_a,{\lambda }_{\tilde{a}}\in [0,1]\},\end{array}$$

we have, by Equation (1), $\ddot{I}(A,A)\subseteq \widehat{I}(A,A)$. Now, given any $A_1,A_2\in U\left([0,1]\right)$ with $A_1\ne A_2$, we have that

$$\begin{array}{ccc}\hfill \ddot{I}(A_1,A_2)& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\}\right]\hfill \\ & =& \left[inf\{I(a_1,a_2)\mid a_i\in A_i\},sup\{I(a_1,a_2)\mid a_i\in A_i\}\right]\hfill \\ & =& \widehat{I}(A_1,A_2).\hfill \end{array}$$

Therefore, $\ddot{I}(A_1,A_2)\subseteq \widehat{I}(A_1,A_2),\forall A_1,A_2\in U\left([0,1]\right)$. □

Remark 5. 

Considering any fuzzy implication I, given $A_1,A_2\in U\left([0,1]\right)$ with $A_1\ne A_2$, we have that $\ddot{I}(A_1,A_2)=\widehat{I}(A_1,A_2)$. So, by [40] (Theorem 6.2), $\ddot{I}(A_1,A_2)=\widehat{I}(A_1,A_2)=\left[I(\overline{a_1},\underline{a_2}),I(\underline{a_1},\overline{a_2})\right]$.

Example 5. 

Consider the Fodor implication $I_{{}_{FD}}(x,y)=\left\{\begin{array}{cc}1,\hfill & ifx\le y\hfill \\ max(1-x,y),\hfill & ifx>y\hfill \end{array}.\right.$ For $A_1=\left[\frac{1}{2},\frac{3}{4}\right]$ and $A_2=\left[0,\frac{1}{4}\right]$, we have that $f_{A_1}\left({\lambda }_{a_1}\right)=\frac{1}{2}+\frac{1}{4}{\lambda }_{a_1}$ and $f_{A_2}\left({\lambda }_{a_2}\right)=\frac{1}{4}{\lambda }_{a_2}$. Then, since $f_{A_1}\left({\lambda }_{a_1}\right)\ge f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$,

$$\begin{array}{c}\hfill I_{{}_{FD}}\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)=max\left(\frac{1}{2}-\frac{1}{4}{\lambda }_{a_1},\frac{1}{4}{\lambda }_{a_2}\right)=\frac{1}{2}-\frac{1}{4}{\lambda }_{a_1},\end{array}$$

for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$. So,

$$\begin{array}{ccc}\hfill {\ddot{I}}_{{}_{FD}}(A_1,A_2)& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{I_{{}_{FD}}\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{I_{{}_{FD}}\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\}\right]\hfill \\ & =& \left[\frac{1}{4},\frac{1}{2}\right].\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\hfill {\ddot{I}}_{{}_{FD}}(A_1,A_2)={\widehat{I}}_{{}_{FD}}(A_1,A_2),\end{array}$$

since $I_{{}_{FD}}(\overline{a_1},\underline{a_2})=I_{{}_{FD}}(\frac{3}{4},0)=\frac{1}{4}$ and $I_{{}_{FD}}(\underline{a_1},\overline{a_2})=I_{{}_{FD}}(\frac{1}{2},\frac{1}{4})=\frac{1}{2}$. Now, for $A_1=A_2=A=\left[0,\frac{1}{4}\right]$, we have $f_A\left({\lambda }_a\right)=\frac{1}{4}{\lambda }_a$, so

$$\begin{array}{ccc}\hfill {\ddot{I}}_{{}_{FD}}(A,A)& =& \left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{I_{{}_{FD}}\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{I_{{}_{FD}}\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\}\right]\hfill \\ & =& \left[1,1\right],\hfill \end{array}$$

since $I_{{}_{FD}}\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)=I_{{}_{FD}}\left(\frac{1}{4}{\lambda }_a,\frac{1}{4}{\lambda }_a\right)=1$. Therefore, ${\ddot{I}}_{{}_{FD}}(A,A)\subset {\widehat{I}}_{{}_{FD}}(A,A)=\left[\frac{3}{4},1\right]$. Note that ${\ddot{I}}_{{}_{FD}}$ is not correct since $I_{{}_{FD}}(\frac{1}{4},0)=\frac{3}{4}\notin {\ddot{I}}_{{}_{FD}}(A,A)$. However, in this case, elements of type $I_{{}_{FD}}(\frac{1}{4},0)$, with $x\ne y$, are not relevant (they are irrelevant); see Remark 6.

We can conclude, then, that this new way of defining interval operators does not always coincide with the best representation. Also, it does not necessarily satisfy Moore’s correctness (see Example 4). However, this does not cause harm; see the following remark:

Example 6. 

(Moore and constraint interval correctness.) It is possible to observe that the loss of the usual correctness (Moore’s correctness) occurs only when we have variable dependence. However, this loss will not cause problems since every element of an interval operates with itself and this operation is contained in the resulting interval, i.e., the correctness with respect to a fixed element is maintained. So, everything “thrown away” is irrelevant. In other words, if the interval A represents a certain number a, we have that

$$\begin{array}{c}\hfill g(a_1,\dots ,a,\dots ,a,\dots ,a_n)\in \ddot{g}(A_1,\dots ,A,\dots ,A,\dots ,A_n),\end{array}$$

then, something like $g(a_1,\dots ,a,\dots ,\tilde{a},\dots ,a_n)$ with $a,\tilde{a}\in A$, and $a\ne \tilde{a}$, is not relevant, since we are only interested in operations with the element in which A represents; that is, operations with the element a. This kind of correctness is called here constraint interval correctness.

The following proposition shows that important properties such as symmetry, (IP), and (NP) are preserved by the constrained interval operators.

Proposition 3. 

Let $g:{[0,1]}^2\to [0,1]$ be a function. Then,

(i) 

If g is symmetric, then $\ddot{g}$ is also symmetric;

(ii) 

If g satisfies identity property (IP), then $\ddot{g}$ satisfies (I-IP);

(iii) 

If g satisfies left neutrality property (NP), then $\ddot{g}$ satisfies (I-NP).

Proof. 

Indeed,

(i)

By the symmetry of g, we have

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,A_2)& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\}\right]\hfill \\ & =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{g\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)\right\}\right]\hfill \\ & =& \ddot{g}(A_2,A_1),\hfill \end{array}$$

for any $A_1,A_2\in U\left([0,1]\right)$.

(ii)

For any $A\in U\left(\right[0,1\left]\right)$,

$$\begin{array}{ccc}\hfill \ddot{g}(A,A)& =& \left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{g\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{g\left(f_A\left({\lambda }_a\right),f_A\left({\lambda }_a\right)\right)\right\}\right]\hfill \\ & =& \left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{1\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{1\right\}\right]=[1,1],\hfill \end{array}$$

since g satisfies (IP).

(iii)

For any $A\in U\left(\right[0,1\left]\right)$,

$$\begin{array}{ccc}\hfill \ddot{g}([1,1],A)& =& \left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{g\left(1,f_A\left({\lambda }_a\right)\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{g\left(1,f_A\left({\lambda }_a\right)\right)\right\}\right]\hfill \\ & =& \left[\underset{0\le {\lambda }_a\le 1}{inf}\left\{f_A\left({\lambda }_a\right)\right\},\underset{0\le {\lambda }_a\le 1}{sup}\left\{f_A\left({\lambda }_a\right)\right\}\right]=A,\hfill \end{array}$$

since g satisfies (NP). □

Proposition 4. 

There is function $g:{[0,1]}^2\to [0,1]$ in which the exchange principle (EP) is satisfied but $\ddot{g}$ does not satisfy (I-EP).

Proof. 

Consider the Łukasiewicz implication $I(x,y)=min(1,1-x+y)$ and the intervals $A_1=[0,1]$ and $A_2=\left[0,\frac{1}{2}\right]$. From Table 1.4. in [36], I satisfies (EP). Now, since $I\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)=min\left(1,1-\frac{{\lambda }_{a_2}}{2}+{\lambda }_{a_1}\right)$, we have

$$\begin{array}{ccc}\hfill \ddot{I}(A_2,A_1)& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{I\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{I\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)\right\}\right]\hfill \\ & =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{min\left(1,1-\frac{{\lambda }_{a_2}}{2}+{\lambda }_{a_1}\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{min\left(1,1-\frac{{\lambda }_{a_2}}{2}+{\lambda }_{a_1}\right)\right\}\right]\hfill \\ & =& \left[\frac{1}{2},1\right].\hfill \end{array}$$

So, $f_{\ddot{I}(A_2,A_1)}\left({\lambda }_{a_{{}_{21}}}\right)=\frac{1}{2}+\frac{{\lambda }_{a_{{}_{21}}}}{2}$ and, for all ${\lambda }_{a_1},{\lambda }_{a_{{}_{21}}}$,

$$\begin{array}{c}\hfill I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{I}(A_2,A_1)}\left({\lambda }_{a_{{}_{21}}}\right)\right)=min\left(1,1-{\lambda }_{a_1}+\frac{1}{2}+\frac{{\lambda }_{a_{21}}}{2}\right)=min\left(1,\frac{3}{2}-{\lambda }_{a_1}+\frac{{\lambda }_{a_{21}}}{2}\right).\end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill \ddot{I}(A_1,\ddot{I}(A_2,A_1))& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{I}(A_2,A_1)}\left({\lambda }_{a_{{}_{21}}}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{I}(A_2,A_1)}\left({\lambda }_{a_{{}_{21}}}\right)\right)\right\}\right]\hfill \\ & =& \left[\frac{1}{2},1\right].\hfill \end{array}$$

On the other hand,

$$\begin{array}{ccc}\hfill \ddot{I}(A_1,A_1)& =& \left[\underset{0\le {\lambda }_{a_1}\le 1}{inf}\left\{I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)\right\},\underset{0\le {\lambda }_{a_1}\le 1}{sup}\left\{I\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_1}\left({\lambda }_{a_1}\right)\right)\right\}\right]\hfill \\ & =& \left[\underset{0\le {\lambda }_{a_1}\le 1}{inf}\left\{min\left(1,1-{\lambda }_{a_1}+{\lambda }_{a_1}\right)\right\},\underset{0\le {\lambda }_{a_1}\le 1}{sup}\left\{min\left(1,1-{\lambda }_{a_1}+{\lambda }_{a_1}\right)\right\}\right]\hfill \\ & =& \left[1,1\right].\hfill \end{array}$$

And, since $f_{\ddot{I}(A_1,A_1)}\left({\lambda }_{a_{{}_{11}}}\right)=1$ for all ${\lambda }_{a_{{}_{11}}}\in [0,1]$,

$$\begin{array}{ccc}\hfill \ddot{I}(A_2,\ddot{I}(A_1,A_1))& =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{I\left(f_{A_1}\left({\lambda }_{a_2}\right),f_{\ddot{I}(A_1,A_1)}\left({\lambda }_{a_{{}_{11}}}\right)\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{I\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{\ddot{I}(A_1,A_1)}\left({\lambda }_{a_{{}_{11}}}\right)\right)\right\}\right]\hfill \\ & =& \left[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{min\left(1,1-\frac{{\lambda }_{a_2}}{2}+1\right)\right\},\underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{min\left(1,1-\frac{{\lambda }_{a_2}}{2}+1\right)\right\}\right]\hfill \\ & =& \left[1,1\right].\hfill \end{array}$$

Therefore, $\ddot{I}(A_1,\ddot{I}(A_2,A_1))\ne \ddot{I}(A_2,\ddot{I}(A_1,A_1))$. □

However, Proposition 5 guarantees that (EP) is preserved by the operator $\ddot{g}$ when g is continuous.

Lemma 1. 

Let g: ${[0,1]}^2\to [0,1]$ be a function. Then,

$$\begin{array}{ccc}\hfill \left\{g\right(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right)& ,& f_{A_3}\left({\lambda }_{a_3}\right)\left)\right)\mid {\lambda }_{a_i}\in [0,1]\}\subseteq \hfill \\ & & \left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{g_{{}_{23}}}\right)\right)\mid {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\in [0,1]\right\},\hfill \end{array}$$

$\forall A_1,A_2,A_3\in U\left([0,1]\right)$. In addition, if g is continuous, then equality holds.

Proof. 

Given $y\in \{g(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\mid {\lambda }_{a_i}\in [0,1]\}$, there are ${\lambda }_{\widetilde{a_1}},{\lambda }_{\widetilde{a_2}},{\lambda }_{\widetilde{a_3}}\in [0,1]$ such that $y=g(f_{A_1}\left({\lambda }_{\widetilde{a_1}}\right),g(f_{A_2}\left({\lambda }_{\widetilde{a_2}}\right),f_{A_3}\left({\lambda }_{\widetilde{a_3}}\right)))$. As $g(f_{A_2}\left({\lambda }_{\widetilde{a_2}}\right),f_{A_3}\left({\lambda }_{\widetilde{a_3}}\right))\in \ddot{g}(A_2,A_3)$, then, by Remark 2, there exists ${\lambda }_{{\tilde{g}}_{{}_{23}}}\in [0,1]$ such that $f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{{\tilde{g}}_{{}_{23}}}\right)=g(f_{A_2}\left({\lambda }_{\widetilde{a_2}}\right),f_{A_3}\left({\lambda }_{\widetilde{a_3}}\right))$. We conclude that

$$\begin{array}{c}\hfill y=g(f_{A_1}\left({\lambda }_{\widetilde{a_1}}\right),f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{{\tilde{g}}_{{}_{23}}}\right))\in \{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{g_{{}_{23}}}\right)\right)\mid {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\in [0,1]\}\end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill \left\{g\right(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right)& ,& f_{A_3}\left({\lambda }_{a_3}\right)\left)\right)\mid {\lambda }_{a_i}\in [0,1]\}\subseteq \hfill \\ & & \left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{g_{{}_{23}}}\right)\right)\mid {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\in [0,1]\right\}.\hfill \end{array}$$

In addition, if g is continuous, then, given

$$\begin{array}{c}\hfill y\in \{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{g_{{}_{23}}}\right)\right)\mid {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\in [0,1]\}\end{array}$$

we have $y=g(f_{A_1}\left({\lambda }_{\widetilde{a_1}}\right),f_{\ddot{g}(A_2,A_3)}({\lambda }_{{\tilde{g}}_{{}_{23}}}))$ for some ${\lambda }_{\widetilde{a_1}},{\lambda }_{{\tilde{g}}_{{}_{23}}}\in [0,1]$. Since $f_{\ddot{g}(A_2,A_3)}({\lambda }_{{\tilde{g}}_{{}_{23}}})\in \ddot{g}(A_2,A_3)$, then, by the continuity of g, there are ${\lambda }_{\widetilde{a_2}},{\lambda }_{\widetilde{a_3}}\in [0,1]$ such that $f_{\ddot{g}(A_2,A_3)}({\lambda }_{{\tilde{g}}_{{}_{23}}})=g(f_{A_2}\left({\lambda }_{\widetilde{a_2}}\right),f_{A_3}\left({\lambda }_{\widetilde{a_3}}\right))$. We conclude that

$$\begin{array}{c}\hfill y=g(f_{A_1}\left({\lambda }_{\widetilde{a_1}}\right),g(f_{A_2}\left({\lambda }_{\widetilde{a_2}}\right),f_{A_3}\left({\lambda }_{\widetilde{a_3}}\right)))\in \{g(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\mid {\lambda }_{a_i}\in [0,1]\}.\end{array}$$

So,

$$\begin{array}{ccc}\hfill \{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}({\lambda }_{g_{{}_{23}}})\right)& \mid & {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\in [0,1]\}\subseteq \hfill \\ & & \left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),g\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)\right)\right)\mid {\lambda }_{a_i}\in [0,1]\right\}.\hfill \end{array}$$

Thus,

$$\begin{array}{ccc}\hfill \{g\left(f_{A_1}\left({\lambda }_{a_1}\right),g\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)\right)\right)& \mid & {\lambda }_{a_i}\in [0,1]\}=\hfill \\ & & \left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}({\lambda }_{g_{{}_{23}}})\right)\mid {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\in [0,1]\right\}.\hfill \end{array}$$

Therefore, the equality holds when g is continuous. □

Remark 7. 

Analogous to the previous lemma, we have that

$$\begin{array}{ccc}\hfill \{g\left(g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)\right),f_{A_3}\left({\lambda }_{a_3}\right)\right)& \mid & {\lambda }_{a_i}\in [0,1]\}\subseteq \hfill \\ & & \left\{g\left(f_{\ddot{g}(A_1,A_2)}\left({\lambda }_{g_{{}_{12}}}\right),f_{A_3}\left({\lambda }_{a_3}\right)\right)\mid {\lambda }_{g_{{}_{12}}},{\lambda }_{a_3}\in [0,1]\right\}.\hfill \end{array}$$

And, if g is continuous, then the equality holds.

Proposition 5. 

Let $g:{[0,1]}^2\to [0,1]$ be a continuous function. Then:

(i)

If g is associative, then $\ddot{g}$ is also associative;

(ii)

If g satisfies the exchange principle (EP), then $\ddot{g}$ satisfies (I-EP).

Proof. 

Indeed,

(i)

For all $A_1,A_2,A_3\in U\left([0,1]\right)$,

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[\underset{0\le {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\le 1}{inf}& \{& g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}({\lambda }_{g_{{}_{23}}})\right)\},\hfill \\ & & \underset{0\le {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\le 1}{sup}\left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}({\lambda }_{g_{{}_{23}}})\right)\right\}].\hfill \end{array}$$

So, by Lemma 1,

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[\underset{0\le {\lambda }_{a_i}\le 1}{inf}& \{& g(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\},\hfill \\ & & \underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\right\}].\hfill \end{array}$$

Now, since g is associative, we have that

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[\underset{0\le {\lambda }_{a_i}\le 1}{inf}\left\{g\right(& g& (f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)),f_{A_3}\left({\lambda }_{a_3}\right)\left)\right\},\hfill \\ & & \underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right)),f_{A_3}\left({\lambda }_{a_3}\right))\right\}].\hfill \end{array}$$

So, by Remark 7,

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[& & \underset{0\le {\lambda }_{g_{{}_{12}}},{\lambda }_{a_3}\le 1}{inf}\left\{g\left(f_{\ddot{g}(A_1,A_2)}\left({\lambda }_{g_{{}_{12}}}\right),f_{A_3}\left({\lambda }_{a_3}\right)\right)\right\},\hfill \\ & & \underset{0\le {\lambda }_{g_{{}_{12}}},{\lambda }_{a_3}\le 1}{sup}\left\{g\left(f_{\ddot{g}(A_1,A_2)}\left({\lambda }_{g_{{}_{12}}}\right),f_{A_3}\left({\lambda }_{a_3}\right)\right)\right\}]\hfill \\ \hfill & =& \ddot{g}(\ddot{g}(A_1,A_2),A_3).\hfill \end{array}$$

Therefore, $\ddot{g}$ is associative.

(ii)

For all $A_1,A_2,A_3\in U\left([0,1]\right)$,

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[\underset{0\le {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\le 1}{inf}& \{& g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}({\lambda }_{g_{{}_{23}}})\right)\},\hfill \\ & & \underset{0\le {\lambda }_{a_1},{\lambda }_{g_{{}_{23}}}\le 1}{sup}\left\{g\left(f_{A_1}\left({\lambda }_{a_1}\right),f_{\ddot{g}(A_2,A_3)}\left({\lambda }_{g_{{}_{23}}}\right)\right)\right\}].\hfill \end{array}$$

So, by Lemma 1,

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[\underset{0\le {\lambda }_{a_i}\le 1}{inf}& \{& g(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\},\hfill \\ & & \underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),g(f_{A_2}\left({\lambda }_{a_2}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\right\}].\hfill \end{array}$$

Therefore, since g satisfies (EP),

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[\underset{0\le {\lambda }_{a_i}\le 1}{inf}& \{& g(f_{A_2}\left({\lambda }_{a_2}\right),g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\},\hfill \\ & & \underset{0\le {\lambda }_{a_i}\le 1}{sup}\left\{g(f_{A_2}\left({\lambda }_{a_2}\right),g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_3}\left({\lambda }_{a_3}\right)))\right\}].\hfill \end{array}$$

So, again by Lemma 1,

$$\begin{array}{ccc}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=[& & \underset{0\le {\lambda }_{a_2},{\lambda }_{g_{{}_{13}}}\le 1}{inf}\left\{g\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{\ddot{g}(A_1,A_3)}({\lambda }_{g_{{}_{13}}})\right)\right\},\hfill \\ & & \underset{0\le {\lambda }_{a_2},{\lambda }_{g_{{}_{13}}}\le 1}{sup}\left\{g\left(f_{A_2}\left({\lambda }_{a_2}\right),f_{\ddot{g}(A_1,A_3)}({\lambda }_{g_{{}_{13}}})\right)\right\}]\hfill \\ & =& \ddot{g}(A_2,\ddot{g}(A_1,A_3)).\hfill \end{array}$$

Therefore, $\ddot{g}$ satisfies (I-EP). □

The binary operator g does not need to be continuous in order for its extension to preserve associativity; it suffices that g is increasing (or decreasing). See the following proposition:

Proposition 6. 

Let $g:{[0,1]}^2\to [0,1]$ be an increasing function. Then, g is associative if and only if $\ddot{g}$ is also associative.

Proof. 

As g is increasing, using Proposition 1, for all $A_i,A_j\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \ddot{g}(A_i,A_j)=[g(\underline{a_i},\underline{a_j}),g(\overline{a_i},\overline{a_j})].\end{array}$$

So, for all $A_1,A_2,A_3\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=\left[g\left(\underline{a_1},\underline{\ddot{g}(A_2,A_3)}\right),g\left(\overline{a_1},\overline{\ddot{g}(A_2,A_3)}\right)\right]=\left[g\left(\underline{a_1},g(\underline{a_2},\underline{a_3})\right),g\left(\overline{a_1},g(\overline{a_2},\overline{a_3})\right)\right].\end{array}$$

Analogously, using Proposition 1, for all $A_1,A_2,A_3\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \ddot{g}(\ddot{g}(A_1,A_2),A_3)=\left[g\left(\underline{\ddot{g}(A_1,A_2)},\underline{a_3}\right),g\left(\overline{\ddot{g}(A_1,A_2)},\overline{a_3}\right)\right]=\left[g\left(g(\underline{a_1},\underline{a_2}),\underline{a_3}\right),g\left(g(\overline{a_1},\overline{a_2}),\overline{a_3}\right)\right].\end{array}$$

So, since g is associative, we conclude that, for all $A_1,A_2,A_3\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \ddot{g}(A_1,\ddot{g}(A_2,A_3))=\ddot{g}(\ddot{g}(A_1,A_2),A_3).\end{array}$$

Conversely, for each $z\in [0,1]$, $z=f_{[z,z]}\left({\lambda }_z\right)$ for all ${\lambda }_z\in [0,1]$; then,

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_x,{\lambda }_y\le 1}{inf}\left\{g(f_{[x,x]}\left({\lambda }_x\right),f_{[y,y]}\left({\lambda }_y\right))\right\}=g(x,y)=\underset{0\le {\lambda }_x,{\lambda }_y\le 1}{sup}\left\{g(f_{[x,x]}\left({\lambda }_x\right),f_{[y,y]}\left({\lambda }_y\right))\right\}\end{array}$$

and, therefore, $\ddot{g}([x,x],[y,y])=[g(x,y),g(x,y)]$. Following a similar reasoning, we have that

$$\begin{array}{c}\hfill \ddot{g}([x,x],\ddot{g}([y,y],[z,z]))=[g(x,g(y,z)),g(x,g(y,z))]\end{array}$$

and

$$\begin{array}{c}\hfill \ddot{g}(\ddot{g}([x,x],[y,y]),[z,z])=[g(g(x,y),z),g(g(x,y),z)].\end{array}$$

So, $g(x,g(y,z\left)\right)=g\left(g\right(x,y),z)$, since $\ddot{g}$ is associative. □

Section 3.1 and Section 3.2 show that the logical connectives studied are extended to their respective intervals logical connectives in relation to the order ${\le }_{{}_{KM}}$, but the same does not occur with Moore’s order ⊴.

3.1. Properties of Some Interval Fuzzy Connectives with Respect to Order ${\le }_{{}_{KM}}$

This section will show that, considering the order of Kulisch–Miranker (see definition below), fuzzy connectives, negation, t-norms, and fuzzy implications are extended to their respective intervals logical connectives, and their main properties are maintained.

Definition 13. 

The Kulisch–Miranker order is defined, for all $A_1,A_2\in U\left([0,1]\right)$, by:

$$\begin{array}{c}\hfill A_1{\le }_{{}_{KM}}{A}_2\iff \underline{a_1}\le \underline{a_2}\mathrm{and}\overline{a_1}\le \overline{a_2}.\end{array}$$

The following proposition shows the connection that exists between the Kulisch–Miranker order and the functional counterpart of the intervals.

Proposition 7. 

Given $A_1,A_2\in U\left([0,1]\right)$, the following are equivalent:

(i) 

$A_1{\le }_{{}_{KM}}{A}_2$;

(ii) 

$f_{A_1}\left(\lambda \right)\le f_{A_2}\left(\lambda \right)$ for all $\lambda \in [0,1]$;

(iii) 

$f_{A_1}^{\prime }\left(\lambda \right)\le f_{A_2}^{\prime }\left(\lambda \right)$ for all $\lambda \in [0,1]$.

Proof. 

$\left(i\right)\iff \left(ii\right)$ Indeed, let $A_1{\le }_{{}_{KM}}{A}_2$ for some $A_1,A_2\in U\left([0,1]\right)$; then, $\underline{a_1}\le \underline{a_2}$ and $\overline{a_1}\le \overline{a_2}$. So, since $\lambda \in [0,1]$, we have $(1-\lambda )\underline{a_1}\le (1-\lambda )\underline{a_2}$ and $\lambda \overline{a_1}\le \lambda \overline{a_2}$ for all $\lambda \in [0,1]$. Thus, $(1-\lambda )\underline{a_1}+\lambda \overline{a_1}\le (1-\lambda )\underline{a_2}+\lambda \overline{a_2}$ for all $\lambda \in [0,1]$, i.e., $f_{A_1}\left(\lambda \right)\le f_{A_2}\left(\lambda \right)$ for all $\lambda \in [0,1]$. Conversely, let $f_{A_1}\left(\lambda \right)\le f_{A_2}\left(\lambda \right)$ for all $\lambda \in [0,1]$; then, in particular, for $\lambda =0$ and $\lambda =1$, we obtain $\underline{a_1}\le \underline{a_2}$ and $\overline{a_1}\le \overline{a_2}$, respectively. So, $A_1{\le }_{{}_{KM}}{A}_2$.

$\left(ii\right)\iff \left(iii\right)$ Straightforward, once $f_A\left(\lambda \right)=f_A^{\prime }(1-\lambda )$. □

Proposition 8. 

There is binary operator g that satisfies (OP), whereas $\ddot{g}$ does not satisfy (I-OP) with respect to ${\le }_{{}_{KM}}$.

Proof. 

Take $A_1,A_2\in U\left([0,1]\right)$ such that $A_1{\le }_{{}_{KM}}{A}_2$ with $\underline{a_1}\ne \underline{a_2}<\overline{a_1}$. So, since g satisfies (OP),

$$\begin{array}{c}\hfill g(f_{A_1}\left(1\right),f_{A_2}\left(0\right))=g(\overline{a_1},\underline{a_2})\ne 1.\end{array}$$

So, $\ddot{g}(A_1,A_2)\ne [1,1]$ since $g(f_{A_1}\left(1\right),f_{A_2}\left(0\right))\in \ddot{g}(A_1,A_2)$. □

The following result shows that the restricted operators preserve the fuzzy implications with respect to the order of Kulisch–Miranker.

Proposition 9. 

Let I be a fuzzy implication. Then, $\ddot{I}$ is an interval fuzzy implication with respect to ${\le }_{{}_{KM}}$.

Proof. 

Indeed, (II1) a for $A_1{\le }_{{}_{KM}}{A}_1$, trivially $\ddot{I}(A_1,C){\le }_{{}_{KM}}\ddot{I}(A_1,C),\forall C\in U\left([0,1]\right)$. On the other hand, if $A_1{\le }_{{}_{KM}}{A}_2$ with $A_1\ne A_2$, then $\underline{a_1}\le \underline{a_2}$ and $\overline{a_1}\le \overline{a_2}$, with $\underline{a_1}\ne \underline{a_2}$ or $\overline{a_1}\ne \overline{a_2}$; so, given any $C\in U\left(\right[0,1\left]\right)$, $I(\underline{a_2},\overline{c})\le I(\underline{a_1},\overline{c})$ and $I(\overline{a_2},\underline{c})\le I(\overline{a_1},\underline{c})$, since I satisfies (I1). So, $\overline{\ddot{I}(A_2,C)}\le \overline{\ddot{I}(A_1,C)}$ and $\underline{\ddot{I}(A_2,C)}\le \underline{\ddot{I}(A_1,C)}$, respectively, since, by Remark 5, $\ddot{I}(A_i,A_j)=\left[I(\overline{a_i},\underline{a_j}),I(\underline{a_i},\overline{a_j})\right]$. Therefore, $\ddot{I}(A_2,C){\le }_{{}_{KM}}\ddot{I}(A_1,C)$. The proof of (II2) is analogous. (II3) Since I satisfies (I3), $I(0,f_{A_2}\left({\lambda }_{a_2}\right))=1,\forall {\lambda }_{a_2}\in [0,1]$, so

$$\begin{array}{c}\hfill \ddot{I}([0,0],A_2)=\left[\underset{0\le {\lambda }_{a_2}\le 1}{inf}\left\{I(0,f_{A_2}\left({\lambda }_{a_2}\right))\right\},\underset{0\le {\lambda }_{a_2}\le 1}{sup}\left\{I(0,f_{A_2}\left({\lambda }_{a_2}\right))\right\}\right]=[1,1].\end{array}$$

(II4) Since I satisfies (I4), then $I(f_{A_1}\left({\lambda }_{a_1}\right),1)=1,\forall {\lambda }_{a_1}\in [0,1]$. So,

$$\begin{array}{c}\hfill \ddot{I}(A_1,[1,1])=\left[\underset{0\le {\lambda }_{a_1}\le 1}{inf}\left\{I(f_{A_1}\left({\lambda }_{a_1}\right),1)\right\},\underset{0\le {\lambda }_{a_1}\le 1}{sup}\left\{I(f_{A_1}\left({\lambda }_{a_1}\right),1)\right\}\right]=[1,1].\end{array}$$

(II5) Since I satisfies (I5), $I(1,0)=0$. So, $\ddot{I}([1,1],[0,0])=[0,0]$. Therefore, $\ddot{I}$ is an interval fuzzy implication. □

Considering the order Kulisch–Miranker, the following result shows that the restricted operators preserve the fuzzy negations.

Proposition 10. 

Let $N:[0,1]\to [0,1]$ be a fuzzy negation. Then, $\ddot{N}$ is an interval fuzzy negation with respect to ${\le }_{{}_{KM}}$.

Proof. 

Indeed,

$$\begin{array}{ccc}\hfill \ddot{N}\left([0,0]\right)& =& [\underset{0\le \lambda \le 1}{inf}\left\{N\left(f_{[0,0]}\left(\lambda \right)\right)\right\},\underset{0\le \lambda \le 1}{sup}\left\{N\left(f_{[0,0]}\left(\lambda \right)\right)\right\}]\hfill \\ & =& [\underset{0\le \lambda \le 1}{inf}\left\{N\left(0\right)\right\},\underset{0\le \lambda \le 1}{sup}\left\{N\left(0\right)\right\}]=[1,1]\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \ddot{N}\left([1,1]\right)& =& [\underset{0\le \lambda \le 1}{inf}\left\{N\left(f_{[1,1]}\left(\lambda \right)\right)\right\},\underset{0\le \lambda \le 1}{sup}\left\{N\left(f_{[1,1]}\left(\lambda \right)\right)\right\}]\hfill \\ & =& [\underset{0\le \lambda \le 1}{inf}\left\{N\left(1\right)\right\},\underset{0\le \lambda \le 1}{sup}\left\{N\left(1\right)\right\}]=[0,0].\hfill \end{array}$$

Now, given $A_1,A_2\in U\left([0,1]\right)$ such that $A_1{\le }_{{}_{KM}}{A}_2$, we have that $\underline{a_1}\le \underline{a_2}$ and $\overline{a_1}\le \overline{a_2}$. So, $N\left(\overline{a_2}\right)\le N\left(\overline{a_1}\right)$ and $N\left(\underline{a_2}\right)\le N\left(\underline{a_1}\right)$, since N is antitonic. By Remark 3, $\underline{\ddot{N}\left(A_2\right)}\le \underline{\ddot{N}\left(A_1\right)}$ and $\overline{\ddot{N}\left(A_2\right)}\le \overline{\ddot{N}\left(A_1\right)}$. Therefore, $\ddot{N}\left(A_2\right){\le }_{{}_{KM}}\ddot{N}\left(A_1\right)$. We conclude that $\ddot{N}$ is an interval fuzzy negation with respect to ${\le }_{{}_{KM}}$. □

Proposition 11. 

Let $N:[0,1]\to [0,1]$ be a fuzzy negation. N is strong if and only if, $\ddot{N}$ is a strong interval fuzzy negation.

Proof. 

Indeed, by Remark 3, $\ddot{N}\left(A\right)=[N\left(\overline{a}\right),N\left(\underline{a}\right)],\forall A\in U\left([0,1]\right)$. So, since N is strong, for all $A_1\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \ddot{N}\left(\ddot{N}\left(A_1\right)\right)=[N\left(\overline{\ddot{N}\left(A_1\right)}\right),N\left(\underline{\ddot{N}\left(A_1\right)}\right)]=[N\left(N\left(\underline{a_1}\right)\right),N\left(N\left(\overline{a_1}\right)\right)]\stackrel{Nstrong}{=}[\underline{a_1},\overline{a_1}]=A_1.\end{array}$$

Conversely, for each $x\in [0,1]$, $x=f_{[x,x]}\left({\lambda }_x\right)$ for all ${\lambda }_x\in [0,1]$, so

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_x\le 1}{inf}\left\{N\left(f_{[x,x]}\left({\lambda }_x\right)\right)\right\}=N\left(x\right)=\underset{0\le {\lambda }_x\le 1}{sup}\left\{N\left(f_{[x,x]}\left({\lambda }_x\right)\right)\right\}\end{array}$$

and then $\ddot{N}\left([x,x]\right)=[N\left(x\right),N\left(x\right)]$. Following a similar reasoning, we have $\ddot{N}\left(\ddot{N}\left([x,x]\right)\right)=[N\left(N\left(x\right)\right),N\left(N\left(x\right)\right)]$ and, since $\ddot{N}$ is strong, we have that $N\left(N\right(x\left)\right)=x$. □

Considering any fuzzy continuous negation, the following proposition guarantees that contrapositive properties are preserved for this negation.

Proposition 12. 

Let N be a continuous fuzzy negation and I be a fuzzy implication. Then,

(i)

I satisfies (CP) with respect to N iff $\ddot{I}$ satisfies (I-CP) with respect to $\ddot{N}$;

(ii)

I satisfies (LCP) with respect to N iff $\ddot{I}$ satisfies (I-LCP) with respect to $\ddot{N}$;

(iii)

I satisfies (RCP) with respect to N iff $\ddot{I}$ satisfies (I-RCP) with respect to $\ddot{N}$.

Proof. 

(i) For all $A_1,A_2\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \left\{I(f_{\ddot{N}\left(A_2\right)}\left({\lambda }_{{}_{N_2}}\right),f_{\ddot{N}\left(A_1\right)}\left({\lambda }_{{}_{N_1}}\right))\mid {\lambda }_{{}_{N_i}}\in [0,1]\right\}=\left\{I(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\mid {\lambda }_{a_i}\in [0,1]\right\}.\end{array}$$

(3)

In fact, $(\subseteq )$ given $y\in \left\{I(f_{\ddot{N}\left(A_2\right)}\left({\lambda }_{{}_{N_2}}\right),f_{\ddot{N}\left(A_1\right)}\left({\lambda }_{{}_{N_1}}\right))\mid {\lambda }_{{}_{N_i}}\in [0,1]\right\}$, there are ${\tilde{\lambda }}_{{}_{N_1}},{\tilde{\lambda }}_{{}_{N_2}}\in [0,1]$ such that $y=I(f_{\ddot{N}\left(A_2\right)}\left({\tilde{\lambda }}_{{}_{N_2}}\right),f_{\ddot{N}\left(A_1\right)}\left({\tilde{\lambda }}_{{}_{N_1}}\right))$. By Remark 3, $\ddot{N}\left(A_i\right)=[N\left(\overline{a_i}\right),N\left(\underline{a_i}\right)]$ for $i\in \{1,2\}$, so, since N is continuous and $f_{\ddot{N}\left(A_i\right)}\left({\tilde{\lambda }}_{{}_{N_i}}\right)\in \ddot{N}\left(A_i\right)$, there exists ${\dot{\lambda }}_{a_i}\in [0,1]$ such that $f_{\ddot{N}\left(A_i\right)}\left({\tilde{\lambda }}_{{}_{N_i}}\right)=N\left(f_{A_i}\left({\dot{\lambda }}_{a_i}\right)\right)$ for $i\in \{1,2\}$. Thus, $y=I(N\left(f_{A_2}\left({\dot{\lambda }}_{a_2}\right)\right),N\left(f_{A_1}\left({\dot{\lambda }}_{a_1}\right)\right))=I(f_{A_1}\left({\dot{\lambda }}_{a_1}\right),f_{A_2}\left({\dot{\lambda }}_{a_2}\right))$, since I satisfies (CP) with respect to N. So, $y\in \left\{I(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\mid {\lambda }_{a_i}\in [0,1]\right\}$. Now, $(\supseteq )$ given $y\in \left\{I(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\mid {\lambda }_{a_i}\in [0,1]\right\}$, there exists ${\tilde{\lambda }}_{a_1},{\tilde{\lambda }}_{a_2}\in [0,1]$ such that $y=I(f_{A_1}\left({\tilde{\lambda }}_{a_1}\right),f_{A_2}\left({\tilde{\lambda }}_{a_2}\right))$. Since I satisfies (CP) with respect to N, $y=I(N\left(f_{A_2}\left({\tilde{\lambda }}_{a_2}\right)\right),N\left(f_{A_1}\left({\tilde{\lambda }}_{a_1}\right)\right))$. So, since $N\left(f_{A_i}\left({\tilde{\lambda }}_{a_i}\right)\right)\in \ddot{N}\left(A_i\right)$, there exists ${\dot{\lambda }}_{{}_{N_i}}\in [0,1]$ such that $f_{\ddot{N}\left(A_i\right)}\left({\dot{\lambda }}_{{}_{N_i}}\right)=N\left(f_{A_i}\left({\tilde{\lambda }}_{a_i}\right)\right)$ for $i\in \{1,2\}$. Therefore, $y=I(f_{\ddot{N}\left(A_2\right)}\left({\dot{\lambda }}_{{}_{N_2}}\right),f_{\ddot{N}\left(A_1\right)}\left({\dot{\lambda }}_{{}_{N_1}}\right))\in \left\{I(f_{\ddot{N}\left(A_2\right)}\left({\lambda }_{{}_{N_2}}\right),f_{\ddot{N}\left(A_1\right)}\left({\lambda }_{{}_{N_1}}\right))\mid {\lambda }_{{}_{N_i}}\in [0,1]\right\}$. We conclude by Equation (3) that $\ddot{I}(\ddot{N}\left(A_2\right),\ddot{N}\left(A_1\right))=\ddot{I}(A_1,A_2)$. Conversely, for each $z\in [0,1]$, we have that $z=f_{[z,z]}\left({\lambda }_z\right)$ for all ${\lambda }_z\in [0,1]$ and $\ddot{N}\left([z,z]\right)=[N\left(z\right),N\left(z\right)]$. Then, for all $x,y\in [0,1]$, we have

$$\begin{array}{c}\hfill I(f_{[x,x]}\left({\lambda }_x\right),f_{[y,y]}\left({\lambda }_y\right))=I(x,y),\end{array}$$

and

$$\begin{array}{c}\hfill I(f_{\ddot{N}\left([y,y]\right)}\left({\lambda }_{{\ddot{N}}_y}\right),f_{\ddot{N}\left([x,x]\right)}\left({\lambda }_{{\ddot{N}}_x}\right))=I(f_{\left[N\right(y),N(y\left)\right]}\left({\lambda }_{{\ddot{N}}_y}\right),f_{\left[N\right(x),N(x\left)\right]}\left({\lambda }_{{\ddot{N}}_x}\right))=I(N\left(y\right),N\left(x\right));\end{array}$$

so, since $\ddot{I}(\ddot{N}\left([y,y]\right),\ddot{N}\left([x,x]\right))=\ddot{I}([x,x],[y,y])$,

$$\begin{array}{c}\hfill \left[I\right(N\left(y\right),N\left(x\right)),I(N\left(y\right),N\left(x\right)\left)\right]=\left[I\right(x,y),I(x,y\left)\right],\end{array}$$

so $I\left(N\right(y),N(x\left)\right)=I(x,y)$.

(ii) For all $A_1,A_2\in U\left([0,1]\right)$,

$$\begin{array}{c}\hfill \left\{I(f_{\ddot{N}\left(A_2\right)}\left({\lambda }_{{}_{N_2}}\right),f_{A_1}\left({\lambda }_{a_1}\right))\mid {\lambda }_{{}_{N_2}},{\lambda }_{a_1}\in [0,1]\right\}\stackrel{(\ast )}{=}\left\{I(f_{\ddot{N}\left(A_1\right)}\left({\lambda }_{{}_{N_1}}\right),f_{A_2}\left({\lambda }_{a_2}\right))\mid {\lambda }_{{}_{N_1}},{\lambda }_{a_2}\in [0,1]\right\}.\end{array}$$

In fact, for $(\subseteq )$, given $y\in \left\{I(f_{\ddot{N}\left(A_2\right)}\left({\lambda }_{{}_{N_2}}\right),f_{A_1}\left({\lambda }_{a_1}\right))\mid {\lambda }_{{}_{N_2}},{\lambda }_{a_1}\in [0,1]\right\}$, there are ${\tilde{\lambda }}_{{}_{N_2}},{\tilde{\lambda }}_{a_1}\in [0,1]$ such that $y=I(f_{\ddot{N}\left(A_2\right)}\left({\tilde{\lambda }}_{{}_{N_2}}\right),f_{A_1}\left({\tilde{\lambda }}_{a_1}\right))$. By Remark 3, $\ddot{N}\left(A_2\right)=[N\left(\overline{a_2}\right),N\left(\underline{a_2}\right)]$; so, since N is continuous and $f_{\ddot{N}\left(A_2\right)}\left({\tilde{\lambda }}_{{}_{N_2}}\right)\in \ddot{N}\left(A_2\right)$, there exists ${\dot{\lambda }}_{a_2}\in [0,1]$ such that $f_{\ddot{N}\left(A_2\right)}\left({\tilde{\lambda }}_{{}_{N_2}}\right)=N\left(f_{A_2}\left({\dot{\lambda }}_{a_2}\right)\right)$. Thus, $y=I(N\left(f_{A_2}\left({\dot{\lambda }}_{a_2}\right)\right),f_{A_1}\left({\tilde{\lambda }}_{a_1}\right))=I(N\left(f_{A_1}\left({\tilde{\lambda }}_{a_1}\right)\right),f_{A_2}\left({\dot{\lambda }}_{a_2}\right))$, since I satisfies (LCP) with respect to N. Now, since $N\left(f_{A_1}\left({\tilde{\lambda }}_{a_1}\right)\right)\in \ddot{N}\left(A_1\right)$, there exists ${\dot{\lambda }}_{{}_{N_1}}\in [0,1]$ such that $N\left(f_{A_1}\left({\tilde{\lambda }}_{a_1}\right)\right)=f_{\ddot{N}\left(A_1\right)}\left({\dot{\lambda }}_{{}_{N_1}}\right)$; so, $y=I(f_{\ddot{N}\left(A_1\right)}\left({\dot{\lambda }}_{{}_{N_1}}\right),f_{A_2}\left({\dot{\lambda }}_{a_2}\right))\in \left\{I(f_{\ddot{N}\left(A_1\right)}\left({\lambda }_{{}_{N_1}}\right),f_{A_2}\left({\lambda }_{a_2}\right))\mid {\lambda }_{{}_{N_1}},{\lambda }_{a_2}\in [0,1]\right\}$. The opposite direction, $(\supseteq )$, follows analogously. Conversely, since, for each $z\in [0,1]$, $z=f_{[z,z]}\left({\lambda }_z\right)$ for all ${\lambda }_z\in [0,1]$ and $\ddot{N}\left([z,z]\right)=[N\left(z\right),N\left(z\right)]$, then, for all $x,y\in [0,1]$, we have

$$\begin{array}{c}\hfill I(f_{\ddot{N}\left([x,x]\right)}\left({\lambda }_{{\ddot{N}}_x}\right),f_{[y,y]}\left({\lambda }_y\right))=I(f_{\left[N\right(x),N(x\left)\right]}\left({\lambda }_{{\ddot{N}}_x}\right),f_{[y,y]}\left({\lambda }_y\right))=I(N\left(x\right),y),\end{array}$$

and

$$\begin{array}{c}\hfill I(f_{\ddot{N}\left([y,y]\right)}\left({\lambda }_{{\ddot{N}}_y}\right),f_{[x,x]}\left({\lambda }_x\right))=I(f_{\left[N\right(y),N(y\left)\right]}\left({\lambda }_{{\ddot{N}}_y}\right),f_{[x,x]}\left({\lambda }_x\right))=I(N\left(y\right),x);\end{array}$$

so, since $\ddot{I}(\ddot{N}\left([y,y]\right),[x,x])=\ddot{I}(\ddot{N}\left([x,x]\right),[y,y])$,

$$\begin{array}{c}\hfill \left[I\right(N\left(y\right),x),I(N\left(y\right),x\left)\right]=\left[I\right(N\left(x\right),y),I(N\left(x\right),y\left)\right],\end{array}$$

so $I\left(N\right(y),x)=I\left(N\right(x),y)$.

(iii) It follows similarly to the previous item. □

The t-norm is also preserved by the constrained interval operators; see the proposition below:

Proposition 13. 

Let T be a t-norm. Then, $\ddot{T}$ is an interval t-norm with respect to ${\le }_{{}_{KM}}$.

Proof. 

The symmetry and associativity are straightforward from Propositions 3$\left(i\right)$ and 6, respectively. Now, to show that $\ddot{T}$ is increasing, consider $A_1,A_2\in U\left([0,1]\right)$ such that $A_1{\le }_{{}_{KM}}{A}_2$; then, $\underline{a_1}\le \underline{a_2}$ and $\overline{a_1}\le \overline{a_2}$. So, for all $C\in U\left(\right[0,1\left]\right)$, by the monotonicity of the t-norm, $T(\underline{a_1},\underline{c})\le T(\underline{a_2},\underline{c})$ and $T(\overline{a_1},\overline{c})\le T(\overline{a_2},\overline{c})$. So, by Proposition 1, we have that $\underline{\ddot{T}(A_1,C)}\le \underline{\ddot{T}(A_2,C)}$ and $\overline{\ddot{T}(A_1,C)}\le \overline{\ddot{T}(A_2,C)}$, respectively. Therefore, $\ddot{T}(A_1,C){\le }_{{}_{KM}}\ddot{T}(A_2,C)$. From symmetry, $\ddot{T}(C,A_1){\le }_{{}_{KM}}\ddot{T}(C,A_2)$ whenever $A_1{\le }_{{}_{KM}}{A}_2$. Finally, by the boundary condition of T, $T(f_{A_1}\left({\lambda }_{a_1}\right),1)=f_A\left({\lambda }_{a_1}\right)$ for all ${\lambda }_a\in [0,1]$, so $\ddot{T}(A_1,[1,1])=A_1$. □

The following section will provide a study with respect to Moore order similar to the one performed with respect to the Kulisch–Miranker order, analyzing their similarities and differences.

3.2. Properties of Constraint Interval Fuzzy Connectives with Respect to Order ⊴

This section addresses the main properties of the constrained interval fuzzy operators with respect to the order of Moore [16], defined below.

Definition 14. 

Moore order is defined by

$$\begin{array}{c}\hfill A_1⊴A_2\iff A_1=A_{2}or\overline{a_1}\le \underline{a_2},\end{array}$$

(4)

$\forall A_1,A_2\in U\left([0,1]\right).$

Proposition 14. 

Given $A_1,A_2\in U\left([0,1]\right)$, the following are equivalent:

(i)

$A_1⊴A_2$;

(ii)

$A_1=A_2$ or $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$;

(iii)

$A_1=A_2$ or $f_{A_1}^{\prime }\left({\lambda }_{a_1}\right)\le f_{A_2}^{\prime }\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$.

Proof. 

(i) ⇔ (ii) Indeed, given $A_1,A_2\in U\left([0,1]\right)$ such that $A_1⊴A_2$, then $A_1=A_2$ or $\overline{a_1}\le \underline{a_2}$. So, it is enough to check the case where $\overline{a_1}\le \underline{a_2}$, since the other follows straight. Therefore, since $f_{A_1}\left({\lambda }_{a_1}\right)\le \overline{a_1}$ and $\underline{a_2}\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$, then $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$. Conversely, if $A_1=A_2$, then, by definition, $A_1⊴A_2$. Now, if $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$, then, in particular, $\overline{a_1}=f_{A_1}\left(1\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_2}\in [0,1]$, so $\overline{a_1}\le f_{A_2}\left(0\right)=\underline{a_2}$; thus, $A_1⊴A_2$.

(ii) ⇔ (iii) Straightforward, once $f_A^{\prime }\left({\lambda }_a\right)=f_A(1-{\lambda }_a)$. □

The following proposition ensures that the order property is preserved by its constrained interval operator.

Proposition 15. 

Let g be a binary function on $[0,1]$. If g satisfies (OP), then $\ddot{g}$ satisfies (I-OP) with respect to ⊴.

Proof. 

Given $A_1,A_2\in U\left([0,1]\right)$ such that $A_1⊴A_2$, we have $A_1=A_2$ or $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$. If $A_1=A_2=A$, then $f_{A_1}\left(\lambda \right)=f_{A_2}\left(\lambda \right)=f_A\left(\lambda \right)$ for all $\lambda \in [0,1]$. So, by Definition 12, $\ddot{g}(A,A)=[1,1]$ since g satisfies (OP). Now, if $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$, then $g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))=1$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$ since g satisfies (OP). Therefore, $\ddot{g}(A_1,A_2)=[1,1]$. Conversely, let $A_1,A_2\in U\left([0,1]\right)$ such that $\ddot{g}(A_1,A_2)=[1,1]$; thus,

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_{a_1},{\lambda }_{a_2}\le 1}{inf}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\right\}=\underset{0\le {\lambda }_{a_1},{\lambda }_{a_2}\le 1}{sup}\left\{g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))\right\}=1.\end{array}$$

So, $g(f_{A_1}\left({\lambda }_{a_1}\right),f_{A_2}\left({\lambda }_{a_2}\right))=1$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$; therefore, $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$ since g satisfies (OP). By Proposition 14, we conclude that $A_1⊴A_2$. □

Unlike Proposition 9, considering Moore’s order, the fuzzy implications are not preserved by constrained interval operators; see the following proposition:

Proposition 16. 

There are fuzzy implications I such that $\ddot{I}$ is not an interval fuzzy implication with respect to ⊴.

Proof. 

Consider the Łukasiewicz implication $I_{LK}$ and the intervals $A_1=\left[\frac{1}{2},\frac{1}{2}\right]$, $A_2=\left[\frac{1}{2},1\right]$ and $A_3=\left[0,\frac{1}{2}\right]$ in $U\left(\right[0,1\left]\right)$. In this case, $A_1⊴A_2$ since $\overline{a_1}=\underline{a_2}$; however, $\ddot{I}(A_2,A_3)⋬\ddot{I}(A_1,A_3)$. Indeed, by Remark 5, $\ddot{I}(A_i,A_j)=\widehat{I}(A_i,A_j)=\left[I(\overline{a_i},\underline{a_j}),I(\underline{a_i},\overline{a_j})\right]$ whenever $A_i\ne A_j$ with $i\ne j$, so $I(\overline{a_1},\underline{a_3})=I(\frac{1}{2},0)=\frac{1}{2}$, $I(\underline{a_1},\overline{a_3})=I(\frac{1}{2},\frac{1}{2})=1$, $I(\overline{a_2},\underline{a_3})=I(1,0)=0$, and $I(\underline{a_2},\overline{a_3})=I(\frac{1}{2},\frac{1}{2})=1$. Therefore,

$$\begin{array}{c}\hfill \ddot{I}(A_2,A_3)=\left[0,1\right]\ne \left[\frac{1}{2},1\right]=\ddot{I}(A_1,A_3)\end{array}$$

and

$$\begin{array}{c}\hfill \overline{\ddot{I}(A_2,A_3)}=1\nleqq \frac{1}{2}=\underline{\ddot{I}(A_1,A_3)}.\end{array}$$

So, $\ddot{I}(A_2,A_3)⋬\ddot{I}(A_1,A_3)$, i.e., $\ddot{I}$ does not satisfy (II1). □

The following proposition shows that, analogous to Proposition 10 with respect to the order of Kulisch–Miranker, the fuzzy negations are also extended to their respective constrained interval operator with respect to Moore’s order.

Proposition 17. 

Let $N:[0,1]\to [0,1]$ be a fuzzy negation. Then, $\ddot{N}$ is an interval fuzzy negation with respect to ⊴.

Proof. 

Indeed, analogous to Proposition 10, $\ddot{N}\left([0,0]\right)=[1,1]$ and $\ddot{N}\left([1,1]\right)=[0,0]$. Now, consider $A_1⊴A_2$ for some $A_1,A_2\in U\left([0,1]\right)$. Then, $A_1=A_2$ or $f_{A_1}\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$. The first case is trivial. Now, if $f_A\left({\lambda }_{a_1}\right)\le f_{A_2}\left({\lambda }_{a_2}\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$, then $N\left(f_{A_2}\left({\lambda }_{a_2}\right)\right)\le N\left(f_A\left({\lambda }_{a_1}\right)\right)$ for all ${\lambda }_{a_1},{\lambda }_{a_2}\in [0,1]$ since N is decreasing. So, in particular, $N\left(\underline{a_2}\right)\le N\left(\overline{a_1}\right)$, i.e.,

$$\begin{array}{c}\hfill \underset{0\le {\lambda }_{a_2}\le 1}{sup}\left\{N\left(f_{A_2}\left({\lambda }_{a_2}\right)\right)\right\}\le \underset{0\le {\lambda }_{a_1}\le 1}{inf}\left\{N\left(f_A\left({\lambda }_{a_1}\right)\right)\right\}.\end{array}$$

Therefore, by Definition 14, $\ddot{N}\left(A_2\right)⊴\ddot{N}\left(A_1\right)$. □

Proposition 18. 

Let $N:[0,1]\to [0,1]$ be a fuzzy negation. N is strong if and only if $\ddot{N}$ is a strong interval fuzzy negation.

Proof. 

Analogous to Proposition 11. □

The contrapositive properties are again preserved for any fuzzy negation.

Proposition 19. 

Let N be a fuzzy negation and I be a fuzzy implication. Then,

(i)

I satisfies (CP) with respect to N iff $\ddot{I}$ satisfies (I-CP) with respect to $\ddot{N}$;

(ii)

I satisfies (LCP) with respect to N iff $\ddot{I}$ satisfies (I-LCP) with respect to $\ddot{N}$;

(iii)

I satisfies (RCP) with respect to N iff $\ddot{I}$ satisfies (I-RCP) with respect to $\ddot{N}$.

Proof. 

The proof follows analogous to Proposition 12. □

Unlike Proposition 13, t-norms are not preserved by the constrained interval operator considering the order ⊴.

Proposition 20. 

There are t-norms T such that $\ddot{T}$ is not an interval t-norm with respect to ⊴.

Proof. 

Consider a t-norm T such that $T(x,y)<T(x,z)$ whenever $x\ne 0$ and $y<z$. Take $A_1$ and $A_2$ in $U\left(\right[0,1\left]\right)$ as not degenerate such that $\overline{a_1}=\underline{a_2}\ne 0$; in this case, $A_1⊴A_2$. Thus, for all $C\in U\left(\right[0,1\left]\right)$ that are not degenerate, we have that $T(\underline{a_2},\underline{c})<T(\overline{a_1},\overline{c})$ since $\underline{c}<\overline{c}$. So,

$$\begin{array}{c}\hfill \overline{\ddot{T}(A_1,C)}\nleqq \underline{\ddot{T}(A_2,C)},\end{array}$$

(5)

since, by Proposition 1, $\ddot{T}(A_i,C)=[T(\underline{a_i},\underline{c}),T(\overline{a_i},\overline{c})]$. And, also, since $A_1$ is not degenerate, then $\underline{a_1}<\underline{a_2}$, so $T(\underline{a_1},\underline{c})<T(\underline{a_2},\underline{c})$. Thus,

$$\begin{array}{c}\hfill \ddot{T}(A_1,C)\ne \ddot{T}(A_2,C).\end{array}$$

(6)

We conclude, by Equations (5) and (6), that $\ddot{T}(A_1,C)⋬\ddot{T}(A_2,C)$. Therefore, $\ddot{T}$ is not increasing. □

It is easy to find in the literature examples of t-norms such that $T(x,y)<T(x,z)$ whenever $x\ne 0$ and $y<z$; for example, the product t-norm, $T_P(x,y)=xy$. Conversely, the t-norm $T_{LK}(x,y)=max(0,x+y-1)$ (Łukasiewicz’s t-norm) does not satisfy this condition, since if $y<z$ and $x+z<1$, then $x+y-1<x+z-1<0$, so $T_{LK}(x,y)=T_{LK}(x,z)=0$.

4. Final Remarks

A well-known problem in interval analysis is that, when there are variable dependencies, the resulting width of the interval is overestimated. This overestimation occurs because each of the two occurrences of one variable is treated as an independent variable.

In order to solve the problem of overestimation, using a reasoning similar to Lodwick’s [28], we have defined the constrained interval fuzzy operators by using the constrained intervals. By Definition 12, we can see that, when there are two or more occurrences of the same interval, it is traversed at the same level. With this, it is possible to state that there is no loss of information; see Remark 6. In view of what was presented, a new notion of correction, called constrained interval correctness, was suggested, in which it is as effective as Moore’s correction and less demanding.

The main objective of this paper is to introduce this new approach to interval-valued fuzzy logic, in which it is possible to soften the problem of overestimation when there are dependencies of variables, and in such a way so that the information of interval connectives is maintained.

In addition, the operators were investigated in terms of correction, algebraic properties, and orders. Furthermore, the authors also proposed a comparative study between the constrained interval operators $\ddot{g}$ and the best interval representation $\widehat{g}$, in which it was shown that $\ddot{g}\subseteq \widehat{g}$. Thus, it is possible to conclude that the operator $\ddot{g}$ improves the overestimation problem and, of the properties studied, it only did not satisfy the order property in relation to Kulisch–Miranker order.