Study of Two Families of Generalized Yager’s Implications for Describing the Structure of Generalized (h,e)-Implications

Abstract

In this study, we analyze the family of generalized $(h,e)$-implications. We determine when this family fulfills some of the main additional properties of fuzzy implication functions and we obtain a representation theorem that describes the structure of a generalized $(h,e)$-implication in terms of two families of fuzzy implication functions. These two families can be interpreted as particular cases of the $(f,g)$ and $(g,f)$-implications, which are two families of fuzzy implication functions that generalize the well-known f and g-generated implications proposed by Yager through a generalization of the internal factors x and $\frac{1}{x}$, respectively. The behavior and additional properties of these two families are also studied in detail.

1. Introduction

In fuzzy logic, fuzzy implication functions play a key role as operators that generalize the classical implications in crisp logic. They are functions $I:{[0,1]}^2\to [0,1]$ fulfilling some boundary properties in order to coincide with the classical implication in ${\{0,1\}}^2$ and some monotonicity properties. These operators are important for both theory and applications. For instance, in the same way classical implications are used in inference schemas such as modus ponens, modus tollens, etc., fuzzy implication functions play a similar role in the generalization of these schemas, which use fuzzy statements whose value is in [0, 1] instead of being in $\{0,1\}$. In addition, fuzzy implication functions also play a different role in other applications such as fuzzy mathematical morphology, fuzzy control or data analysis [1]. Depending on the context and the proper rule and its behavior, various fuzzy implication functions with different properties can be adequate. This fact has motivated the study and definition of many families of fuzzy implication functions, and more than 100 families have been defined until now. For further information, consult the surveys [1,2,3] and the books [4,5].

Although apparently there is a huge amount of fuzzy implication functions available, these families can present intersection or even coincide with others already known [6]. For this reason, it is of the utmost importance to study the additional properties that the operators of certain families satisfy and to provide an axiomatic characterization of the new operators in the literature in order to find its possible relation with respect to those already known. In this respect, the characterization of several families of fuzzy implication functions have already been achieved: $(S,N)$-implications with a continuous negation [7], R-implications obtained from left-continuous t-norms [5,8,9], some $QL$-implications [10], Yager’s implications [11], h-implications [12], probabilistic and survival S-implications [6]; among others [13,14].

In this paper, we are interested in the study of the family of $(h,e)$-implications. These fuzzy implication functions were defined for the first time in [15] under the motivation of generalizing h-implications to a new family of functions satisfying the property $I(e,y)=y$ for all $y\in [0,1]$ and some $e\in (0,1)$. The family of $(h,e)$-implications has presented very interesting properties such as the controlled increase in the second variable by the parameter e or that they constitute a whole family of fuzzy implication functions fulfilling the exchange principle, but not the law of importation for any t-norm T [16]. Additionally, they have been applied on image processing for edge detection (see [17]), obtaining good results with respect to other families of fuzzy implication functions. Although the properties of $(h,e)$-implications were analyzed for the first time in [15], in [18] it was pointed out that a more general definition was possible. Therefore, our first contribution is to adapt all the existing results to this more general definition and to provide all the proofs.

In [12], it was proved that h-implications were characterized by the fact that their structure is determined by two Yager’s implications through the threshold horizontal method [12], more specifically, the structure of an h-implication is like an adequately scaled f-implication whenever $y\le e$ and like an adequately scaled g-implication whenever $y>e$. In this paper, as a second and main contribution, we provide a similar result for $(h,e)$-implications but, in our case, the structure of $(h,e)$-implications is described in terms of two new families of fuzzy implication functions that are generalizations of Yager’s implications, called $(f,g)$ and $(g,f)$-implications [19].

To date, there have been several proposals of generalizations of Yager’s implications by considering different approaches: generalizing the inner factors x and $\frac{1}{x}$, in the expression of f- and g-generated implications, respectively, [19,20,21,22], considering a different internal function from the product [23,24,25]; among others [26,27]. The families of $(f,g)$ and $(g,f)$-implications are of the first kind since they consider the internal factor x as a continuous and strictly decreasing function $g:[0,1]\to [0,+\infty ]$ with $g\left(0\right)=0$ and $\frac{1}{x}$ as a strictly decreasing function $f:[0,1]\to [0,+\infty ]$ with $f\left(0\right)=+\infty$. This approach is different from all the other generalization proposals, and then the interest in these two families is twofold: to describe the structure of $(h,e)$-implications and to study these families in order to compare its properties with other generalizations considered in the literature. Although these two families were preliminary studied in [19], the results were provided without any proof and some of them were partially erroneous. Our third contribution is to provide the proofs of all the results and to rectify the mistakes in some statements.

The paper is structured as follows. In Section 2, we list essential results on fuzzy implication functions. In Section 3, the family of generalized $(h,e)$-implications and its main properties are recalled, and a proof of the representation theorem is given. In Section 4, the families of $(f,g)$ and $(g,f)$-implications are presented and their properties are studied. Finally, in Section 5, we gather concluding remarks and we outline some future work.

2. Preliminaries

In this section, we recollect some concepts and results that are useful throughout the paper. First, we recall the definition of fuzzy negation, t-norm and t-conorm, which are generalizations of the classical negation, conjunction and disjunction, respectively, to fuzzy logic.

Definition 1

([8]). A decreasing function $N:[0,1]\to [0,1]$ is called a fuzzy negation if $N\left(0\right)=1$ and $N\left(1\right)=0$.

Definition 2

([28]). A triangular norm or t-norm is a binary function $T:{[0,1]}^2\to [0,1]$ that is commutative, associative, increasing in both variables and 1 is its neutral element. A triangular conorm or t-conorm is a binary function $S:{[0,1]}^2\to [0,1]$ that is commutative, associative, increasing in both variables and 0 is its neutral element.

Next, we introduce the definition of fuzzy implication function.

Definition 3

([5,8]). A binary operator $I:{[0,1]}^2\to [0,1]$ is said to be afuzzy implication functionif it satisfies:

(I1)

$I(x,z)\ge I(y,z)$ when $x\le y$, for all $z\in [0,1]$.

(I2)

$I(x,y)\le I(x,z)$ when $y\le z$, for all $x\in [0,1]$.

(I3)

$I(0,0)=I(1,1)=1$ and $I(1,0)=0.$

From this definition, it is straightforward to prove that a fuzzy implication function is such that $I(x,1)=1$ and $I(0,x)=1$ for all $x\in [0,1]$. However, the 0-horizontal section $I(x,0)$ and 1-vertical section $I(1,x)$ for $x\in (0,1)$ cannot be deduced from the definition.

Since the definition of fuzzy implication function is rather general, additional properties on these operators are considered. Although there exist many additional properties, we recall here the ones that have been more studied.

• The identity principle

  $$\left(\mathbf{IP}\right)I(x,x)=1,x\in [0,1].$$
• The ordering property

  $$\left(\mathbf{OP}\right)I(x,y)=1\iff x\le y,x,y\in [0,1].$$
• The exchange principle

  $$\left(\mathbf{EP}\right)I(x,I(y,z\left)\right)=I(y,I(x,z\left)\right),x,y,z\in [0,1].$$
• The law of importation with respect to a t-norm T

  $$\left({\mathbf{LI}}_{\mathbf{T}}\right)I(T(x,y),z)=I(x,I(y,z)),x,y,z\in [0,1].$$
• The left neutrality principle

  $$\left(\mathbf{NP}\right)I(1,y)=y,y\in [0,1].$$
• The left neutrality principle with $e\in (0,1)$

  $$\left({\mathbf{NP}}_{\mathbf{e}}\right)I(e,y)=y,y\in [0,1].$$
• The contrapositive symmetry with respect to a fuzzy negation N,

  $$\left(\mathbf{CP}\right(\mathbf{N}\left)\right)I(x,y)=I\left(N\right(y),N(x\left)\right),x,y\in [0,1].$$

From the values $I(x,0)$ of a fuzzy implication function I, we obtain a fuzzy negation called the natural negation of I.

Definition 4

([5]). Let I be a fuzzy implication function. The function $N_I$ defined by $N_I\left(x\right)=I(x,0)$ for all $x\in [0,1]$ is called the natural negation of I.

The wide definition of fuzzy implication function allows the existence of many families with different additional properties. Depending on the construction method of a family, we can distinguish between three classes: those obtained by the combination of other logical operators such as fuzzy negations, t-norms or t-conorms; those generated by univaluated functions in the interval [0, 1]; and those generated from other fuzzy implication functions. In [29], the reader can find an overview of construction methods of fuzzy implication functions.

Among those fuzzy implication functions whose definition is based on the use generator functions, we can highlight Yager’s implications, called f and g-generated implications. These fuzzy implication functions are generated from additive generators of continuous Archimedean t-norms and t-conorms, respectively.

Definition 5

([5], Definition 3.1.1). Let $f:[0,1]\to [0,+\infty ]$ be a strictly decreasing and continuous function with $f\left(1\right)=0$. The function $I_f:{[0,1]}^2\to [0,1]$ defined by

$$I_f(x,y)=f^{-1}(x·f\left(y\right)),x,y\in [0,1],$$

with the understanding $0·(+\infty )=0$, is called an f-generated implication. The function f itself is called an f-generator.

Definition 6

([5], Definition 3.2.3). Let $g:[0,1]\to [0,+\infty ]$ be a strictly increasing and continuous function with $g\left(0\right)=0$. The function $I_g:{[0,1]}^2\to [0,1]$ defined by

$$I_g(x,y)=g^{(-1)}\left(\frac{1}{x}g\left(y\right)\right),x,y\in [0,1],$$

with the understanding $\frac{1}{0}=+\infty$ and $+\infty ·0=+\infty$ is called a g-generated implication, where the function $g^{(-1)}$ is the pseudo-inverse of g given by

$$g^{(-1)}\left(x\right)=\left\{\begin{array}{cc}{g}^{-1}\left(x\right)\hfill & ifx\in [0,g(1\left)\right],\hfill \\ 1\hfill & ifx\in \left[g\right(1),+\infty ].\hfill \end{array}\right.$$

In this case, g is called the g-generator of the fuzzy implication function I defined as above.

Another family of fuzzy implication functions of this kind is the h-implications, introduced in [15] following the idea behind the definition of Yager’s implications, but considering additive generators of representable uninorms.

Definition 7

([15], Definition 7). Fix an $e\in (0,1)$ and let $h:[0,1]\to [-\infty ,+\infty ]$ be a strictly increasing and continuous function with $h\left(0\right)=-\infty$, $h\left(e\right)=0$ and $h\left(1\right)=+\infty$. The function $I^h:{[0,1]}^2\to [0,1]$ defined by

$$I^h(x,y)=\left\{\begin{array}{cc}1\hfill & ifx=0,\hfill \\ h^{-1}(x·h\left(y\right))\hfill & ifx>0\mathit{and}y\le e,\hfill \\ h^{-1}(\frac{1}{x}·h\left(y\right))\hfill & ifx>0\mathit{and}y>e.\hfill \end{array}\right.$$

is called an h-implication. The function h itself is called an h-generator (with respect to e) of the fuzzy implication function $I^h$ defined as above.

These three families of fuzzy implication functions were characterized in [11,12]. Moreover, it was proved that the characterization of h-implications could be derived from the characterization of Yager’s implications since the structure of an h-implication is given by an f and a g-generated implication through the horizontal threshold generation method. The horizontal threshold generation method is a construction method that generates a fuzzy implication function from two given ones and it consists in an appropriate scaling of the second variable of the two fuzzy implication functions.

Theorem 1

([12], Theorem 3). Let $I_1$, $I_2$ be two fuzzy implication functions and $e\in (0,1)$. Then, the binary function $I_{I_1-I_2}:{[0,1]}^2\to [0,1]$, called the e-horizontal threshold generated implication from $I_1$ and $I_2$, defined as

$$I_{I_1-I_2}(x,y)=\left\{\begin{array}{cc}1\hfill & ifx=0,\hfill \\ e·I_1\left(x,\frac{y}{e}\right)\hfill & ifx>0\mathit{and}y\le e,\hfill \\ e+(1-e)·I_2\left(x,\frac{y-e}{1-e}\right)\hfill & ifx>0\mathit{and}y>e,\hfill \end{array}\right.$$

is a fuzzy implication function.

3. Generalized $(\mathit{h},\mathit{e})$-Implications

In [15], a new class of fuzzy implications was presented, the family of $(h,e)$-implications. The motivation behind its definition was modifying the h-implications towards fulfilling the property $\left({\mathbf{NP}}_{\mathbf{e}}\right)$. Indeed, the family of h-implications presented an unexpected behavior. Most of the families of fuzzy implication functions generated from uninorms such as RU-implications [30], which are generalizations of R-implications, tend to satisfy $\left({\mathbf{NP}}_{\mathbf{e}}\right)$ instead of $\left(\mathbf{NP}\right)$. Being h-generators the additive generators of representable uninorms, it would be expected that this family of fuzzy implications functions would satisfy $\left({\mathbf{NP}}_{\mathbf{e}}\right)$ instead of $\left(\mathbf{NP}\right)$. This is achieved with a slight modification in the definition leading to the so-called $(h,e)$-implications.

Although $(h,e)$-implications were first defined in [15], in [18] it was pointed out that a more general definition was possible. The latter is the one we recall here.

Definition 8

([18], Definition 11). Fix an $e\in (0,1)$ and let $h:[0,1]\to [-\infty ,+\infty ]$ be a strictly increasing and continuous function with $h\left(e\right)=0$ and $h\left(1\right)=+\infty$. The function $I^{h_g,e}:{[0,1]}^2\to [0,1]$ defined by

$$I^{h_g,e}(x,y)=\left\{\begin{array}{ccc}1\hfill & if& x=0,\hfill \\ h^{(-1)}\left(\frac{x}{e}h\left(y\right)\right)\hfill & if& x>0,y\le e,\hfill \\ h^{-1}\left(\frac{e}{x}h\left(y\right)\right)\hfill & if& x>0,y>e,\hfill \end{array}\right.$$

where the function $h^{(-1)}$ is the pseudo-inverse of h given by

$$h^{(-1)}\left(x\right)=\left\{\begin{array}{ccc}{h}^{-1}\left(x\right)\hfill & if& x\in \left[h\right(0),+\infty ),\hfill \\ 0\hfill & if& x\in (-\infty ,h(0\left)\right),\hfill \end{array}\right.$$

is called a generalized $(h,e)$-implication. The function h itself is called a generalized h-generator (with respect to e) of the implication function $I^{h_g,e}$ defined as above.

Although the above definition was proposed in [18] and the properties that were already studied for $(h,e)$-implications in [15] were reconsidered for this more general definition, the results in [18] were announced without any proof. Therefore, hereafter we recall those results, but provide the corresponding proof. Having said this, the next proposition ensures that generalized $(h,e)$-implications are indeed fuzzy implication functions.

Proposition 1

([18], Proposition 9). If h is a generalized h-generator with respect to a fixed $e\in (0,1)$, then $I^{h_g,e}$ is a fuzzy implication function.

Proof. 

-

Let $x_1,x_2,y\in [0,1]$ with $x_1<x_2$. Since h is strictly increasing, we have $h\left(x_1\right)<h\left(x_2\right)$ and it holds that $h^{(-1)}$ is an increasing function. Now, we have to distinguish three cases:

-

If $x_1=0$ then $I^{h_g,e}(0,y)=1\ge I^{h_g,e}(x_2,y)$.

-

If $x_1\ne 0$ and $y\le e$ then $h\left(y\right)\le 0$ and $\frac{x_1}{e}h\left(y\right)\ge \frac{x_2}{e}h\left(y\right)$. Consequently,

$$I^{h_g,e}(x_1,y)=h^{(-1)}\left(\frac{x_1}{e}h\left(y\right)\right)\ge h^{(-1)}\left(\frac{x_2}{e}h\left(y\right)\right)=I^{h_g,e}(x_2,y).$$

-

If $x_1\ne 0$ and $y>e$ then $h\left(y\right)>0$ and $\frac{e}{x_1}h\left(y\right)>\frac{e}{x_2}h\left(y\right)$. Therefore,

$$I^{h_g,e}(x_1,y)=h^{(-1)}\left(\frac{e}{x_1}h\left(y\right)\right)\ge h^{(-1)}\left(\frac{e}{x_2}h\left(y\right)\right)=I^{h_g,e}(x_2,y).$$

-

Let $x,y_1,y_2\in [0,1]$ with $y_1<y_2$. Then, similarly to the previous item, we have $h\left(y_1\right)<h\left(y_2\right)$ and we have to consider four different cases:

-

If $x=0$ then $I^{h_g,e}(0,y_1)=1=I^{h_g,e}(0,y_2)$.

-

If $x\ne 0$ and $y_1<y_2\le e$ we have that $h\left(y_1\right)<h\left(y_2\right)\le h\left(e\right)=0$ and

$$I^{h_g,e}(x,y_1)=h^{(-1)}\left(\frac{x}{e}h\left(y_1\right)\right)\le h^{(-1)}\left(\frac{x}{e}h\left(y_2\right)\right)=I^{h_g,e}(x,y_1).$$

-

If $x\ne 0$ and $y_1\le e<y_2$, we have that $h\left(y_1\right)\le 0<h\left(y_2\right)$ so $\frac{x}{e}h\left(y_1\right)\le 0<\frac{e}{x}h\left(y_2\right)$ and we get that

$$I^{h_g,e}(x,y_1)=h^{(-1)}\left(\frac{x}{e}h\left(y_1\right)\right)\le h^{(-1)}\left(0\right)=e<h^{-1}\left(\frac{e}{x}h\left(y_2\right)\right)=I^{h_g,e}(x,y_2).$$

-

If $x\ne 0$ and $e<y_1<y_2$ then we have that $0<h\left(y_1\right)<h\left(y_2\right)$. Thus,

$$I^{h_g,e}(x,y_1)=h^{-1}\left(\frac{e}{x}h\left(y_1\right)\right)\le h^{-1}\left(\frac{e}{x}h\left(y_2\right)\right)=I^{h_g,e}(x,y_2).$$

-

Finally, $I^{h_g,e}$ satisfies the boundary conditions since

-

$I^{h_g,e}(0,0)=1$ by construction.

-

$I^{h_g,e}(1,1)=h^{-1}\left(\frac{1}{e}h\left(1\right)\right)=h^{-1}(+\infty )=1$.

-

$I^{h_g,e}(1,0)=h^{(-1)}\left(\frac{1}{e}h\left(0\right)\right)=0$.

□

Moreover, like in the case of h-implications, the generator of a generalized $(h,e)$-implication is unique up to a positive multiplicative constant piecewise.

Proposition 2

([18], Proposition 10). Let $h_1,h_2:[0,1]\to [-\infty ,+\infty ]$ be two generalized h-generators with respect to a fixed $e\in (0,1)$. Then, the following statements are equivalent:

(i)

$I^{h_{1,g},e}=I^{h_{2,g},e}$.

(ii)

There exist constants $k,c\in (0,+\infty )$ such that

$$h_2\left(x\right)=\left\{\begin{array}{cc}k·h_1\left(x\right)\hfill & ifx\in [0,e),\hfill \\ c·h_1\left(x\right)\hfill & ifx\in [e,1].\hfill \end{array}\right.$$

Proof. 

The proof is identical to the proof of Theorem 17 in [15]. □

Hereunder, we recall some of the basic properties of generalized $(h,e)$-implications. First of all, the next result studies the natural negation. Notice that in contrast with $(h,e)$-implications, the presence of $h^{(-1)}$ in the more general definition implies that the behavior of the natural negation depends on the value of h in zero.

Proposition 3

([18], Proposition 11). Let h be a generalized h-generator. Then,

-

If $h\left(0\right)=-\infty$, then the natural negation $N_{I^{h_g,e}}$ is the Gödel negation or least negation $N_{D_1}$, given by

$$N_{D_1}\left(x\right)=\left\{\begin{array}{cc}1\hfill & ifx=0,\hfill \\ 0\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

-

If $h\left(0\right)>-\infty$, then the natural negation $N_{I^{h_g,e}}$ is given by

$$N_{I^{h_g,e}}\left(x\right)=I^{h_g,e}(x,0)=\left\{\begin{array}{ccc}1\hfill & if& x=0,\hfill \\ h^{-1}\left(\frac{x}{e}h\left(0\right)\right)\hfill & if& x\le e,\hfill \\ 0\hfill & if& x>e.\hfill \end{array}\right.$$

Proof. 

Let h be a generalized h-generator. Then,

-

If $h\left(0\right)=-\infty$, then $h^{(-1)}=h^{-1}$ and for every $x\in [0,1]$ we get

$$N_{I^{h_g,e}}\left(x\right)=I^{h_g,e}(x,0)=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}& x=0,\hfill \\ h^{-1}\left(-\infty \right)\hfill & \mathrm{if}& x\in (0,1],\hfill \end{array}\right.=\left\{\begin{array}{cc}1\hfill & \mathrm{if}x=0,\hfill \\ 0\hfill & \mathrm{if}x\in (0,1].\hfill \end{array}\right.$$

-

If $h\left(0\right)>-\infty$ then we have

$$\begin{array}{ccc}\hfill N_{I^{h_g,e}}\left(x\right)& =& I^{h_g,e}(x,0)=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}& x=0,\hfill \\ h^{-1}\left(\frac{x}{e}h\left(0\right)\right)\hfill & \mathrm{if}& \frac{x}{e}h\left(0\right)\ge h\left(0\right),\hfill \\ 0\hfill & \mathrm{if}& \frac{x}{e}h\left(0\right)<h\left(0\right),\hfill \end{array}\right.\hfill \\ & =& \left\{\begin{array}{cc}1\hfill & \mathrm{if}x=0,\hfill \\ h^{-1}\left(\frac{x}{e}h\left(0\right)\right)\hfill & \mathrm{if}x\le e,\hfill \\ 0\hfill & \mathrm{if}x>e.\hfill \end{array}\right.\hfill \end{array}$$

□

The following proposition studies when generalized $(h,e)$-implications fulfill the additional properties of fuzzy implication functions considered in this paper.

Theorem 2

([18], Theorem 22). Let h be a generalized h-generator and $e\in (0,1)$. The following properties hold:

(i)

$I^{h_g,e}(x,y)\le e$ if and only if $(x>0\mathrm{and}y\le e)$. Moreover, $I^{h_g,e}(x,e)=e$ for all $x>0$.

(ii)

$I^{h_g,e}$ satisfies(EP)if and only if $h\left(0\right)=-\infty$.

(iii)

$I^{h_g,e}(x,y)=1$ if and only if $x=0$ or $y=1$. Thus, $I^{h_g,e}$ does not satisfy either(OP)or(IP).

(iv)

$I^{h_g,e}$ is continuous, except at the points $(0,y)$ with $y\le e$.

(v)

$I^{h_g,e}$ satisfies $\left({\mathbf{NP}}_{\mathbf{e}}\right)$, but does not satisfy(NP).

(vi)

$I^{h_g,e}$ does not satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to any t-norm T.

(vii)

$I^{h_g,e}$ does not satisfy (CP(N)) with any fuzzy negation N.

Proof. 

(i)

It is clear that if $x>0$ and $y\le e$, then $I^{h_g,e}(x,y)=h^{(-1)}\left(\frac{x}{e}h\left(y\right)\right)\le e$ since $h\left(y\right)\le 0$. Otherwise, if $x=0$, $I^{h_g,e}(0,y)=1$ for all $y\in [0,1]$ and if $x>0$ and $y>e$, $I^{h_g,e}(x,y)=h^{-1}\left(\frac{e}{x}h\left(y\right)\right)>e$ because $h\left(y\right)>0$. Moreover, we have that $I^{h_g,e}(x,e)=h^{(-1)}\left(\frac{x}{e}h\left(e\right)\right)=h^{(-1)}\left(0\right)=e$ for all $x>0$.

(ii)

Assume that $I^{h_g,e}$ satisfies (EP). Now, let us consider $h\left(0\right)>-\infty$ and we will get a contradiction. On the one hand, if $0<x_0<e^2$, we have

$$I^{h_g,e}(x_0,I^{h_g,e}(1,0))=I^{h_g,e}(x_0,0)=h^{(-1)}\left(\frac{x_0}{e}h\left(0\right)\right)=h^{-1}\left(\frac{x_0}{e}h\left(0\right)\right).$$

On the other hand, let us compute $I^{h_g,e}(1,I^{h_g,e}(x_0,0))$. First, we have

$$x_0<e\Rightarrow \frac{x_0}{e}h\left(0\right)>h\left(0\right)\Rightarrow I^{h_g,e}(x_0,0)=h^{-1}\left(\frac{x_0}{e}h\left(0\right)\right).$$

Now, since $x_0<e^2$ and by item (i) $I^{h_g,e}(x_0,0)\le e$ we get that

$$I^{h_g,e}(1,I^{h_g,e}(x_0,0))=I^{h_g,e}\left(1,h^{-1}\left(\frac{x_0}{e}h\left(0\right)\right)\right)=h^{(-1)}\left(\frac{x_0}{e^2}h\left(0\right)\right)=h^{-1}\left(\frac{x_0}{e^2}h\left(0\right)\right).$$

Since we have that $h^{-1}$ is strictly increasing in $\left[h\right(0),+\infty ]$, we get

$$h^{-1}\left(\frac{x_0}{e^2}h\left(0\right)\right)<h^{-1}\left(\frac{x_0}{e}h\left(0\right)\right).$$

Hence, $I^{h_g,e}(1,I^{h_g,e}(x_0,0))<I^{h_g,e}(x_0,I^{h_g,e}(1,0))$, in contradiction with the fact that $I^{h_g,e}$ satisfies (EP).

For the reverse implication, if $h\left(0\right)=-\infty$, we know that in this case $h^{(-1)}=h^{-1}$ and $N_{I^{h_g,e}}=N_{D_1}$. For any $x,y,z\in [0,1]$, let us distinguish five cases:

-

If $x=0$, then for all $y,z\in [0,1]$ we have

$$I^{h_g,e}(0,I^{h_g,e}(y,z))=1=I^{h_g,e}(y,1)=I^{h_g,e}(y,I^{h_g,e}(0,z)).$$

-

If $y=0$ for all $x,z\in [0,1]$ we have

$$I^{h_g,e}(x,I^{h_g,e}(0,z))=I^{h_g,e}(x,1)=1=I^{h_g,e}(0,I^{h_g,e}(x,z)).$$

-

If $x\ne 0$, $y\ne 0$ and $z=0$, we obtain

$$\begin{array}{ccc}\hfill I^{h_g,e}(x,I^{h_g,e}(y,0))& =& I^{h_g,e}(x,N_{I^{h_g,e}}\left(y\right))=I^{h_g,e}(x,0)=N_{I^{h_g,e}}\left(x\right)=0\hfill \\ & =& N_{I^{h_g,e}}\left(y\right)=I^{h_g,e}(y,0)=I^{h_g,e}(y,N_{I^{h_g,e}}\left(x\right))\hfill \\ & =& I^{h_g,e}(y,I^{h_g,e}(x,0)).\hfill \end{array}$$

-

If $x\ne 0$, $y\ne 0$ and $z\le e$, by the item (i), $I^{h_g,e}(y,z)\le e$ and $I^{h_g,e}(x,z)\le e$ and consequently

$$I^{h_g,e}(x,I^{h_g,e}(y,z))=I^{h_g,e}\left(x,h^{-1}\left(\frac{y}{e}h\left(z\right)\right)\right)=h^{-1}\left(\frac{xy}{e^2}h\left(z\right)\right).$$

Similarly,

$$I^{h_g,e}(y,I^{h_g,e}(x,z))=I^{h_g,e}\left(y,h^{-1}\left(\frac{x}{e}h\left(z\right)\right)\right)=h^{-1}\left(\frac{xy}{e^2}h\left(z\right)\right).$$

-

Finally, if $x\ne 0$, $y\ne 0$ and $e<z\le 1$, then again by (i) we have $I^{h_g,e}(y,z)>e$ and $I^{h_g,e}(x,z)>e$ and thus

$$I^{h_g,e}(x,I^{h_g,e}(y,z))=I^{h_g,e}\left(x,h^{-1}\left(\frac{e}{y}h\left(z\right)\right)\right)=h^{-1}\left(\frac{e^2}{xy}h\left(z\right)\right),$$

$$I^{h_g,e}(y,I^{h_g,e}(x,z))=I^{h_g,e}\left(y,h^{-1}\left(\frac{e}{x}h\left(z\right)\right)\right)=h^{-1}\left(\frac{e^2}{xy}h\left(z\right)\right).$$

(iii)

It is obvious that if $x=0$ or $y=1$, $I^{h_g,e}(x,y)=1$ since $I^{h_g,e}$ is a fuzzy implication function. If $y\le e$, by item (i), $I^{h_g,e}(x,y)\le e<1$ and if $e<y<1$, $0<h\left(y\right)<+\infty$ and we have

$$I^{h_g,e}(x,y)=h^{-1}\left(\frac{e}{x}h\left(y\right)\right)<h^{-1}(+\infty )=1.$$

(iv)

By definition, the implication $I^{h_g,e}$ is continuous for all $(x,y)\in (0,1]\times [0,e)$ and for all $(x,y)\in (0,1]\times (e,1]$. Further, the vertical sections with a fixed $x>0$ are continuous since

$$I^{h_g,e}(x,e)=h^{(-1)}\left(\frac{x}{e}h\left(e\right)\right)=h^{(-1)}\left(0\right)=e,$$

and

$$\underset{y\to e^{-}}{lim}{I}^{h_g,e}(x,y)=\underset{y\to e^{-}}{lim}{h}^{(-1)}\left(\frac{x}{e}h\left(y\right)\right)=h^{(-1)}\left(0\right)=e,$$

$$\underset{y\to e^{+}}{lim}{I}^{h_g,e}(x,y)=\underset{y\to e^{+}}{lim}{h}^{-1}\left(\frac{e}{x}h\left(y\right)\right)=h^{-1}\left(0\right)=e.$$

On the other hand, the horizontal sections with $y>e$ are continuous since $I^{h_g,e}(0,y)=1$ and

$$\underset{x\to 0^{+}}{lim}{I}^{h_g,e}(x,y)=\underset{x\to 0^{+}}{lim}{h}^{-1}\left(\frac{e}{x}h\left(y\right)\right)=h^{-1}(+\infty )=1.$$

However, fixed $0<y\le e$, $-\infty <h\left(y\right)\le 0$ and we know that $I^{h_g,e}(0,y)=1$, but

$$\underset{x\to 0^{+}}{lim}{I}^{h_g,e}(x,y)=\underset{x\to 0^{+}}{lim}{h}^{(-1)}\left(\frac{x}{e}h\left(y\right)\right)=h^{(-1)}\left(0\right)=e,$$

thus $I^{h_g,e}$ horizontal sections with $y\le e$ are continuous except at the points $(0,y)$ with $0<y\le e$. Finally, by Proposition 3, $I^{h_g,e}$ is also not continuous at the point $(0,0)$. Now, applying ([5] Theorem A.0.4) adequately, we can prove that $I^{h_g,e}$ is continuous, except at the points $(0,y)$ with $y\le e$.

(v)

For all $y\in [0,1]$ we have that

$$I^{h_g,e}(e,y)=\left\{\begin{array}{ccc}{h}^{(-1)}\left(h\left(y\right)\right)\hfill & \mathrm{if}& y\le e,\hfill \\ h^{-1}\left(h\left(y\right)\right)\hfill & \mathrm{if}& y>e,\hfill \end{array}\right.=y.$$

Thus, $I^{h_g,e}$ satisfies $\left({\mathbf{NP}}_{\mathbf{e}}\right)$. On the other hand, for all $y>e$, we have

$$I^{h_g,e}(1,y)=y\iff h^{-1}\left(eh\left(y\right)\right)=y\iff eh\left(y\right)=h\left(y\right)\iff e=1.$$

Thus, $I^{h_g,e}$ does not satisfy (NP).

(vi)

Suppose that $I^{h_g,e}$ fulfills $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to a t-norm T, then we know that it also fulfills (EP), and by item (ii), $h\left(0\right)=-\infty$. Now, taking $x=y=1$ and $e<z<1$, since $T(1,1)=1$ we find that

$$I^{h_g,e}(T(1,1),z)=I^{h_g,e}(1,I^{h_g,e}(1,z))\iff h^{-1}\left(eh\left(z\right)\right)=h^{-1}\left(e^{2}h\left(z\right)\right)\iff e=1.$$

Thus, $I^{h_g,e}$ does not satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to any t-norm T.

(vii)

Suppose that $I^{h_g,e}$ satisfies (CP(N)) with a fuzzy negation N. So, we have $I^{h_g,e}(x,y)=I^{h_g,e}(N\left(y\right),N\left(x\right))$ for all $x,y\in [0,1]$. Taking $x=1$ and $y=e$, we know by item (i) that $I^{h_g,e}(1,e)=e$ and then

$$e=I^{h_g,e}(1,e)=I^{h_g,e}(N\left(e\right),N\left(1\right))=I^{h_g,e}(N\left(e\right),0)=N_{I^{h_g,e}}\circ N\left(e\right).$$

If $h\left(0\right)=-\infty$ then $N_{I^{h_g,e}}=N_{D_1}$ and we obtain a contradiction. If $h\left(0\right)>-\infty$ then

$$e=N_{I^{h_g,e}}\circ N\left(e\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}N\left(e\right)=0\hfill \\ h^{-1}\left(\frac{N\left(e\right)}{e}h\left(0\right)\right)\hfill & \mathrm{if}N\left(e\right)\le e,\hfill \\ 0\hfill & \mathrm{if}N\left(e\right)>e,\hfill \end{array}\right.$$

and the only feasible case is $N\left(e\right)\in (0,e]$ and then $N\left(e\right)h\left(0\right)=h\left(e\right)·e=0$, which is also a contradiction.

□

Perhaps one of the main properties of $(h,e)$-implications is given by (i) in the previous theorem. It reflects that these operators have a controlled increase with respect to the second variable produced by the insertion of the parameter e, as we can graphically see in Figure 1. Observe that the fuzzy implication functions generated by the horizontal threshold method in Theorem 1 had a similar property, so it is intuitive to think that generalized $(h,e)$-implications are related in some way with this method. Furthermore, notice that the family of generalized $(h,e)$-implications provides an example of fuzzy implication functions that do not satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to any t-norm T and yet satisfy (EP) when $h\left(0\right)=-\infty$, providing another argument of the fact that $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ is stronger than (EP) [16].

Example 1.

Let us consider the generalized h-generator

$$h\left(x\right)=\left\{\begin{array}{cc}ln\left(\frac{x}{e}\right)\hfill & ifx\le e,\hfill \\ -ln\left(\frac{1-x}{1-e}\right)\hfill & ifx>e,\hfill \end{array}\right.$$

with $e\in (0,1)$. The corresponding $(h,e)$-implication is the following one

$$I^{h_1,e}(x,y)=\left\{\begin{array}{cc}1\hfill & ifx=0,\hfill \\ e{\left(\frac{y}{e}\right)}^{\frac{x}{e}}\hfill & ifx>0,y\le e,\hfill \\ 1-(1-e)·{\left(\frac{1-y}{1-e}\right)}^{\frac{e}{x}}\hfill & ifx>0,y>e.\hfill \end{array}\right.$$

(1)

The plot of this $(h,e)$-implication for different values of the parameter e can be seen in Figure 1.

Figure 1. Plot of the $(h,e)$-implication given by Equation (1) for different values of e. (a) e = 0.25, (b) e = 0.5, (c) e = 0.75.

Figure 1. Plot of the $(h,e)$-implication given by Equation (1) for different values of e. (a) e = 0.25, (b) e = 0.5, (c) e = 0.75.

Representation Theorem

Although there is no axiomatic characterization of $(h,e)$-implications, in [19] a representation theorem for this family was presented without the corresponding proof. In this article, we provide a proof to that result and we adjust it to the case of generalized $(h,e)$-implications.

Theorem 3.

Let $I:{[0,1]}^2\to [0,1]$ be a binary function and $e\in (0,1)$. Then, I is a generalized $(h,e)$-implication with respect to e if and only if there exist an f-generator and a g-generator with $g\left(1\right)=+\infty$ such that I is given by

$$I(x,y)=\left\{\begin{array}{ccc}1\hfill & if& x=0,\hfill \\ e·f^{(-1)}\left(\frac{x}{e}·f\left(\frac{y}{e}\right)\right)\hfill & if& x>0,y\le e,\hfill \\ e+(1-e)·g^{-1}\left(\frac{e}{x}·g\left(\frac{y-e}{1-e}\right)\right)\hfill & if& x>0,y>e.\hfill \end{array}\right.$$

(2)

Moreover, in this case generators h, f and g are related in the following way:

$$f\left(x\right)=-h(e·x)\mathrm{for}\mathrm{all}x\in [0,1],$$

$$g\left(x\right)=h(e+(1-e)·x)\mathrm{for}\mathrm{all}x\in [0,1],$$

$$h\left(x\right)=\left\{\begin{array}{ccc}-f\left(\frac{x}{e}\right)\hfill & \mathrm{if}& x\le e,\hfill \\ g\left(\frac{x-e}{1-e}\right)\hfill & \mathrm{if}& x>e.\hfill \end{array}\right.$$

Proof. 

Let I be a generalized $(h,e)$-implication with respect to e. We know that h is a continuous and strictly increasing function with $h\left(e\right)=0$ and $h\left(1\right)=+\infty$. First of all, note that $f\left(x\right)=-h\left(ex\right)$ and $g\left(x\right)=h(e+(1-e\left)x\right)$ are f and g-generators, respectively, since f is a continuous and strictly decreasing function with $f\left(1\right)=-h\left(e\right)=0$ and g is a continuous and strictly increasing function with $g\left(0\right)=h\left(e\right)=0$. Note that since $h^{-1}$ is well-defined on $\left[h\right(0),+\infty )$ with $h\left(0\right)<0$ then we have for all $x\in [0,+\infty )$ that

$$f^{(-1)}\left(x\right)=\frac{h^{(-1)}(-x)}{e},g^{-1}\left(x\right)=\frac{h^{-1}\left(x\right)-e}{1-e}.$$

We will split the proof in two cases:

-

If $x>0$ and $y\le e$, then

$$e{f}^{(-1)}\left(\frac{x}{e}f\left(\frac{y}{e}\right)\right)=e{f}^{(-1)}\left(-\frac{x}{e}h\left(y\right)\right)=h^{(-1)}\left(\frac{x}{e}h\left(y\right)\right)=I(x,y).$$

-

If $x>0$ and $y>e$ then

$$\begin{array}{ccc}\hfill e+(1-e)·g^{-1}\left(\frac{e}{x}g\left(\frac{y-e}{1-e}\right)\right)& =& e+(1-e)·\frac{h^{-1}\left(\frac{e}{x}h\left(e+(1-e)\frac{y-e}{1-e}\right)\right)-e}{1-e}\hfill \\ & =& h^{-1}\left(\frac{e}{x}h\left(y\right)\right)=I(x,y).\hfill \end{array}$$

For the reverse implication, let us consider f and g-generators such that I is given by Equation (2). Consider

$$h\left(x\right)=\left\{\begin{array}{ccc}-f\left(\frac{x}{e}\right)\hfill & \mathrm{if}& x\le e,\\ g\left(\frac{x-e}{1-e}\right)\hfill & \mathrm{if}& x>e.\end{array}\right.$$

This function is continuous, strictly increasing, $h\left(e\right)=-f\left(1\right)=0$ and $h\left(1\right)=g\left(1\right)=+\infty$. Now, let us prove that $I=I^{h_g,e}$. Notice that

$$\begin{array}{ccc}\hfill h^{(-1)}\left(x\right)& =& \left\{\begin{array}{ccc}{h}^{-1}\left(x\right)\hfill & \mathrm{if}& x\in \left[h\right(0),+\infty ),\\ 0\hfill & \mathrm{if}& x\in (-\infty ,h(0\left)\right),\end{array}\right.\hfill \\ & =& \left\{\begin{array}{ccc}0\hfill & \mathrm{if}& x\in (-\infty ,-f(0\left)\right),\\ e·f^{-1}(-x)\hfill & \mathrm{if}& x\in [-f(0),0],\\ e+(1-e)·g^{-1}\left(x\right)\hfill & \mathrm{if}& x\in (0,+\infty ).\end{array}\right.\hfill \end{array}$$

Then, studying again two cases we have that

-

If $x>0$ and $y\le e$ then

$$I^{h_g,e}(x,y)=h^{(-1)}\left(\frac{x}{e}h\left(y\right)\right)=h^{(-1)}\left(-\frac{x}{e}f\left(\frac{y}{e}\right)\right)=e·f^{(-1)}\left(\frac{x}{e}f\left(\frac{y}{e}\right)\right)=I(x,y).$$

-

If $x>0$ and $y>e$ then

$$\begin{array}{ccc}\hfill I^{h_g,e}(x,y)& =& h^{-1}\left(\frac{e}{x}h\left(y\right)\right)=h^{-1}\left(\frac{e}{x}g\left(\frac{y-e}{1-e}\right)\right)=e+(1-e)·g^{-1}\left(\frac{e}{x}g\left(\frac{y-e}{1-e}\right)\right)\hfill \\ & =& I(x,y).\hfill \end{array}$$

□

The next example provides the construction of an $(h,e)$-implication by using the threshold horizontal method given an f-generator and g-generator with $g\left(1\right)=+\infty$.

Example 2.

Take, for instance, $e=\frac{1}{2}$ and the subsequent f and g-generators

$$f\left(x\right)=-ln\left(\frac{x}{2-x}\right),g\left(x\right)=ln\left(\frac{1+x}{1-x}\right).$$

Then, it is easy to check that the following functions are fuzzy implication functions

$$I_1(x,y)=f^{(-1)}\left(\frac{x}{e}f\left(y\right)\right)=\frac{2y^{2x}}{{(2-y)}^{2x}+y^{2x}},$$

$$I_2(x,y)=g^{-1}\left(\frac{e}{x}g\left(y\right)\right)=\frac{{(1+y)}^{\frac{1}{2x}}-{(1-y)}^{\frac{1}{2x}}}{{(1+y)}^{\frac{1}{2x}}+{(1-y)}^{\frac{1}{2x}}}.$$

Then, an $(h,e)$-implication is constructed from $I_1$ and $I_2$ by using the threshold horizontal method as Theorem 3 shows. Concretely, the h-generator corresponds to

$$h\left(x\right)=ln\left(\frac{x}{1-x}\right).$$

We can see the construction method graphically in Figure 2.

Although Theorem 3 gives a useful description of the family of generalized $(h,e)$-implications, it is not an axiomatic characterization of this family, i.e., a characterization in terms of their own properties. For providing such results, a deeper study of this family is needed.

Let us recall that the characterization of h-implications presented in [12] was written in terms of the threshold horizontal method, in particular h-implications are characterized by the fact that they are generated by an f-implication and a g-implication through the horizontal threshold method. In this case, the axiomatic characterization was not provided, but it can be easily obtained by using the characterizations of Yager’s implications presented in [11]. For the case of generalized $(h,e)$-implications, it is straightforward to prove that if we consider an f-generator, the function $I_{f,e}:{[0,1]}^2\to [0,1]$ defined by

$$I_{f,e}(x,y)=f^{(-1)}\left(\frac{x}{e}f\left(y\right)\right),x,y\in [0,1],$$

with the understanding $+\infty ·0=0$ is a fuzzy implication function and if we consider a g-generator with $g\left(1\right)=+\infty$, the function $I_{g,e}:{[0,1]}^2\to [0,1]$ defined by

$$I_{g,e}(x,y)=g^{(-1)}\left(\frac{e}{x}g\left(y\right)\right),x,y\in [0,1],$$

with the understanding $\frac{1}{0}=+\infty$ and $+\infty ·0=+\infty$ is also a fuzzy implication function. Notice that Theorem 3 discloses that generalized $(h,e)$-implications are also characterized by the fact that they can be generated through the horizontal threshold method by the two new families of fuzzy implication functions just introduced. Therefore, in order to obtain a characterization of $(h,e)$-implications, we have to study and characterize the two families of fuzzy implication functions defined.

In particular, $I_{f,e}$ and $I_{g,e}$ are fuzzy implication functions that belong to two families which are generalizations of the well-known Yager’s implications. In the next section, we deeply study these two families.

4. Generalized Yager’s Implications

In [19], two new families of fuzzy implication functions, called the $(f,g)$ and $(g,f)$-generated implications, were defined as a generalization of the well-known Yager’s f and g-generated implications, respectively. In the definition of the f-generated implications, one can consider the function x as a particular case of a family of strictly increasing and continuous functions defined as $g:[0,1]\to [0,+\infty ]$ such that $g\left(0\right)=0$. The same happens to the role of $\frac{1}{x}$ as a concrete case of a continuous, strictly decreasing function $f:[0,1]\to [0,+\infty ]$ such that $f\left(0\right)=+\infty$. In this section, we recall the definitions and properties of these two families of fuzzy implication functions published in [19], providing the corresponding proofs and rectifying some wrongly stated results.

4.1. Generalization of f-Generated Implications

First, we will study a generalization of the f-generated implications, generalizing the function x in its definition as a strictly increasing function $g:[0,1]\to [0,+\infty ]$ with $g\left(0\right)=0$.

Definition 9.

Let $f:[0,1]\to [0,+\infty ]$ be a strictly decreasing and continuous function with $f\left(1\right)=0$ and $g:[0,1]\to [0,+\infty ]$ be a continuous and strictly increasing function with $g\left(0\right)=0$. The function $I_{f,g}:{[0,1]}^2\to [0,1]$ defined by

$$I_{f,g}(x,y)=f^{(-1)}\left(g\left(x\right)f\left(y\right)\right),x,y\in [0,1],$$

(3)

with the understanding $0·(+\infty )=0$, is called an $(f,g)$-generated operation.

Remark 1.

An initial difference between the family of f-generated implications and its generalization is that we need to consider the pseudo-inverse of f. This is because when $f\left(0\right)<+\infty$, $g\left(x\right)·f\left(y\right)$ may be bigger than the initial value $f\left(0\right)$. Nevertheless, notice that Equation (3) can also be written in the following form without explicitly using the pseudo-inverse of f:

$$I(x,y)=f^{-1}\left(min\left(f\left(x\right)g\left(y\right),f\left(0\right)\right)\right),x,y\in [0,1].$$

(4)

An $(f,g)$-generated operation may not fulfill all the conditions in Definition 3, and then it is not always a fuzzy implication function.

Theorem 4.

An $(f,g)$-operation $I_{f,g}$ is a fuzzy implication function if and only if one of the following conditions hold:

(i)

$f\left(0\right)=+\infty$.

(ii)

$f\left(0\right)<+\infty$ and $g\left(1\right)\ge 1$.

Proof. 

First, we will consider that $I_{f,g}$ is a fuzzy implication function with $f\left(0\right)<+\infty$. In this case,

$$I_{f,g}(1,0)=f^{(-1)}\left(g\left(1\right)f\left(0\right)\right)=\left\{\begin{array}{ccc}{f}^{-1}\left(g\left(1\right)f\left(0\right)\right)\hfill & \mathrm{if}& g\left(1\right)<1,\\ \\ 0\hfill & \mathrm{if}& g\left(1\right)\ge 1.\end{array}\right.$$

Since $I_{f,g}$ is a fuzzy implication function, then it holds that $I_{f,g}(1,0)=0$ and hence, $g\left(1\right)\ge 1$.

Now, let us consider an $(f,g)$-operation satisfying (i) or (ii). The fact that $I_{f,g}$ is a fuzzy implication function can be seen from the following:

-

Let $x_1,x_2,y\in [0,1]$ with $x_1\le x_2$. Since g is strictly increasing, we have that $g\left(x_1\right)\le g\left(x_2\right)$. Now, since f is strictly decreasing, $f^{(-1)}$ is decreasing, and we find that

$$I_{f,g}(x_1,y)=f^{(-1)}\left(g\left(x_1\right)f\left(y\right)\right)\ge f^{(-1)}\left(g\left(x_2\right)f\left(y\right)\right)=I_{f,g}(x_2,y),$$

and $I_{f,g}$ satisfies (I1).

-

Consider $x,y_1,y_2\in [0,1]$ with $y_1\le y_2$ then, again by the strictly decreasing nature of f, $f^{(-1)}$ is decreasing, and hence, we have

$$\begin{array}{ccc}\hfill f\left(y_1\right)\ge f\left(y_2\right)& \Rightarrow & g\left(x\right)·f\left(y_1\right)\ge g\left(x\right)·f\left(y_2\right)\hfill \\ & \Rightarrow & f^{(-1)}(g\left(x\right)·f\left(y_1\right))\le f^{(-1)}(g\left(x\right)·f\left(y_2\right))\hfill \\ & \Rightarrow & I_{f,g}(x,y_1)\le I_{f,g}(x,y_2),\hfill \end{array}$$

and $I_{f,g}$ satisfies (I2).

-

$I_{f,g}(0,0)=f^{(-1)}\left(g\left(0\right)f\left(0\right)\right)=f^{(-1)}\left(0\right)=1$.

-

$I_{f,g}(1,1)=f^{(-1)}\left(g\left(1\right)f\left(1\right)\right)=f^{(-1)}\left(0\right)=1$.

-

$I_{f,g}(1,0)=f^{(-1)}\left(g\left(1\right)f\left(0\right)\right)$ and we have two cases. If $f\left(0\right)=+\infty$ then $I_{f,g}(1,0)=f^{-1}(+\infty )=0$. Otherwise, if $f\left(0\right)<+\infty$ and $g\left(1\right)\ge 1$ then, $f\left(0\right)g\left(1\right)\ge f\left(0\right)$ and $I_{f,g}(1,0)=0$.

□

Remark 2.

In [21], a similar approach to provide a generalization of Yager’s f-implications was considered. In this case, the authors consider the fuzzy implication function given by $I_{f,g}(x,y)=f^{(-1)}\left(g\left(x\right)f\left(y\right)\right)$ where f is an f-generator and $g:[0,1]\to [0,1]$ is an increasing function satisfying $g\left(0\right)=0$ and $g\left(1\right)=1$. In this case, they consider functions g, which are not necessarily continuous, but with $g\left(1\right)=1$. Our approach restricts to the case when g is continuous, but allows any value in $(0,+\infty )$ of $g\left(1\right)$ whenever $f\left(0\right)=+\infty$ and any value $g\left(1\right)\ge 1$ whenever $f\left(0\right)<+\infty$. Clearly, the two families intersect when we consider a continuous, strictly decreasing function g with $g\left(1\right)=1$. Moreover, by Remark 2.1 in [21], in this particular case the resulting $(f,g)$-implications are in fact ϕ-conjugated of f-generated implications with f generator given by $f\circ g^{-1}$ and $\phi =g$.

When an $(f,g)$-operation fulfills Definition 3, we will use the nomenclature $(f,g)$-implication and we will call an admissible pair of generators to the pair of functions $(f,g)$.

The next result shows that it is enough to consider the pairs $(f,g)$ of admissible generators such that $f\left(0\right)=+\infty$ or $f\left(0\right)=1$.

Proposition 4.

Let $I_{f,g}$ be a fuzzy implication function with $f\left(0\right)<+\infty$, then there exists a function $f_1$ with $f_1\left(0\right)=1$ such that $(f_1,g)$ is an admissible pair of generators and $I_{f,g}=I_{f_1,g}$.

Proof. 

Let $I_{f,g}$ be a fuzzy implication function with $f\left(0\right)<+\infty$ and consider $f_1\left(x\right)=\frac{f\left(x\right)}{f\left(0\right)}$. Then, $(f_1,g)$ is an admissible pair of generators with $f_1\left(0\right)=\frac{f\left(0\right)}{f\left(0\right)}=1$ and since $f_1^{-1}\left(x\right)=f^{-1}\left(xf\left(0\right)\right)$ then

$$\begin{array}{ccc}\hfill I_{f_1,g}(x,y)& =& f_1^{(-1)}\left(g\left(x\right)f_1\left(y\right)\right)=f_1^{(-1)}\left(g\left(x\right)\frac{f\left(y\right)}{f\left(0\right)}\right)=f_1^{-1}\left(min\left\{g\left(x\right)\frac{f\left(y\right)}{f\left(0\right)},f_1\left(0\right)\right\}\right)\hfill \\ & =& f^{-1}\left(min\left\{g\right(x\left)f\right(y),f(0\left)\right\}\right)=I_{f,g}(x,y).\hfill \end{array}$$

□

Then next proposition shows that the $(f,g)$-generated implications have non-trivial zero region for some choice of generators.

Proposition 5.

Let $(f,g)$ be an admissible pair of generators. Then, the following statements hold:

(i)

If $g\left(1\right)<f\left(0\right)=+\infty$, then $I_{f,g}(x,y)=0$ if and only if $y=0<x$.

(ii)

If $g\left(1\right)=f\left(0\right)=+\infty$, then $I_{f,g}(x,y)=0$ if and only if $y<x=1$ or $y=0<x$.

(iii)

If $f\left(0\right)<+\infty$, then $I_{f,g}(x,y)=0$ if and only if $g\left(x\right)\ge 1$ and $y\le f^{-1}\left(\frac{f\left(0\right)}{g\left(x\right)}\right)$.

Proof. 

(i)

Let us assume that $g\left(1\right)<f\left(0\right)=+\infty$ then $f^{(-1)}=f^{-1}$. Hence, for every $x,y\in [0,1]$, we find that

$$I_{f,g}(x,y)=f^{-1}\left(g\left(x\right)f\left(y\right)\right)=0\iff g\left(x\right)f\left(y\right)=f\left(0\right)=+\infty .$$

However, we know that $g\left(x\right)\le g\left(1\right)<+\infty$, then the only possibility is $f\left(y\right)=+\infty$ and $g\left(x\right)\ne 0$. Consequently, $y=0<x$.

(ii)

Again we have that $f^{(-1)}=f^{-1}$ and then,

$$I_{f,g}(x,y)=0\iff g\left(x\right)f\left(y\right)=+\infty .$$

Therefore, $g\left(x\right)=+\infty$ and $f\left(y\right)>0$ or, $g\left(x\right)>0$ and $f\left(y\right)=+\infty$. Hence, the results follows.

(iii)

Consider $x,y\in [0,1]$ then

$$I_{f,g}(x,y)=0\iff f^{(-1)}\left(g\left(x\right)f\left(y\right)\right)=0\iff g\left(x\right)f\left(y\right)\in [f\left(0\right),+\infty ).$$

Now, since f is strictly decreasing, $f\left(y\right)\le f\left(0\right)$ for all $y\in [0,1]$ and then necessarily $g\left(x\right)\ge 1$. Finally,

$$g\left(x\right)f\left(y\right)\ge f\left(0\right)\iff f\left(y\right)\ge \frac{f\left(0\right)}{g\left(x\right)}\iff y\le f^{-1}\left(\frac{f\left(0\right)}{g\left(x\right)}\right).$$

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On the other hand, the next proposition shows that the region where the $(f,g)$-generated implications take the value 1 is independent of their generators.

Proposition 6.

Let $(f,g)$ be an admissible pair of generators. Then $I_{f,g}(x,y)=1$ if and only if $x=0$ or $y=1$.

Proof. 

Let $(f,g)$ be an admissible pair of generators and $x,y\in [0,1]$. Then,

$$\begin{array}{ccc}\hfill I_{f,g}(x,y)=1& \iff & f^{(-1)}\left(g\left(x\right)f\left(y\right)\right)=1\iff g\left(x\right)f\left(y\right)=0\hfill \\ & \iff & g\left(x\right)=0\mathrm{or}f\left(y\right)=0\iff x=0\mathrm{or}y=1.\hfill \end{array}$$

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The determination of the one region obtained in the previous proposition is a property of fuzzy implication functions deeply studied in [31] where it is explained that the property $I(x,y)=1\iff x=0\mathrm{or}y=1$ is very important for the definition of strong equality indices. Consequently, $(f,g)$-implications could be used to generate strong equality indices. Furthermore, in [11], this property plays a crucial role in the characterization of f-generated implications.

From the previous proposition, the following result is straightforward.

Corollary 1.

Let $(f,g)$ be an admissible pair of generators. Then, the $(f,g)$-implication $I_{f,g}$ does not satisfy either(IP)or(OP).

The next proposition studies under which conditions (NP) is satisfied by the $(f,g)$-generated implications. This property is satisfied by many of the most well-known families and therefore, to determine when $(f,g)$-implications fulfill (NP) is a necessary step for forthcoming studies on the intersections of this family with other existing families.

Proposition 7.

Let $(f,g)$ be an admissible pair of generators. Then, $I_{f,g}$ satisfies(NP)if and only if $g\left(1\right)=1$.

Proof. 

Consider $(f,g)$ an admissible pair of generators. It is clear from the definition of fuzzy implication function that $I_{f,g}(1,0)=0$ and $I_{f,g}(1,1)=1$. Otherwise, since f is strictly decreasing and continuous with $f\left(1\right)=0$, we can choose $y\in (0,1)$ such that $g\left(1\right)f\left(y\right)<f\left(0\right)$ and then

$$I_{f,g}(1,y)=f^{(-1)}\left(g\left(1\right)f\left(y\right)\right)=y\iff g\left(1\right)f\left(y\right)=f\left(y\right)\iff g\left(1\right)=1.$$

□

The following result studies the natural negation of these fuzzy implication functions. The properties of the natural negation play an important role in many characterization results of fuzzy implication functions.

Proposition 8.

Let $(f,g)$ be an admissible pair of generators. Then, the following properties hold:

(i)

If $f\left(0\right)=+\infty$, then the natural negation $N_{I_{f,g}}$ is the Gödel or least negation $N_{D_1}$.

(ii)

If $f\left(0\right)<+\infty$, then the natural negation $N_{I_{f,g}}$ is given by

$$N_{I_{f,g}}(x,y)=\left\{\begin{array}{ccc}{f}^{-1}\left(g\left(x\right)f\left(0\right)\right)\hfill & \mathrm{if}& g\left(x\right)<1,\\ 0\hfill & \mathrm{if}& g\left(x\right)\ge 1.\end{array}\right.$$

(iii)

The natural negation $N_{I_{f,g}}$ is continuous and strictly increasing if and only if $f\left(0\right)<+\infty$ and $g\left(1\right)=1$.

Proof. 

(i)

If $f\left(0\right)=+\infty$, then $f^{(-1)}=f^{-1}$ and for every $x\in [0,1]$ we get

$$\begin{array}{ccc}\hfill N_{I_{f,g}}\left(x\right)& =& f^{-1}\left(g\left(x\right)f\left(0\right)\right)=\left\{\begin{array}{ccc}{f}^{-1}(+\infty )\hfill & \mathrm{if}& g\left(x\right)\ne 0,\\ f^{-1}\left(0\right)\hfill & \mathrm{if}& g\left(x\right)=0,\end{array}\right.=\left\{\begin{array}{ccc}0\hfill & \mathrm{if}& x>0,\\ 1\hfill & \mathrm{if}& x=0,\end{array}\right.\hfill \\ & =& N_{D_1}\left(x\right).\hfill \end{array}$$

(ii)

If $f\left(0\right)<+\infty$ then we have

$$\begin{array}{ccc}\hfill N_{I_{f,g}}\left(x\right)& =& f^{(-1)}\left(g\left(x\right)f\left(0\right)\right)=\left\{\begin{array}{ccc}{f}^{-1}\left(g\left(x\right)f\left(0\right)\right)\hfill & \mathrm{if}& g\left(x\right)f\left(0\right)<f\left(0\right),\\ 0\hfill & \mathrm{if}& g\left(x\right)f\left(0\right)\ge f\left(0\right),\end{array}\right.\hfill \\ & =& \left\{\begin{array}{ccc}{f}^{-1}\left(g\left(x\right)f\left(0\right)\right)\hfill & \mathrm{if}& g\left(x\right)<1,\\ 0\hfill & \mathrm{if}& g\left(x\right)\ge 1.\end{array}\right.\hfill \end{array}$$

(iii)

If $f\left(0\right)<+\infty$ and $g\left(1\right)=1$, it is straightforward from point (ii) that $N_{I_{f,g}}$ is a continuous function since it is the composition of real continuous functions. Consider $x_1<x_2$, by the strictly increasing nature of g, we have that $g\left(x_1\right)f\left(0\right)<g\left(x_2\right)f\left(0\right)$. Now, since f is strictly decreasing, $f^{-1}$ is strictly decreasing in $[0,f(0\left)\right]$, and we get that

$$N_{I_{f,g}}\left(x_1\right)=f^{-1}\left(g\left(x_1\right)f\left(0\right)\right)>f^{-1}\left(g\left(x_2\right)f\left(0\right)\right)=N_{I_{f,g}}\left(x_2\right).$$

Hence, $N_{I_{f,g}}$ is strictly decreasing. Reciprocally, items (i) and (ii) prove that the obtained natural negations are not strictly decreasing when $f\left(0\right)=+\infty$ or when $f\left(0\right)<+\infty$ and $g\left(1\right)>1$.

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At this point, we analyze the discontinuity points of the $(f,g)$-generated implications, which can be $(0,0)$ or $(1,1)$.

Proposition 9.

Let $(f,g)$ be an admissible pair of generators. Then $I_{f,g}$ is continuous everywhere except at point $(0,0)$ when $f\left(0\right)=+\infty$ or at point $(1,1)$ when $g\left(1\right)=+\infty$.

Proof. 

Let $(f,g)$ be an admissible pair of generators, then by definition $I_{f,g}$ is continuous at each $(x,y)\in {[0,1]}^2$ by being the composition of real continuous functions except for the cases when ($g\left(x\right)=0$ and $f\left(y\right)=+\infty$) or ($g\left(x\right)=+\infty$ or $f\left(y\right)$ =0 ), since in these situations we have considered the convention $0·(+\infty )=0$. These two situations correspond to the following two cases:

• If $x=y=0$ and $f\left(0\right)=+\infty$, then because of (i) in Proposition 8, the natural negation of $I_{f,g}$ is not continuous at $x=0$, and therefore $I_{f,g}$ is non-continuous at $(0,0)$.
• If $x=y=1$ and $g\left(1\right)=+\infty$, then

  $$I_{f,g}(1,y)=f^{(-1)}(+\infty ·f\left(y\right))=\left\{\begin{array}{ccc}{f}^{-1}\left(0\right)\hfill & \mathrm{if}& y=1,\\ 0\hfill & \mathrm{if}& y<1,\end{array}\right.=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}& y=1,\\ 0\hfill & \mathrm{if}& y<1.\end{array}\right.$$
  Hence, we have that

  $$\underset{y\to 1^{-}}{lim}{I}_{f,g}(1,y)=0\ne 1=I_{f,g}(1,1),$$

  and $I_{f,g}$ is not continuous at $(1,1)$.

□

Consequently, there are members of the family that are continuous in the whole domain, and therefore, feasible to be applied in several fields.

Finally, we present two results that determine completely when the $(f,g)$-generated implications satisfy (EP) or $\left({\mathbf{LI}}_{\mathbf{T}}\right)$.

Proposition 10.

Let $(f,g)$ be an admissible pair of generators. Then, the following statements are equivalent:

(i)

$I_{f,g}$ satisfies(EP).

(ii)

$f\left(0\right)=+\infty$ or ($f\left(0\right)<+\infty$ and $g\left(1\right)=1$).

Proof. 

Assume that $I_{f,g}$ satisfies (EP). Now, let us consider $f\left(0\right)<+\infty$ and $g\left(1\right)>1$ and get a contradiction. On the one hand, we have that

$$I_{f,g}(x,I_{f,g}(1,0))=I_{f,g}(x,0)=f^{(-1)}\left(g\left(x\right)f\left(0\right)\right).$$

Now, since g is strictly increasing and continuous with $g\left(0\right)=0$, we can find some $x_0\in (0,1)$ small enough such that

$$0<g\left(1\right)g\left(x_0\right)<1\mathrm{and}0<g\left(x_0\right)<1.$$

Then, $I_{f,g}(x_0,0)=f^{-1}\left(g\left(x_0\right)f\left(0\right)\right)$ and we get that

$$I_{f,g}(1,I_{f,g}(x_0,0))=I_{f,g}(1,f^{-1}\left(g\left(x_0\right)f\left(0\right)\right))=f^{-1}\left(g\left(1\right)g\left(x_0\right)f\left(0\right)\right).$$

However, since $g\left(1\right)g\left(x_0\right)f\left(0\right)>g\left(x_0\right)f\left(0\right)$ and $f^{-1}$ is strictly decreasing in $[0,f(0\left)\right]$, we have that

$$f^{-1}\left(g\left(1\right)g\left(x_0\right)f\left(0\right)\right)<f^{-1}\left(g\left(x_0\right)f\left(0\right)\right).$$

Hence, $I_{f,g}(x_0,I_{f,g}(1,0))>I_{f,g}(1,I_{f,g}(x_0,0))$. Contradiction with the fact that $I_{f,g}$ satisfies (EP).

For the reverse implication we have two cases:

-

If $f\left(0\right)=+\infty$ then $f^{(-1)}=f^{-1}$ and we obtain

$$\begin{array}{ccc}\hfill I_{f,g}(x,I_{f,g}(y,z))& =& f^{-1}(g\left(x\right)·(f\circ f^{-1})\left(g\left(y\right)f\left(z\right)\right))=f^{-1}\left(g\left(x\right)g\left(y\right)f\left(z\right)\right)\hfill \\ & =& f^{-1}(g\left(y\right)·(f\circ f^{-1})\left(g\left(x\right)f\left(z\right)\right)=I_{f,g}(y,I_{f,g}(x,z)).\hfill \end{array}$$

Hence, $I_{f,g}$ satisfies (EP).

-

If $f\left(0\right)<+\infty$ and $g\left(1\right)=1$, since g is strictly increasing and f strictly decreasing then

$$g\left(x\right)f\left(y\right)\le f\left(y\right)\le f\left(0\right)\mathrm{for}\mathrm{all}x,y\in [0,1].$$

Thus, in this case we have that $I_{f,g}(x,y)=f^{-1}\left(g\left(x\right)f\left(y\right)\right)$ for all $x,y\in [0,1]$ and, similarly to the previous point, we can prove that $I_{f,g}$ satisfies (EP).

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Proposition 11.

Let $(f,g)$ be an admissible pair of generators and T a t-norm. Then the following statements are equivalent:

(i)

The couple of functions $I_{f,g}$ and T satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$.

(ii)

$g\left(1\right)=1$ and $T={\left(T_P\right)}_g$, i.e., $T(x,y)=g^{-1}\left(g\left(x\right)g\left(y\right)\right)$ for all $x,y\in [0,1]$.

Proof. 

First, let us consider $g\left(1\right)=1$ and $T(x,y)=g^{-1}\left(g\left(x\right)g\left(y\right)\right)$, then $g\left(x\right)\in [0,1]$ for all $x\in [0,1]$ and we have that

$$I_{f,g}(T(x,y),z)=f^{(-1)}\left(g\left(x\right)g\left(y\right)f\left(z\right)\right)=f^{-1}\left(g\left(x\right)g\left(y\right)f\left(z\right)\right).$$

On the other hand, by the strictly decreasing nature of f we have that

$$g\left(x\right)f\left(y\right)\le f\left(y\right)\le f\left(0\right)\mathrm{for}\mathrm{all}x,y\in [0,1].$$

Then, $I_{f,g}(x,y)=f^{-1}\left(g\left(x\right)f\left(y\right)\right)$ for all $x,y\in [0,1]$ and we get

$$I_{f,g}(x,I_{f,g}(y,z))=f^{-1}\left(g\left(x\right)g\left(y\right)f\left(z\right)\right).$$

Hence, $I_{f,g}$ satisfies $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to ${\left(T_P\right)}_g$. Now, let us assume that $I_{f,g}$ satisfies $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to a certain t-norm T, we know that $I_{f,g}$ also satisfies (EP). Then, by Proposition 10$f\left(0\right)=+\infty$ or ($g\left(1\right)=1$ and $f\left(0\right)<+\infty$) and we have two cases:

-

If $f\left(0\right)=+\infty$ we know that $f^{(-1)}=f^{-1}$. Therefore, for all $x,y,z\in [0,1]$, we find that

$$\begin{array}{ccc}\hfill I_{f,g}(T(x,y),z)=I_{f,g}(x,I_{f,g}(x,y))& \iff & f^{-1}\left(g\left(T(x,y)\right)f\left(z\right)\right)=f^{-1}\left(g\left(x\right)g\left(y\right)f\left(z\right)\right)\hfill \\ & \iff & g\left(T\right(x,y\left)\right)=g\left(x\right)g\left(y\right).\hfill \end{array}$$

Now, since T is a t-norm, for all $y\in (0,1)$ we have

$$g\left(y\right)=g\left(T\right(1,y\left)\right)=g\left(1\right)g\left(y\right),$$

hence, $g\left(1\right)=1$. Then $g\left(x\right)g\left(y\right)\in [0,1]$ and $T(x,y)=g^{-1}\left(g\left(x\right)g\left(y\right)\right)$ for all $x,y\in [0,1]$.

-

On the other hand, if $f\left(0\right)<+\infty$ and $g\left(1\right)=1$, then we know from the proof of Proposition 10 that in this case $I_{f,g}(x,y)=f^{-1}\left(g\left(x\right)f\left(y\right)\right)$ and from the equality

$$I_{f,g}(T(x,y),z)=I_{f,g}(x,I_{f,g}(y,z))\iff g\left(T(x,y)\right)f\left(z\right)=g\left(x\right)g\left(y\right)f\left(z\right),$$

we obtain the result.

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It is worth noting that if $g\left(1\right)\ne 1$, then these implications do not satisfy the law of importation with any t-norm, which is a huge difference from the particular case of the f-generated implications whose characterization in [11] is based on this property. Moreover, note that this family of fuzzy implication functions provides new examples of functions satisfying the exchange principle, but not the law of importation with respect to any t-norm. The main consequence of these previous results is that while $(f,g)$-implications are not adequate for simplifying the process of applying the compositional rule of inference (CRI) of Zadeh (see [32]), some of them can be applied in approximate reasoning.

4.2. Generalization of g-Generated Implications

Now, we introduce a similar generalization for the g-generated implications by replacing the role of $\frac{1}{x}$ in their definition for a strictly decreasing function $f:[0,1]\to [0,+\infty ]$ with $f\left(0\right)=+\infty$.

Definition 10.

Let $g:[0,1]\to [0,+\infty ]$ be a strictly increasing and continuous function with $g\left(0\right)=0$ and $f:[0,1]\to [0,+\infty ]$ be a continuous and strictly decreasing function with $f\left(0\right)=+\infty$. The function $I:{[0,1]}^2\to [0,1]$ defined by

$$I_{g,f}(x,y)=g^{(-1)}\left(f\left(x\right)g\left(y\right)\right),x,y\in [0,1],$$

(5)

with the understanding $0·(+\infty )=+\infty$ and $\frac{1}{0}=+\infty$, is called a $(g,f)$-generated operation.

Remark 3.

The use of the pseudo-inverse in Equation (5) can be avoided using the following expression

$$I_{g,f}(x,y)=g^{-1}(min\{f\left(x\right)g\left(y\right),g\left(1\right)\}),x,y\in [0,1].$$

(6)

As in the case of the $(f,g)$-operations, not for all pairs of functions $(g,f)$ under the previous conditions we obtain a fuzzy implication function.

Theorem 5.

A $(g,f)$-operation $I_{g,f}$ is a fuzzy implication function if and only if, one of the following conditions hold:

(i)

$g\left(1\right)=+\infty$.

(ii)

$g\left(1\right)<+\infty$ and $f\left(1\right)\ge 1$.

Proof. 

First, let us assume that $I_{g,f}$ is a fuzzy implication function such that $g\left(1\right)<+\infty$. In this case, we have

$$1=I_{g,f}(1,1)=g^{(-1)}\left(f\left(1\right)g\left(1\right)\right)=\left\{\begin{array}{ccc}{g}^{-1}\left(f\left(1\right)g\left(1\right)\right)\hfill & \mathrm{if}& f\left(1\right)<1,\\ 1\hfill & \mathrm{if}& f\left(1\right)\ge 1.\end{array}\right.$$

Hence, necessarily $f\left(1\right)\ge 1$.

On the other hand, consider a $(g,f)$-operation satisfying (i) or (ii).

-

Let $x_1,x_2,y\in [0,1]$ with $x_1\le x_2$. Since f is strictly decreasing, then $f\left(x_1\right)\ge f\left(x_2\right)$. Now, since g is strictly increasing, $g^{(-1)}$ is increasing and we find

$$I_{g,f}(x_1,y)=g^{(-1)}\left(f\left(x_1\right)g\left(y\right)\right)\ge g^{(-1)}\left(f\left(x_2\right)g\left(y\right)\right)=I_{g,f}(x_2,y),$$

and $I_{g,f}$ satisfies (I1).

-

Consider $x,y_1,y_2\in [0,1]$ with $y_1\le y_2$, since g and $g^{(-1)}$ are increasing, then $g\left(y_1\right)\le g\left(y_2\right)$ and we find that

$$I_{g,f}(x,y_1)=g^{(-1)}\left(f\left(x\right)g\left(y_1\right)\right)\le g^{(-1)}\left(f\left(x\right)g\left(y_2\right)\right)=I_{g,f}(x,y_2),$$

then $I_{g,f}$ satisfies (I2).

-

$I_{g,f}(0,0)=g^{(-1)}\left(f\left(0\right)g\left(0\right)\right)=g^{(-1)}(+\infty ·0)=g^{(-1)}(+\infty )=1$.

-

$I_{g,f}(1,1)=g^{(-1)}\left(g\left(1\right)f\left(1\right)\right)$ and we need to distinguish two cases. If $g\left(1\right)=+\infty$, then we have that $I_{g,f}(1,1)=g^{-1}(+\infty )=1$. On the other hand, if $g\left(1\right)<+\infty$ and $f\left(1\right)\ge 1$, then $g\left(1\right)f\left(1\right)\ge g\left(1\right)$ and we get that $I_{g,f}(1,1)=1$.

-

$I_{g,f}(1,0)=g^{(-1)}\left(f\left(1\right)g\left(0\right)\right)=g^{(-1)}(f\left(1\right)·0)=g^{(-1)}\left(0\right)=0$.

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Remark 4.

In [22], a similar approach was considered. The authors define the family of fuzzy implication functions given by $I(x,y)=g^{(-1)}\left(f\left(x\right)g\left(y\right)\right)$ where $f:[0,1]\to [1,+\infty ]$ is a continuous, decreasing function satisfying $f\left(0\right)=+\infty$, $f\left(1\right)=1$ and g is a g-generator. In this case, they consider functions f not necessarily strictly decreasing, but with $f\left(1\right)=1$. In our case, we consider functions f, which are strictly decreasing, but we allow any value in $(0,+\infty )$ of $f\left(1\right)$ whenever $g\left(1\right)=+\infty$ and $f\left(1\right)\ge 1$ when $g\left(1\right)<+\infty$. Then, the two families are not equivalent, but they have intersection when we consider a continuous, strictly decreasing function with $f\left(1\right)=1$. However, since the two families are very similar, one can verify that the conditions that ensure that the two families fulfill a certain property are very similar in the two cases.

Whenever a $(g,f)$-operation satisfies the properties given in Definition 3, we will call it a $(g,f)$-implication with their associated admissible pair of generators $(g,f)$.

In a similar way as in the $(f,g)$-implications, for $(g,f)$-implications, it is only necessary to consider those pairs of admissible generators $(g,f)$ such that $g\left(1\right)=1$ or $g\left(1\right)=+\infty$, as it is shown in the following result.

Proposition 12.

Let $I_{g,f}$ be a fuzzy implication function with $g\left(1\right)<+\infty$, then there exists a function $g_1$ with $g_1\left(1\right)=1$ such that $(g_1,f)$ is an admissible pair of generators and $I_{g,f}=I_{g_1,f}$.

Proof. 

Let $I_{g,f}$ be a fuzzy implication function with $g\left(1\right)<+\infty$. If we consider $g_1\left(x\right)=\frac{g\left(x\right)}{g\left(1\right)}$, then $(g_1,f)$ is also an admissible pair of generators with $g_1\left(1\right)=1$. Moreover, $g_1^{-1}\left(x\right)=g^{-1}\left(xg\left(1\right)\right)$ and we find that

$$\begin{array}{ccc}\hfill I_{g_1,f}\left(x\right)& =& g_1^{(-1)}\left(f\left(x\right)g_1\left(y\right)\right)=g_1^{(-1)}\left(f\left(x\right)\frac{g\left(y\right)}{g\left(1\right)}\right)=g_1^{-1}\left(min\left\{f\left(x\right)\frac{g\left(y\right)}{g\left(1\right)},g_1\left(1\right)\right\}\right)\hfill \\ & =& g^{-1}(min\{f\left(x\right)g\left(y\right),g\left(1\right)\})=I_{g,f}(x,y).\hfill \end{array}$$

□

Now, we will follow a similar approach to the previous section in order to study the properties of these fuzzy implication functions. Let us start by studying the region where the $(g,f)$-implications take value 1. Notice that $(g,f)$-implications may have a non-trivial 1 region.

Proposition 13.

Let $(g,f)$ be an admissible pair of generators. Then, the following statements hold:

(i)

If $g\left(1\right)=+\infty$, then $I_{g,f}(x,y)=1\iff x=0$ or $y=1$.

(ii)

If $g\left(1\right)<+\infty$, then $I_{g,f}(x,y)=1\iff y\ge g^{-1}\left(\frac{g\left(1\right)}{f\left(x\right)}\right)$.

Proof. 

(i)

If $g\left(1\right)=+\infty$, we know that $g^{(-1)}=g^{-1}$ and then for any $x,y\in [0,1]$ we have

$$\begin{array}{ccc}\hfill I_{g,f}(x,y)=g^{-1}\left(f\left(x\right)g\left(y\right)\right)=1& \iff & f\left(x\right)g\left(y\right)=g\left(1\right)=+\infty \hfill \\ & \iff & f\left(x\right)=+\infty \mathrm{or}g\left(y\right)=+\infty \hfill \\ & \iff & x=0\mathrm{or}y=1.\hfill \end{array}$$

(ii)

If $g\left(1\right)<+\infty$, then by the definition of $g^{(-1)}$, for every $x,y\in [0,1]$ we know that

$$I_{g,f}(x,y)=g^{(-1)}\left(f\left(x\right)g\left(y\right)\right)=1\iff f\left(x\right)g\left(y\right)\ge g\left(1\right)\iff y\ge g^{-1}\left(\frac{g\left(1\right)}{f\left(x\right)}\right).$$

□

From the previous result we can see that unlike the $(f,g)$-implications, $(g,f)$-implications satisfy the identity principle in certain cases.

Corollary 2.

Let $(g,f)$ be an admissible pair of generators. Then $I_{g,f}$ satisfies(IP)if and only if $g\left(1\right)<+\infty$ and $f\left(x\right)\ge \frac{g\left(1\right)}{g\left(x\right)}$ for all $x\in [0,1]$.

On the other hand, in the following proposition study, the region where these fuzzy implication functions value zero. This result has been corrected since in ([19], Proposition 9) the case when $f\left(1\right)=0$ was not contemplated.

Proposition 14.

Let $(g,f)$ be an admissible pair of generators. Then the following statements hold:

(i)

If $f\left(1\right)>0$, then $I_{g,f}(x,y)=0$ if and only if $x>0$ and $y=0$.

(ii)

If $f\left(1\right)=0$ and $g\left(1\right)=+\infty$, then $I_{g,f}(x,y)=0$ if and only if ($x=1$ and $y<1$) or ($x>0$ and $y=0$).

Proof. 

Consider $x,y\in [0,1]$ then

$$I_{g,f}(x,y)=g^{(-1)}\left(f\left(x\right)g\left(y\right)\right)=0\iff f\left(x\right)g\left(y\right)=0.$$

Taking into account that either $g\left(1\right)=+\infty$ or $g\left(1\right)<+\infty$ and $f\left(1\right)\ge 1$ and the understanding $0·(+\infty )=+\infty$ we obtain the result. □

From the last result, it is straightforward that these fuzzy implication functions satisfy that their natural negation is $N_{D_1}$. Then, the natural negation of $(g,f)$-implications is independent of their generators.

Corollary 3.

Let $(g,f)$ be an admissible pair of generators. Then, the natural negation $N_{I_{g,f}}$ is the Gödel or least fuzzy negation $N_{D_1}$.

The next result reflects that, as in the case of (IP), the property (OP) can be satisfied by $(g,f)$-implications under certain conditions of their generators. This result has been corrected with respect to the original one ([19], [Proposition 11]) since the expression in (iii) was not correct. Moreover, we explicitly find the constant in item (ii).

Proposition 15.

Let $(g,f)$ be an admissible pair of generators. Then, the following statements are equivalent:

(i)

$I_{g,f}$ satisfies(OP).

(ii)

$g\left(1\right)<+\infty$ and $f\left(x\right)=\frac{g\left(1\right)}{g\left(x\right)}$.

(iii)

$f\left(1\right)=1$ and $I_{g,f}(x,y)=f^{-1}\left(max\left\{1,\frac{f\left(y\right)}{f\left(x\right)}\right\}\right)$.

Proof. 

(i)⇒ (ii). Let us assume $I_{g,f}$ satisfies (OP). By Proposition 13 we know that in this case $g\left(1\right)=+\infty$ is not possible. Considering $g\left(1\right)<+\infty$ we know from Proposition 12 that considering the function $g_1\left(x\right)=\frac{g\left(x\right)}{g\left(1\right)}$ we have that $I_{g_1,f}=I_{g,f}$ with $g_1\left(1\right)=1$. Then,

$$I_{g_1,f}(x,y)=1\iff y\ge g_1^{-1}\left(\frac{g_1\left(1\right)}{f\left(x\right)}\right)\iff f\left(x\right)\ge \frac{1}{g_1\left(y\right)}.$$

Thus, if $I_{g,f}$ satisfies (OP), we have that

$$x\le y\iff f\left(x\right)\ge \frac{1}{g_1\left(y\right)}.$$

(7)

We will prove that $f\left(x\right)=\frac{1}{g_1\left(x\right)}$ for all $x\in [0,1]$. For $x=0$ we have that

$$\frac{1}{g_1\left(0\right)}=\frac{1}{0}=+\infty =f\left(0\right).$$

For $x=1$, by Equation (7) we obtain that

$$f\left(1\right)<\frac{1}{g_1\left(y\right)},\mathrm{for}\mathrm{all}0\le y<1.$$

Taking limits we get that

$$f\left(1\right)\le \underset{y\to 1^{-}}{lim}\frac{1}{g_1\left(y\right)}=1,$$

and since we already had that $f\left(1\right)\ge 1$ it holds that $f\left(1\right)=1=\frac{1}{g_1\left(1\right)}$. Finally, suppose that for some $x_0\in (0,1)$ the equality does not hold. By Equation (7), we have that

$$f\left(x_0\right)\ge \frac{1}{g_1\left(x_0\right)},$$

then let us assume that $f\left(x_0\right)>\frac{1}{g_1\left(x_0\right)}$. We consider the following continuous function

$$h_1\left(y\right)=f\left(x_0\right)g_1\left(y\right),$$

then, we have that $h_1\left(0\right)=f\left(x_0\right)g_1\left(0\right)=0$ and $h_1\left(x_0\right)=f\left(x_0\right)g_1\left(x_0\right)>1$. However, then there exists a $y_0\in (0,x_0)$ such that $f\left(x_0\right)g_1\left(y_0\right)=1$. Contradiction with Equation (7). Then, we have proved that $f\left(x\right)=\frac{1}{g_1\left(x\right)}=\frac{g\left(1\right)}{g\left(x\right)}$ for all $x\in [0,1]$.

(ii) ⇒ (iii) If $g\left(1\right)<+\infty$ and $f\left(x\right)=\frac{g\left(1\right)}{g\left(x\right)}$, then the result follows replacing $g^{-1}\left(x\right)$ by $f^{-1}\left(\frac{g\left(1\right)}{x}\right)$ in Equation (6).

(iii) ⇒ (i) Let us prove that $I_{g,f}(x,y)<1\iff y<x$. Consider $x,y\in [0,1]$ then

$$I_{g,f}(x,y)=f^{-1}\left(max\left\{1,\frac{f\left(y\right)}{f\left(x\right)}\right\}\right)<1\iff max\left\{1,\frac{f\left(y\right)}{f\left(x\right)}\right\}>f\left(1\right)=1\iff \frac{f\left(y\right)}{f\left(x\right)}>1.$$

Since f is strictly decreasing we get that

$$I_{g,f}(x,y)<1\iff f\left(y\right)>f\left(x\right)\iff y<x.$$

□

Consecutively, the next result deals with the continuity of $(g,f)$-implications. We already know from Proposition 3 that these implications are never continuous at $(0,0)$, since their natural negation is $N_{D_1}$. Similarly from $(f,g)$-implications, the next result shows that $(0,0)$ and $(1,1)$ are the only possible points of discontinuity.

Proposition 16.

Let $(g,f)$ be an admissible pair of generators. Then, the following properties hold:

(i)

$I_{g,f}$ is continuous everywhere except at the point $(0,0)$ if and only if $g\left(1\right)<+\infty$ or $\left(g\right(1)=+\infty$ and $f\left(1\right)>0$).

(ii)

$I_{g,f}$ is continuous everywhere except at the points $(0,0)$ and $(1,1)$ if and only if $g\left(1\right)=+\infty$ and $f\left(1\right)=0$.

Proof. 

Let $(g,f)$ be an admissible pair of generators, then by definition $I_{g,f}$ is continuous at each $(x,y)\in {[0,1]}^2$ by being the composition of real continuous functions except for the cases when ($f\left(x\right)=0$ and $g\left(y\right)=+\infty$) or ($g\left(y\right)=0$ and $f\left(x\right)=+\infty$), since in these situations we have considered the convention $0·(+\infty )=+\infty$. These two situations correspond to the following two cases:

• If $x=y=1$ and $f\left(1\right)=0$ then $g\left(1\right)<+\infty$ and

  $$\underset{y\to 1^{-}}{lim}{I}_{g,f}(1,y)=\underset{y\to 1^{-}}{lim}{g}^{-1}\left(f\left(1\right)g\left(y\right)\right)=g^{-1}\left(0\right)=0\ne 1=I_{g,f}(1,1).$$
  Thus, $I_{g,f}$ is discontinuous at $(1,1)$.
• $x=y=0$ then we know by Corollary 3 that $I_{g,f}$ is not continuous at $(0,0)$ for any choice of its generators.

□

Finally, we study the properties (EP) and $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ for these fuzzy implication functions.

Proposition 17.

Let $(g,f)$ be an admissible pair of generators. Then, $I_{g,f}$ always satisfies(EP).

Proof. 

We distinguish between two cases:

• If $g\left(1\right)=+\infty$, then for each $x,y,z\in [0,1]$ we find that $g^{(-1)}=g^{-1}$ and then

  $$I_{g,f}(x,I_{g,f}(y,z))=g^{-1}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right)=I_{g,f}(y,I_{g,f}(x,z)).$$
• If $g\left(1\right)<+\infty$ and $f\left(1\right)\ge 1$, consider $x,y,z\in [0,1]$ and let us distinguish two cases:

  -

  If $f\left(x\right)f\left(y\right)g\left(z\right)\le g\left(1\right)$ then, since $f\left(x\right)\ge f\left(1\right)\ge 1$, we find that

  $$g\left(1\right)\ge f\left(x\right)g\left(z\right)\mathrm{and}g\left(1\right)\ge f\left(y\right)g\left(z\right).$$

  Then, we get

  $$\begin{array}{ccc}\hfill I_{g,f}(x,I_{g,f}(y,z))& =& g^{(-1)}\left(f\left(x\right)(g\circ g^{(-1)})\left(f\left(y\right)g\left(z\right)\right)\right)=g^{(-1)}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right)\hfill \\ & =& g^{(-1)}\left(f\left(y\right)(g\circ g^{(-1)})\left(f\left(x\right)g\left(z\right)\right)\right)=I_{g,f}(y,I_{g,f}(x,z)).\hfill \end{array}$$

  -

  If $f\left(x\right)f\left(y\right)g\left(z\right)>g\left(1\right)$, then we find that

  $$I_{g,f}(x,I_{g,f}(y,z))=\left\{\begin{array}{ccc}{g}^{(-1)}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right)\hfill & \mathrm{if}& f\left(y\right)g\left(z\right)\le g\left(1\right),\\ g^{(-1)}\left(f\left(x\right)g\left(1\right)\right)\hfill & \mathrm{if}& f\left(y\right)g\left(z\right)>g\left(1\right).\end{array}\right.$$

  In any case, since $f\left(x\right)f\left(y\right)g\left(z\right)>g\left(1\right)$ and $f\left(x\right)g\left(1\right)\ge g\left(1\right)$, it is always $I_{g,f}(x,I_{g,f}(y,z))=1$. An analogous argument proves that $I_{g,f}(y,I_{g,f}(x,z))=1$.

□

Proposition 18.

Let $(g,f)$ be an admissible pair of generators and T a t-norm. Then, the following statements are equivalent:

(i)

The couple of functions $I_{g,f}$ and T satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$.

(ii)

$f\left(1\right)=1$ and $T(x,y)=f^{-1}\left(f\left(x\right)f\left(y\right)\right)$ for all $x,y\in [0,1]$.

Proof. 

(i) ⇒ (ii) Let us distinguish between two cases:

• If $g\left(1\right)=+\infty$ then

  $$\begin{array}{ccc}\hfill I\left(T\right(x,y),z)=I(x,I(y,z\left)\right)& \iff & g^{-1}\left(f\left(T(x,y)\right)g\left(z\right)\right)=g^{-1}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right)\hfill \\ & \iff & f\left(T\right(x,y\left)\right)=f\left(x\right)f\left(y\right).\hfill \end{array}$$
  Since T is a t-norm we have that for all $y\in (0,1)$, $f\left(y\right)=f\left(T\right(1,y\left)\right)=f\left(1\right)f\left(y\right)$. Thus, $f\left(1\right)=1$ and $T(x,y)=f^{-1}\left(f\left(x\right)f\left(y\right)\right)$.
• If $g\left(1\right)<+\infty$ and $f\left(1\right)\ge 1$, from the proof of Proposition 17, we deduce the following equality

  $$I_{g,f}(x,I_{g,f}(y,z))=\left\{\begin{array}{ccc}{g}^{-1}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right)\hfill & \mathrm{if}& f\left(x\right)f\left(y\right)g\left(z\right)<g\left(1\right),\\ 1\hfill & \mathrm{if}& f\left(x\right)f\left(y\right)g\left(z\right)\ge g\left(1\right).\end{array}\right.$$
  On the other hand, we have that

  $$I_{g,f}(T(x,y),z)=\left\{\begin{array}{ccc}{g}^{-1}\left(f\left(T(x,y)\right)g\left(z\right)\right)\hfill & \mathrm{if}& f\left(T\right(x,y\left)\right)g\left(z\right)<g\left(1\right),\\ 1\hfill & \mathrm{if}& f\left(T\right(x,y\left)\right)g\left(z\right)\ge g\left(1\right).\end{array}\right.$$
  Now, first let us prove that $f\left(1\right)=1$. Since g is continuous with $g\left(0\right)=0$, for all $y\in (0,1]$ we can find some $z\in (0,1)$ such that $f\left(1\right)f\left(y\right)g\left(z\right)\le g\left(1\right)$ and by $f\left(1\right)\ge 1$ we have that $f\left(y\right)g\left(z\right)\le g\left(1\right)$. Since $I_{g,f}$ and T satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$, we find that

  $$g^{-1}\left(f\left(1\right)f\left(y\right)g\left(z\right)\right)=I_{g,f}(1,I_{g,f}(y,z))=I_{g,f}(T(1,y),z)=I_{g,f}(y,z)=g^{-1}\left(f\left(y\right)g\left(z\right)\right).$$
  Thus, $f\left(1\right)=1$. If $x,y\in [0,1]\setminus (0,1)$ is straightforward to see that $T(x,y)=f^{-1}\left(f\left(x\right)f\left(y\right)\right)$. Let us assume $f\left(T\right(x,y\left)\right)\ne f\left(x\right)f\left(y\right)$ for some $x,y\in (0,1)$ and get a contradiction. We have two cases:

  -

  If $f\left(T\right(x,y\left)\right)<f\left(x\right)f\left(y\right)$ then we choose $z=g^{-1}\left(\frac{g\left(1\right)}{f\left(x\right)f\left(y\right)}\right)$ and we get that

  $$f\left(x\right)f\left(y\right)g\left(z\right)=f\left(x\right)f\left(y\right)\frac{g\left(1\right)}{f\left(x\right)f\left(y\right)}=g\left(1\right),$$

  then $I_{g,f}(x,I_{g,f}(y,z))=1$. However, on the other hand,

  $$f\left(T(x,y)\right)g\left(z\right)=\frac{f\left(T\right(x,y\left)\right)}{f\left(x\right)f\left(y\right)}g\left(1\right)<g\left(1\right),$$

  and then $I_{g,f}(T(x,y),z)<1$.

  -

  If $f\left(T\right(x,y\left)\right)>f\left(x\right)f\left(y\right)$, let us consider $z=g^{-1}\left(\frac{g\left(1\right)}{f\left(T\right(x,y\left)\right)}\right)$. Then, we have that

  $$f\left(T\right(x,y\left)\right)g\left(z\right)=g\left(1\right),$$

  and then $I_{g,f}(T(x,y),z)=1$. Otherwise,

  $$f\left(x\right)f\left(y\right)g\left(z\right)=\frac{f\left(x\right)f\left(y\right)}{f\left(T\right(x,y\left)\right)}g\left(1\right)<g\left(1\right),$$

  and $I_{g,f}(x,I_{g,f}(y,z))<1$.

(ii) ⇒ (i) Let us consider an admissible pair of generators $(g,f)$ with $f\left(1\right)=1$ and $T(x,y)=f^{-1}\left(f\left(x\right)f\left(y\right)\right)$, then

$$I_{g,f}(T(x,y),z)=g^{(-1)}\left((f\circ f^{-1})\left(f\left(x\right)f\left(y\right)\right)g\left(z\right)\right)=g^{(-1)}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right).$$

On the other hand, since f is strictly decreasing and g strictly increasing then $f\left(y\right)g\left(z\right)\le f\left(1\right)g\left(z\right)=g\left(z\right)\le g\left(1\right)$ and

$$I_{g,f}(x,I_{g,f}(y,z))=I_{g,f}(x,g^{(-1)}\left(f\left(y\right)g\left(z\right)\right))=I_{g,f}(x,g^{-1}\left(f\left(y\right)g\left(z\right)\right))=g^{(-1)}\left(f\left(x\right)f\left(y\right)g\left(z\right)\right).$$

□

In a similar way to $(f,g)$-implications, although $(g,f)$-implications always satisfy (EP) if $f\left(1\right)>1$, they do not satisfy $\left({\mathbf{LI}}_{\mathbf{T}}\right)$ with respect to any t-norm.

5. Conclusions and Future Work

In this paper, our main interest has been to study the family of generalized $(h,e)$-implications. Although these fuzzy implication functions were first defined in [15], we have considered the more general definition given in [18] and hence, all the results have been adapted to this definition. Then, we have recalled the representation theorem for $(h,e)$-implications presented in [19]. This theorem fully determines the structure of a generalized $(h,e)$-implication given an f and g-generator with $g\left(1\right)=+\infty$. However, in order to properly study this structure, the necessity of defining two novel families of fuzzy implication functions which are generalizations of Yager’s implications has been exposed. In this sense, the additional properties of these two families of fuzzy implication functions, namely $(f,g)$ and $(g,f)$-implications, have been fully investigated. Although these properties were already presented in [19] without proof, in this paper we have proved all the results. Moreover, some of the results in [19] were not fully correct and therefore, these results have been corrected.

As future work, we want to complete our study by providing an axiomatic characterization of generalized $(h,e)$-implications. In order to provide this characterization, we want to follow the same approach as in the case of h-implications, i.e., we want to provide axiomatic characterizations of the subfamilies of $(f,g)$ and $(g,f)$-implications involved in the representation theorem of generalized $(h,e)$-implications and to use these results to derive the axiomatic characterization of generalized $(h,e)$-implications.