On distributivity equations of implications over overlap functions and contrapositive symmetry equations of implications

1 Introduction

Probability theory, evidence theory, fuzzy set theory and rough set theory are most important mathematical tools for dealing with vague, imprecise, inconsistent and uncertain knowledge. In recent years, researches and applications on fuzzy set theory and rough set theory have attracted more and more researchers’ attention. And they are also hot issues in the artificial intelligence field [36–38].

In 2009, Bustince et al. introduced the concept of overlap functions in [10] for the purpose of dealing with the frequent overlapping problem of classification in image processing, and the main use of overlap functions is to measure the overlapping degree between the two functions standing for the object and background, respectively. Later, it was Elkano et al. who applied overlap functions into classification. In 2015, they adapted the inference system of Fuzzy Association Rule-based Classification model for High-Dimensional problems by replacing the product t-norm with n-dimensional overlap functions [20]. In 2016, they studied the influence of the usage of n-dimensional overlap functions to model the conjunction in several Fuzzy Rule-Based Classification Systems in order to enhance their performance in multi-class classification problems applying decomposition techniques [19]. At the same time, overlap functions were also applied to many other fields, such as fuzzy preference modeling and decision making [12, 21].

Not only in application but also in theory have overlap functions developed quickly. In 2013, some important properties of overlap functions were discussed by Bedregal et al. in [9]. Then Dimuro and Bedregal studied Archimedean overlap functions and idempotent overlap functions, and gave the ordinal sum of overlap functions in [16]. In 2015, the concept of residual implication derived from overlap function was proposed by Dimuro and Bedregal in [14] and the laws of contraposition for these residual implications were analyzed in [15]. In 2016, Dimuro et al. introduced overlap functions with additive generator pairs in [17]. Two years later, Qiao and Hu proposed overlap and grouping functions with multiplicative generator pairs in [25]. Some other studies related to overlap functions can be found also in [11, 31].

The investigations of distributive laws of fuzzy implications over different fuzzy logic connectives have attracted the attention of many authors. In order to achieve the desired accuracy better in fuzzy approximation, a large rule base is commonly needed. And, in many cases, the problem of rule explosion hinders the further application and development of fuzzy system. Thus, there arise many works to construct fuzzy systems having specified performance by using as less as possible fuzzy rules. For instance, to avoid the problem of combinatorial rule explosion, Combs and Andrews required of the classical tautology (p ∧ q) → r ≡ (p → r) ∨ (q → r) in their inference mechanism towards reduction in the complexity of fuzzy “If-then” rule in [13]. After that, from the point of fuzzy logic, Trillas and Alsina firstly introduced the following generalized version of the above law in [32]: I (T (x, y), z) = S (I (x, z), I (y, z)), where T is a t-norm, S is a t-conorm and I is a fuzzy implication, and got all solutions of T when I is R-implication, or S-implication, or QL-implication. Along the line, the other three functional equations were also discussed by Balasubramaniam and Rao in [8]. On the other hand, Türksen et al. in [33] gave the necessary conditions for a fuzzy implication I satisfying the equation

I ( x , T ( y , z ) ) = T ( I ( x , y ) , I ( x , z ) ) ,(1) when T is the product t-norm. And then, Baczyński investigated Equation (1) while T is a strict t-norm in [1] and presented the sufficient and necessary conditions of functional equations composed of Eq.(1) and the equation I (x, I (y, z)) = I (T (x, y), z). Subsequently, he gave a further study in [4] of the system of functional equations consisting of Eq.(1) and the following equation

I ( x , y ) = I ( N ( y ) , N ( x ) ) .(2) Eq.(2), called contrapositive symmetry equation of implication, was introduced by Fodor in [6] to construct a new family of t-norms, and it has important application in fuzzy preference modeling and expert systems with incomplete information [4, 33]. However, the full characterization of the system of Equations (1) and (2) for strict t-norm was given by Yang and Qin in [35]. And then later, the distributive laws of fuzzy implications over nilpotent or strict or t-conorms were studied in [2, 34]. Finally, Qin and Baczyński generalized Eq.(1) into the following equation

I ( x , T 1 ( y , z ) ) = T 2 ( I ( x , y ) , I ( x , z ) ) ,(3) and presented the sufficient and necessary conditions of solutions for Eq.(3) and the system of functional equations composed of Eq.(2) and Eq.(3), when T₁ is a continuous t-norm, T₂ is a continuous Archimedean t-norm, I is an unknown function and N is a strong negation, all these related results have been summarized in [5].

OPEN IN VIEWER

Fig.1

The relation between t-norms and overlap function.

Overlap function, as a relatively new non-associative fuzzy logic connective, has close relationship with t-norm, as shown in Fig. 1. So it is natural to consider the distributivity of fuzzy implication functions over overlap functions. The pioneer work in this framework was started by Qiao and Hu, who introduced the distributive equation in [27] as follows:

I ( x , O 1 ( y , z ) ) = O 2 ( I ( x , y ) , I ( x , z ) ) ,(4) where O₁ and O₂ are overlap functions, I is one of the residual implications derived from overlap functions, or (G, N)-implications for grouping functions and fuzzy negations, or QL-operations for overlap functions, grouping functions and fuzzy negations. Recently, Qiao and Hu gave the characterization of Eq.(4) when O₁ and O₂ are additively generated by strict additive generator pairs and I is an unknown function in [28]. The purpose of this paper is to explore and give solutions to the system of functional equations consisting of Eq.(2) and Eq.(4) when O₁ and O₂ have multiplicative generator pairs and I is a continuous binary function. What’s more, another possibility is considered and some solutions to the system consisting of Eq.(2) and Eq.(4) are also presented at the situation that O₁ is an idempotent overlap function and O₂ is an Archimedean overlap function.

This paper is organized as follows: Section 2 presents a specific formalization of the main concepts used throughout the paper, such as multiplicatively generated overlap functions, idempotent overlap functions, Archimedean overlap functions, fuzzy implications, fuzzy negations and so on. Section 3 gives characterizations of solutions to Eq.(4) when O₁ and O₂ are multiplicatively generated overlap functions and I is a continuous function, and shows that no continuous fuzzy implications satisfy Eq.(4). But sufficient and necessary conditions are given when fuzzy implication I is continuous except for the point (0, 0). Subsequently, section 4 characterizes solutions to the functional equations composed of Eq.(2) and Eq.(4) at the situation that I is a fuzzy implication and is continuous expect at the points (0, 0) and (1, 1), N is a strong fuzzy negation. Section 5 considers some other possibilities and gives sufficient and necessary conditions of solutions to Eq.(4) when O₁ is an idempotent overlap function, O₂ is an Archimedean overlap function, I is an unknown function, and discusses some solutions to the system of Eq.(2) and Eq.(4).

This section recollects some main concepts and results used throughout the subsequent sections. For a further reading about such concepts, we recommend [4, 25].

Definition 2.1. ([16]) A binary function O: [0, 1] ² ⟶ [0, 1] is said to be an overlap function if it satisfies the following conditions:

(O₁) O is commutative;

Definition 2.2. ([16]) Let O be an overlap function, an element a ∈ [0, 1] is called an idempotent element of O if O (a, a) = a. If each a ∈ [0, 1] is an idempotent element of O, then O is said to be idempotent. The numbers 0 and 1 are called trivial idempotent elements of O, each idempotent element in (0, 1) will be called a non-trivial idempotent element of O. And the set of idempotent elements of O is denoted by Idem (O).

The power of the overlap function O is defined as: x O ( 1 ) = x a n d x O ( n + 1 ) = O ( x , x O ( n ) ) with n ∈ ℕ * .

Definition 2.3. ([16]) An overlap function is said to be Archimedean if for each (x, y) ∈ (0, 1) ² there is an n ∈ ℕ * such that x O ( n ) < y .

Proposition 2.4. ([16]) Let O be an Archimedean overlap function. Then O has only trivial idempotent elements.

Example 2.5. ([16]) (1) Any positive continuous t-norm is an overlap function. (2) The overlap function O _p : [0, 1] ² ⟶ [0, 1] with p > 0 and p ≠ 1, given by O _p (x, y) = x ^p y ^p , is neither associative nor with 1 as its neutral element. Specially, when p = 1 2 , O is an idempotent overlap function; when p > 1, O _p is an Archimedean overlap function.

Definition 2.6. ([25]) Let g, h: [0, 1] ⟶ [0, 1] be two continuous and increasing functions. If the binary function O_g,h: [0, 1] ² ⟶ [0, 1] given by O g , h ( x , y ) = g ( h ( x ) h ( y ) ) is an overlap function, then (g, h) is called a multiplicative generator pair of O_g,h and O_g,h is said to be multiplicatively generated by the pair (g, h).

Theorem 2.7. ([25]) Let g, h: [0, 1] ⟶ [0, 1] be two continuous and increasing functions such that

(a) h (x) = 0 if and only if x = 0;

Then (g, h) multiplicatively generates an overlap function.

Proposition 2.8. ([25]) Let g, h: [0, 1] ⟶ [0, 1] be two continuous and increasing functions such that

g (x) = 0 if and only if x = 0;

Then the following statements hold:

h (x) = 0 if and only if x = 0;

Proposition 2.9. ([25]) ıLet O be an overlap function multiplicatively generated by the pair (g, h). Then the following statements hold: (1) h (x) = 0 if and only if x = 0; (2) g (x) = 0 if and only if x = 0.

Example 2.10. ([25]) (1) Consider O _p (p > 0, p ≠ 1), its multiplicative generator pair (g, h) can be given by g (x) = x and h (x) = x ^p . (2) The overlap function O ( x , y ) = 2 xy 1 + xy , x, y ∈ [0, 1] can be multiplicatively generated by the pair (g, h), where g ( x ) = 2 x 1 + x and h (x) = x.

Definition 2.11. ([6]) A function I: [0, 1] ² ⟶ [0, 1] is called a fuzzy implication, if it fulfills the following conditions: (I₁) I is decreasing with respect to the first variable; (I₂) I is increasing with respect to the second one; (I₃) I (0, 0) = I (1, 1) = 1, I (1, 0) = 0.

Remark 2.12. ([6]) Clearly, fuzzy implication I satisfies the following property: (RB) I (x, 1) = 1 for all x ∈ [0, 1](right boundary condition).

Definition 2.13. ([6]) A function N: [0, 1] ⟶ [0, 1] is a fuzzy negation if the following properties are satisfied: for all x, y ∈ [0, 1], (1) If x ≤ y then N (y) ≤ N (x); (2) N (0) = 1 and N (1) = 0. Moreover, a fuzzy negation N is strong if it also satisfies the involution property, i. e. N (N (x)) = x for each x ∈ [0, 1].

Theorem 2.14. ([24]) A function f : D ⟶ ℝ is a continuous solution of the multiplicative Cauchy functional equation

f ( x y ) = f ( x ) f ( y ) , x , y ∈ D(5) if and only if either f = 0, or f = 1, or there exists a constant c ∈ ℝ such that f ( x ) = x c , x ∈ D , where D is one of sets (0, 1), [0, 1), (0, 1]. If 0 ∈ D, then c > 0.

In this section, Eq.(4) is investigated and its continuous solutions are presented for multiplicatively generated overlap functions. To start with, we show some properties of multiplicative generator pair of overlap function.

Lemma 3.1. Let (g, h) be a multiplicative generator pair of the overlap function O_g,h. Then g (1) = 1.

Proposition 3.2. Let g, h: [0, 1] ⟶ [0, 1] be two continuous and increasing functions. If g is strict, then the following statements are equivalent: (1) (g, h) is a multiplicative generator pair of the overlap function O_g,h; (2) g and h satisfy the following conditions: (a) g (x) = 0 if and only if x = 0; (b) g (x) = 1 if and only if x = 1; (c) h (x) = 0 if and only if x = 0; (d) h (x) = 1 if and only if x = 1.

Theorem 3.3. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing and I: [0, 1] ² ⟶ [0, 1] be a continuous binary function satisfying (RB). Then the following statements are equivalent: (1) the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) I = 1 or there exists a continuous function c: [0, 1] ⟶ (0, ∞) such that for all x, y ∈ [0, 1],

I ( x , y ) = h 2 − 1 ( ( h 1 ( y ) ) c ( x ) ) = g 2 ( ( g 1 − 1 ( y ) ) c ( x ) ) .(6)

Proof. (1)⇒(2) Let the triple of functions (O₁, O₂, I) satisfy Eq.(4). Since O₁, O₂ are generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂), so Eq.(4) has the following form: g 2 − 1 ( I ( x , g 1 ( h 1 ( y ) h 1 ( z ) ) ) ) = h 2 ( I ( x , y ) ) h 2 ( I ( x , z ) ) For any fixed x ∈ [0, 1], define I _x : [0, 1] ⟶ [0, 1] by I _x (y) = I (x, y), y ∈ [0, 1]. Moreover, take f x = g 2 - 1 ∘ I x ∘ g 1 and k x = h 2 ∘ I x ∘ h 1 - 1 , then both f _x and k _x are unary function on [0, 1]. Let u = h₁ (y) and v = h₁ (z), Eq.(4) can be rewritten as follows:

f x ( u v ) = k x ( u ) k x ( v ) , u , v ∈ [ 0 , 1 ] .(7) By a simple calculation, it shows that f _x (1) = k _x (1) = 1. Let u = 1, then it holds that f _x (v) = k _x (1) k _x (v) = k _x (v) for all v ∈ [0, 1]. That is to say, Eq.(7) can be seen as

k x ( u v ) = k x ( u ) k x ( v ) , u , v ∈ [ 0 , 1 ] .(8) Thus, from Theorem 2.14, it leads to three possibilities:

f x = k x = 0 , v ∈ [ 0 , 1 ) ,(9) or

f x ( v ) = k x ( v ) = v c x , v ∈ [ 0 , 1 ) .(10) or there exists a constant c _x ∈ (0, ∞) such that

f x ( v ) = k x ( v ) = v c x , v ∈ [ 0 , 1 ) .(11) Combined with the continuity of I and f _x (1) = k _x (1) = 1, it can be concluded that f _x = k _x = 1, or there exists a constant c _x ∈ (0, ∞) such that f x ( v ) = k x ( v ) = v c x v ∈ [ 0 , 1 ] . Thus, by definitions of f _x and k _x , it shows that

I ( x , y ) = 1 , y ∈ [ 0 , 1 ] ,(12) or there exists a constant c _x ∈ (0, ∞) such that for all y ∈ [0, 1],

I ( x , y ) = g 2 ( ( g 1 − 1 ( y ) ) c x ) = h 2 − 1 ( ( h 1 ( y ) ) c x ) .(13) Let us assume that there exists x₀ ∈ [0, 1] such that I (x₀, y) = 1 for all y ∈ [0, 1]. Particularly, I (x₀, 0) = 1. But for another possible vertical section, the I (x, 0) = 0 always holds. Since I is continuous, so the only possible solution in this case is I = 1. So the first solution is obtained. Finally, assume that the vertical sections I _x is Eq.(13) for all x ∈ [0, 1]. Define function c: [0, 1] ⟶ (0, ∞) by c (x) = c _x , Eq.(13) means that for all x, y ∈ [0, 1], I ( x , y ) = g 2 ( ( g 1 - 1 ( y ) ) c ( x ) ) = h 2 - 1 ( ( h 1 ( y ) ) c ( x ) ) . The function c is continuous since for any fixed y ∈ (0, 1), it is the composition of continuous functions.

(2)⇒(1) If I = 1 for all x, y ∈ [0, 1], it is obvious that the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. If I is the form (6), then for any x, y, z ∈ [0, 1], Eq.(4) can be verified as follows: I ( x , O 1 ( y , z ) ) = I ( x , g 1 ( h 1 ( y ) h 1 ( z ) ) ) = g 2 ( ( h 1 ( y ) ) c ( x ) ( h 1 ( z ) ) c ( x ) ) = g 2 ( h 2 ( I ( x , y ) ) h 2 ( I ( x , z ) ) ) = O 2 ( I ( x , y ) , I ( x , z ) ) . □

Example 3.4. Consider O 1 = O 2 = O in Eq.(4) and take its multiplicative generator pair (g, h) as g ( x ) = x and h (x) = x. If c (x) = 1 + x, x ∈ [0, 1], then the solution for Eq.(4) is as follows: I ( x , y ) = y x + 1 , x , y ∈ [ 0 , 1 ] .

Lemma 3.5. Let I be a binary function on [0, 1], and O₁, O₂ be two overlap functions, x ∈ [0, 1]. If the vertical section I (x, ·) has one of the following forms: I ( x , y ) = { 0 if y = 0 , 1 if y ∈ ( 0 , 1 ] , I ( x , y ) = { 0 if y ∈ [ 0 , 1 ) , 1 if y = 1 . then the triple of functions (O₁, O₂, I (x, ·)) satisfies Eq.(4) for all y, z ∈ [0, 1].

Since Eq.(4) is the generalization of a tautology from the classical logic involving Boolean implication, it is reasonable to expect that the solution I of Eq.(4) is also a fuzzy implication.

Corollary 3.6. If O₁, O₂ are two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing, then there is no continuous solution I of Eq.(4) that is fuzzy implication.

Proof. Suppose that fuzzy implication I is a continuous solution of Eq.(4). Since the triple of functions (O₁, O₂, I) satisfies Eq.(4), by Theorem 3.3, I must be I = 1 or Eq.(6). Obviously, I = 1 doesn’t satisfy (I₃). If I is Eq.(6), then I (0, 0) = 0, it is an obvious contradiction with (I₃). Thus, there is no continuous solution I of Eq.(4) that is fuzzy implication.□

For Eq.(6), the conditions of (I₁) and (I₂) can be reached easily. However, from Corollary 3.6, it shows that Eq.(6) fails to be a fuzzy implication only because I (0, 0) = 1 doesn’t hold. So it is reasonable to explore solutions that I is not continuous at point (0, 0).

Lemma 3.7. Let O₁, O₂ be two overlap functions and I: [0, 1] ² ⟶ [0, 1] be a binary function satisfying (I₃). If the triple of functions (O₁, O₂, I) satisfies Eq.(4), then I (0, y) = 1 for all y ∈ [0, 1].

Proof. Let the triple of functions (O₁, O₂, I) satisfy Eq.(4). Substitute x = 0 and z = 0 in Eq.(4), it follows that I (0, O₁ (y, 0)) = O₂ ((I (0, y), I (0, 0)) for all y ∈ [0, 1]. That is I (0, 0) = O₂ ((I (0, y), 1). By (I₃), it holds that O₂ ((I (0, y), 1)=1. Since O₂ satisfies (O₃), so I (0, y) = 1 holds for all y ∈ [0, 1].□

Theorem 3.8. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing and I: [0, 1] ² ⟶ [0, 1] be continuous expect at the point (0, 0) which satisfies (RB) and (I₃). Then the following statements are equivalent: (1) the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) there exists a continuous function c: [0, 1] ⟶ [0, ∞) with c (x) >0 for x ∈ (0, 1] and c (0) = 0 such that I has the form

I ( x , y ) = { 1 if x = y = 0 , h 2 - 1 ( ( h 1 ( y ) ) c ( x ) ) = g 2 ( ( g 1 - 1 ( y ) ) c ( x ) ) otherwise _(14) for all x, y ∈ [0, 1].

(1)⇒(2) Let the triple of functions (O₁, O₂, I) satisfy Eq.(4) for all x, y, z ∈ [0, 1]. Since I is continuous on (0, 1] × [0, 1], similar to the proof of Theorem 3.3, it shows that either I = 1 or there exists a continuous function c′: (0, 1] ⟶ (0, ∞) such that I has the form I ( x , y ) = h 2 - 1 ( ( h 1 ( y ) ) c ′ ( x ) ) = g 2 ( ( g 1 - 1 ( y ) ) c ′ ( x ) ) for all (x, y) ∈ (0, 1] × [0, 1]. Obviously, the solution I = 1 doesn’t satisfy I (1, 0) = 0, so the only possible solution is the latter. And by Lemma 3.7, I (0, y) = 1 holds for all y ∈ [0, 1]. Fix arbitrarily y ∈ (0, 1), from the continuity of I, it holds that lim x → 0 + I ( x , y ) = lim x → 0 + h 2 - 1 ( ( h 1 ( y ) ) c ′ ( x ) ) = 1 . Moreover, h₁, h₂ are bijections, so it follows that h 2 - 1 ( ( h 1 ( y ) ) lim x → 0 + c ′ ( x ) ) = 1 . Since h₁ (y) ∈ (0, 1), so lim x → 0 + c ′ ( x ) = 0 . Define function c: [0, 1] ⟶ [0, ∞) by c ( x ) = { 0 if x = 0 , c ′ ( x ) if y ∈ ( 0 , 1 ] . Obviously, the function c is continuous and I has the form (14).

(2)⇒(1) It can be proved easily.□

The previous proof shows that the function given by Eq.(14) with a continuous function c satisfies condition (I₃) and is increasing with respect to the second variable. Unfortunately, nothing can be said about its monotonicity with respect to the first one. The next result solves this by showing the necessary and sufficient condition.

Theorem 3.9. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing and I be a fuzzy implication that is continuous expect at the point (0, 0). Then the following statements are equivalent: (1) the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) there exists a continuous and increasing function c: [0, 1] ⟶ [0, ∞) with c (x) >0 for x ∈ (0, 1] and c (0) = 0 such that I has the form I ( x , y ) = { 1 if x = y = 0 , h 2 - 1 ( ( h 1 ( y ) ) c ( x ) ) = g 2 ( ( g 1 - 1 ( y ) ) c ( x ) ) otherwise for all x, y ∈ [0, 1].

In this section, solutions to the system of functional equations composed of Eq.(2) and Eq.(4) are discussed. At this situation, if I is discontinuous at the point (0, 0), then I must be discontinuous at point (1, 1). Therefore, the discussion about the solutions to Eq.(4) under the assumption that I is not continuous at the points (0, 0) and (1, 1) is necessary.

Proposition 4.1. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing and I: [0, 1] ² ⟶ [0, 1] be continuous expect at the points (0, 0) and (1, 1) which satisfies (RB) and (I₃). Then the following statements are equivalent: (1) the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) there exists a continuous function c: [0, 1] ⟶ [0, ∞] with 0< c (x) < ∞ for x ∈ (0, 1) and c (0) = 0, c (1) = ∞ such that I has the form I ( x , y ) = { 1 if (x,y)=(0,0) , 0 if (x,y)=(1,0) , h 2 - 1 ( ( h 1 ( y ) ) c ( x ) ) = g 2 ( ( g 1 - 1 ( y ) ) c ( x ) ) otherwise(15) for all x, y ∈ [0, 1].

Proof. (1)⇒(2) Let us assume that I is the solution of Eq.(4) satisfying the required properties. Since I is continuous on (0, 1) × [0, 1], again from the proof of Theorem 3.8, there exists a continuous function c′: [0, 1) ⟶ [0, ∞) with c′ (x) >0 for x ∈ (0, 1) and c′ (0) = 0 such that I ( x , y ) = { 1 i f x = y = 0 h 2 − 1 ( ( h 1 ( y ) ) c ′ ( x ) ) = g 2 ( ( g 1 − 1 ( y ) ) c ′ ( x ) ) o t h e r w i s e for all (x, y) ∈ [0, 1) × [0, 1]. For x = 1, using argumentation for the function I₁ (y) = I (1, y), y ∈ [0, 1), and similar to the proof of Theorem 3.3, it can be obtained that I ( 1 , y ) = { 0 if y ∈ [ 0 , 1 ) , 1 if y = 1 . Finally, fix arbitrarily y ∈ (0, 1), since lim x → 1 - I ( x , y ) = I ( 1 , y ) = 0 , so it holds that lim x → 1 - c ′ ( x ) = ∞ . Define function c: [0, 1] ⟶ [0, ∞] by c ( x ) = { c ′ ( x ) if x ∈ [ 0 , 1 ) , ∞ if x = 1 . Obviously, the function c is continuous and I has the form (15).

(2)⇒(1) It is clear from Theorem 3.3, Lemma 3.5 and Lemma 3.7.□

As a direct consequence of Proposition 4.1, the following result about fuzzy implication is given.

Corollary 4.2. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing and I be a fuzzy implication that is continuous expect at the points (0, 0) and (1, 1). Then the following statements are equivalent: (1) the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) there exists a continuous and increasing function c: [0, 1] ⟶ [0, ∞] with 0< c (x) < ∞ for x ∈ (0, 1) and c (0) = 0, c (1) = ∞ such that I has the form I ( x , y ) = { 1 if (x,y)=(0,0) , 0 if (x,y)=(1,0) , h 2 - 1 ( ( h 1 ( y ) ) c ( x ) ) = g 2 ( ( g 1 - 1 ( y ) ) c ( x ) ) otherwise(16) for all x, y ∈ [0, 1].

Now, the solutions to the system consisting of Eq.(2) and Eq.(4) are given as follows:

Theorem 4.3. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing, I: [0, 1] ² ⟶ [0, 1] be continuous expect at the points (0, 0) and (1, 1) which satisfies (RB) and (I₃), and N be strong fuzzy negation. Then the following statements are equivalent: (1) the quaternion of functions (O₁, O₂, I, N) satisfies Eq.(2) and Eq.(4) for all x, y, z ∈ [0, 1]. (2) there exist constants r₁, r₂ ∈ (- ∞, 0) such that

( h 1 ( x ) ) r 1 = ( g 1 − 1 ( x ) ) r 2 , x ∈ [ 0 , 1 ] ,(17) and I has the form I ( x , y ) = { 1 if ( x , y ) = ( 0 , 0 ) , 0 if ( x , y ) = ( 1 , 0 ) , h 2 − 1 ( ( h 1 ( y ) ) r 1 ln ( h 1 ( N ( x ) ) ) ) = g 2 ( ( g 1 − 1 ( y ) ) r 2 ln ( g 1 − 1 ( N ( x ) ) ) ) otherwise(18) for all x, y ∈ [0, 1].

Proof. (1)⇒(2) Let I be the solution of Eq.(4) satisfying the required properties. Then by Proposition 4.1, I is the form (15). Since I and N satisfy Eq.(2), it can be obtained that for any (x, y) ∈ (0, 1)², h 2 - 1 ( ( h 1 ( y ) ) c ( x ) ) = h 2 - 1 ( ( h 1 ( N ( x ) ) ) c ( N ( y ) ) ) , where c: (0, 1) ⟶ (0, ∞) is a continuous function. So it holds that (h₁ (y)) ^c(x) = (h₁ (N (x))) ^c(N(y)). Take natural logarithms of both sides, i.e.c (x) ln(h₁ (y)) = c (N (y)) ln(h₁ (N (x))). This implies that for all (x, y) ∈ (0, 1) ², c ( x ) ln ( h 1 ( N ( x ) ) ) = c ( N ( y ) ) ln ( h 1 ( y ) ) . Obviously, two sides of this equation are both unary function and independent with each other. Since c (x), ln(h₁ (N (x))), c (N (y)), ln(h₁ (y)) ∉ {0, - ∞ + ∞}, so exists a constant r₁ ∈ (- ∞, 0) such that c (x) = r₁ ln(h₁ (N (x))), x ∈ (0, 1). By the similar way, it can be obtained that there exists a constant r₂ ∈ (- ∞, 0) such that c ( x ) = r 2 ln ( g 1 - 1 ( N ( x ) ) ) , x ∈ ( 0 , 1 ) . So it must hold that ( h 1 ( N ( x ) ) ) r 1 = ( g 1 - 1 ( N ( x ) ) ) r 2 , x ∈ ( 0 , 1 ) . Since N is strong, so it means that ( h 1 ( x ) ) r 1 = ( g 1 - 1 ( x ) ) r 2 for all x ∈ (0, 1). Moreover, x = 0 and x = 1 naturally hold, so Eq.(17) is obtained. Furthermore, by the assumption that c (0) = 0, c (1) = ∞, I is the form (18).

(2)⇒(1) Obviously, (O₁, O₂, I) satisfies Eq.(4). Since N is strong, so we prove the left part through the following cases: If x, y ∈ (0, 1), then this law follows from our construction. If x = 1 and y ∈ (0, 1), then I ( 1 , y ) = h 2 - 1 ( ( h 1 ( y ) ) r 1 ln ( h 1 ( N ( 1 ) ) ) ) = h 2 - 1 ( 0 ) = 0 , I ( N ( y ) , 0 ) = h 2 - 1 ( ( h 1 ( 0 ) ) r 1 ln ( h 1 ( y ) ) ) = 0 . If x = 0 and y ∈ (0, 1], then I ( 0 , y ) = h 2 - 1 ( ( h 1 ( y ) ) r 1 ln ( h 1 ( N ( 0 ) ) ) ) = h 2 - 1 ( 1 ) = 1 , I ( N ( y ) , 1 ) = h 2 - 1 ( ( h 1 ( 1 ) ) r 1 ln ( h 1 ( y ) ) ) = 1 . If x = 1 and y = 0, thenI (N (0), N (1)) = I (1, 0) = 0. If x = 1 and y = 1, then I ( 1 , 1 ) = h 2 - 1 ( ( h 1 ( 1 ) ) r 1 ln ( h 1 ( N ( 1 ) ) ) ) = h 2 - 1 ( 1 ) = 1 , I (N (1), N (1)) = I (0, 0) = 1. Consequently, I satisfies Eq.(2).□

In fact, by a simple analysis, it can seen that Eq.(18) is actually a fuzzy implication, that is the following result:

Corollary 4.4. Let O₁, O₂ be two overlap functions generated, respectively, by the multiplicative generator pairs (g₁, h₁) and (g₂, h₂) with g₁, h₁, g₂, h₂ being strictly increasing, I be a fuzzy implication which is continuous expect at the points (0, 0) and (1, 1), and N be strong fuzzy negation. Then the following statements are equivalent: (1) the quaternion of functions (O₁, O₂, I, N) satisfies Eq.(2) and Eq.(4) for all x, y, z ∈ [0, 1]. (2) there exist two constants r₁, r₂ ∈ (- ∞, 0) such that ( h 1 ( x ) ) r 1 = ( g 1 - 1 ( x ) ) r 2 , x ∈ [0, 1], and I has the form I ( x , y ) = { 1 if ( x , y ) = ( 0 , 0 ) , 0 if ( x , y ) = ( 1 , 0 ) , h 2 − 1 ( ( h 1 ( y ) ) r 1 ln ( h 1 ( N ( x ) ) ) ) = g 2 ( ( g 1 − 1 ( y ) ) r 2 ln ( g 1 − 1 ( N ( x ) ) ) ) otherwise(15) for all x, y ∈ [0, 1].

Not every overlap function can be multiplicatively generated, such as T _M and O ( x , y ) = xy ( x + y ) 2 3 , x, y ∈ [0, 1]. So it is natural to consider whether there exist any other cases that also have solutions for Eq.(2) and Eq.(4). As we all know, either idempotent overlap functions or Archimedean overlap functions have many effective properties, which may be helpful for us to solve some problems. Fortunately, solutions to Eq.(2) and Eq.(4) can be given under the condition that O₁ is an idempotent overlap function and O₂ is an Archimedean overlap function. Firstly, a full characterization for Eq.(4) under this assumption should be given. In this paper, sup  ∅ = 0 and inf  ∅ = 1 are stipulated.

Lemma 5.1. Let O₁ be an overlap function, O₂ be an Archimedean overlap function and I be a binary function on [0, 1]. If the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1], then either I (x, y) = 0 or I (x, y) = 1 holds for every x ∈ [0, 1] and every y ∈ Idem (O₁).

Proof. For every x ∈ [0, 1] and y ∈ Idem (O₁), by virtue of the assumption that the triple of functions (O₁, O₂, I) satisfies Eq.(4), it shows that I (x, y) = I (x, O₁ (y, y)) = O₂ (I (x, y), I (x, y)). That is to say, I (x, y) is an idempotent element of O₂. Since O₂ is an Archimedean overlap function, so by Proposition 2.4, it shows that I (x, y) = 0 or I (x, y) = 1.□

Lemma 5.2. Let O₁, O₂ be overlap functions, I: [0, 1] ² ⟶ [0, 1] be a binary function and y₀ be a fixed idempotent element of O₁. If I (x, y₀) = 0 for some x ∈ [0, 1] and the triple of functions (O₁, O₂, I (x, ·)) satisfies Eq.(4) for all y, z ∈ [0, 1], then it follows I (x, t) = 0 for all t ≤ y₀.

Proof. Let I (x, y₀) = 0 and t ≤ y₀. Because O₁ (·, y₀) is continuous on [0, y₀], and O₁ (0, y₀) = 0, O₁ (y₀, y₀) = y₀, so there exists u ∈ [0, y₀] such that t = O₁ (u, y₀). Since (O₁, O₂, I (x, ·)) satisfies Eq.(4), it holds that I (x, t) = I (x, O₁ (u, y₀)) = O₂ (I (x, u), I (x, y₀)) = O₂ (I (x, u), 0) = 0.

Lemma 5.3. Let O₁ be an overlap function, O₂ be an Archimedean overlap function, I be a binary function on [0, 1] and y₀ be a fixed idempotent element of O₁. If I (x, y₀) = 1 for some x ∈ [0, 1] and the triple of functions (O₁, O₂, I (x, ·)) satisfies Eq.(4) for all y, z ∈ [0, 1], then it follows that I (x, t) = 1 for every t ∈ (↑ y₀) ⋂ Idem (O₁).

Proof. For any t ∈ (↑ y₀) ⋂ Idem (O₁), then t is an idempotent element of O₁, by Lemma 5.1, it shows that I (x, t) = 0 or I (x, t) = 1. Suppose that I (x, t) = 0. Because of t ≥ y₀, then by Lemma 5.2, it can be concluded that I (x, y₀) = 0, which is a contradiction. Consequently, I (x, t) = 1.

Proposition 5.4. Let O₁, O₂ be overlap functions, I: [0, 1] ² ⟶ [0, 1] be a binary function, y₁, y₂ be two different idempotent elements of O₁ and assume that the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. If both I (x, y₁) = 0 and I (x, y₂) = 1 hold for some x ∈ [0, 1], then y₁ < y₂.

Proof. Let I (x, y₁) = 0 and I (x, y₂) = 1 hold for some x ∈ [0, 1]. Suppose that y₁ > y₂, then by Lemma 5.2, I (x, y₂) = 0 holds, which is a contradiction.□

Theorem 5.5. Let O₁ be an idempotent overlap function, O₂ be an Archimedean overlap function and I be a binary function on [0, 1]. Then the following statements are equivalent: (1) the triple of functions (O₁, O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) For any x ∈ [0, 1], the vertical section I (x, ·) belongs to one of the following statements: (a) There exists a constant c _x ∈ [0, 1] with O₁ (c _x , 1) = c _x such that for all y ∈ [0, 1], I ( x , y ) = { 0 if y ≤ c x , 1 if y > c x . (b) There exists a constant c _x ∈ [0, 1] satisfying the condition that y < c _x if and only if O₁ (y, 1) < c _x , such that for all y ∈ [0, 1], I ( x , y ) = { 0 if y < c x , 1 if y ≥ c x .

Proof. (1)⇒(2) Let the triple of functions (O₁, O₂, I) satisfy Eq.(4). For x ∈ [0, 1], since O₁ is idempotent, so by Lemma 5.1, it follows that I (x, y) = 0 or I (x, y) = 1 for all y ∈ [0, 1]. Definec _x = ⋁ {y ∈ [0, 1] ∣ I (x, y) = 0}d _x = ⋀ {y ∈ [0, 1] ∣ I (x, y) = 1}. Clearly, c _x = d _x . (a) If I (x, c _x ) = 0, then by Lemma 5.2, Lemma 5.3 and Proposition 5.4, it shows that I ( x , y ) = { 0 if y ≤ c x , 1 if y > c x . On the one hand, since the triple (O₁, O₂, I (x, ·)) satisfies Eq.(4), so it follows that I (x, O₁ (c _x , 1)) = O₂ (I (x, c _x ), I (x, 1)) = O₂ (0, I (x, 1)) = 0. Furthermore, O₁ (c _x , 1) ≤ c _x holds. On the other hand, by virtue of the monotonicity and idempotency of O₁, it can be obtained that O₁ (c _x , 1) ≥ O₁ (c _x , c _x ) = c _x . Consequently, O₁ (c _x , 1) = c _x . (b) If I (x, c _x ) = 1, similarly, it holds that I ( x , y ) = { 0 if y < c x , 1 if y ≥ c x . Now we show that y < c _x if and only if O₁ (y, 1) < c _x . On the one hand, when y < c _x holds, by Eq.(4), it follows that I (x, O₁ (y, 1)) = O₂ (I (x, y), I (x, 1)) = O₂ (0, I (x, 1)) = 0. Obviously, O₁ (y, 1) < c _x . On the other hand, assume that O₁ (y, 1) < c _x , let us now prove that y < c _x . Assume to the contrary, then by the monotonicity and idempotency of O₁, it holds that O₁ (y, 1) ≥ O₁ (c _x , 1) ≥ O₁ (c _x , c _x ) = c _x , which is a contradiction. Consequently, y < c _x if and only if O₁ (y, 1) < c _x .

(2)⇒(1) For each fixed x ∈ [0, 1]. Two cases are considered.

If I (x, ·) belongs to the case (a). Fix arbitrarily y, z ∈ [0, 1]. If max (y, z) ≤ c _x , then by the monotonicity of O₁, it shows that O₁ (y, z) ≤ O₁ (max (y, z), max (y, z)) = max (y, z) ≤ c _x . So the left side of Eq.(4) is I (x, O₁ (y, z)) = 0 and the right side of Eq.(4) is O₂ (I (x, y), I (x, z)) = O₂ (0, 0) = 0. If min (y, z) ≤ c _x < max (y, z), similarly, it can be obtained that O₁ (y, z) ≤ O₁ (c _x , 1) = c _x . Thus, the left side of Eq.(4) is I (x, O₁ (y, z)) = 0 and the right side of Eq.(4) is O₂ (I (x, y), I (x, z)) = O₂ (I (x, min (y, z)), I (x, max (y, z))) = O₂ (0, 1)   =  0. If c _x < min (y, z), it shows that O₁ (y, z) ≥ O₁ (min (y, z), min (y, z)) = min (y, z) > c _x . So the left side of Eq.(4) is I (x, O₁ (y, z)) = 1 and the right side of Eq.(4) is O₂ (I (x, y), I (x, z)) = O₂ (1, 1) = 1.

If I (x, ·) belongs to the case (b). Fix arbitrarily y, z ∈ [0, 1]. If max (y, z) < c _x , then O₁ (y, z) < c _x . So the left side of Eq.(4) is I (x, O₁ (y, z)) = 0 and the right side of Eq.(4) is O₂ (I (x, y), I (x, z)) = O₂ (0, 0) = 0. If min (y, z) < c _x ≤ max (y, z), then it holds that O₁ (y, z) ≤ O₁ (min (y, z), 1) < c _x . Thus, the left side of Eq.(4) is I (x, O₁ (y, z)) = 0 and the right side of Eq.(4) is O₂ (I (x, y), I (x, z)) = O₂ (0, 1) = 0. If c _x ≤ min (y, z), then it holds that O₁ (y, z) ≥ min (y, z) ≥ c _x . So the left side of Eq.(4) is I (x, O₁ (y, z)) = 1 and the right side of Eq.(4) is O₂ (I (x, y), I (x, z)) = O₂ (1, 1) = 1.□

Corollary 5.6. Let O₁ be the idempotent t-norm T _M , O₂ be an Archimedean overlap function and I be a binary function on [0, 1]. Then the following statements are equivalent: (1) the triple of functions (T _M , O₂, I) satisfies Eq.(4) for all x, y, z ∈ [0, 1]. (2) For any x ∈ [0, 1], there exists a constant c _x ∈ [0, 1] such that the vertical section I (x, ·) has one of the following forms: I ( x , y ) = { 0 if y ≤ c x , 1 if y > c x , I ( x , y ) = { 0 if y < c x , 1 if y ≥ c x .

Remark 5.7. 2 If the set of all continuous Archimedean t-norms and the set of all Archimedean overlap functions are denoted by AT and AO, respectively, then it can be obtained that AT ⊈ AO, AO ⊈ AT and AO⋂ AT ≠ ∅. For instance, O _p (p > 1) ∈ AO, but O _p ∉ AT; the Łukasiewicz t-norm T _LK ∈ AT, but T _LK ∉ AO; the product t-norm T _P ∈ AO ⋂ AT. According to Corollary 3.9 obtained by Qin and Baczyński [29], O₂ should be a continuous Archimedean t-norm, while in Corollary 5.6, we take O₂ as an Archimedean overlap function, so two results are essentially different. However, after a carefully comparison, we can find that the two equations have identical solutions.

Example 5.8. Consider O 1 ( x , y ) = xy ( x + y ) 2 3 , x, y ∈ [0, 1] and O₂ = O _p for p > 1 in Eq.(4), then solutions for Eq.(4) can be given: for any x ∈ [0, 1], the vertical section I (x, ·) has one of the following forms: I(x, y) = 0, y ∈ [0, 1], I(x, y) = 1, y ∈ [0, 1] , I ( x , y ) = { 0 i f y = 0 1 i f y ∈ ( 0 , 1 ] , I ( x , y ) = { 0 i f y ∈ [ 0 , 1 ) , 1 i f y = 1

Corollary 5.9. If O₁ is an idempotent overlap function, O₂ is an Archimedean overlap function, then there is no continuous solution I of Eq.(4) that is fuzzy implication.

Proof. Assume that fuzzy implication I is a continuous solution of Eq.(4). For any x ∈ [0, 1], since the triple of functions (O₁, O₂, I (x, ·)) satisfies Eq.(4), by Theorem 5.5, it holds that I (x, ·) = 0 or I (x, ·) = 1. And the continuity of I also implies I = 0 or I = 1, which is a contradiction with (I₃). Thus, there is no continuous solution I of Eq.(4) that is fuzzy implication.□

Theorem 5.10. Let O₁ be an idempotent overlap function, O₂ be an Archimedean overlap function, I be a fuzzy implication which is continuous expect at the point (1, 0) and N be strong fuzzy negation. Then the following statements are equivalent: (1) the quaternion of functions (O₁, O₂, I, N) satisfies Eq.(2) and Eq.(4) for all x, y, z ∈ [0, 1]. (2) I is the greatest fuzzy implication.

Theorem 5.11. Let O₁ be an idempotent overlap function, O₂ be an Archimedean overlap function, I be a fuzzy implication which is continuous expect at the boundary x = 0, y = 1, and N be strong fuzzy negation. Then the following statements are equivalent: (1) the quaternion of functions (O₁, O₂, I, N) satisfies Eq.(2) and Eq.(4) for all x, y, z ∈ [0, 1]. (2) I is the least fuzzy implication.

6 Conclusion

This work is a further study of the distributivity equation of fuzzy implication over overlap functions, it concentrates on giving solutions for the system composed of this kind of distributivity equation and contraposition symmetry equation. The main contributions of this paper are listed as follows:

(1) Characterizations of continuous binary function I which satisfies Eq.(4) for multiplicatively generated overlap functions are given. Although there is no continuous fuzzy implication fulfilling Eq.(4), sufficient and necessary conditions for fuzzy implication I satisfying Eq.(4) when I is continuous expect for the point (0, 0) are presented.

(2) Equivalent characterizations of the system consisting of Eq.(2) and Eq.(4) are given under the assumption that O₁, O₂ are multiplicatively generated overlap functions, I is a fuzzy implication which is continuous expect at the points (0, 0) and (1, 1), and N is a strong fuzzy negation.

(3) Eq.(4) under the assumption that O₁ is an idempotent overlap function and O₂ is an Archimedean overlap function is studied, and its equivalent characterization is presented. At the same time, solutions to the system composed of Eq.(2) and Eq.(4) are also characterized.

For the further work, we tend to discuss the distributive laws and conditional distributive laws between overlap functions and some other special aggregation functions, such as t-norms, t-conorms, uninorms, nullnorms, semi-norms, semi-t-operators and so on.