On the distributivity of fuzzy implications over representable uninorms

Fuzzy Sets and Systems 161 (2010) 2256 – 2275 www.elsevier.com/locate/fss

On the distributivity of fuzzy implications over representable uninorms夡 Michał Baczy´nski Institute of Mathematics, University of Silesia, 40-007 Katowice, ul. Bankowa 14, Poland Received 12 March 2009; received in revised form 26 February 2010; accepted 1 March 2010 Available online 19 March 2010

Abstract Recently, many works have appeared dealing with the distributivity of fuzzy implications over t-norms, t-conorms and uninorms. These equations have a very important role to play in efficient inferencing in approximate reasoning, especially fuzzy control systems. In this work we deal with the following two distributive equations I (x, U1 (y, z))=U2 (I (x, y), I (x, z)) and I (U1 (x, y), z)= U2 (I (x, z), I (y, z)), where U1 , U2 are given representable uninorms and I is an unknown function, in particular a fuzzy implication. Using the obtained characterizations we show that the first equation does not hold when U1 , U2 are both representable uninorms and I is a continuous fuzzy implication. Therefore, we present solutions of the first equation which are fuzzy implications with continuous vertical sections. For the second equation the situation is quite different—when U1 , U2 are both representable uninorms, then there are no solutions which are fuzzy implications. As a byproduct result we have obtained solutions of the additive Cauchy functional equation for an unknown function f : [−∞, ∞] → [−∞, ∞] with different meaning of additions ∞ + (−∞) and (−∞) + ∞. © 2010 Elsevier B.V. All rights reserved. MSC: 03B52; 03E72; 39B52 Keywords: Fuzzy connectives; Fuzzy implication; Functional equations; Uninorm

1. Introduction Distributivity of fuzzy implications over different fuzzy logic connectives has been studied in the recent past by many authors. This interest, perhaps, was kick started by Combs and Andrews in [9] wherein they exploit the following classical tautology: ( p ∧ q) → r ≡ ( p → r ) ∨ (q → r ) in their inference mechanism towards reduction in the complexity of fuzzy “If–Then” rules. Subsequently, there were many discussions (see [10,16]), most of them pointing out the need for a theoretical investigation required for employing such equations in practice. 夡 Expanded version of a talk presented at the 12th International Conference on Information Processing and Management of Uncertainty in Knowledge-Based Systems, IPMU 2008 (Málaga (Torremolinos), Spain, June 22–27, 2008). E-mail address: [email protected]. 0165-0114/$ - see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2010.03.005

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It was Trillas and Alsina [19], who were the first to investigate the following generalized version of the above law: I (T (x,y),z) = S(I (x,z),I (y,z)), x,y,z ∈ [0,1],

(1)

where T, S are a t-norm and a t-conorm, respectively, and I is a fuzzy implication. Using similar techniques as above, Balasubramaniam and Rao [7] considered the following dual equations of (1): I (S(x,y),z) = T (I (x,z),I (y,z)),

(2)

I (x,T1 (y,z)) = T2 (I (x,y),I (x,z)),

(3)

I (x,S1 (y,z)) = S2 (I (x,y),I (x,z)),

(4)

where again T, T1 , T2 and S, S1 , S2 are t-norms and t-conorms, respectively, and I is a fuzzy implication. In both papers it was shown that when I is either an R-implication obtained from a left-continuous t-norm or an S-implication, in almost all the cases the distributivity equations (1)–(4) hold if and only if T1 = T2 = T = min and S1 = S2 = S = max. In fact, Eq. (4) for the case when I is an R-implication obtained from a strict t-norm was left unsolved in [7], but this situation was considered by Baczy´nski and Jayaram in [6], where they characterized functions I, which satisfy the functional equation (4), when S1 ,S2 are either both strict, or nilpotent t-conorms. These results have been generalized in [4] for continuous and Archimedean triangular conorms. Meanwhile, the author in [2,3] considered the functional equation (3) and characterized fuzzy implications I in the case when T1 =T2 is a strict t-norm. In this paper we investigate the other possible generalization, i.e., we concentrate on the following two distributive equations: I (x,U1 (y,z)) = U2 (I (x,y),I (x,z)),

(D1)

I (U1 (x,y),z) = U2 (I (x,z),I (y,z)),

(D2)

by characterizing functions I (in particular fuzzy implications), which satisfy the above equations for all x, y, z ∈ [0, 1], when U1 and U2 are given representable uninorms. It should be noted that Eq. (D2) was studied by Ruiz-Aguilera and Torrens in [17] for the major part of known classes of uninorms with continuous underlying t-norm and t-conorm and for strong implications derived from uninorms, while in [18], they also studied (D2), but with the assumption that I is a residual implication derived from a given uninorm. Observe that the general solutions of the distributive equation F(x,G(y,z)) = G(F(x,z),F(y,z)), where F is continuous and G is assumed to be continuous, strictly increasing and associative were presented by Aczél (see [1, Theorem 6, p. 319]). Our results can be seen as a generalization of the above-mentioned result without any assumptions on the function F and less assumptions on the function G. 2. Basic fuzzy logic connectives We assume that the reader is familiar with the classical results concerning uninorms, so we only recall basic definitions and facts which will be useful in the sequel. Uninorms were introduced by Yager and Rybalov in 1996 (see [20]) as a generalization of triangular norms and conorms. For the recent overview of this family of operations see [11]. Definition 2.1. An associative, commutative and increasing operation U : [0, 1]2 → [0, 1] is called a uninorm, if there exists e ∈ [0, 1], called the neutral element of U, such that U (e, x) = U (x, e) = x, for all x ∈ [0, 1]. Remark 2.2. (i) If e = 0, then U is a t-conorm and if e = 1, then U is a t-norm. (ii) The neutral element e corresponding to a uninorm U is unique. (iii) For any uninorm U we have U(0,0) = 0 and U(1,1) = 1 and U (0, 1) ∈ {0, 1}. A uninorm U such that U(0,1) = 0 is called conjunctive and if U(0,1) = 1, then it is called disjunctive.

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In the literature one can find several different classes of uninorms (see [13]). We only mention relevant details and results, which will be useful in the sequel, connected with the one class. Uninorms that can be represented as in Theorem 2.3 are called representable uninorms. Theorem 2.3 (Fodor et al. [13, Theorem 3]). For a function U : [0, 1]2 → [0, 1] the following statements are equivalent: (i) U is a strictly increasing and continuous on ]0,1[2 uninorm with the neutral element e ∈]0, 1[ such that U is self-dual, except in points (0,1) and (1,0), with respect to a strong negation N with the fixed point e, i.e., U (x,y) = N (U (N (x),N (y))), x,y ∈ [0,1]2 \ {(0,1),(1,0)}. (ii) U has a continuous additive generator, i.e., there exists a continuous and strictly increasing function h: [0, 1] → [−∞, ∞], such that h(0) = −∞, h(e) = 0 for e ∈]0, 1[ and h(1) = ∞, which is uniquely determined up to a positive multiplicative constant, such that either  0 i f (x,y) ∈ {(0,1),(1,0)}, U (x,y) = x,y ∈ [0,1], (5) −1 h (h(x) + h(y)) other wise, when U is conjunctive, or  1 i f (x,y) ∈ {(0,1),(1,0)}, U (x,y) = −1 h (h(x) + h(y)) other wise,

x,y ∈ [0,1],

(6)

when U is disjunctive. Example 2.4. For a continuous additive generator h(x) = ln(x/(1 − x)) we get the following conjunctive representable uninorm: ⎧ if (x,y) ∈ {(0,1),(1,0)}, ⎨0 xy x,y ∈ [0,1], U (x,y) = ⎩ otherwise, (1 − x)(1 − y) + x y and the following disjunctive representable uninorm: ⎧ if (x,y) ∈ {(0,1),(1,0)}, ⎨1 xy U (x,y) = ⎩ otherwise, (1 − x)(1 − y) + x y In both situations e =

1 2

x,y ∈ [0,1].

and N(x) = 1−x.

Remark 2.5. One can easily observe that if a representable uninorm U is conjunctive, then the representation U(x,y)=h−1 (h(x)+h(y)) holds for all x, y ∈ [0, 1] with the assumption that (−∞) + ∞ = ∞ + (−∞) = −∞.

(A−)

Similarly, if a representable uninorm U is disjunctive, then U(x,y)=h−1 (h(x)+h(y)) holds for all x, y ∈ [0, 1] with the assumption that (−∞) + ∞ = ∞ + (−∞) = ∞.

(A+)

In the literature we can find several diverse definitions of fuzzy implications. In this article we will use the following one, which is equivalent to the well accepted definition proposed by Fodor and Roubens [12, Definition 1.15] (see also Kitainik [14, p. 50], Bustince et al. [8] and Baczy´nski and Jayaram [5]).

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Definition 2.6. A function I : [0, 1]2 → [0, 1] is called a fuzzy implication if it satisfies the following conditions: I is decreasing in the first variable,

(I1)

I is increasing in the second variable,

(I2)

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

(I3)

Directly from the above definition we see that each fuzzy implication I satisfies the following left and right boundary conditions, respectively: I (0,y) = 1, y ∈ [0,1],

(LB)

I (x,1) = 1, x ∈ [0,1].

(RB)

Therefore, I satisfies also the normality condition I (0,1) = 1,

(NC)

and, consequently, every fuzzy implication restricted to the set {0,1}2 coincides with the classical implication. 3. Preliminary results Our main goal in this paper is to present the representations of some classes of fuzzy implications that satisfy Eqs. (D1) and (D2) when U1 , U2 are representable uninorms. Within this context, we firstly show that only some cases are needed to be investigated (cf. [17, Remark 3]). Lemma 3.1. Let a function I : [0, 1]2 → [0, 1] satisfy (I3) and also (D1) with some uninorms U1 , U2 . Then U1 is conjunctive if and only if U2 is conjunctive. Proof. Firstly, substituting x=y = 1 and z = 0 into (D1), we get I (1,U1 (1,0)) = U2 (I (1,1),I (1,0)).

(7)

Now, if U1 is conjunctive, then U1 (1,0) = 0 and by (I3) we get I(1,0)=U2 (1,0), thus 0=U2 (1,0), i.e., U2 is also a conjunctive uninorm. Instead, if U1 is disjunctive, then U1 (1,0) = 1 and we get from (7) and (I3) that 1=U2 (1,0), i.e., U2 is also a disjunctive uninorm.  However, quite contrastingly, we have the following fact which is easy to prove as the above one. Lemma 3.2. Let a function I : [0, 1]2 → [0, 1] satisfy (I3) and also (D2) with some uninorms U1 , U2 . Then U1 is conjunctive if and only if U2 is disjunctive. Proof. Firstly, substituting x=z = 0 and y = 1 into (D2), we get I (U1 (0,1),0) = U2 (I (0,0),I (1,0)).

(8)

Now, if U1 is conjunctive, then U1 (0,1) = 0 and by (I3) we get I(0,0)=U2 (1,0), thus 1=U2 (1,0), i.e., U2 is a disjunctive uninorm. Instead, if U1 is disjunctive, then U1 (0,1) = 1 and we get from (8) and (I3) that 0=U2 (1,0), i.e., U2 is a conjunctive uninorm.  By the above results it is enough, in our context, to consider the functional equation (D1) only when both uninorms U1 , U2 are either conjunctive or disjunctive uninorms, and the functional equation (D2) only when either U1 is conjunctive and U2 is disjunctive, or U1 is disjunctive and U2 is conjunctive.

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4. Some new results pertaining to functional equations Here we show some new results related to the additive Cauchy functional equation: f (x + y) = f (x) + f (y)

(C)

for a function f : [−∞, ∞] → [−∞, ∞] and x, y ∈ [−∞, ∞]. It should be noted that the main problem in this context is with the adequate definition of the additions ∞ + (−∞) and (−∞) + ∞. Therefore we consider all possible situations, which will be important for applications in uninorms. The presented facts, which are new and crucial in the proofs of the main theorems, can be seen as the generalizations of the some theorems from the theory of functional equations (see [1,15]). Proposition 4.1. Let X = Y = [−∞, ∞]. For a function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A−), i.e., (−∞) + ∞ = ∞ + (−∞) = −∞ in both sets X and Y. (ii) Either f = −∞, or f = 0, or f = ∞, or  f (x) = or

0 

f (x) = or



f (x) =

i f x ∈] − ∞,∞],

−∞ i f x = −∞, ∞



i f x ∈] − ∞,∞],

∞ i f x = −∞, 0

f (x) = or

−∞ i f x = −∞,

i f x ∈] − ∞,∞],

−∞ i f x ∈ {−∞,∞}, ∞

i f x ∈ R,

or there exists a unique additive function g: R → R such that  −∞ i f x ∈ {−∞,∞}, f (x) = g(x) i f x ∈ R, or f (x) =



∞

i f x ∈ {−∞,∞},

g(x) i f x ∈ R,

(9)

(10)

(11)

(12)

(13)

(14)

or

⎧ ⎪ ⎨ −∞ i f x = −∞, f (x) = g(x) i f x ∈ R, ⎪ ⎩ ∞ i f x = ∞.

(15)

Proof. (ii) ⇒ (i) It is a direct calculation that all the above functions satisfy Eq. (C) with the assumption (A−) in the domain and codomain.

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(i) ⇒ (ii) Let X = Y = [−∞, ∞] and let f : X → Y satisfy (C) for all x, y ∈ X with the assumption (A−) in X and Y. Firstly, observe that the situation when both f (−∞) = ∞ and f (∞) = −∞ is not possible. Indeed, if we assume the above, then by (A−) we get ∞ = f (−∞) = f ((−∞) + ∞) = f (−∞) + f (∞) = ∞ + (−∞) = −∞, a contradiction. Setting x = y = −∞ in (C) we get f (−∞) = f (−∞) + f (−∞), therefore f (−∞) = −∞, or f (−∞) = 0, or f (−∞) = ∞. If f (−∞) = 0, then for any x ∈ [−∞, ∞] we have 0 = f (−∞) = f ((−∞) + x) = f (−∞) + f (x) = 0 + f (x) = f (x), thus we obtain the first possible solution f = 0. Setting x = y = ∞ in (C) we get f (∞) = f (∞) + f (∞), therefore f (∞) = −∞, or f (∞) = 0, or f (∞) = ∞. If f (∞) = 0, then for any x ∈] − ∞, ∞] we have 0 = f (∞) = f (∞ + x) = f (∞) + f (x) = 0 + f (x) = f (x), thus considering the remaining values for −∞ (obtained earlier) we get next possible solutions (9) or (10). Setting x = y = 0 in (C) we get f(0) = f(0) + f(0), so f (0) = −∞, or f(0) = 0, or f (0) = ∞. If f (0) = −∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + (−∞) = −∞, thus we obtain the solution f = −∞. If f (0) = ∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + ∞, so for every fixed x ∈ [−∞, ∞] we get that either f (x) = −∞ or f (x) = ∞. But for all x ∈ R we have ∞ = f (0) = f (x + (−x)) = f (x) + f (−x), hence, by our assumption (A−) we obtain that f (x) = ∞ for all x ∈ R. By the first step of our proof and our remaining values for −∞ and ∞ we get the next three possible solutions: f = ∞, or (11), or (12). Let us assume now that f(0) = 0. See that for any x ∈ R we have 0 = f (0) = f (x + (−x)) = f (x) + f (−x), so f (x) ∈ R for every x ∈ R. Therefore there exists a unique additive function g: R → R such that f(x) = g(x) for x ∈ R. Taking again into account the previous steps of our proof we obtain the last three possible solutions, i.e., f has either the form (13), or (14), or (15).  By the well known continuous solutions of the additive Cauchy functional equation for real numbers (see [1,15, Theorem 5.2.1]) we get Corollary 4.2. Let X = Y = [−∞, ∞]. For a continuous function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A−), i.e., (−∞) + ∞ = ∞ + (−∞) = −∞ in both sets X and Y. (ii) Either f = −∞, or f = 0, or f = ∞, or there exists a unique constant c ∈]0, ∞[ such that f (x) = cx, for all x ∈ [−∞, ∞]. Proof. (ii) ⇒ (i) Obvious.

(16)

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(i) ⇒ (ii) Let us assume that f is a continuous solution of (C). From Proposition 4.1 we see that the only possible continuous solutions are the constant one or that given by (15). But in this case g is continuous, so g(x) = cx, for x ∈ R, with some c ∈ R. Since f is continuous we get that c > 0.  Proposition 4.3. Let X = Y = [−∞, ∞]. For a function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A+), i.e., (−∞) + ∞ = ∞ + (−∞) = ∞ in both sets X and Y. (ii) Either f = −∞, or f = 0, or f = ∞, or  0 i f x ∈ [−∞,∞[, f (x) = (17) −∞ i f x = ∞, or



f (x) = or

f (x) =

i f x ∈ [−∞,∞[,

∞ i f x = ∞, 

f (x) = or

0

−∞ i f x ∈ [−∞,∞[, ∞



i f x = ∞,

−∞ i f x ∈ R, ∞

i f x ∈ {−∞,∞},

(18)

(19)

(20)

or there exists a unique additive function g: R → R such that f has the form (13), or (14), or (15). Proof. (ii) ⇒ (i) It is a direct calculation that all the above functions satisfy Eq. (C) with the assumption (A+) in the domain and codomain. (i) ⇒ (ii) Let X = Y = [−∞, ∞] and let f : X → Y satisfy (C) for all x, y ∈ X with the assumption (A+) in X and Y. Firstly, observe that the situation when both f (−∞) = ∞ and f (∞) = −∞ is not possible. Indeed, if we assume the above, we get −∞ = f (∞) = f ((−∞) + ∞) = f (−∞) + f (∞) = ∞ + (−∞) = ∞, a contradiction. Setting x = y = ∞ in (C) we get f (∞) = f (∞) + f (∞), therefore f (∞) = −∞, or f (∞) = 0, or f (∞) = ∞. If f (∞) = 0, then for any x ∈ [−∞, ∞] we have 0 = f (∞) = f (∞ + x) = f (∞) + f (x) = 0 + f (x) = f (x), thus we obtain the first possible solution f = 0. Setting x = y = −∞ in (C) we get f (−∞) = f (−∞) + f (−∞), therefore f (−∞) = −∞, or f (−∞) = 0, or f (−∞) = ∞. If f (−∞) = 0, then for any x ∈] − ∞, ∞] we have 0 = f (−∞) = f (−∞ + x) = f (−∞) + f (x) = 0 + f (x) = f (x), thus considering the remaining values for ∞ (obtained earlier) we get next possible solutions (17) or (18). Setting x = y = 0 in (C) we get f(0) = f(0) + f(0), so f (0) = −∞, or f(0) = 0, or f (0) = ∞. If f (0) = ∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + ∞ = ∞, thus we obtain the solution f = ∞. If f (0) = −∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + (−∞),

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so for every fixed x ∈ [−∞, ∞] we get that either f (x) = −∞ or f (x) = ∞. But for all x ∈ R we have −∞ = f (0) = f (x + (−x)) = f (x) + f (−x), hence, by our assumption (A+) we obtain that f (x) = −∞ for all x ∈ R. By the first step of our proof and our remaining values for −∞ and ∞ we get the next three possible solutions: f = −∞, or (19) or (20). Let us assume now that f(0) = 0. See that for all x ∈ R we have 0 = f (0) = f (x + (−x)) = f (x) + f (−x), so f (x) ∈ R for every x ∈ R. Therefore there exists a unique additive function g: R → R such that f(x) = g(x) for x ∈ R. Taking again into account the previous steps of our proof we obtain the last three possible solutions, i.e., f has either the form (13), or (14), or (15).  Corollary 4.4. Let X = Y = [−∞, ∞]. For a continuous function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A+), i.e., (−∞) + ∞ = ∞ + (−∞) = ∞ in both sets X and Y. (ii) Either f = −∞, or f = 0, or f = ∞, or there exists a unique constant c ∈]0, ∞[ such that f has the representation (16) for all x ∈ [−∞, ∞]. Proposition 4.5. Let X = Y = [−∞, ∞]. For a function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A−), i.e., (−∞) + ∞ = ∞ + (−∞) = −∞ in the set X and the assumption (A+), i.e., (−∞) + ∞ = ∞ + (−∞) = ∞ in the set Y. (ii) Either f = −∞, or f = 0, or f = ∞, or f has the form (9), or (10), or (20), or  ∞ i f x = −∞, f (x) = (21) −∞ i f x ∈] − ∞,∞], or there exists a unique additive function g: R → R such that f has the form (13), or (14), or ⎧ ⎪ ⎨ ∞ i f x = −∞, f (x) = g(x) i f x ∈ R, ⎪ ⎩ −∞ i f x = ∞.

(22)

Proof. (ii) ⇒ (i) It is a direct calculation that all the above functions satisfy Eq. (C) with the assumption (A−) in the domain and the assumption (A+) in the codomain. (i) ⇒ (ii) Let f : X → Y satisfy (C) for all x, y ∈ X with the assumption (A−) in the set X and the assumption (A+) in the set Y. Firstly, observe that the situation when both f (−∞) = −∞ and f (∞) = ∞ is not possible. Indeed, if we assume the above, we get −∞ = f (−∞) = f ((−∞) + ∞) = f (−∞) + f (∞) = (−∞) + ∞ = ∞, a contradiction. Setting x = y = −∞ in (C) we get f (−∞) = f (−∞) + f (−∞), therefore f (−∞) = −∞, or f (−∞) = 0, or f (−∞) = ∞. If f (−∞) = 0, then for any x ∈ [−∞, ∞] we have 0 = f (−∞) = f ((−∞) + x) = f (−∞) + f (x) = 0 + f (x) = f (x), thus we obtain the first possible solution f = 0. Setting x = y = ∞ in (C) we get f (∞) = f (∞) + f (∞), therefore f (∞) = −∞, or f (∞) = 0, or f (∞) = ∞. If f (∞) = 0, then for any x ∈] − ∞, ∞] we have 0 = f (∞) = f (∞ + x) = f (∞) + f (x) = 0 + f (x) = f (x),

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thus considering the remaining values for −∞ (obtained earlier) we get next possible solutions (9) or (10). Setting x = y = 0 in (C) we get f(0) = f(0) + f(0), so f (0) = −∞, or f(0) = 0, or f (0) = ∞. If f (0) = ∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + ∞ = ∞, thus we obtain the next solution f = ∞. If f (0) = −∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + (−∞), so for every fixed x ∈ [−∞, ∞] we get that either f (x) = −∞ or f (x) = ∞. But for all x ∈ R we have −∞ = f (0) = f (x + (−x)) = f (x) + f (−x), hence, by our assumption (A+) in the set Y we obtain that f (x) = −∞ for all x ∈ R. By the first step of our proof and our remaining values for −∞ and ∞ we get the next three possible solutions: f = −∞, (20) or (21). Let us assume now that f(0) = 0. See that for all x ∈ R we have 0 = f (0) = f (x + (−x)) = f (x) + f (−x), so f (x) ∈ R for every x ∈ R. Therefore there exists a unique additive function g: R → R such that f(x) = g(x) for x ∈ R. Taking again into account the previous steps of our proof we obtain the last three possible solutions, i.e., f has either the form (13), or (14), or (22).  Corollary 4.6. Let X = Y = [−∞, ∞]. For a continuous function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A−), i.e., (−∞)+∞ = ∞ + (−∞) = −∞ in the set X and the assumption (A+), i.e., (−∞) + ∞ = ∞ + (−∞) = ∞ in the set Y. (ii) Either f = −∞, or f = 0, or f = ∞, or there exists a unique constant c ∈]−∞, 0[ such that f has the representation (16) for all x ∈ [−∞, ∞]. Proposition 4.7. Let X = Y = [−∞, ∞]. For a function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A+), i.e., (−∞) + ∞ = ∞ + (−∞) = ∞ in the set X and the assumption (A−), i.e., (−∞) + ∞ = ∞ + (−∞) = −∞ in the set Y. (ii) Either f = −∞, or f = 0, or f = ∞, or f has the form (12), or (17), or (18), or  ∞ i f x = −∞, f (x) = (23) −∞ i f x ∈] − ∞,∞], or there exists a unique additive function g: R → R such that f has the form (13), or (14), or (22). Proof. (ii) ⇒ (i) It is a direct calculation that all the above functions satisfy Eq. (C) with the assumption (A+) in the domain and the assumption (A−) in the codomain. (i) ⇒ (ii) Let f : X → Y satisfy (C) for all x, y ∈ X with the assumption (A+) in the set X and the assumption (A−) in the set Y. Firstly, observe that the situation when both f (−∞) = −∞ and f (∞) = ∞ is not possible. Indeed, if we assume the above, we get ∞ = f (∞) = f ((−∞) + ∞) = f (−∞) + f (∞) = (−∞) + ∞ = −∞, a contradiction. Setting x = y = ∞ in (C) we get f (∞) = f (∞) + f (∞), therefore f (∞) = −∞, or f (∞) = 0, or f (∞) = ∞. If f (∞) = 0, then for any x ∈ [−∞, ∞] we have 0 = f (∞) = f (∞ + x) = f (∞) + f (x) = 0 + f (x) = f (x), thus we obtain the first possible solution f = 0.

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Setting x = y = −∞ in (C) we get f (−∞) = f (−∞) + f (−∞), therefore f (−∞) = −∞, or f (−∞) = 0, or f (−∞) = ∞. If f (−∞) = 0, then for any x ∈] − ∞, ∞] we have 0 = f (−∞) = f (−∞ + x) = f (−∞) + f (x) = 0 + f (x) = f (x), thus considering the remaining values for ∞ (obtained earlier) we get next possible solutions (17) or (18). Setting x = y = 0 in (C) we get f(0) = f(0) + f(0), so f (0) = − ∞, or f(0) = 0, or f (0) = ∞. If f (0) = −∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + (−∞) = −∞, thus we obtain the next solution f = −∞. If f (0) = ∞, then for any x ∈ [−∞, ∞] we get f (x) = f (x + 0) = f (x) + f (0) = f (x) + ∞, so for every fixed x ∈ [−∞, ∞] we get that either f (x) = −∞ or f (x) = ∞. But for all x ∈ R we have ∞ = f (0) = f (x + (−x)) = f (x) + f (−x), hence, by our assumption (A−) in the set Y we obtain that f (x) = ∞ for all x ∈ R. By the first step of our proof and our remaining values for −∞ and ∞ we get the next three possible solutions: f = ∞, (12) or (23). Let us assume now that f(0) = 0. See that for all x ∈ R we have 0 = f (0) = f (x + (−x)) = f (x) + f (−x), so f (x) ∈ R for every x ∈ R. Therefore there exists a unique additive function g: R → R such that f(x) = g(x) for x ∈ R. Taking again into account the previous steps of our proof we obtain the last three possible solutions, i.e., f has either the form (13), or (14), or (22).  Corollary 4.8. Let X = Y = [−∞, ∞]. For a continuous function f : X → Y the following statements are equivalent: (i) f satisfies the additive Cauchy functional equation (C) for all x, y ∈ X , with the assumption (A+), i.e., (−∞) + ∞ = ∞ + (−∞) = ∞ in the set X and the assumption (A−), i.e., (−∞) + ∞ = ∞ + (−∞) = −∞ in the set Y. (ii) Either f = −∞, or f = 0, or f = ∞, or there exists a unique constant c ∈]−∞, 0[ such that f has the representation (16) for all x ∈ [−∞, ∞]. 5. On Eq. (D1) for representable uninorms 5.1. General solutions of (D1) when U1 , U2 are both conjunctive representable uninorms In this part we will show main results for the first equation (D1) when U1 , U2 are both conjunctive representable uninorms. Theorem 5.1. For conjunctive representable uninorms U1 , U2 with the neutral elements e1 , e2 ∈]0, 1[ and a function I : [0, 1]2 → [0, 1] the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D1) for all x, y, z ∈ [0, 1]. (ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative constants, such that U1 , U2 admit the representation (5) with h1 , h2 , respectively, and for every fixed x ∈ [0, 1], the vertical section I (x, ·) has, for all y ∈ [0, 1], one of the following forms: I (x,y) = 0,

(24)

I (x,y) = e2 ,

(25)

I (x,y) = 1,

(26)

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 I (x,y) =  I (x,y) =  I (x,y) =  I (x,y) =  I (x,y) =  I (x,y) =

0

i f y = 0,

(27)

e2 i f y ∈]0,1], 1

i f y = 0,

(28)

e2 i f y ∈]0,1], 0 i f y = 0,

(29)

1 i f y ∈]0,1], 0 i f y = 0 or y = 1, 1 i f y ∈]0,1[, i f y = 0 or y = 1,

0

h −1 2 (gx (h 1 (y))) i f y ∈]0,1[, 1

i f y = 0 or y = 1,

h −1 2 (gx (h 1 (y)))

i f y ∈]0,1[,

⎧ 0 i f y = 0, ⎪ ⎨ −1 I (x,y) = h 2 (gx (h 1 (y))) i f y ∈]0,1[, ⎪ ⎩ 1 i f y = 1,

(30)

(31)

(32)

(33)

with a unique additive function gx : R → R. Moreover, if h 1 (y) = ah 1 (y), h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[, and we assume that

−1

h −1 2 (gx (h 1 (y))) = h 2 (gx (h 1 (y))), y ∈]0,1[,

for some x ∈ [0, 1], then gx (u) = bgx (u/a), for all u ∈ R. Proof. (ii) ⇒ (i) The proof in this direction can be checked by a direct substitution. (i) ⇒ (ii) Let us assume that uninorms U1 , U2 and a function I are solutions of the functional equation (D1) satisfying the required properties. By the definition of a representable uninorm, U1 and U2 admit the representation (5) for some continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞. Moreover, both generators are uniquely determined up to positive multiplicative constants. Now, by Remark 2.5, Eq. (D1) becomes −1 I (x,h −1 1 (h 1 (y) + h 1 (z))) = h 2 (h 2 (I (x,y)) + h 2 (I (x,z))), x,y,z ∈ [0,1],

with the assumption (A−) in the codomains of h1 and h2 . Let x ∈ [0, 1] be arbitrarily fixed. Define a function I x : [0, 1] → [0, 1] by the formula I x (y) = I (x,y), y ∈ [0,1]. By routine substitutions, f x = h 2 ◦ I x ◦ h −1 1 , u = h1 (y), v = h1 (z), for y, z ∈ [0, 1], we obtain the following additive Cauchy functional equation: f x (u + v) = f x (u) + f x (v), u,v ∈ [−∞,∞], where f x : [−∞, ∞] → [−∞, ∞] is an unknown function, with the assumption (A−) in its domain and its codomain. Proposition 4.1 describes all solutions fx . Because of the definition of the function fx we get all our formulas. We show now the second part of point (ii). Let us assume that for some x ∈ [0, 1] the vertical section is given by the function with gx . Let h 1 (y) = ah 1 (y) and h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[. If we assume that

−1

h −1 2 (gx (h 1 (y))) = h 2 (gx (h 1 (y))), y ∈]0,1[,

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with some additive functions gx and gx , then using the formula for the inverse of the bijection h 2 we get  

−1 −1 gx (h 1 (y)) , y ∈]0,1[, h 2 (gx (h 1 (y))) = h 2 b so gx (h 1 (y)) =

gx (ah 1 (y)) , y ∈]0,1[, b

thus, for every u ∈ R we obtain u

.  gx (u) = bgx a Since we are interested in finding solutions of (D1) in the fuzzy logic context, we can easily obtain an infinite number of solutions which are fuzzy implications. It should be noted that with this assumption only the vertical sections (26), (29) and (33) are possible, while for x = 0 the vertical section should be (26). We should also remember that a fuzzy implication is decreasing in the first variable while it is increasing in the second one. Example 5.2. (i) If U1 and U2 are conjunctive representable uninorms, then the greatest solution of (D1) which is a fuzzy implication is the greatest fuzzy implication (see [5]):  0 if x = 1 and y = 0, I1 (x,y) = 1 otherwise. The vertical sections are the following: for x ∈ [0, 1[ it is the solution (26) and for x = 1 it is the solution (29) in Theorem 5.1. (ii) If U1 and U2 are both conjunctive representable uninorms with the neutral elements e1 and e2 , then the following fuzzy implication: ⎧ ⎪ ⎨ 1 if x = 0 or y = 1, I (x,y) = e2 if x > 0 and y ∈]0,1[, ⎪ ⎩ 0 otherwise, satisfies (D1) with U1 and U2 . The vertical sections are the following: for x = 0 it is the solution (26) and for x ∈]0, 1] it is the solution (33) with gx = 0. 5.2. General solutions of (D1) when U1 , U2 are both disjunctive representable uninorms In this part we will show main results for the first equation (D1) when U1 , U2 are both disjunctive representable uninorms. By Proposition 4.3 and using similar deduction as in the proof of Theorem 5.1 we get Theorem 5.3. For disjunctive representable uninorms U1 , U2 with the neutral elements e1 , e2 ∈]0, 1[ and a function I : [0, 1]2 → [0, 1] the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D1) for all x, y, z ∈ [0, 1]. (ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative constants, such that U1 , U2 admit the representation (6) with h1 , h2 , respectively, and for every fixed x ∈ [0, 1], the vertical section I (x, ·) has, for all y ∈ [0, 1], one of the following forms: I (x,y) = 0,

(34)

I (x,y) = e2 ,

(35)

I (x,y) = 1,

(36)

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 I (x,y) =

0 

I (x,y) = I (x,y) =  I (x,y) =  I (x,y) = 

(37)

i f y = 1,

e2 i f y ∈ [0,1[, 1



I (x,y) =

e2 i f y ∈ [0,1[,

(38)

i f y = 1,

0 i f y ∈ [0,1[,

(39)

1 i f y = 1, 0 i f y ∈]0,1[, 1 i f y = 0 or y = 1, i f y = 0 or y = 1,

0

h −1 2 (gx (h 1 (y))) i f y ∈]0,1[, 1,

i f y = 0 or y = 1,

h −1 2 (gx (h 1 (y))),

i f y ∈]0,1[,

⎧ 0 i f y = 0, ⎪ ⎨ −1 I (x,y) = h 2 (gx (h 1 (y))) i f y ∈]0,1[, ⎪ ⎩ 1 i f y = 1,

(40)

(41)

(42)

(43)

with a unique additive function gx : R → R. Moreover, if h 1 (y) = ah 1 (y), h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[, and we assume that

−1

h −1 2 (gx (h 1 (y))) = h 2 (gx (h 1 (y))), y ∈]0,1[,

for some x ∈ [0, 1], then gx (u) = bgx (u/a), for all u ∈ R. Here again we can present an infinite number of solutions of (D1) which are fuzzy implications. It should be noted that with this assumption only the vertical sections (36), (38), (39) and (43) are possible, while for x = 0 the vertical section should be (36). We should also remember that a fuzzy implication is decreasing in the first variable while it is increasing in the second one. Example 5.4. If U1 and U2 are disjunctive representable uninorms, then the least solution which is a fuzzy implication is the least fuzzy implication (see [5]):  1 if x = 0 or y = 1, I0 (x,y) = 0 otherwise. The vertical sections are the following: for x = 0 it is the solution (36) and for x ∈]0, 1] it is the solution (39) in Theorem 5.3. 5.3. Continuous solutions of (D1) Observe firstly that continuous solutions of the additive Cauchy functional equation (C) are the same in Corollaries 4.2 and 4.4. Thus, by Theorems 5.1 and 5.3, we are in a position to describe the continuous solutions I of (D1) for representable uninorms. Theorem 5.5. Let U1 , U2 be both conjunctive (or disjunctive) representable uninorms with the neutral elements e1 , e2 ∈]0, 1[. For a continuous function I : [0, 1]2 → [0, 1] the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D1) for all x, y, z ∈ [0, 1].

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(ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative constants, such that U1 , U2 admit the representation (5) (or (6)) with h1 , h2 , respectively, and either I = 0, or I = e2 , or I = 1, or there exists a unique continuous function c: [0, 1] →]0, ∞[, such that I has the following form: I (x,y) = h −1 2 (c(x)h 1 (y)), x,y ∈ [0,1].

(44)

Moreover, if h 1 (y) = ah 1 (y), h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[, and we assume that

−1

h −1 2 (c(x)h 1 (y)) = h 2 (c (x)h 1 (y)), y ∈ [0,1],

then c (x) = (b/a)c(x), for all x ∈ [0, 1]. Proof. (ii) ⇒ (i) It is obvious that all presented solutions I are continuous and the result follows from the general solutions presented in Theorems 5.1 and 5.3. (i) ⇒ (ii) Let us assume, without loss of generality, that U1 and U2 are both conjunctive representable uninorms. When both uninorms are disjunctive, then the reasoning is similar. From Theorem 5.1 we know which vertical sections are possible for a fixed x ∈ [0, 1]. Since I is continuous, for every x ∈ [0, 1] the vertical sections are also continuous and consequently the vertical sections (27)–(30) are not possible. Let us assume that there exists some x0 ∈ [0, 1] such that I has the form (31) for some additive function gx0 : R → R. This implies, in particular, that gx0 is continuous, so there exists cx0 ∈ R such that gx0 (u) = cx0 u for all u ∈ R. Hence ⎧ −1 ⎨ h 2 (cx0 h 1 (0)) if cx0  0 −1 lim I (x0 ,y) = lim h −1 (g (h (y))) = lim h (c h (y)) = −1 x0 1 x0 1 2 2 ⎩ lim h 2 (0) if cx0 = 0 y→0+ y→0+ y→0+ y→1−

⎧ ⎧ −1 ⎪ ⎪ ⎨ 0 if cx0 > 0, ⎨ h 2 (−∞) if cx0 > 0 −1 = h 2 (∞) if cx0 < 0 = 1 if cx0 < 0, ⎪ ⎪ ⎩ ⎩ e2 if cx0 = 0, e2 if cx0 = 0 and −1 lim I (x0 ,y) = lim h −1 2 (gx0 (h 1 (y))) = lim h 2 (cx0 h 1 (y)) =

y→1−

y→1−

y→1−

⎧ −1 ⎨ h 2 (cx0 h 1 (1)) if cx0  0 −1 ⎩ lim− h 2 (0) y→1

if cx0 = 0

⎧ ⎧ −1 if cx0 > 0 ⎪ ⎪ ⎨ 1 if cx0 > 0, ⎨ h 2 (∞) = 0 if cx0 < 0, = h −1 2 (−∞) if cx0 < 0 ⎪ ⎪ ⎩ ⎩ e2 if cx0 = 0, e2 if cx0 = 0 while I (x0 , 0) = I (x0 , 1) = 0 in this case from (31). Thus we see that in every case this vertical section is not continuous, a contradiction. Hence the vertical section (31) is not possible. Let us assume that there exists some x0 ∈ [0, 1] such that I has the form (32) for some additive function gx0 : R → R. This implies, in particular, that gx0 is continuous, so there exists cx0 ∈ R such that gx0 (u) = cx0 u, for all u ∈ R. Now I (x0 , 0) = I (x0 , 1) = 1 and using similar reasoning as above we see that this vertical section is not continuous, a contradiction. Hence the vertical section (32) is not possible. Let us assume that there exists some x0 ∈ [0, 1] such that the vertical section for it is (24). In particular I(x0 ,1) = 0, but for the other possible vertical sections we always have I (x, 1)  0, so the only possibility in this case is I = 0. Analogously, let us assume that there exists some x0 ∈ [0, 1] such that the vertical section for it is (25). In particular I(x0 ,0) = e2 , but for the other possible vertical sections we always have I (x, 0)  e2 , so the only possibility in this case is I = e2 . Similarly, let us assume that there exists some x0 ∈ [0, 1] such that the vertical section for it is (26). In particular I(x0 ,0) = 1, but for the other possible vertical section (33) we have I(x,0) = 0, so the only possibility in this case is I = 1. Finally, let us assume that for all x ∈ [0, 1] the vertical section is (33). This implies, in particular, that gx is continuous for all x ∈ [0, 1], thus gx (u) = cx u with cx ∈] − ∞, ∞[ and for u ∈ R. Since I should be continuous

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and I(x,0) = 0 and I(x,1) = 1 we get that cx > 0. Indeed, for any fixed x ∈ [0, 1] we have  −1   h 2 (cx h 1 (0)) if cx  0 −1 0 = I (x,0) = I x, lim y = lim I (x,y) = lim h 2 (cx h 1 (y)) = + + + y→0 y→0 y→0 h −1 if cx = 0 2 (0) ⎧ −1 ⎧  ⎪ ⎪ ⎨ h 2 (−∞) if cx0 > 0 ⎨ 0 if cx0 > 0, h −1 2 (cx (−∞)) if cx  0 −1 = h 2 (∞) = if cx0 < 0 = 1 if cx0 < 0, ⎪ ⎪ e2 if cx = 0 ⎩ ⎩ e2 if cx0 = 0 e2 if cx0 = 0 and

 1 = I (x,1) = I

 x, lim y y→1−

 = lim I (x,y) = lim h −1 2 (cx h 1 (y) = y→1−

y→1−

h −1 2 (cx h 1 (1)) if cx  0 h −1 2 (0)

⎧ ⎧ −1  h 2 (∞) if cx0 > 0 ⎪ ⎪ −1 ⎨ 1 if cx0 > 0, ⎨ h 2 (cx (∞)) if cx  0 −1 = h 2 (−∞) if cx0 < 0 = 0 if cx0 < 0, = ⎪ ⎪ e2 if cx = 0 ⎩ ⎩ e2 if cx0 = 0, e2 if cx0 = 0

if cx = 0

thus in both cases cx > 0. Therefore there exists a function c: [0, 1] →]0, ∞[ such that I has the form (44). This function c is continuous since for any fixed y ∈]0, 1[\{e1 } it is a composition of continuous functions: c(x) =

h 2 (I (x,y)) , x ∈ [0,1]. h 1 (y)

From the above formula one can immediately see that the function c is uniquely determined. The second part of this theorem follows from Theorem 5.1.  Example 5.6. (i) If we assume that U1 = U2 are both conjunctive or disjunctive representable uninorms and the function c = 1 in (44), then our solution is trivial: I (x,y) = h −1 (c(x)h(y)) = h −1 (h(y)) = y, x,y ∈ [0,1]. (ii) Let us assume that c(x) = −x+2 for all x ∈ [0, 1] and consider the conjunctive (or disjunctive) representable uninorm U given in Example 2.4. Then the continuous function which satisfies (D1) with U1 = U2 = U has the following form: −x+2  y 1−y I (x,y) = −x+2 , x,y ∈ [0,1],  y 1+ 1−y with the standard (limit) calculations for y = 1. Since Eq. (D1) is the generalization of a tautology from the classical logic involving Boolean implication, it is reasonable to expect that the continuous solution I of (D1) is also a fuzzy implication. But from Theorem 5.5 and Lemma 3.1 we get Corollary 5.7. If U1 , U2 are both representable uninorms, then there are no continuous solutions I of (D1) which satisfy (I3). Proof. Let a continuous function I satisfy (I3) and (D1) with some representable t-conorms U1 , U2 with continuous additive generators h1 , h2 , respectively. Then U1 , U2 are both conjunctive or disjunctive. Hence I has to have the form (44) with a continuous function c: [0, 1] →]0, ∞[, but in this case we get −1 −1 I (0,0) = h −1 2 (c(0)h 1 (0)) = h 2 (c(0)(−∞)) = h 2 (−∞) = 0,

so I does not satisfy the first condition in (I3). 

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From Corollary 5.7 it is obvious that in our situation we need to look for solutions I of (D1) which are fuzzy implications non-continuous at the point (0,0). But it is not possible to obtain a fuzzy implication which is noncontinuous only in this point or, generally, only in one point. To see this assume that I is a fuzzy implication with continuous vertical sections, i.e., the only possible vertical sections are those given by (26) or (33) (or equivalently, by (36) or (43)). Firstly see that I satisfies (LB), so I(0,y) = 1 for all y ∈ [0, 1]—this means that for x = 0 we have the vertical section (26). Next, for x = 1 the vertical section is (33), because I(1,0) = 0. Now, let us assume that for some x1 ∈]0, 1[ the vertical section is also (33), i.e., I (x1 ,y) = h −1 2 (cx1 h 1 (y)), y ∈ [0,1], with cx1 ∈]0, ∞[, since I is increasing with respect to the second variable. Therefore for all x ∈]x1 , 1] it is also given by (33). So let us take any x2 ∈]x1 , 1]. Since I is increasing with respect to the second variable, cx2 ∈]0, ∞[. Now, since I is decreasing with respect to second variable, we get −1 h −1 2 (cx1 h 1 (y))≥h 2 (cx2 h 1 (y)),

thus cx1 h 1 (y)≥cx2 h 1 (y). If h 1 (y) > 0, i.e., y > e1 , then we get that cx1 ≥cx2 . If h 1 (y) < 0, i.e., y < e1 , then we get that cx1 ≤cx2 . Therefore we have obtained the fact that cx1 = cx2 = c is some constant in ]0, ∞[. Thus we have shown the following result. Theorem 5.8. Let U1 , U2 be representable uninorms with the neutral elements e1 , e2 ∈]0, 1[. For a fuzzy implication I : [0, 1]2 → [0, 1], which has continuous vertical sections, the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D1) for all x, y, z ∈ [0, 1]. (ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative constants, such that U1 , U2 admit simultaneously the representation (5) or (6) with h1 , h2 , respectively, and there exists a unique constant c ∈]0, ∞[ such that I has the form  I (x,y) =

1

i f x < x0 ,

h −1 2 (ch 1 (y)) i f x≥x 0 ,

x,y ∈ [0,1],

where x 0 ∈]0, 1], or  I (x,y) =

1

i f x≤x0 ,

h −1 2 (ch 1 (y)) i f x > x 0 ,

x,y ∈ [0,1],

where x 0 ∈ [0, 1[. In both cases I is not continuous with respect to the second variable for x = x0 . Example 5.9. Let us consider the conjunctive (or disjunctive) representable uninorm U given in Example 2.4. In the following we show three different fuzzy implications which satisfy the distributive equation (D1) with U1 = U2 = U: ⎧ 1 if x < 1 or y = 1, ⎪ ⎪ ⎪ ⎪   ⎪ 2 ⎪ y ⎨ (i) I (x, y) = x, y ∈ [0, 1], 1−y ⎪ otherwise, ⎪   ⎪ 2 ⎪ y ⎪ ⎪ ⎩ 1+ 1−y

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1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 X

0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

Y

Fig. 1. Plot of fuzzy implication from Example 5.9(ii).

(ii) I (x, y) =

⎧ ⎪ ⎪ ⎪1 ⎪ ⎪ ⎪ ⎪ ⎨  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 

(iii) I (x, y) =

if x <

1 or y = 1, 2

5 y 1−y 5 otherwise,  y 1+ 1−y

1 if x = 0, y otherwise,

x, y ∈ [0, 1],

x, y ∈ [0, 1].

In Fig. 1 we have presented the plot of the fuzzy implication from point (ii).

6. On Eq. (D2) for representable uninorms 6.1. General solutions of (D2) when U1 is a conjunctive and U2 is a disjunctive representable uninorms In this section we will show main results for the second equation (D2) when U1 is conjunctive and U2 is disjunctive representable uninorm. By Proposition 4.5 and using similar deduction as in the proof of Theorem 5.1 we get Theorem 6.1. For a conjunctive representable uninorm U1 with the neutral element e1 ∈]0, 1[, disjunctive representable uninorm U2 with the neutral element e2 ∈]0, 1[ and a function I : [0, 1]2 → [0, 1] the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D2) for all x, y, z ∈ [0, 1]. (ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative constants, such that U1 admit the representation (5) with h1 , U2 admit the representation (6) with h2 and for every

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fixed x ∈ [0, 1], the vertical section I (x, ·) has, for all y ∈ [0, 1], one of the following forms: I (x,y) = 0, I (x,y) = e2 , I (x,y) = 1,  I (x,y) =

i f y = 0,

e2 i f y ∈]0,1], 

I (x,y) =

1

i f y = 0,

e2 i f y ∈]0,1], 

I (x,y) =

0 i f y ∈]0,1[, 1 i f y = 0 or y = 1,

 I (x,y) =

1 i f y = 0, 0 i f y ∈]0,1],

 I (x,y) =

i f y = 0 or y = 1,

0

h −1 2 (gx (h 1 (y))) i f y ∈]0,1[, 

I (x,y) =

0

i f y = 0 or y = 1,

1

h −1 2 (gx (h 1 (y))) i f y ∈]0,1[,

⎧ 1 i f y = 0, ⎪ ⎪ ⎨ I (x,y) = h −1 2 (gx (h 1 (y))) i f y ∈]0,1[, ⎪ ⎪ ⎩ 0 i f y = 1, with a unique additive function gx : R → R. Moreover, if h 1 (y) = ah 1 (y), h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[, and we assume that

−1

h −1 2 (gx (h 1 (y))) = h 2 (gx (h 1 (y))), y ∈]0,1[,

for some x ∈ [0, 1], then gx (u) = bgx (u/a), for all u ∈ R. Remark 6.2. Observe that only the third vertical section in above theorem can be considered for fuzzy implications, but in general it is not possible to obtain that I(1,0) = 0 and I(1,1) = 1 and I (1, ·) is an increasing function of one variable. 6.2. General solutions of (D2) when U1 is a disjunctive and U2 is a conjunctive representable uninorms In this section we will show main results for the second equation (D2) when U1 is disjunctive and U2 is conjunctive representable uninorm. By Proposition 4.7 and using similar deduction as in the proof of Theorem 5.1 we get Theorem 6.3. For a disjunctive representable uninorm U1 with the neutral element e1 ∈]0, 1[, conjunctive representable uninorm U2 with the neutral element e2 ∈]0, 1[ and a function I : [0, 1]2 → [0, 1] the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D2) for all x, y, z ∈ [0, 1]. (ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative

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constants, such that U1 admit the representation (6) with h1 , U2 admit the representation (5) with h2 and for every fixed x ∈ [0, 1], the vertical section I (x, ·) has, for all y ∈ [0, 1], one of the following forms: I (x,y) = 0, I (x,y) = e2 , I (x,y) = 1,  I (x,y) =  I (x,y) =

1 i f y ∈]0,1[, e2 i f y ∈ [0,1[, 0

 I (x,y) = I (x,y) =  I (x,y) = 

i f y = 1,

e2 i f y ∈ [0,1[, 1



I (x,y) =

0 i f y = 0 or y = 1,

i f y = 1,

1 i f y = 0, 0 i f y ∈]0,1], 0

i f y = 0 or y = 1,

h −1 2 (gx (h 1 (y)))

i f y ∈]0,1[,

1

i f y = 0 or y = 1,

h −1 2 (gx (h 1 (y))) i f y ∈]0,1[,

⎧ 1 i f y = 0, ⎪ ⎨ −1 I (x,y) = h 2 (gx (h 1 (y))) i f y ∈]0,1[, ⎪ ⎩ 0 i f y = 1, with a unique additive function gx : R → R. Moreover, if h 1 (y) = ah 1 (y), h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[, and we assume that

−1

h −1 2 (gx (h 1 (y))) = h 2 (gx (h 1 (y))), y ∈]0,1[,

for some x ∈ [0, 1], then gx (u) = bgx (u/a), for all u ∈ R. Remark 6.4. Observe that only the third and the sixth vertical sections can be considered for fuzzy implications, but in general it is not possible to obtain that I(1,0) = 0 and I(1,1) = 1 and I (1, ·) is an increasing function of one variable. Corollary 6.5. If U1 , U2 are both representable uninorms, then there are no solutions I of (D2) which are fuzzy implications in the sense of Definition 2.6. 6.3. Continuous solutions of (D2) We know that there are no fuzzy implications which satisfy Eq. (D2) with representable uninorms, but at the end of this article we show the continuous solutions of (D2) for representable uninorms. Maybe the formulas of these solutions can also be useful in some other applications. Theorem 6.6. Let U1 , U2 be both representable uninorms with the neutral elements e1 , e2 ∈]0, 1[, one conjunctive and second disjunctive. For a continuous function I : [0, 1]2 → [0, 1] the following statements are equivalent: (i) The triple of functions U1 , U2 , I satisfies the functional equation (D2) for all x, y, z ∈ [0, 1].

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(ii) There exist continuous, strictly increasing functions h 1 , h 2 : [0, 1] → [−∞, ∞] with h 1 (0) = h 2 (0) = −∞, h 1 (e1 ) = h 2 (e2 ) = 0 and h 1 (1) = h 2 (1) = ∞, which are uniquely determined up to positive multiplicative constants, such that U1 , U2 admit the representation (5) and (6) (or inversely) with h1 and h2 , respectively, and either I = 0, or I = e2 , or I = 1, or there exists a unique continuous function c: [0, 1] →] − ∞, 0[, such that I has the form (44). Moreover, if h 1 (y) = ah 1 (y), h 2 (y) = bh 2 (y) for all y ∈ [0, 1] and some a, b ∈]0, ∞[, and we assume that

−1

h −1 2 (c(x)h 1 (y)) = h 2 (c (x)h 1 (y)), y ∈ [0,1],

then c (x) = (b/a)c(x), for all x ∈ [0, 1]. 7. Summary In this paper we have investigated the distributivity of fuzzy implications over representable uninorms. To obtain all solutions, we have firstly shown, in Section 3, that only some cases are important in our context. Next, in Section 4, we have obtained new results dealing with the additive Cauchy functional equation (C), when the domain and range of the function f is equal to the extended real line [−∞, ∞]. In Section 5 we have presented the general solutions of Eq. (D1) for representable uninorms and some solutions which are fuzzy implications. Finally, in Section 6 we have presented the general solutions of Eq. (D2) for representable uninorms. In this case we have obtained the fact that there are no fuzzy implications which satisfy Eq. (D2) with representable uninorms. References [1] J. Aczél, Lectures on Functional Equations and their Applications, Academic Press, New York, 1966. [2] M. Baczy´nski, On a class of distributive fuzzy implications, Int. J. Uncertainty Fuzziness Knowl.-Based Syst. 9 (2001) 229–238. [3] M. Baczy´nski, Contrapositive symmetry of distributive fuzzy implications, Int. J. Uncertainty Fuzziness Knowl.-Based Syst. 10 (2002) 135–147. [4] M. Baczy´nski, On the distributivity of fuzzy implications over continuous and Archimedean triangular conorms, Fuzzy Sets and Systems 161 (2010) 1406–1419. [5] M. Baczy´nski, B. Jayaram, Fuzzy implications, Studies in Fuzziness and Soft Computing, Vol. 231, Springer, Berlin, Heidelberg, 2008. [6] M. Baczy´nski, B. Jayaram, On the distributivity of fuzzy implications over nilpotent or strict triangular conorms, IEEE Trans. Fuzzy Syst. 17 (2009) 590–603. [7] J. Balasubramaniam, C.J.M. Rao, On the distributivity of implication operators over T- and S-norms, IEEE Trans. Fuzzy Syst. 12 (2004) 194–198. [8] H. Bustince, P. Burillo, F. Soria, Automorphisms, negation and implication operators, Fuzzy Sets and Systems 134 (2003) 209–229. [9] W.E. Combs, J.E. Andrews, Combinatorial rule explosion eliminated by a fuzzy rule configuration, IEEE Trans. Fuzzy Syst. 6 (1998) 1–11. [10] S. Dick, A. Kandel, Comments on “Combinatorial rule explosion eliminated by a fuzzy rule configuration”, IEEE Trans. Fuzzy Syst. 7 (1999) 475–477. [11] J. Fodor, B. DeBaets, Uninorm basics, in: P.P. Wang, D. Ruan, E.E. Kerre (Eds.), Fuzzy Logic: A Spectrum of Theoretical and Practical Issues, Studies in Fuzziness and Soft Computing, Vol. 215, Springer, Berlin, Heidelberg, 2007, pp. 49–64. [12] J.C. Fodor, M. Roubens, Fuzzy Preference Modelling and Multicriteria Decision Support, Kluwer, Dordrecht, 1994. [13] J.C. Fodor, R.R. Yager, A. Rybalov, Structure of uninorms, Int. J. Uncertainty Fuzziness Knowl.-Based Syst. 5 (1997) 411–427. [14] L. Kitainik, Fuzzy Decision Procedures With Binary Relations, Kluwer, Dordrecht, 1993. [15] M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities: Cauchy’s Equation and Jensen’s Inequality, PWN-Polish Scientific Publishers & Silesian University, Warszawa, Katowice, 1985. [16] J.M. Mendel, Q. Liang, Comments on “Combinatorial rule explosion eliminated by a fuzzy rule configuration”, IEEE Trans. Fuzzy Syst. 7 (1999) 369–371. [17] D. Ruiz-Aguilera, J. Torrens, Distributivity of strong implications over conjunctive and disjunctive uninorms, Kybernetika 42 (2005) 319–336. [18] D. Ruiz-Aguilera, J. Torrens, Distributivity of residual implications over conjunctive and disjunctive uninorms, Fuzzy Sets and Systems 158 (2007) 23–37. [19] E. Trillas, C. Alsina, On the law [ p ∧ q → r ] = [( p → r ) ∨ (q → r )] in fuzzy logic, IEEE Trans. Fuzzy Syst. 10 (2002) 84–88. [20] R.R. Yager, A. Rybalov, Uninorm aggregation operators, Fuzzy Sets and Systems 80 (1996) 111–120.