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S- and R-implications from uninorms continuous in ]0,1[2 and their distributivity over uninorms
Fuzzy Sets and Systems 160 (2009) 832 – 852 www.elsevier.com/locate/fss
S- and R-implications from uninorms continuous in ]0, 1[2 and their distributivity over uninorms D. Ruiz-Aguilera, J. Torrens∗ Department of Mathematics and Computer Science, University of the Balearic Islands, Cra. de Valldemossa, Km. 7,5, 07122 Palma de Mallorca, Spain Available online 20 June 2008
Abstract This work deals with two kinds of implications derived from uninorms continuous in ]0, 1[2 : S-implications and R-implications. In both cases, the general expression of such implications is found and several properties are studied. In particular, the distributivity of the S- and R-implications over conjunctive and disjunctive uninorms is investigated. © 2008 Elsevier B.V. All rights reserved. Keywords: Uninorm; S-implication; R-implication; Contrapositive symmetry; Exchange principle; Distributivity
1. Introduction The most usual kinds of implication functions used in fuzzy logic are strong implications or S-implications given by I (x, y) = S(N (x), y) for all x, y ∈ [0, 1],
(1)
where S is a t-conorm and N a strong negation, and residual implications or R-implications given by I (x, y) = sup{z ∈ [0, 1] | T (x, z) y} for all x, y ∈ [0, 1],
(2)
where T is a t-norm (usually left-continuous). Commonly, these implications are performed by t-norms and t-conorms (see for instance [12]) and they are successfully used in several problems, like fuzzy logic, aggregation of fuzzy relations, mathematical morphology and others. On the other hand, uninorms are a special kind of aggregation functions that have proved to be useful in many fields like expert systems, neural networks, aggregation, fuzzy system modelling, measure theory, mathematical morphology, etc. They are specially interesting because of their structure as a special combination of a t-norm and a t-conorm (see [10] or [12]), and because they must be conjunctive (U (1, 0) = 0) or disjunctive (U (1, 0) = 1). This allows to define fuzzy implication functions from uninorms and, in fact, some studies have been made in this direction in [3,18]. In these references only three classes of uninorms are used, namely, uninorms in Umin and Umax , representable uninorms and idempotent ones. However, another important class, that includes representable uninorms, is the class of uninorms continuous in ]0, 1[2 introduced and characterized in [11]. This work deals with S- and R-implications derived from this kind of ∗ Corresponding author.
E-mail address: [email protected] (J. Torrens). 0165-0114/$ - see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2008.05.015
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uninorms. For S-implications the uninorm must be disjunctive, whereas for R-implications we need a uninorm U such that U (x, 0) = 0 for all x < 1. In both cases, two possible kinds of uninorms continuous in ]0, 1[2 can be used and we give the expression of S- and R-implications, respectively, for both kinds of continuous uninorms. Moreover, several properties are studied like contrapositive symmetry, exchange principle, and specially the distributivity of such implications over conjunctive and disjunctive uninorms. The interest of this last property was clear from [1] where the authors use the classical tautology ( p ∧ q) → r ≡ ( p → r ) ∨ (q → r )
(3)
to avoid combinatorial rule explosion in their inference mechanism. This article produced some discussions around it [5,16] that pointed out the interest of a theoretical investigation necessary for using such equation in practice. Eq. (3), translated to the framework of fuzzy logic, was already studied by the same authors in [21,22] using the above mentioned classes of uninorms, but not with uninorms continuous in ]0, 1[2 . In this sense, the current work can be viewed as a completion of these papers. 2. Preliminaries We suppose the reader to be familiar with basic results concerning t-norms and t-conorms that can be found in [12]. Definition 1 (Fodor et al. [10]). A uninorm is a two-place function U : [0, 1] × [0, 1] −→ [0, 1] which is associative, commutative, increasing in each place and such that there exists some element e ∈ [0, 1], called neutral element, such that U (e, x) = x for all x ∈ [0, 1]. It is clear that the function U becomes a t-norm when e = 1 and a t-conorm when e = 0. For any uninorm we have U (0, 1) ∈ {0, 1} and a uninorm U is called conjunctive when U (1, 0) = 0 and disjunctive when U (1, 0) = 1. The structure of any uninorm U with neutral element e ∈]0, 1[ is always as follows. It works like a t-norm T in the interval [0, e], like a t-conorm S in the interval [e, 1] and it takes values between the minimum and the maximum in all other cases. For this reason, such a uninorm is usually denoted by U = (e, T, S) although this notation is ambiguous because there are many different uninorms with the same e, T and S. For all these results and more details on uninorms see [10]. The known classes of uninorms most commonly used are: • Uninorms in Umin and Umax , those with the 0-section and 1-section continuous except perhaps at the point x = e (see [10]). • Idempotent uninorms, those satisfying U (x, x) = x for all x ∈ [0, 1] (see [2,13] and also [8]). • The class Umnx given by those uninorms with neutral element e satisfying U (x, y) ∈ {x, y} for all x, y in A(e) = [0, e] × [e, 1] ∪ [e, 1] × [0, e] (see [7]). Note that this class includes both previous classes. • Representable uninorms, see Definition 2 below (characterized in [10]). • Continuous in ]0, 1[2 , characterized in [11] (see also Theorem 1). Again this class contains the previous one. Because we will mainly use the last two classes of uninorms in the paper, we recall them. Definition 2 (Fodor et al. [10]). A uninorm U with neutral element e ∈]0, 1[ is said to be representable if there is an increasing continuous mapping h : [0, 1] → [−∞, +∞] (called an additive generator of U), with h(0) = −∞, h(e) = 0 and h(1) = +∞ such that U is given by U (x, y) = h −1 (h(x) + h(y)) for all (x, y) ∈ [0, 1]2 \ {(0, 1), (1, 0)} and either U (0, 1) = U (1, 0) = 0 or U (0, 1) = U (1, 0) = 1. Remark 1. Representable uninorms were initially introduced under another name (see [6]). A representable uninorm is clearly continuous in [0, 1]2 \{(0, 1), (1, 0)}, and strictly increasing in ]0, 1[2 . Moreover, there exists a strong negation N with fixed point e such that for all (x, y) ∈ [0, 1]2 \ {(0, 1), (1, 0)}, U (x, y) = N (U (N (x), N (y))). This negation N is given by N (x) = h −1 (−h(x)), where h is an additive generator of U.
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Uninorms continuous in ]0, 1[2 were characterized in [11] as follows. Theorem 1 (Hu and Li [11]). Suppose U is a uninorm continuous in ]0, 1[2 with neutral element e ∈]0, 1[. Then either one of the following cases is satisfied:
(a) There exist ∈ [0, e[, ∈ [0, ], two continuous t-norms T and T and a representable uninorm R such that U can be represented as ⎧
x, y ⎪ T if x, y ∈]0, [ ⎪ ⎪ ⎪ ⎪ ⎪ x − y−
⎪ ⎪ , if x, y ∈], [ + ( − )T ⎪ ⎪ ⎪ ⎨ − − U (x, y) = + (1 − )R x − , y − (4) if x, y ∈], 1[ ⎪ ⎪ 1 − 1 − ⎪ ⎪ ⎪ ⎪ 1 if min(x, y) ∈], 1] and max(x, y) = 1 ⎪ ⎪ ⎪ ⎪ or 1 if (x, y) ∈ {(, 1), (1, )} ⎪ ⎩ min(x, y) otherwise
(b) There exist ∈]e, 1], ∈ [, 1], two continuous t-conorms S and S and a representable uninorm R such that U can be represented as ⎧ x y ⎪ ⎪ R , x, y ∈]0, [ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
x − , y − ⎪ if x, y ∈], [ ⎪ + ( − )S ⎪ ⎨ − − (5) U (x, y) = x − y−
⎪ + (1 − )S , if x, y ∈], 1[ ⎪ ⎪ 1 − 1 − ⎪ ⎪ ⎪ ⎪ 0 if max(x, y) ∈ [0, [ and min(x, y) = 0 ⎪ ⎪ ⎪ ⎪ or 0 if (x, y) ∈ {(0, ), (, 0)} ⎪ ⎩ max(x, y) otherwise Remark 2. Note that formula (4) appears in [11] with the t-norm TU on the square [0, ]2 . But in the approach of [11], it is proved that both and must be idempotent elements of U and consequently, by continuity, the t-norm TU must
be an ordinal sum of two t-norms T and T on [0, ]2 in such a way that formula (4) follows. Similarly, formula (5) appears also in [11] with the t-conorm SU in the square [, 1]2 , but , are idempotents of U and SU is an ordinal sum
of two t-conorms S and S on [, 1]2 such that (5) holds. Denote by Ucos the class of these uninorms and, particularly, by Ucos,min the class of uninorms with form (4) and by Ucos,max the class of uninorms with form (5). A uninorm U in Ucos,min (or in Ucos,max ) will be denoted as
U ≡ T , , T , , (R, e) cos,min (or U ≡ (R, e), , S , , S cos,max ) to represent its parameters. Remark 3. Any uninorm U in Ucos,min with = 0 or U in Ucos,max with = 1 is a representable uninorm. The distributivity condition between uninorms has been studied for several kinds of these operations (see [14,15,17, 20,21]). The following result will be used in the paper and can be found in [21]. Proposition 1. Let U and Ud be two uninorms with the same neutral element e ∈]0, 1[. If U is distributive over Ud , then Ud is idempotent and Ud (x, y) = U (x, y) for all (x, y) ∈ A(e) = [0, e] × [e, 1] ∪ [e, 1] × [0, e]. Definition 3 (Fodor and Roubers [9]). A binary operation I : [0, 1] × [0, 1] → [0, 1] is said to be an implication operation, or an implication, if it satisfies: (I1) I is nonincreasing in the first variable and nondecreasing in the second one. (I2) I (0, 0) = I (1, 1) = 1 and I (1, 0) = 0.
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Note that, any implication satisfies I (0, x) = 1 and I (x, 1) = 1 for all x ∈ [0, 1] whereas the symmetrical values I (x, 0) and I (1, x) are not derived from the definition. Strong and residual implications, obtained by Eqs. (1) and (2), respectively, are studied in the literature not only from t-norms and t-conorms, but also from conjunctive and disjunctive uninorms. With respect to the residual implication obtained from a t-norm T, usually denoted by IT , we have the following result (see for instance [4]). Proposition 2. Let T be a continuous t-norm. • IT (x, y) = 1 if and only if x y, known as the ordering property. • If T = min, then IT is the well-known Gödel implication given by
1 if x y IT (x, y) = y otherwise • If T is Archimedean with additive generator f, then IT (x, y) = f −1 (max(0, f (y) − f (x))) x, y ∈ [0, 1] • If T is an ordinal sum T = ( k , k , Tk )k∈K ⎧ 1 ⎪ ⎪ ⎨ −1 f k (ITk ( f k (x), f k (y))) IT (x, y) = ⎪ ⎪ ⎩ y
with K finite or countably infinite, then if x y if x > y and (x, y) ∈ [k , k ]2 , for some k ∈ K otherwise
where ITk is the R-implication derived from the Archimedean t-norm Tk for all k ∈ K , and f k (x) = (x − k )/(k − k ). Remark 4. Note that, from expressions above, it is easy to see that when T is nonstrict Archimedean (that is, f (0) < +∞), IT is continuous, whereas for strict t-norms ( f (0) = +∞), (0, 0) is a point of discontinuity of IT . In any case, when T has an Archimedean ordinal summand Tk in an interval [k , k ], fixed any c ∈]k , k [⊂]0, 1[, the section IT (−, c) is strictly decreasing in the interval [c, k ] taking all values from lim IT (x, c) = k to IT (k , c) = c
x→c+
but IT is not continuous when k < 1 because IT (c, c) = 1. The following proposition can be found in [3]. Proposition 3. Let U be a representable uninorm with neutral element e ∈]0, 1[ and additive generator h. Let U ∗ be the disjunctive representable uninorm with the same additive generator h, and N be the strong negation given by N (x) = h −1 (−h(x)). Then, (i) The residual implicator IU is given by
−1 h (h(y) − h(x)) if (x, y)(0, 0), (1, 1) IU (x, y) = 1 otherwise (ii) IU (x, y) = IU ∗ (x, y) = U ∗ (N (x), y) for all x, y ∈ [0, 1]. 3. Strong implications Definition 4. Given a disjunctive uninorm U and a strong negation N, the operation defined by IU,N (x, y) = U (N (x), y) for all x, y ∈ [0, 1] will be called the strong implication of U and N.
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1
R
N (x)
y
T
0 0
N ()
1
Fig. 1. IU,N with U ∈ Ucos,min , where operations in the figure are applied to the pairs (N (x), y).
With the previous assumptions IU,N is always an implication as its name suggests. Note that, for a uninorm U ∈ Ucos , U is disjunctive if and only if one of the following items holds:
(a) U ≡ (R, e), , S , , S cos,max , with U (1, 0) = 1. (b) A uninorm U in Ucos,min with = 0 (consequently only a t-norm T is necessary in its expression), and U (1, 0) = 1. We will denote such uninorm by U ≡ 0, T, , (R, e) cos,min . In both cases, the general structure of the strong implication derived from such uninorms can be easily derived and they are given in the following two propositions, respectively. The proofs of both propositions are straightforward. Proposition 4. If U ≡ 0, T, , (R, e) cos,min is a disjunctive uninorm, then IU,N is given by ⎧ N (x) y ⎪ ⎪ T , ⎪ ⎪ ⎪ ⎪ ⎨ N (x) − y − IU,N (x, y) = + (1 − )R , ⎪ 1− 1− ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎩ min(N (x), y)
if x ∈]N (), 1[ and y ∈]0, [ if x ∈]0, N ()[ and y ∈], 1[ if x = 0 or y = 1 otherwise
Proposition 5. If U ≡ (R, e), , S , , S cos,max is a disjunctive uninorm in Ucos,max , then IU,N is given by ⎧ N (x) − y −
⎪ ⎪ , ⎪ + (1 − )S ⎪ ⎪ 1 − ⎪ 1− ⎪ ⎪ y − N (x) − ⎪ ⎪ , ⎪ + ( − )S
⎪ ⎨ − − N (x) y IU,N (x, y) = , ⎪ R ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ 0 or max(N (x), y) ⎪ ⎪ ⎩ max(N (x), y)
if x ∈]0, N ()[ and y ∈], 1[ if x ∈]N (), N ()[ and y ∈], [ if x ∈]N (), 1[ and y ∈]0, [ if x = 1, y ∈ [0, [, or y = 0, x ∈]N (), 1] if (x, y) = (1, ) or (x, y) = (N (), 0) otherwise
We can see the general structure of IU,N being U ∈ Ucos,min and being U ∈ Ucos,max in Figs. 1 and 2, respectively. From its definition, it is clear that strong implications IU,N always satisfy contrapositive symmetry with respect to N, I (N (y), N (x)) = I (x, y) for all x, y ∈ [0, 1]
(6)
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1 S ′′ y •
S′ N (x) 0 0
• N ()
R
1
N ()
Fig. 2. IU,N with U ∈ Ucos,max , where operations in the figure are applied to the pairs (N (x), y).
because U is commutative and N involutive, and the exchange principle I (x, I (y, z)) = I (y, I (x, z)) for all x, y, z ∈ [0, 1], because U is commutative and associative. Other interesting property is the distributivity of such implications over conjunctive and disjunctive uninorms. In others words, the translation of Eq. (3) to our framework, that is, IU,N (Uc (x, y), z) = Ud (IU,N (x, z), IU,N (y, z))
(7)
for all x, y, z ∈ [0, 1], where Uc and Ud are uninorms in one of the classes considered, such that Uc is conjunctive, Ud is disjunctive and their underlying t-norms and t-conorms are continuous. This property was already studied for strong implications derived from other types of uninorms in [21]. Thus, we complete here that study for uninorms continuous in ]0, 1[2 . First of all, we have the following result, which proof is not included because it is the same as the one given in Theorem 1 of [21]. Theorem 2. Let U , Ud be disjunctive uninorms, Uc a conjunctive uninorm and N a strong negation. Then IU,N , Uc and Ud satisfy the distributivity equation (7) if and only if Uc and Ud are N-dual and U is distributive over Ud . i.e., U (x, Ud (y, z)) = Ud (U (x, y), U (x, z))
(8)
for all x, y, z ∈ [0, 1]. To solve Eq. (8) for uninorms in Ucos , we will distinguish two cases depending on which class is in the uninorm U, Ucos,max or Ucos,min . However, we give first the following lemma that is true in a more general setting. Lemma 1. Let U, Ud be two disjunctive uninorms with neutral elements e, ed ∈ [0, 1[, respectively, such that U is distributive over Ud . Then U (ed , ed ) = ed , that is, ed is an idempotent element of U. Proof. Take y = e and z = ed and x = ed in (8), then we obtain ed = U (ed , e) = U (ed , Ud (e, ed )) = Ud (U (ed , e), U (ed , ed )) = Ud (ed , U (ed , ed )) = U (ed , ed ).
3.1. Distributivity when U ∈ Ucos,max Now we consider the case when U ∈ Ucos,max . Recall that we assume for the distributivity property that the underlying t-norm and t-conorm of Ud are continuous.
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Lemma 2. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then e < ed or ed = 0. In both cases Ud (e, e) > 0. Proof. We divide the proof in several steps. (i) If ed < , as U is in Ucos,max , their only idempotent elements less than , are e and 0. By Lemma 1, ed is an idempotent element of U. But if ed = e and U is distributive over Ud , we know by Proposition 1 that Ud must be idempotent and Ud (x, y) = U (x, y) for all (x, y) ∈ A(e), that gives a contradiction with the definition of U. Thus, we have e < ed or ed = 0. (ii) If e < ed and Ud (e, e) = 0, then we have U (x, Ud (e, e)) = U (x, 0) for all x ∈ [0, 1] but by the other side, since U is distributive over Ud , U (x, Ud (e, e)) = Ud (U (x, e), U (x, e)) = Ud (x, x) and then, Ud (x, x) = U (x, 0) which contradicts the continuity of TUd and SUd . Then, we can conclude that e < ed and Ud (e, e) > 0. In the case ed = 0, Ud (e, e) > 0 trivially because then Ud is a t-conorm. Now, taking into account the results in Lemma 2, we distinguish two cases according to the possible values of ed , that is ed = 0 or ed > 0. In the case ed = 0 we have that Ud = S is a t-conorm and then, the results are already known (see [19, Theorem 9]) and they are as follows.
Proposition 6. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud = S be a t-conorm. Then U is distributive over S if and only if • S = max or • S is an ordinal sum of the form S = (0, , S1 ) with S1 strict, and the additive generator s of S1 satisfying s(e/) = 1, is also a multiplicative generator of R. In the case ed > 0 we will reach new solutions through several lemmas.
Lemma 3. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed > 0, such that U is distributive over Ud . Then Ud (x, x) = x for all x ∈ [, 1]. Proof. We know by Lemma 2 that 0 < Ud (e, e) and e < ed . Thus, Ud (e, e) is given by the underlying t-norm of Ud and consequently we have 0 < Ud (e, e) e < ed . Then for all x , we have, by definition of U ∈ Ucos,max : U (x, Ud (e, e)) = max(x, Ud (e, e)) = x but also, because U is distributive over Ud , U (x, Ud (e, e)) = Ud (x, x) and then Ud (x, x) = x for all x ∈ [, 1].
Lemma 4. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed > 0, such that U is distributive over Ud . Then, Ud cannot be continuous in ]0, 1[2 and, for all 0 < y < ed < z, we have that Ud (y, z)y. Proof. The first assertion is an immediate consequence of Lemma 3. On the other hand, suppose that there exists z > ed such that for some 0 < y < ed < z we have Ud (y, z) = y. We can take x such that 0 < y < ed < x < z, and then, using Lemma 1, and the structure of U, we have, by one side: U (x, Ud (y, z)) = U (x, y) = max(x, y) = x,
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but by the other, we have Ud (U (x, y), U (x, z)) = Ud (x, U (x, z))Ud (x, z) z > x that gives a contradiction, and then the result is proven. Thus, when ed > 0 solutions of the distributivity equation can only appear when Ud is in Umnx . To reach all of them we distinguish two cases according that ed < or ed .
Lemma 5. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm in Umnx with neutral element ed , such that U is distributive over Ud , and 0 < ed < . Then, Ud (x, 0) = U (x, 0) for all x ∈ [0, 1]. Proof. We divide our proof in several cases: • If x ed , by definition of U, we have Ud (x, 0) = U (x, 0) = 0. • If ed < x < . Take z such that 0 < z < ed < x < . Then we have, U (x, 0) = U (x, Ud (z, 0)) = Ud (U (x, z), U (x, 0)) = Ud (max(x, z), 0) = Ud (x, 0) where in the third equality we have used that U ∈ Ucos,max . • If < x, we take some y satisfying ed < y < < x, and then we have Ud (x, 0) = Ud (U (x, 0), U (y, 0)) = U (Ud (x, y), 0) = U (max(x, y), 0) = U (x, 0) • When x = , we distinguish two subcases: ◦ When U (0, ) = , take y with 0 < y < ed < . Using the previous lemma, Ud (, y)y, and since Ud ∈ Umnx , Ud (, y) = . Now we have U (0, ) = U (0, Ud (, y)) = Ud (U (0, ), U (0, y)) = Ud (, 0) ◦ When U (0, ) = 0. Now, take 0 < z ed < : U (, 0) = U (, Ud (z, 0)) = Ud (U (, z), U (, 0)) = Ud (, 0).
Corollary 1. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud , and 0 < ed < . Then, if Ud is in Umnx , there exists a continuous t-norm T such that Ud is given by ⎧ x y ⎪ ⎪ , if (x, y) ∈ [0, ]2 ⎪ T ⎪ ⎨ if max(x, y) > ed and min(x, y) > 0 Ud (x, y) = max(x, y) (9) ⎪ ⎪ U (x, y) if min(x, y) = 0 ⎪ ⎪ ⎩ min(x, y) otherwise Proof. From Lemmas 2 and 3 we know that ed and that Ud (x, x) = x for all x . This implies that the underlying t-conorm of Ud must be the maximum and the underlying t-norm must be an ordinal sum of the form ( 0, /ed , T ) for a continuous t-norm T. The result follows now from the previous lemmas.
Lemma 6. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud , and ed . Then Ud (x, 0) = x for all x > ed , that is, Ud must be in Umax .
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Proof. Suppose that there is x > ed such that Ud (0, x) = 0, then we can take ed < z < x, and we can write, by one hand U (z, Ud (0, x)) = U (z, 0) = z and by the other: Ud (U (z, 0), U (z, x)) = Ud (z, U (z, x))Ud (z, x) = x that gives a contradiction. Then, Ud (0, x) = x for all x > ed .
Corollary 2. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud , and ed . Then, if Ud is in Umnx , then U ∈ Umax and there exists a continuous t-norm T such that Ud must be given by ⎧ x y ⎪ ⎪ , if (x, y) ∈ [0, ]2 ⎨ T Ud (x, y) = (10) max(x, y) if max(x, y)ed ⎪ ⎪ ⎩ min(x, y) otherwise Proof. From the previous lemma it is clear that Ud must be in Umax and the result follows from Lemmas 2 and 3 similarly as in Corollary 1.
Theorem 3. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm in one of the classes considered in the Preliminaries. Then, U is distributive over Ud if and only if • Ud = S is a t-conorm as in Proposition 6, or • ed , U (ed , ed ) = ed , there exists a continuous t-norm T such that R is distributive over T (that is, T = min or T is strict and, if t is the additive generator of T satisfying t(e/) = 1 then 1/t is also a multiplicative generator of R), and one of the following two cases holds: (i) ed < and then Ud is given by Eq. (9), or (ii) ed and then Ud is given by Eq. (10). Proof. If U is distributive over Ud we know by the previous lemmas that Ud is a t-conorm as in Proposition 6, or ed , U (ed , ed ) = ed , and there exists a continuous t-norm T such that Ud is given by Eq. (9) or Eq. (10), depending on ed < or ed , respectively. Moreover, since U is distributive over Ud we have in particular that U (x, Ud (y, z)) = Ud (U (x, y), U (x, z)) for all x, y, z and consequently we will have that R is distributive over T and the result follows from the solutions of the distributivity equation for representable uninorms given in [21]. Conversely, it is a straightforward computation to show that when U and Ud are as in the enunciate they are distributive, by considering several cases depending on where x, y, z are placed with respect to and ed . The general structure of U and Ud such that U is distributive over Ud is depicted in Fig. 3 for case (i), and Fig. 4 for case (ii). 3.2. Distributivity when U ∈ Ucos,min We start with the case that U ∈ Ucos,min . Recall that in this case, to be U disjunctive we necessarily have = 0. Lemma 7. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm in Ucos,min , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then ed and Ud (e, e) < 1. Proof. Let us suppose first that ed > . By Lemma 1, ed is an idempotent element of U which is in Ucos,min . Thus, ed = e. But if ed = e and U is distributive over Ud , we know by Proposition 1 that Ud must be idempotent and
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Fig. 3. U ∈ Ucos,max (left) distributive over Ud ∈ Umnx (right), with < ed < .
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Fig. 4. U ∈ Ucos,max (left) distributive over Ud ∈ Umax (right), with < ed .
Ud (x, y) = U (x, y) for all (x, y) ∈ [0, e] × [e, 1] ∪ [e, 1] × [0, e], that gives a contradiction with the definition of U and consequently ed < e. Let us suppose now that Ud (e, e) = 1, then we have U (0, Ud (e, e)) = U (0, 1) = 1, whereas Ud (U (0, e), U (0, e)) = Ud (0, 0) = 0. But this is a contradiction since U is distributive over Ud . Then, we can conclude that Ud (e, e) < 1. Lemma 8. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm in Ucos,min , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then Ud (x, x) = x for all x ∈ [0, ]. Proof. We know by the previous lemma, that ed < e and Ud (e, e) < 1. Then for all x < e Ud (e, e) < 1, we have, by definition of U ∈ Ucos,min : U (x, Ud (e, e)) = x, but also, because U is distributive over Ud , U (x, Ud (e, e)) = Ud (x, x), and then Ud (x, x) = x for all x ∈ [0, ].
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Corollary 3. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm, and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then Ud is neither representable, nor in Ucos . Lemma 9. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm in Ucos,min , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . For all y < ed < z < 1 we have that Ud (y, z)z. Proof. Suppose on the contrary that there exists z such that for some y < ed < z < 1 we have Ud (y, z) = z. Now, we can take x such that y < x < ed < z < 1, and then, using Lemma 1, and the structure of U, we have, by one side U (x, Ud (y, z)) = U (x, z) = min(x, z) = x, but by the other, we have Ud (U (x, y), U (x, z)) = Ud (U (x, y), x)Ud (y, x) y < x that gives a contradiction, and then the result is proven. Corollary 4. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm, and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then Ud is not in Umax and, if it is in Umnx , there exists a continuous t-conorm S such that Ud must be given by ⎧ min(x, y) ⎪ ⎪ if min(x, y) < ed and max(x, y) < 1 ⎨ x − y− Ud (x, y) = + (1 − )S , if (x, y) ∈ [, 1]2 (11) ⎪ 1− 1− ⎪ ⎩ max(x, y) otherwise Proof. Similarly as in Corollary 1 we deduce from Lemma 8 that the underlying t-norm of Ud is the minimum and the underlying t-conorm is an ordinal sum of the form ( − ed /1 − ed , 1, S ) for a continuous t-conorm S. Now, the result follows from Lemma 9. Theorem 4. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm, and let Ud be a disjunctive uninorm in one of the classes considered in the Preliminaries. Then, U is distributive over Ud if and only if ed and
(i) U (ed , ed ) = ed , that is, there exist T and T t-norms such that T = ( 0, ed /, T , ed /, 1, T ). (ii) There exists a t-conorm S such that Ud is given by Eq. (11). (iii) R is distributive over S, that is, S = max or S is strict and the additive generator s of S satisfying s((e − )/(1 − )) = 1, is also a multiplicative generator of R. Proof. By the previous lemmata, if U is distributive over Ud , then (i) and (ii) are satisfied. Moreover, we also have that R must be distributive over S and, from the solutions of the distributivity equation for representable uninorms and t-conorms (see [19] and also [21]) we conclude (iii). Conversely, if U and Ud satisfy (i)–(iii) it is a straightforward computation to check that U is distributive over Ud by distinguish several cases, depending on the values of x, y and z in (8). Remark 5. Note that in the extreme case ed = 0, Ud is a t-conorm and Eq. (11) reduces to ⎧ x − y− ⎨ + (1 − )S , if (x, y) ∈ [, 1]2 Ud (x, y) = 1− 1− ⎩ max(x, y) otherwise Thus, the theorem above contains, for the case ed = 0, the results in [19] about distributivity of a uninorm in Ucos,min over a continuous t-conorm. The general structure of U ∈ Ucos,min and Ud such that U is distributive over Ud is depicted in Fig. 5.
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Fig. 5. U ∈ Ucos,min (left) distributive over Ud ∈ Umnx (right).
3.3. Other related distributivities Related to Eq. (7) there are several distributivity equations involving implications and uninorms. Namely, IU,N (x, Ud (y, z)) = Ud (IU,N (x, y), IU,N (x, z))
(12)
IU,N (Ud (x, y), z) = Uc (IU,N (x, z), IU,N (y, z))
(13)
IU,N (x, Uc (y, z)) = Uc (IU,N (x, y), IU,N (x, z))
(14)
where uninorms with subscript c are conjunctive and those with subscript d are disjunctive. All of these properties were studied for strong implications derived from other types of uninorms in [21]. As in such reference, it can be easily proved using contrapositive symmetry of strong implications that Eq. (12) is equivalent to Eq. (7) taking Uc the N-dual of Ud and consequently we have the following theorem. Theorem 5. Let U be a disjunctive uninorm in Ucos , and Ud and Ud two disjunctive uninorms. Then IU,N , Ud and Ud
satisfy Eq. (12) if and only if Ud = Ud and U is distributive over Ud . Thus, Theorems 3 and 4 give all the solutions of Eq. (12). Analogously Eq. (13) is equivalent to Eq. (14) using this duality, and so, we only need to solve this last equation. Again, we easily have the following result. Lemma 10. Let U be a disjunctive uninorm in Ucos , and Uc , Uc two conjunctive uninorms. Then, IU,N , Uc and Uc
satisfy Eq. (14) if and only if Uc = Uc and U is distributive over Uc . Thus, we need to solve the distributivity equation of U over Uc . However, in this case the solutions among the four classes of uninorms considered, appears only when ec = 1, that is, when Uc is a t-norm. Let us prove this fact following the lemmas below. Lemma 11. Let U be a disjunctive uninorm in Ucos , and Uc a conjunctive uninorm with neutral element ec < 1. If U is distributive over Uc , necessarily U ∈ Ucos,max . Proof. Suppose that U ∈ Ucos,min , then, as U is distributive over Uc , we have by one side U (0, Uc (1, ec )) = U (0, 1) = 1
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and by the other, since ec < 1: Uc (U (0, 1), U (0, ec )) = Uc (1, 0) = 0 that is a contradiction, and therefore U ∈ Ucos,max . However, even when U is in Ucos,max , there are no solutions with ec < 1, as the following lemma shows.
Lemma 12. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Uc be a conjunctive uninorm with neutral element ec . If U is distributive over Uc , then (i) e < ec , U (ec , ec ) = ec and Uc (e, e) > 0. (ii) Uc (x, x) = x for all x ∈ [, 1] and consequently Uc cannot be in Ucos . (iii) If Uc ∈ Umnx then ec = 1, that is Uc is a t-norm. Proof. The proofs of parts (i) and (ii) are very similar to the ones given in Lemmas 2 and 3, respectively. With respect to part (iii), take Uc ∈ Umnx with U distributive over Uc and suppose ec < 1. We will reach a contradiction in several steps: • First we prove that Uc (1, y) = 1 for all 0 < y < ec . Suppose on the contrary that there is such a y with Uc (1, y) = y. Taking x such that ec < x < 1 we have x = U (x, y) = U (x, Uc (y, 1)) = Uc (U (x, y), U (x, 1)) = Uc (x, 1) = 1 obtaining contradiction. • Now, for all 0 < y < e < ec we have, using the previous step, that 1 = U (0, 1) = U (0, Uc (1, y)) = Uc (U (0, 1), U (0, y)) = Uc (1, 0) = 0 obtaining again contradiction. Thus, ec = 1 and Uc must be a t-norm. Finally when ec = 1, we have Uc = T is a t-norm and the distributivity of U ∈ Ucos over T is already known (see [19]). In spite of that we include here the solutions for the sake of completeness.
Proposition 7. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Uc = T be a t-norm. Then U is distributive over T if and only if one of the following cases is satisfied • T = min. • U ≡ 0, T, , (R, e) cos,min and T is the ordinal sum given by T = (, 1, T ∗ ) , where T ∗ is a strict t-norm which additive generator t with t((e − )/(1 − )) = 1 satisfies that 1/t is also a multiplicative generator of R.
• U ≡ (R, e), , S , , S cos,max and T is the ordinal sum given by T = (0, , 1, T ∗∗ ) , where T ∗∗ is a strict t-norm in which additive generator t with t(e/) = 1 satisfies that 1/t is also a multiplicative generator of R. 4. R-implications from uninorms continuous in ]0, 1[2 Now, we want to deal with residual implications defined from uninorms in Ucos by IU (x, y) = sup{z ∈ [0, 1] | U (x, z) y}. In this case, to be IU an implication it is necessary (and sufficient) that the uninorm U satisfies U (x, 0) = 0 for all x < 1 (see [3]). Thus we can consider uninorms U in Ucos,min , but also uninorms in Ucos,max with = 1. In both cases, the general structure of the residual implication from such uninorm U can be easily derived.
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Fig. 6. Structure of IU , being U inbb= 0 0 134 133 Ucos,min .
Theorem 6. If U ≡ T , , T , , (R, e) cos,min is a conjunctive uninorm in Ucos,min , then IU is given by ⎧ 1 ⎪ ⎪ ⎪ x y ⎪ ⎪
I , ⎪ T ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ( − )I
x − , y − ⎪ ⎨ T − − IU (x, y) = x − y− ⎪ , + (1 − )I R ⎪ ⎪ ⎪ 1− 1− ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ y
if x ∈ [0, ] and y x if x ∈ [0, [ and y < x if x ∈ [, ] and y < x if (x, y) ∈ ], 1[2
(15)
if y = 1 if x = 1 and y > . otherwise
The structure of these implications can be seen in Fig. 6. Now we will study the second case, in which U ∈ Ucos,max . Theorem 7. If U ≡ (R, e), , S, 1 cos,max , then IU ⎧ x y ⎪ ⎪ I R , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x − y− ⎪ ⎪ , ⎨ + (1 − )R S 1− 1− IU (x, y) = 0 ⎪ ⎪ ⎪ ⎪ y ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎩ 0
is given by if (x, y) ∈]0, ]2 if (x, y) ∈], 1[2 and y x if x ∈], 1[ and y < x if (x, y) ∈]0, ] × [, 1[ if y = 1 or x = 0 if y = 0 and x0
(16)
The structure of such implications can be viewed in Fig. 7. Now we can study two properties of implications that we have recalled before: the exchange principle and contrapositive symmetry, for these implications. For the exchange principle, we know that all left-continuous uninorms are such that its residual implication will satisfy the exchange principle, and in the case of uninorms in Ucos we have the following result. Theorem 8. Let U be a uninorm in Ucos such that U (0, x) = 0 for all x < 1. If we have any of the following cases:
(i) U ≡ T , , T , , (R, e) cos,min , with = and U (, 1) = U (1, ) = . (ii) U ≡ (R, e), , S, 1 cos,max , conjunctive, i.e., U (0, 1) = U (1, 0) = 0. Then U is left-continuous and IU will satisfy the exchange principle.
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1 RS y 0
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Fig. 7. Structure of IU , with U ∈ Ucos,max .
On the contrapositive symmetry property, we get the next theorem. Theorem 9. Let U be a uninorm in Ucos such that U (0, x) = 0 for all x < 1. Then IU satisfy the contrapositive symmetry respect to a strong negation N if and only if U is representable and N = NU . Proof. First of all, it was proven in [3] that if U is representable, then it satisfy contrapositive symmetry with respect to NU . To prove from left to right, we will distinguish two different cases.
(i) If U ≡ T , , T , , (R, e) cos,min , let us suppose that there exists a strong negation N for which IU satisfy contrapositive symmetry. Then, by applying x = y in the equation of contrapositive symmetry (6) we will obtain IU (x, x) = IU (N (x), N (x))
(17)
Now we calculate, using Eq. (15), the value of IU (x, x), knowing that uninorm R has neutral element (e − )/(1 − ), ⎧ ⎨ 1 if x IU (x, x) = e if < x < 1 ⎩ 1 if x = 1 As N is a strong negation, there exists a fix point of N, that we will call z, N (z) = z. Now we will see that if 0, there exists x such that x < < N (x) by distinguishing three cases: • If z = , then we can take 0 < x < 1 such that x < = N () < N (x). • If z < , then we can take 0 < x < 1 such that x < N () < z < < N (x). • If < z, then we can take 0 < x < 1 such that x < < z < N () < N (x). Then, in any case, there exists x such that x < < N (x), and if contrapositive symmetry with respect to N is satisfied, by Eq. (17) it would have to be 1 = e, but this is a contradiction, because then U would be a t-norm and not from Ucos,min . Therefore, = 0 and U is representable.
(ii) If U ≡ (R, e), , S , , S cos,max , following a similar reasoning as in the previous case, now using Eq. (16) and taking into account that the neutral element of R is e/, we can deduce that ⎧ e if x < ⎪ ⎪ ⎪ ⎪ ⎨ if x = x − x − IU (x, x) = + (1 − )R S , if < x < 1 ⎪ ⎪ ⎪ 1− 1− ⎪ ⎩ 1 if x = 1
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And similarly to the previous case, if 1 then there exists x0 such that x0 < < N (x0 ), and if (17) is satisfied, then we have e = + (1 − )R S ((x0 − )/(1 − ), (x0 − )/(1 − )), but this is a contradiction, because e < . Then = 1 and U is representable. Finally, for this kind of implications we are also interested in solving the distributivity equation as in the case of strong implications. That is, we want to study the equation IU (Uc (x, y), z) = Ud (IU (x, z), IU (y, z))
(18)
for all x, y, z ∈ [0, 1]. As in the case of strong implications, Uc and Ud are a conjunctive and a disjunctive uninorm, respectively, and we assume that their underlying t-norms and t-conorms are continuous. On the other hand, in this case U is a uninorm in Ucos and IU its residual implication. We distinguish again two cases, when U is in Ucos,min , and when it is in Ucos,max and we do this in the following subsections. 4.1. Distributivity when U ∈ Ucos,min Lemma 13. Consider Uc a conjunctive uninorm with neutral element ec , Ud a disjunctive uninorm,
U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then Uc is a t-norm. Proof. • First of all, if we suppose that ec , then we would have, taking x = ec and y = z = e in (18), Ud (IU (ec , e), IU (e, e)) = IU (Uc (ec , e), e) that means Ud (1, e) = IU (e, e), because ec < e. This last equality implies e = 1, that is a contradiction. Then < ec . • Now, suppose that ec < 1. Then we can take y0 such that ec < y0 < 1 with IU (ec , z) > IU (y0 , z) for all z > , because < ec < y0 , by the previous reasoning. As IU is continuous in [, 1]2 , there exists z 0 such that IU (ec , z 0 ) > IU (y0 , z 0 ) ed , and therefore we have, by putting x = ec , y = y0 , and z = z 0 in (18): Ud (IU (ec , z 0 ), IU (y0 , z 0 )) = IU (Uc (ec , y0 ), z 0 ) = IU (y0 , z 0 ) but we also have Ud (IU (ec , z 0 ), IU (y0 , z 0 )) max(IU (ec , z 0 ), IU (y0 , z 0 )) > IU (y0 , z 0 ) that is a contradiction. Then we can say that ec = 1, that is, Uc is a t-norm. From now on we will consider Uc be a (continuous) t-norm.
Lemma 14. Let Uc be a t-norm, Ud a disjunctive uninorm with neutral element ed , U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then ed and Ud (z, z) = z for all z . Proof. Taking x = 1, y = e in (18) we have z = IU (e, z) = IU (T (1, e), z) = Ud (IU (1, z), IU (e, z)) =
Ud (, z) if z Ud (z, z) if z
Now, if < ed we would have Ud (, ed ) = ed = , that is a contradiction. Then, Ud (z, z) = z for all z , and ed .
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Lemma 15. Let Uc be a t-norm, Ud a disjunctive uninorm with neutral element ed , U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then U (x, x) = x for all x ed , and consequently IU (x, y) = y for all x, y such that x < y ed . Proof. Suppose that there is an x ed such that U (x, x) < x, that is, T (x/e, x/e) < x/e and by continuity there will exist an interval [a, b] such that a < x < b and the restriction of T to the interval [a/e, b/e] is Archimedean. Now, fixed z 0 such that a < z 0 < x and taking into account Remark 4, IU (−, z 0 ) takes all values from z 0 to b. Thus, let x0 be such that a < z 0 < IU (x0 , z 0 ) < min(ed , b), we have on one hand IU (Uc (x0 , 1), z 0 ) = IU (x0 , z 0 ) whereas, on the other hand, Ud (IU (x0 , z 0 ), IU (1, z 0 )) = Ud (IU (x0 , z 0 ), z 0 ) = min(IU (x0 , z 0 ), z 0 ) = z 0 obtaining contradiction.
Lemma 16. Let Uc be a t-norm, Ud = (ed , Td , Sd ) a disjunctive uninorm, U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then Uc (e, e) > and Ud (x, x) = Uc (x, x) = x for all x . That is, Uc must be an ordinal sum Uc = (, 1, T0 ) for some continuous t-norm T0 , Td = min and Sd must be also an ordinal sum Sd = (( − ed )/(1 − ed ), 1, S0 ) . Proof. Let us give the proof in several steps. • First we prove that Uc (e, e) > . Suppose on the contrary that Uc (e, e) , then 1 = IU (Uc (e, e), ) = Ud (IU (e, ), IU (e, )) = Ud (, ). Thus, since e > we have Ud (e, e) = 1. Since Uc is continuous, we can take an x such that 1 > x > with 1 > Uc (x, x) > , but then IU (Uc (x, x), x) = Ud (IU (x, x), IU (x, x)) = Ud (e, e) = 1 obtaining a contradiction because IU is strictly monotone in [, 1]2 (in this region, U is given by the representable uninorm R and then we can apply Proposition 3). • Second we prove that Ud (x, x) = x for all x . Just taking x = y = e in (18) we obtain IU (Uc (e, e), z) = Ud (IU (e, z), IU (e, z)) = Ud (z, z). Now, for all z , IU (Uc (e, e), z) = z by the previous step and consequently Ud (z, z) = z for all z . This directly implies that Td = min and that Sd is an ordinal sum of the form Sd = (( − ed )/(1 − ed ), 1, S0 ) . • Finally, we prove that Uc (x, x) = x for all x . Suppose on the contrary that there is some x < such that Uc (x, x) < x and take z such that Uc (x, x) < z < x < . For these values we have IU (x, z) and then, using the step above, 1 = IU (Uc (x, x), z) = Ud (IU (x, z), IU (x, z)) = IU (x, z) obtaining again a contradiction. Thus, Uc (x, x) = x for all x and consequently Uc is an ordinal sum of the form Uc = (, 1, T0 ) . Theorem 10. Let Uc and Ud = (ed , Td , Sd ) be a conjunctive and a disjunctive uninorm, respectively,
U ≡ T , , T , , (R, e) cos,min and IU its residual implication. Then, Uc , Ud and IU satisfy (18) if and only there exist a continuous t-norm T0 and a continuous t-conorm S0 such that (i) Uc is the t-norm given by the ordinal sum Uc = (, 1, T0 ) , (ii) ed , Td = min and Sd = (( − ed )/(1 − ed ), 1, S0 ) ,
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e
Fig. 8. Structure of Uc (top left), Ud (top right) and IU (bottom), with U ∈ Ucos,min , that satisfy Eq. (18).
(iii) U (x, x) = x for all x ed , IU (x, y) = y for all x < y ed and (iv) T0 , S0 and I R satisfy Eq. (18), that is, T0 and S0 are N R -dual and: • T0 = min and S0 = max or, • S0 is strict and if s is its additive generator with s((e − )/(1 − )) = 1 then s is also a multiplicative generator of R. Proof. We first suppose that Uc , Ud and IU satisfy (18). Then, (i)–(iii) follows directly from the previous lemmas. On the other hand, since Uc , Ud and IU satisfy Eq. (18) in particular for all x, y, z , we derive that T0 , S0 and R must also satisfy Eq. (18) and the remaining items follows from the results of such equation for representable uninorms given in [22]. Conversely, due to their symmetry we only need to prove Eq. (18) for values x y and we can do this by distinguishing several cases: • If x , then Uc (x, y) = min(x, y) = x. On the other hand, ◦ When z x then I (x, z) = 1 and Eq. (18) follows trivially because Ud is disjunctive. ◦ When z < x, ed we have I (x, z) = I (y, z) = z and (18) follows because Ud (z, z) = z. ◦ When ed z < x we have I (x, z), I (y, z) ∈]ed , ]. But Ud is given by the maximum for these values and again (18) is satisfied. • If x, Eq. (18) is satisfied because T0 S0 and I R also satisfy it. The structure of solutions of the previous theorem can be viewed in Fig. 8.
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4.2. Distributivity when U ∈ Ucos,max Recall that in this case we must have a uninorm U with = 0, that is U ≡ (R, e), , S, 1 cos,max . We obtain new solutions with similar reasonings as in the subsection above. Lemma 17. Let Uc and Ud be a conjunctive and a disjunctive uninorm, respectively, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. If they satisfy (18), then ed = 0, that is Ud must be a t-conorm, and Ud (x, x) = x for all x . Proof. Let us prove first that ed = 0. Suppose on the contrary that ed > 0, then: • If ec we have IU (Uc (ec , e), z) = IU (e, z) = z, whereas for all z < ec , Ud (IU (ec , z), IU (e, z)) = Ud (0, z). Thus, we obtain Ud (0, z) = z for all z < ec which is a contradiction. • If ec < we can take an y0 such that 0 < y0 < ec < . But then there will exist a z 0 < such that IU (ec , z 0 ) < IU (y0 , z 0 ) < ed and then IU (Uc (ec , y0 ), z 0 ) = IU (y0 , z 0 ) whereas Ud (IU (ec , z 0 ), IU (y0 , z 0 )) min(IU (ec , z 0 ), IU (y0 , z 0 )) < IU (y0 , z 0 ) obtaining again contradiction. Thus, ed = 0 and finally, for all z we have z = IU (0, z) = IU (Uc (0, 0), z) = Ud (IU (0, z), IU (0, z)) = Ud (z, z).
Lemma 18. Let Uc be a conjunctive uninorm and Ud a t-conorm, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. If they satisfy (18), then ec > and Uc is in Umin with Uc (x, x) = x for all x . Proof. We give the proof in several steps: • First we prove that ec > . Suppose that ec and take z such that < z < 1. Then on one hand IU (Uc (ec , 1), z) = IU (1, z) = 0, and on the other hand Ud (IU (ec , z), IU (1, z)) = Ud (z, 0) = z, which gives a contradiction. • Now we prove that Uc (1, x) = x for all x > . Suppose that there is an x > such that Uc (1, x) > x > , we can take z such that x < z < Uc (1, x) and for this z we have 0 = IU (Uc (1, x), z) = Ud (IU (1, z), IU (x, z)) = Ud (0, IU (x, z)) = IU (x, z) > 0 obtaining contradiction. • From the step above we have by increasingness that Uc is in Umin . • Finally we prove that Uc (x, x) = x for all x . We prove it by distinguishing two cases: ◦ Suppose there is an x such that Uc (x, x) < x and take z 0 such that Uc (x, x) < z 0 < x. Then IU (Uc (x, x), z 0 ) = Ud (IU (x, z 0 ), IU (x, z 0 )) = Ud (0, 0) = 0 which is a contradiction since Uc (x, x) < z 0 . ◦ Suppose now that there is an x such that Uc (x, x) > x. Similarly, taking z 0 such that x < z 0 < Uc (x, x) we have 0 = IU (Uc (x, x), z 0 ) = Ud (IU (x, z 0 ), IU (x, z 0 )) = IU (x, z 0 ) which again is a contradiction because x < z 0 and consequently IU (x, z 0 ) . Lemma 19. Let Uc be a conjunctive uninorm and Ud a t-conorm, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. If they satisfy (18), then U (x, x) = x for all x ec .
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max
min ec
ec min
max
max
S0
max
min
min
e
T0
e
e
min
e
ec
ec
y ec
y RS
0
e
IR
e
0
ec
Fig. 9. Structure of Uc (top left), Ud (top right), IU (bottom), with U ∈ Ucos,max , that satisfy Eq. (18).
Proof. Suppose U (x, x) > x for some x > ec . There is an interval [a, b] containing x and U (x, x) in which IU is strictly monotone. But then, taking z such that ec < x < z < U (x, x), we have IU (Uc (x, U (x, x)), z) = IU (U (x, x), z) = 0 and also Ud (IU (x, z), IU (U (x, x), z)) = Ud (IU (x, z), 0) = IU (x, z) obtaining a contradiction. Theorem 11. Let Uc = (ec , Tc , Sc ) and Ud be a conjunctive and a disjunctive uninorm, respectively, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. Then, Uc , Ud and IU satisfy (18) if and only there exist a continuous t-norm T0 and a continuous t-conorm S0 such that (i) (ii) (iii) (iv)
Ud is the t-conorm given by the ordinal sum Ud = (0, , S0 ) , Uc is in Umin with ec > , Sc = max and Tc = (0, /ec , T0 ) , U (x, x) = x for all x ec and T0 , S0 and I R satisfy Eq. (18), that is, T0 and S0 are N R -dual and: • T0 = min and S0 = max or, • S0 is strict and if s is its additive generator with s(/ec ) = 1 then s is also a multiplicative generator of R.
Proof. Completely similar to the proof of Theorem 10. The structure of solutions given in the previous theorem can be viewed in Fig. 9.
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5. Conclusion Implication functions are essential not only in fuzzy logic, but also in many applications like approximate reasoning and fuzzy control. Among other approaches S-implications and R-implications are the most widely used in this context. In this paper, these kinds of implications have been derived from uninorms continuous in ]0, 1[2 pointing out their structure and studying some of their properties. One of them, particularly important, is the distributivity of such implications over conjunctive and disjunctive uninorms, that has been successfully used in avoiding combinatorial rule explosion in fuzzy systems (see [1]). In this paper, new solutions of this equation have been found among S- and R-implications derived from uninorms continuous in ]0, 1[2 . Acknowledgments The authors want to thank the referees for their valuable comments. This work has been partially supported by the Spanish Grant MTM2006-05540 and the Government of the Balearic Islands Project PRIB-2004-9250. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
[21] [22]
W.E. Combs, Combinatorial rule explosion eliminated by a fuzzy rule configuration, IEEE Trans. Fuzzy Systems 6 (1998) 1–11. B. De Baets, Idempotent uninorms, European J. Oper. Res. 118 (1999) 631–642. B. De Baets, J.C. Fodor, Residual operators of uninorms, Soft Comput. 3 (1999) 89–100. B. De Baets, R. Mesiar, Residual implicators of continuous t-norms, in: Proc. EUFIT-96 Aachen, Germany, 1996, pp. 27–31. S. Dick, A. Kandel, Comments on “Combinatorial rule explosion eliminated by a fuzzy rule configuration”, IEEE Trans. Fuzzy Systems 7 (1999) 475–477. J. Dombi, Basic concepts for the theory of evaluation: the aggregative operator, European J. Oper. Res. 10 (1982) 282–293. P. Dryga´s, Discussion of the structure of uninorms, Kybernetika 41 (2005) 213–226. P. Dryga´s, Remarks about idempotent uninorms, J. Elec. Eng. 57 (2006) 92–94. J.C. Fodor, M. Roubens, Fuzzy preference modelling and multicriteria decision support, Theory and Decision Library, Serie D: System Theory, Knowledge Engineering and Problem Solving, Kluwer Academic Publishers, Dordrecht, 1994. J.C. Fodor, R.R. Yager, A. Rybalov, Structure of uninorms, Int. J. Uncertainty Fuzziness Knowledge-based Systems 5 (1997) 411–427. S. Hu, Z. Li, The structure of continuous uni-norms, Fuzzy Sets and Systems 124 (2001) 43–52. E.P. Klement, R. Mesiar, E. Pap, Triangular Norms, Kluwer Academic Publishers, Dordrecht, 2000. J. Martín, G. Mayor, J. Torrens, On locally internal monotonic operations, Fuzzy Sets and Systems 137 (2003) 27–42. M. Mas, G. Mayor, J. Torrens, The distributivity condition for uninorms and t-operators, Fuzzy Sets and Systems 128 (2002) 209–225. M. Mas, G. Mayor, J. Torrens, Corrigendum to “The distributivity condition for uninorms and t-operators” [Fuzzy Sets and Systems 128 (2002) 209–225], Fuzzy Sets and Systems 153 (2005) 297–299. J.M. Mendel, Q. Liang, Comments on “Combinatorial rule explosion eliminated by a fuzzy rule configuration”, IEEE Trans. Fuzzy Systems 7 (1999) 369–371. D. Ruiz, J. Torrens, Distributive idempotent uninorms, Internat. J. Uncertainty, Fuzziness Knowledge-Based Systems 11 (2003) 413–428. D. Ruiz, J. Torrens, Residual implications and co-implications from idempotent uninorms, Kybernetika 40 (2004) 21–38. D. Ruiz, J. Torrens, Distributivity and conditional distributivity of a uninorm and a continuous t-conorm, IEEE Trans. Fuzzy Systems 14 (2006) 180–190. D. Ruiz-Aguilera, Contribució a l’estudi de les uninormes en el marc de les equacions funcionals, Aplicacions a la morfologia matemàtica, Ph.D. Thesis (in catalan) University of the Balearic Islands, Palma de Mallorca, Spain, 2007, Available at: http://www.tesisenxarxa.net/ TDX-1121107-124008/ . D. Ruiz-Aguilera, J. Torrens, Distributivity of strong implications over conjunctive and disjunctive uninorms, Kybernetika 42 (2005) 319–336. D. Ruiz-Aguilera, J. Torrens, Distributivity of residual implications over conjunctive and disjunctive uninorms, Fuzzy Sets and Systems 158 (2007) 23–37.
S- and R-implications from uninorms continuous in ]0, 1[2 and their distributivity over uninorms D. Ruiz-Aguilera, J. Torrens∗ Department of Mathematics and Computer Science, University of the Balearic Islands, Cra. de Valldemossa, Km. 7,5, 07122 Palma de Mallorca, Spain Available online 20 June 2008
Abstract This work deals with two kinds of implications derived from uninorms continuous in ]0, 1[2 : S-implications and R-implications. In both cases, the general expression of such implications is found and several properties are studied. In particular, the distributivity of the S- and R-implications over conjunctive and disjunctive uninorms is investigated. © 2008 Elsevier B.V. All rights reserved. Keywords: Uninorm; S-implication; R-implication; Contrapositive symmetry; Exchange principle; Distributivity
1. Introduction The most usual kinds of implication functions used in fuzzy logic are strong implications or S-implications given by I (x, y) = S(N (x), y) for all x, y ∈ [0, 1],
(1)
where S is a t-conorm and N a strong negation, and residual implications or R-implications given by I (x, y) = sup{z ∈ [0, 1] | T (x, z) y} for all x, y ∈ [0, 1],
(2)
where T is a t-norm (usually left-continuous). Commonly, these implications are performed by t-norms and t-conorms (see for instance [12]) and they are successfully used in several problems, like fuzzy logic, aggregation of fuzzy relations, mathematical morphology and others. On the other hand, uninorms are a special kind of aggregation functions that have proved to be useful in many fields like expert systems, neural networks, aggregation, fuzzy system modelling, measure theory, mathematical morphology, etc. They are specially interesting because of their structure as a special combination of a t-norm and a t-conorm (see [10] or [12]), and because they must be conjunctive (U (1, 0) = 0) or disjunctive (U (1, 0) = 1). This allows to define fuzzy implication functions from uninorms and, in fact, some studies have been made in this direction in [3,18]. In these references only three classes of uninorms are used, namely, uninorms in Umin and Umax , representable uninorms and idempotent ones. However, another important class, that includes representable uninorms, is the class of uninorms continuous in ]0, 1[2 introduced and characterized in [11]. This work deals with S- and R-implications derived from this kind of ∗ Corresponding author.
E-mail address: [email protected] (J. Torrens). 0165-0114/$ - see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2008.05.015
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uninorms. For S-implications the uninorm must be disjunctive, whereas for R-implications we need a uninorm U such that U (x, 0) = 0 for all x < 1. In both cases, two possible kinds of uninorms continuous in ]0, 1[2 can be used and we give the expression of S- and R-implications, respectively, for both kinds of continuous uninorms. Moreover, several properties are studied like contrapositive symmetry, exchange principle, and specially the distributivity of such implications over conjunctive and disjunctive uninorms. The interest of this last property was clear from [1] where the authors use the classical tautology ( p ∧ q) → r ≡ ( p → r ) ∨ (q → r )
(3)
to avoid combinatorial rule explosion in their inference mechanism. This article produced some discussions around it [5,16] that pointed out the interest of a theoretical investigation necessary for using such equation in practice. Eq. (3), translated to the framework of fuzzy logic, was already studied by the same authors in [21,22] using the above mentioned classes of uninorms, but not with uninorms continuous in ]0, 1[2 . In this sense, the current work can be viewed as a completion of these papers. 2. Preliminaries We suppose the reader to be familiar with basic results concerning t-norms and t-conorms that can be found in [12]. Definition 1 (Fodor et al. [10]). A uninorm is a two-place function U : [0, 1] × [0, 1] −→ [0, 1] which is associative, commutative, increasing in each place and such that there exists some element e ∈ [0, 1], called neutral element, such that U (e, x) = x for all x ∈ [0, 1]. It is clear that the function U becomes a t-norm when e = 1 and a t-conorm when e = 0. For any uninorm we have U (0, 1) ∈ {0, 1} and a uninorm U is called conjunctive when U (1, 0) = 0 and disjunctive when U (1, 0) = 1. The structure of any uninorm U with neutral element e ∈]0, 1[ is always as follows. It works like a t-norm T in the interval [0, e], like a t-conorm S in the interval [e, 1] and it takes values between the minimum and the maximum in all other cases. For this reason, such a uninorm is usually denoted by U = (e, T, S) although this notation is ambiguous because there are many different uninorms with the same e, T and S. For all these results and more details on uninorms see [10]. The known classes of uninorms most commonly used are: • Uninorms in Umin and Umax , those with the 0-section and 1-section continuous except perhaps at the point x = e (see [10]). • Idempotent uninorms, those satisfying U (x, x) = x for all x ∈ [0, 1] (see [2,13] and also [8]). • The class Umnx given by those uninorms with neutral element e satisfying U (x, y) ∈ {x, y} for all x, y in A(e) = [0, e] × [e, 1] ∪ [e, 1] × [0, e] (see [7]). Note that this class includes both previous classes. • Representable uninorms, see Definition 2 below (characterized in [10]). • Continuous in ]0, 1[2 , characterized in [11] (see also Theorem 1). Again this class contains the previous one. Because we will mainly use the last two classes of uninorms in the paper, we recall them. Definition 2 (Fodor et al. [10]). A uninorm U with neutral element e ∈]0, 1[ is said to be representable if there is an increasing continuous mapping h : [0, 1] → [−∞, +∞] (called an additive generator of U), with h(0) = −∞, h(e) = 0 and h(1) = +∞ such that U is given by U (x, y) = h −1 (h(x) + h(y)) for all (x, y) ∈ [0, 1]2 \ {(0, 1), (1, 0)} and either U (0, 1) = U (1, 0) = 0 or U (0, 1) = U (1, 0) = 1. Remark 1. Representable uninorms were initially introduced under another name (see [6]). A representable uninorm is clearly continuous in [0, 1]2 \{(0, 1), (1, 0)}, and strictly increasing in ]0, 1[2 . Moreover, there exists a strong negation N with fixed point e such that for all (x, y) ∈ [0, 1]2 \ {(0, 1), (1, 0)}, U (x, y) = N (U (N (x), N (y))). This negation N is given by N (x) = h −1 (−h(x)), where h is an additive generator of U.
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Uninorms continuous in ]0, 1[2 were characterized in [11] as follows. Theorem 1 (Hu and Li [11]). Suppose U is a uninorm continuous in ]0, 1[2 with neutral element e ∈]0, 1[. Then either one of the following cases is satisfied:
(a) There exist ∈ [0, e[, ∈ [0, ], two continuous t-norms T and T and a representable uninorm R such that U can be represented as ⎧
x, y ⎪ T if x, y ∈]0, [ ⎪ ⎪ ⎪ ⎪ ⎪ x − y−
⎪ ⎪ , if x, y ∈], [ + ( − )T ⎪ ⎪ ⎪ ⎨ − − U (x, y) = + (1 − )R x − , y − (4) if x, y ∈], 1[ ⎪ ⎪ 1 − 1 − ⎪ ⎪ ⎪ ⎪ 1 if min(x, y) ∈], 1] and max(x, y) = 1 ⎪ ⎪ ⎪ ⎪ or 1 if (x, y) ∈ {(, 1), (1, )} ⎪ ⎩ min(x, y) otherwise
(b) There exist ∈]e, 1], ∈ [, 1], two continuous t-conorms S and S and a representable uninorm R such that U can be represented as ⎧ x y ⎪ ⎪ R , x, y ∈]0, [ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
x − , y − ⎪ if x, y ∈], [ ⎪ + ( − )S ⎪ ⎨ − − (5) U (x, y) = x − y−
⎪ + (1 − )S , if x, y ∈], 1[ ⎪ ⎪ 1 − 1 − ⎪ ⎪ ⎪ ⎪ 0 if max(x, y) ∈ [0, [ and min(x, y) = 0 ⎪ ⎪ ⎪ ⎪ or 0 if (x, y) ∈ {(0, ), (, 0)} ⎪ ⎩ max(x, y) otherwise Remark 2. Note that formula (4) appears in [11] with the t-norm TU on the square [0, ]2 . But in the approach of [11], it is proved that both and must be idempotent elements of U and consequently, by continuity, the t-norm TU must
be an ordinal sum of two t-norms T and T on [0, ]2 in such a way that formula (4) follows. Similarly, formula (5) appears also in [11] with the t-conorm SU in the square [, 1]2 , but , are idempotents of U and SU is an ordinal sum
of two t-conorms S and S on [, 1]2 such that (5) holds. Denote by Ucos the class of these uninorms and, particularly, by Ucos,min the class of uninorms with form (4) and by Ucos,max the class of uninorms with form (5). A uninorm U in Ucos,min (or in Ucos,max ) will be denoted as
U ≡ T , , T , , (R, e) cos,min (or U ≡ (R, e), , S , , S cos,max ) to represent its parameters. Remark 3. Any uninorm U in Ucos,min with = 0 or U in Ucos,max with = 1 is a representable uninorm. The distributivity condition between uninorms has been studied for several kinds of these operations (see [14,15,17, 20,21]). The following result will be used in the paper and can be found in [21]. Proposition 1. Let U and Ud be two uninorms with the same neutral element e ∈]0, 1[. If U is distributive over Ud , then Ud is idempotent and Ud (x, y) = U (x, y) for all (x, y) ∈ A(e) = [0, e] × [e, 1] ∪ [e, 1] × [0, e]. Definition 3 (Fodor and Roubers [9]). A binary operation I : [0, 1] × [0, 1] → [0, 1] is said to be an implication operation, or an implication, if it satisfies: (I1) I is nonincreasing in the first variable and nondecreasing in the second one. (I2) I (0, 0) = I (1, 1) = 1 and I (1, 0) = 0.
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Note that, any implication satisfies I (0, x) = 1 and I (x, 1) = 1 for all x ∈ [0, 1] whereas the symmetrical values I (x, 0) and I (1, x) are not derived from the definition. Strong and residual implications, obtained by Eqs. (1) and (2), respectively, are studied in the literature not only from t-norms and t-conorms, but also from conjunctive and disjunctive uninorms. With respect to the residual implication obtained from a t-norm T, usually denoted by IT , we have the following result (see for instance [4]). Proposition 2. Let T be a continuous t-norm. • IT (x, y) = 1 if and only if x y, known as the ordering property. • If T = min, then IT is the well-known Gödel implication given by
1 if x y IT (x, y) = y otherwise • If T is Archimedean with additive generator f, then IT (x, y) = f −1 (max(0, f (y) − f (x))) x, y ∈ [0, 1] • If T is an ordinal sum T = ( k , k , Tk )k∈K ⎧ 1 ⎪ ⎪ ⎨ −1 f k (ITk ( f k (x), f k (y))) IT (x, y) = ⎪ ⎪ ⎩ y
with K finite or countably infinite, then if x y if x > y and (x, y) ∈ [k , k ]2 , for some k ∈ K otherwise
where ITk is the R-implication derived from the Archimedean t-norm Tk for all k ∈ K , and f k (x) = (x − k )/(k − k ). Remark 4. Note that, from expressions above, it is easy to see that when T is nonstrict Archimedean (that is, f (0) < +∞), IT is continuous, whereas for strict t-norms ( f (0) = +∞), (0, 0) is a point of discontinuity of IT . In any case, when T has an Archimedean ordinal summand Tk in an interval [k , k ], fixed any c ∈]k , k [⊂]0, 1[, the section IT (−, c) is strictly decreasing in the interval [c, k ] taking all values from lim IT (x, c) = k to IT (k , c) = c
x→c+
but IT is not continuous when k < 1 because IT (c, c) = 1. The following proposition can be found in [3]. Proposition 3. Let U be a representable uninorm with neutral element e ∈]0, 1[ and additive generator h. Let U ∗ be the disjunctive representable uninorm with the same additive generator h, and N be the strong negation given by N (x) = h −1 (−h(x)). Then, (i) The residual implicator IU is given by
−1 h (h(y) − h(x)) if (x, y)(0, 0), (1, 1) IU (x, y) = 1 otherwise (ii) IU (x, y) = IU ∗ (x, y) = U ∗ (N (x), y) for all x, y ∈ [0, 1]. 3. Strong implications Definition 4. Given a disjunctive uninorm U and a strong negation N, the operation defined by IU,N (x, y) = U (N (x), y) for all x, y ∈ [0, 1] will be called the strong implication of U and N.
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1
R
N (x)
y
T
0 0
N ()
1
Fig. 1. IU,N with U ∈ Ucos,min , where operations in the figure are applied to the pairs (N (x), y).
With the previous assumptions IU,N is always an implication as its name suggests. Note that, for a uninorm U ∈ Ucos , U is disjunctive if and only if one of the following items holds:
(a) U ≡ (R, e), , S , , S cos,max , with U (1, 0) = 1. (b) A uninorm U in Ucos,min with = 0 (consequently only a t-norm T is necessary in its expression), and U (1, 0) = 1. We will denote such uninorm by U ≡ 0, T, , (R, e) cos,min . In both cases, the general structure of the strong implication derived from such uninorms can be easily derived and they are given in the following two propositions, respectively. The proofs of both propositions are straightforward. Proposition 4. If U ≡ 0, T, , (R, e) cos,min is a disjunctive uninorm, then IU,N is given by ⎧ N (x) y ⎪ ⎪ T , ⎪ ⎪ ⎪ ⎪ ⎨ N (x) − y − IU,N (x, y) = + (1 − )R , ⎪ 1− 1− ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎩ min(N (x), y)
if x ∈]N (), 1[ and y ∈]0, [ if x ∈]0, N ()[ and y ∈], 1[ if x = 0 or y = 1 otherwise
Proposition 5. If U ≡ (R, e), , S , , S cos,max is a disjunctive uninorm in Ucos,max , then IU,N is given by ⎧ N (x) − y −
⎪ ⎪ , ⎪ + (1 − )S ⎪ ⎪ 1 − ⎪ 1− ⎪ ⎪ y − N (x) − ⎪ ⎪ , ⎪ + ( − )S
⎪ ⎨ − − N (x) y IU,N (x, y) = , ⎪ R ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ 0 or max(N (x), y) ⎪ ⎪ ⎩ max(N (x), y)
if x ∈]0, N ()[ and y ∈], 1[ if x ∈]N (), N ()[ and y ∈], [ if x ∈]N (), 1[ and y ∈]0, [ if x = 1, y ∈ [0, [, or y = 0, x ∈]N (), 1] if (x, y) = (1, ) or (x, y) = (N (), 0) otherwise
We can see the general structure of IU,N being U ∈ Ucos,min and being U ∈ Ucos,max in Figs. 1 and 2, respectively. From its definition, it is clear that strong implications IU,N always satisfy contrapositive symmetry with respect to N, I (N (y), N (x)) = I (x, y) for all x, y ∈ [0, 1]
(6)
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1 S ′′ y •
S′ N (x) 0 0
• N ()
R
1
N ()
Fig. 2. IU,N with U ∈ Ucos,max , where operations in the figure are applied to the pairs (N (x), y).
because U is commutative and N involutive, and the exchange principle I (x, I (y, z)) = I (y, I (x, z)) for all x, y, z ∈ [0, 1], because U is commutative and associative. Other interesting property is the distributivity of such implications over conjunctive and disjunctive uninorms. In others words, the translation of Eq. (3) to our framework, that is, IU,N (Uc (x, y), z) = Ud (IU,N (x, z), IU,N (y, z))
(7)
for all x, y, z ∈ [0, 1], where Uc and Ud are uninorms in one of the classes considered, such that Uc is conjunctive, Ud is disjunctive and their underlying t-norms and t-conorms are continuous. This property was already studied for strong implications derived from other types of uninorms in [21]. Thus, we complete here that study for uninorms continuous in ]0, 1[2 . First of all, we have the following result, which proof is not included because it is the same as the one given in Theorem 1 of [21]. Theorem 2. Let U , Ud be disjunctive uninorms, Uc a conjunctive uninorm and N a strong negation. Then IU,N , Uc and Ud satisfy the distributivity equation (7) if and only if Uc and Ud are N-dual and U is distributive over Ud . i.e., U (x, Ud (y, z)) = Ud (U (x, y), U (x, z))
(8)
for all x, y, z ∈ [0, 1]. To solve Eq. (8) for uninorms in Ucos , we will distinguish two cases depending on which class is in the uninorm U, Ucos,max or Ucos,min . However, we give first the following lemma that is true in a more general setting. Lemma 1. Let U, Ud be two disjunctive uninorms with neutral elements e, ed ∈ [0, 1[, respectively, such that U is distributive over Ud . Then U (ed , ed ) = ed , that is, ed is an idempotent element of U. Proof. Take y = e and z = ed and x = ed in (8), then we obtain ed = U (ed , e) = U (ed , Ud (e, ed )) = Ud (U (ed , e), U (ed , ed )) = Ud (ed , U (ed , ed )) = U (ed , ed ).
3.1. Distributivity when U ∈ Ucos,max Now we consider the case when U ∈ Ucos,max . Recall that we assume for the distributivity property that the underlying t-norm and t-conorm of Ud are continuous.
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Lemma 2. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then e < ed or ed = 0. In both cases Ud (e, e) > 0. Proof. We divide the proof in several steps. (i) If ed < , as U is in Ucos,max , their only idempotent elements less than , are e and 0. By Lemma 1, ed is an idempotent element of U. But if ed = e and U is distributive over Ud , we know by Proposition 1 that Ud must be idempotent and Ud (x, y) = U (x, y) for all (x, y) ∈ A(e), that gives a contradiction with the definition of U. Thus, we have e < ed or ed = 0. (ii) If e < ed and Ud (e, e) = 0, then we have U (x, Ud (e, e)) = U (x, 0) for all x ∈ [0, 1] but by the other side, since U is distributive over Ud , U (x, Ud (e, e)) = Ud (U (x, e), U (x, e)) = Ud (x, x) and then, Ud (x, x) = U (x, 0) which contradicts the continuity of TUd and SUd . Then, we can conclude that e < ed and Ud (e, e) > 0. In the case ed = 0, Ud (e, e) > 0 trivially because then Ud is a t-conorm. Now, taking into account the results in Lemma 2, we distinguish two cases according to the possible values of ed , that is ed = 0 or ed > 0. In the case ed = 0 we have that Ud = S is a t-conorm and then, the results are already known (see [19, Theorem 9]) and they are as follows.
Proposition 6. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud = S be a t-conorm. Then U is distributive over S if and only if • S = max or • S is an ordinal sum of the form S = (0, , S1 ) with S1 strict, and the additive generator s of S1 satisfying s(e/) = 1, is also a multiplicative generator of R. In the case ed > 0 we will reach new solutions through several lemmas.
Lemma 3. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed > 0, such that U is distributive over Ud . Then Ud (x, x) = x for all x ∈ [, 1]. Proof. We know by Lemma 2 that 0 < Ud (e, e) and e < ed . Thus, Ud (e, e) is given by the underlying t-norm of Ud and consequently we have 0 < Ud (e, e) e < ed . Then for all x , we have, by definition of U ∈ Ucos,max : U (x, Ud (e, e)) = max(x, Ud (e, e)) = x but also, because U is distributive over Ud , U (x, Ud (e, e)) = Ud (x, x) and then Ud (x, x) = x for all x ∈ [, 1].
Lemma 4. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed > 0, such that U is distributive over Ud . Then, Ud cannot be continuous in ]0, 1[2 and, for all 0 < y < ed < z, we have that Ud (y, z)y. Proof. The first assertion is an immediate consequence of Lemma 3. On the other hand, suppose that there exists z > ed such that for some 0 < y < ed < z we have Ud (y, z) = y. We can take x such that 0 < y < ed < x < z, and then, using Lemma 1, and the structure of U, we have, by one side: U (x, Ud (y, z)) = U (x, y) = max(x, y) = x,
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but by the other, we have Ud (U (x, y), U (x, z)) = Ud (x, U (x, z))Ud (x, z) z > x that gives a contradiction, and then the result is proven. Thus, when ed > 0 solutions of the distributivity equation can only appear when Ud is in Umnx . To reach all of them we distinguish two cases according that ed < or ed .
Lemma 5. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm in Umnx with neutral element ed , such that U is distributive over Ud , and 0 < ed < . Then, Ud (x, 0) = U (x, 0) for all x ∈ [0, 1]. Proof. We divide our proof in several cases: • If x ed , by definition of U, we have Ud (x, 0) = U (x, 0) = 0. • If ed < x < . Take z such that 0 < z < ed < x < . Then we have, U (x, 0) = U (x, Ud (z, 0)) = Ud (U (x, z), U (x, 0)) = Ud (max(x, z), 0) = Ud (x, 0) where in the third equality we have used that U ∈ Ucos,max . • If < x, we take some y satisfying ed < y < < x, and then we have Ud (x, 0) = Ud (U (x, 0), U (y, 0)) = U (Ud (x, y), 0) = U (max(x, y), 0) = U (x, 0) • When x = , we distinguish two subcases: ◦ When U (0, ) = , take y with 0 < y < ed < . Using the previous lemma, Ud (, y)y, and since Ud ∈ Umnx , Ud (, y) = . Now we have U (0, ) = U (0, Ud (, y)) = Ud (U (0, ), U (0, y)) = Ud (, 0) ◦ When U (0, ) = 0. Now, take 0 < z ed < : U (, 0) = U (, Ud (z, 0)) = Ud (U (, z), U (, 0)) = Ud (, 0).
Corollary 1. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud , and 0 < ed < . Then, if Ud is in Umnx , there exists a continuous t-norm T such that Ud is given by ⎧ x y ⎪ ⎪ , if (x, y) ∈ [0, ]2 ⎪ T ⎪ ⎨ if max(x, y) > ed and min(x, y) > 0 Ud (x, y) = max(x, y) (9) ⎪ ⎪ U (x, y) if min(x, y) = 0 ⎪ ⎪ ⎩ min(x, y) otherwise Proof. From Lemmas 2 and 3 we know that ed and that Ud (x, x) = x for all x . This implies that the underlying t-conorm of Ud must be the maximum and the underlying t-norm must be an ordinal sum of the form ( 0, /ed , T ) for a continuous t-norm T. The result follows now from the previous lemmas.
Lemma 6. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud , and ed . Then Ud (x, 0) = x for all x > ed , that is, Ud must be in Umax .
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Proof. Suppose that there is x > ed such that Ud (0, x) = 0, then we can take ed < z < x, and we can write, by one hand U (z, Ud (0, x)) = U (z, 0) = z and by the other: Ud (U (z, 0), U (z, x)) = Ud (z, U (z, x))Ud (z, x) = x that gives a contradiction. Then, Ud (0, x) = x for all x > ed .
Corollary 2. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud , and ed . Then, if Ud is in Umnx , then U ∈ Umax and there exists a continuous t-norm T such that Ud must be given by ⎧ x y ⎪ ⎪ , if (x, y) ∈ [0, ]2 ⎨ T Ud (x, y) = (10) max(x, y) if max(x, y)ed ⎪ ⎪ ⎩ min(x, y) otherwise Proof. From the previous lemma it is clear that Ud must be in Umax and the result follows from Lemmas 2 and 3 similarly as in Corollary 1.
Theorem 3. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Ud be a disjunctive uninorm in one of the classes considered in the Preliminaries. Then, U is distributive over Ud if and only if • Ud = S is a t-conorm as in Proposition 6, or • ed , U (ed , ed ) = ed , there exists a continuous t-norm T such that R is distributive over T (that is, T = min or T is strict and, if t is the additive generator of T satisfying t(e/) = 1 then 1/t is also a multiplicative generator of R), and one of the following two cases holds: (i) ed < and then Ud is given by Eq. (9), or (ii) ed and then Ud is given by Eq. (10). Proof. If U is distributive over Ud we know by the previous lemmas that Ud is a t-conorm as in Proposition 6, or ed , U (ed , ed ) = ed , and there exists a continuous t-norm T such that Ud is given by Eq. (9) or Eq. (10), depending on ed < or ed , respectively. Moreover, since U is distributive over Ud we have in particular that U (x, Ud (y, z)) = Ud (U (x, y), U (x, z)) for all x, y, z and consequently we will have that R is distributive over T and the result follows from the solutions of the distributivity equation for representable uninorms given in [21]. Conversely, it is a straightforward computation to show that when U and Ud are as in the enunciate they are distributive, by considering several cases depending on where x, y, z are placed with respect to and ed . The general structure of U and Ud such that U is distributive over Ud is depicted in Fig. 3 for case (i), and Fig. 4 for case (ii). 3.2. Distributivity when U ∈ Ucos,min We start with the case that U ∈ Ucos,min . Recall that in this case, to be U disjunctive we necessarily have = 0. Lemma 7. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm in Ucos,min , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then ed and Ud (e, e) < 1. Proof. Let us suppose first that ed > . By Lemma 1, ed is an idempotent element of U which is in Ucos,min . Thus, ed = e. But if ed = e and U is distributive over Ud , we know by Proposition 1 that Ud must be idempotent and
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Fig. 3. U ∈ Ucos,max (left) distributive over Ud ∈ Umnx (right), with < ed < .
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Fig. 4. U ∈ Ucos,max (left) distributive over Ud ∈ Umax (right), with < ed .
Ud (x, y) = U (x, y) for all (x, y) ∈ [0, e] × [e, 1] ∪ [e, 1] × [0, e], that gives a contradiction with the definition of U and consequently ed < e. Let us suppose now that Ud (e, e) = 1, then we have U (0, Ud (e, e)) = U (0, 1) = 1, whereas Ud (U (0, e), U (0, e)) = Ud (0, 0) = 0. But this is a contradiction since U is distributive over Ud . Then, we can conclude that Ud (e, e) < 1. Lemma 8. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm in Ucos,min , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then Ud (x, x) = x for all x ∈ [0, ]. Proof. We know by the previous lemma, that ed < e and Ud (e, e) < 1. Then for all x < e Ud (e, e) < 1, we have, by definition of U ∈ Ucos,min : U (x, Ud (e, e)) = x, but also, because U is distributive over Ud , U (x, Ud (e, e)) = Ud (x, x), and then Ud (x, x) = x for all x ∈ [0, ].
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Corollary 3. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm, and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then Ud is neither representable, nor in Ucos . Lemma 9. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm in Ucos,min , and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . For all y < ed < z < 1 we have that Ud (y, z)z. Proof. Suppose on the contrary that there exists z such that for some y < ed < z < 1 we have Ud (y, z) = z. Now, we can take x such that y < x < ed < z < 1, and then, using Lemma 1, and the structure of U, we have, by one side U (x, Ud (y, z)) = U (x, z) = min(x, z) = x, but by the other, we have Ud (U (x, y), U (x, z)) = Ud (U (x, y), x)Ud (y, x) y < x that gives a contradiction, and then the result is proven. Corollary 4. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm, and let Ud be a disjunctive uninorm with neutral element ed , such that U is distributive over Ud . Then Ud is not in Umax and, if it is in Umnx , there exists a continuous t-conorm S such that Ud must be given by ⎧ min(x, y) ⎪ ⎪ if min(x, y) < ed and max(x, y) < 1 ⎨ x − y− Ud (x, y) = + (1 − )S , if (x, y) ∈ [, 1]2 (11) ⎪ 1− 1− ⎪ ⎩ max(x, y) otherwise Proof. Similarly as in Corollary 1 we deduce from Lemma 8 that the underlying t-norm of Ud is the minimum and the underlying t-conorm is an ordinal sum of the form ( − ed /1 − ed , 1, S ) for a continuous t-conorm S. Now, the result follows from Lemma 9. Theorem 4. Let U ≡ 0, T, , (R, e) cos,min be a disjunctive uninorm, and let Ud be a disjunctive uninorm in one of the classes considered in the Preliminaries. Then, U is distributive over Ud if and only if ed and
(i) U (ed , ed ) = ed , that is, there exist T and T t-norms such that T = ( 0, ed /, T , ed /, 1, T ). (ii) There exists a t-conorm S such that Ud is given by Eq. (11). (iii) R is distributive over S, that is, S = max or S is strict and the additive generator s of S satisfying s((e − )/(1 − )) = 1, is also a multiplicative generator of R. Proof. By the previous lemmata, if U is distributive over Ud , then (i) and (ii) are satisfied. Moreover, we also have that R must be distributive over S and, from the solutions of the distributivity equation for representable uninorms and t-conorms (see [19] and also [21]) we conclude (iii). Conversely, if U and Ud satisfy (i)–(iii) it is a straightforward computation to check that U is distributive over Ud by distinguish several cases, depending on the values of x, y and z in (8). Remark 5. Note that in the extreme case ed = 0, Ud is a t-conorm and Eq. (11) reduces to ⎧ x − y− ⎨ + (1 − )S , if (x, y) ∈ [, 1]2 Ud (x, y) = 1− 1− ⎩ max(x, y) otherwise Thus, the theorem above contains, for the case ed = 0, the results in [19] about distributivity of a uninorm in Ucos,min over a continuous t-conorm. The general structure of U ∈ Ucos,min and Ud such that U is distributive over Ud is depicted in Fig. 5.
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Fig. 5. U ∈ Ucos,min (left) distributive over Ud ∈ Umnx (right).
3.3. Other related distributivities Related to Eq. (7) there are several distributivity equations involving implications and uninorms. Namely, IU,N (x, Ud (y, z)) = Ud (IU,N (x, y), IU,N (x, z))
(12)
IU,N (Ud (x, y), z) = Uc (IU,N (x, z), IU,N (y, z))
(13)
IU,N (x, Uc (y, z)) = Uc (IU,N (x, y), IU,N (x, z))
(14)
where uninorms with subscript c are conjunctive and those with subscript d are disjunctive. All of these properties were studied for strong implications derived from other types of uninorms in [21]. As in such reference, it can be easily proved using contrapositive symmetry of strong implications that Eq. (12) is equivalent to Eq. (7) taking Uc the N-dual of Ud and consequently we have the following theorem. Theorem 5. Let U be a disjunctive uninorm in Ucos , and Ud and Ud two disjunctive uninorms. Then IU,N , Ud and Ud
satisfy Eq. (12) if and only if Ud = Ud and U is distributive over Ud . Thus, Theorems 3 and 4 give all the solutions of Eq. (12). Analogously Eq. (13) is equivalent to Eq. (14) using this duality, and so, we only need to solve this last equation. Again, we easily have the following result. Lemma 10. Let U be a disjunctive uninorm in Ucos , and Uc , Uc two conjunctive uninorms. Then, IU,N , Uc and Uc
satisfy Eq. (14) if and only if Uc = Uc and U is distributive over Uc . Thus, we need to solve the distributivity equation of U over Uc . However, in this case the solutions among the four classes of uninorms considered, appears only when ec = 1, that is, when Uc is a t-norm. Let us prove this fact following the lemmas below. Lemma 11. Let U be a disjunctive uninorm in Ucos , and Uc a conjunctive uninorm with neutral element ec < 1. If U is distributive over Uc , necessarily U ∈ Ucos,max . Proof. Suppose that U ∈ Ucos,min , then, as U is distributive over Uc , we have by one side U (0, Uc (1, ec )) = U (0, 1) = 1
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and by the other, since ec < 1: Uc (U (0, 1), U (0, ec )) = Uc (1, 0) = 0 that is a contradiction, and therefore U ∈ Ucos,max . However, even when U is in Ucos,max , there are no solutions with ec < 1, as the following lemma shows.
Lemma 12. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Uc be a conjunctive uninorm with neutral element ec . If U is distributive over Uc , then (i) e < ec , U (ec , ec ) = ec and Uc (e, e) > 0. (ii) Uc (x, x) = x for all x ∈ [, 1] and consequently Uc cannot be in Ucos . (iii) If Uc ∈ Umnx then ec = 1, that is Uc is a t-norm. Proof. The proofs of parts (i) and (ii) are very similar to the ones given in Lemmas 2 and 3, respectively. With respect to part (iii), take Uc ∈ Umnx with U distributive over Uc and suppose ec < 1. We will reach a contradiction in several steps: • First we prove that Uc (1, y) = 1 for all 0 < y < ec . Suppose on the contrary that there is such a y with Uc (1, y) = y. Taking x such that ec < x < 1 we have x = U (x, y) = U (x, Uc (y, 1)) = Uc (U (x, y), U (x, 1)) = Uc (x, 1) = 1 obtaining contradiction. • Now, for all 0 < y < e < ec we have, using the previous step, that 1 = U (0, 1) = U (0, Uc (1, y)) = Uc (U (0, 1), U (0, y)) = Uc (1, 0) = 0 obtaining again contradiction. Thus, ec = 1 and Uc must be a t-norm. Finally when ec = 1, we have Uc = T is a t-norm and the distributivity of U ∈ Ucos over T is already known (see [19]). In spite of that we include here the solutions for the sake of completeness.
Proposition 7. Let U ≡ (R, e), , S , , S cos,max be a disjunctive uninorm in Ucos,max , and let Uc = T be a t-norm. Then U is distributive over T if and only if one of the following cases is satisfied • T = min. • U ≡ 0, T, , (R, e) cos,min and T is the ordinal sum given by T = (, 1, T ∗ ) , where T ∗ is a strict t-norm which additive generator t with t((e − )/(1 − )) = 1 satisfies that 1/t is also a multiplicative generator of R.
• U ≡ (R, e), , S , , S cos,max and T is the ordinal sum given by T = (0, , 1, T ∗∗ ) , where T ∗∗ is a strict t-norm in which additive generator t with t(e/) = 1 satisfies that 1/t is also a multiplicative generator of R. 4. R-implications from uninorms continuous in ]0, 1[2 Now, we want to deal with residual implications defined from uninorms in Ucos by IU (x, y) = sup{z ∈ [0, 1] | U (x, z) y}. In this case, to be IU an implication it is necessary (and sufficient) that the uninorm U satisfies U (x, 0) = 0 for all x < 1 (see [3]). Thus we can consider uninorms U in Ucos,min , but also uninorms in Ucos,max with = 1. In both cases, the general structure of the residual implication from such uninorm U can be easily derived.
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1
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Fig. 6. Structure of IU , being U inbb= 0 0 134 133 Ucos,min .
Theorem 6. If U ≡ T , , T , , (R, e) cos,min is a conjunctive uninorm in Ucos,min , then IU is given by ⎧ 1 ⎪ ⎪ ⎪ x y ⎪ ⎪
I , ⎪ T ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ( − )I
x − , y − ⎪ ⎨ T − − IU (x, y) = x − y− ⎪ , + (1 − )I R ⎪ ⎪ ⎪ 1− 1− ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ y
if x ∈ [0, ] and y x if x ∈ [0, [ and y < x if x ∈ [, ] and y < x if (x, y) ∈ ], 1[2
(15)
if y = 1 if x = 1 and y > . otherwise
The structure of these implications can be seen in Fig. 6. Now we will study the second case, in which U ∈ Ucos,max . Theorem 7. If U ≡ (R, e), , S, 1 cos,max , then IU ⎧ x y ⎪ ⎪ I R , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x − y− ⎪ ⎪ , ⎨ + (1 − )R S 1− 1− IU (x, y) = 0 ⎪ ⎪ ⎪ ⎪ y ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎩ 0
is given by if (x, y) ∈]0, ]2 if (x, y) ∈], 1[2 and y x if x ∈], 1[ and y < x if (x, y) ∈]0, ] × [, 1[ if y = 1 or x = 0 if y = 0 and x0
(16)
The structure of such implications can be viewed in Fig. 7. Now we can study two properties of implications that we have recalled before: the exchange principle and contrapositive symmetry, for these implications. For the exchange principle, we know that all left-continuous uninorms are such that its residual implication will satisfy the exchange principle, and in the case of uninorms in Ucos we have the following result. Theorem 8. Let U be a uninorm in Ucos such that U (0, x) = 0 for all x < 1. If we have any of the following cases:
(i) U ≡ T , , T , , (R, e) cos,min , with = and U (, 1) = U (1, ) = . (ii) U ≡ (R, e), , S, 1 cos,max , conjunctive, i.e., U (0, 1) = U (1, 0) = 0. Then U is left-continuous and IU will satisfy the exchange principle.
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1 RS y 0
IR
0
0
1
Fig. 7. Structure of IU , with U ∈ Ucos,max .
On the contrapositive symmetry property, we get the next theorem. Theorem 9. Let U be a uninorm in Ucos such that U (0, x) = 0 for all x < 1. Then IU satisfy the contrapositive symmetry respect to a strong negation N if and only if U is representable and N = NU . Proof. First of all, it was proven in [3] that if U is representable, then it satisfy contrapositive symmetry with respect to NU . To prove from left to right, we will distinguish two different cases.
(i) If U ≡ T , , T , , (R, e) cos,min , let us suppose that there exists a strong negation N for which IU satisfy contrapositive symmetry. Then, by applying x = y in the equation of contrapositive symmetry (6) we will obtain IU (x, x) = IU (N (x), N (x))
(17)
Now we calculate, using Eq. (15), the value of IU (x, x), knowing that uninorm R has neutral element (e − )/(1 − ), ⎧ ⎨ 1 if x IU (x, x) = e if < x < 1 ⎩ 1 if x = 1 As N is a strong negation, there exists a fix point of N, that we will call z, N (z) = z. Now we will see that if 0, there exists x such that x < < N (x) by distinguishing three cases: • If z = , then we can take 0 < x < 1 such that x < = N () < N (x). • If z < , then we can take 0 < x < 1 such that x < N () < z < < N (x). • If < z, then we can take 0 < x < 1 such that x < < z < N () < N (x). Then, in any case, there exists x such that x < < N (x), and if contrapositive symmetry with respect to N is satisfied, by Eq. (17) it would have to be 1 = e, but this is a contradiction, because then U would be a t-norm and not from Ucos,min . Therefore, = 0 and U is representable.
(ii) If U ≡ (R, e), , S , , S cos,max , following a similar reasoning as in the previous case, now using Eq. (16) and taking into account that the neutral element of R is e/, we can deduce that ⎧ e if x < ⎪ ⎪ ⎪ ⎪ ⎨ if x = x − x − IU (x, x) = + (1 − )R S , if < x < 1 ⎪ ⎪ ⎪ 1− 1− ⎪ ⎩ 1 if x = 1
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847
And similarly to the previous case, if 1 then there exists x0 such that x0 < < N (x0 ), and if (17) is satisfied, then we have e = + (1 − )R S ((x0 − )/(1 − ), (x0 − )/(1 − )), but this is a contradiction, because e < . Then = 1 and U is representable. Finally, for this kind of implications we are also interested in solving the distributivity equation as in the case of strong implications. That is, we want to study the equation IU (Uc (x, y), z) = Ud (IU (x, z), IU (y, z))
(18)
for all x, y, z ∈ [0, 1]. As in the case of strong implications, Uc and Ud are a conjunctive and a disjunctive uninorm, respectively, and we assume that their underlying t-norms and t-conorms are continuous. On the other hand, in this case U is a uninorm in Ucos and IU its residual implication. We distinguish again two cases, when U is in Ucos,min , and when it is in Ucos,max and we do this in the following subsections. 4.1. Distributivity when U ∈ Ucos,min Lemma 13. Consider Uc a conjunctive uninorm with neutral element ec , Ud a disjunctive uninorm,
U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then Uc is a t-norm. Proof. • First of all, if we suppose that ec , then we would have, taking x = ec and y = z = e in (18), Ud (IU (ec , e), IU (e, e)) = IU (Uc (ec , e), e) that means Ud (1, e) = IU (e, e), because ec < e. This last equality implies e = 1, that is a contradiction. Then < ec . • Now, suppose that ec < 1. Then we can take y0 such that ec < y0 < 1 with IU (ec , z) > IU (y0 , z) for all z > , because < ec < y0 , by the previous reasoning. As IU is continuous in [, 1]2 , there exists z 0 such that IU (ec , z 0 ) > IU (y0 , z 0 ) ed , and therefore we have, by putting x = ec , y = y0 , and z = z 0 in (18): Ud (IU (ec , z 0 ), IU (y0 , z 0 )) = IU (Uc (ec , y0 ), z 0 ) = IU (y0 , z 0 ) but we also have Ud (IU (ec , z 0 ), IU (y0 , z 0 )) max(IU (ec , z 0 ), IU (y0 , z 0 )) > IU (y0 , z 0 ) that is a contradiction. Then we can say that ec = 1, that is, Uc is a t-norm. From now on we will consider Uc be a (continuous) t-norm.
Lemma 14. Let Uc be a t-norm, Ud a disjunctive uninorm with neutral element ed , U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then ed and Ud (z, z) = z for all z . Proof. Taking x = 1, y = e in (18) we have z = IU (e, z) = IU (T (1, e), z) = Ud (IU (1, z), IU (e, z)) =
Ud (, z) if z Ud (z, z) if z
Now, if < ed we would have Ud (, ed ) = ed = , that is a contradiction. Then, Ud (z, z) = z for all z , and ed .
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Lemma 15. Let Uc be a t-norm, Ud a disjunctive uninorm with neutral element ed , U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then U (x, x) = x for all x ed , and consequently IU (x, y) = y for all x, y such that x < y ed . Proof. Suppose that there is an x ed such that U (x, x) < x, that is, T (x/e, x/e) < x/e and by continuity there will exist an interval [a, b] such that a < x < b and the restriction of T to the interval [a/e, b/e] is Archimedean. Now, fixed z 0 such that a < z 0 < x and taking into account Remark 4, IU (−, z 0 ) takes all values from z 0 to b. Thus, let x0 be such that a < z 0 < IU (x0 , z 0 ) < min(ed , b), we have on one hand IU (Uc (x0 , 1), z 0 ) = IU (x0 , z 0 ) whereas, on the other hand, Ud (IU (x0 , z 0 ), IU (1, z 0 )) = Ud (IU (x0 , z 0 ), z 0 ) = min(IU (x0 , z 0 ), z 0 ) = z 0 obtaining contradiction.
Lemma 16. Let Uc be a t-norm, Ud = (ed , Td , Sd ) a disjunctive uninorm, U ≡ T , , T , , (R, e) cos,min and IU its residual implication. If they satisfy (18), then Uc (e, e) > and Ud (x, x) = Uc (x, x) = x for all x . That is, Uc must be an ordinal sum Uc = (, 1, T0 ) for some continuous t-norm T0 , Td = min and Sd must be also an ordinal sum Sd = (( − ed )/(1 − ed ), 1, S0 ) . Proof. Let us give the proof in several steps. • First we prove that Uc (e, e) > . Suppose on the contrary that Uc (e, e) , then 1 = IU (Uc (e, e), ) = Ud (IU (e, ), IU (e, )) = Ud (, ). Thus, since e > we have Ud (e, e) = 1. Since Uc is continuous, we can take an x such that 1 > x > with 1 > Uc (x, x) > , but then IU (Uc (x, x), x) = Ud (IU (x, x), IU (x, x)) = Ud (e, e) = 1 obtaining a contradiction because IU is strictly monotone in [, 1]2 (in this region, U is given by the representable uninorm R and then we can apply Proposition 3). • Second we prove that Ud (x, x) = x for all x . Just taking x = y = e in (18) we obtain IU (Uc (e, e), z) = Ud (IU (e, z), IU (e, z)) = Ud (z, z). Now, for all z , IU (Uc (e, e), z) = z by the previous step and consequently Ud (z, z) = z for all z . This directly implies that Td = min and that Sd is an ordinal sum of the form Sd = (( − ed )/(1 − ed ), 1, S0 ) . • Finally, we prove that Uc (x, x) = x for all x . Suppose on the contrary that there is some x < such that Uc (x, x) < x and take z such that Uc (x, x) < z < x < . For these values we have IU (x, z) and then, using the step above, 1 = IU (Uc (x, x), z) = Ud (IU (x, z), IU (x, z)) = IU (x, z) obtaining again a contradiction. Thus, Uc (x, x) = x for all x and consequently Uc is an ordinal sum of the form Uc = (, 1, T0 ) . Theorem 10. Let Uc and Ud = (ed , Td , Sd ) be a conjunctive and a disjunctive uninorm, respectively,
U ≡ T , , T , , (R, e) cos,min and IU its residual implication. Then, Uc , Ud and IU satisfy (18) if and only there exist a continuous t-norm T0 and a continuous t-conorm S0 such that (i) Uc is the t-norm given by the ordinal sum Uc = (, 1, T0 ) , (ii) ed , Td = min and Sd = (( − ed )/(1 − ed ), 1, S0 ) ,
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e
e
T0
min
min
849
max
S0
max
max
min ed
ed
min ed
ed
e
e
IR
1
e
IT ′′
1
• y
IT ′
ed
y y
ed
e
Fig. 8. Structure of Uc (top left), Ud (top right) and IU (bottom), with U ∈ Ucos,min , that satisfy Eq. (18).
(iii) U (x, x) = x for all x ed , IU (x, y) = y for all x < y ed and (iv) T0 , S0 and I R satisfy Eq. (18), that is, T0 and S0 are N R -dual and: • T0 = min and S0 = max or, • S0 is strict and if s is its additive generator with s((e − )/(1 − )) = 1 then s is also a multiplicative generator of R. Proof. We first suppose that Uc , Ud and IU satisfy (18). Then, (i)–(iii) follows directly from the previous lemmas. On the other hand, since Uc , Ud and IU satisfy Eq. (18) in particular for all x, y, z , we derive that T0 , S0 and R must also satisfy Eq. (18) and the remaining items follows from the results of such equation for representable uninorms given in [22]. Conversely, due to their symmetry we only need to prove Eq. (18) for values x y and we can do this by distinguishing several cases: • If x , then Uc (x, y) = min(x, y) = x. On the other hand, ◦ When z x then I (x, z) = 1 and Eq. (18) follows trivially because Ud is disjunctive. ◦ When z < x, ed we have I (x, z) = I (y, z) = z and (18) follows because Ud (z, z) = z. ◦ When ed z < x we have I (x, z), I (y, z) ∈]ed , ]. But Ud is given by the maximum for these values and again (18) is satisfied. • If x, Eq. (18) is satisfied because T0 S0 and I R also satisfy it. The structure of solutions of the previous theorem can be viewed in Fig. 8.
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4.2. Distributivity when U ∈ Ucos,max Recall that in this case we must have a uninorm U with = 0, that is U ≡ (R, e), , S, 1 cos,max . We obtain new solutions with similar reasonings as in the subsection above. Lemma 17. Let Uc and Ud be a conjunctive and a disjunctive uninorm, respectively, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. If they satisfy (18), then ed = 0, that is Ud must be a t-conorm, and Ud (x, x) = x for all x . Proof. Let us prove first that ed = 0. Suppose on the contrary that ed > 0, then: • If ec we have IU (Uc (ec , e), z) = IU (e, z) = z, whereas for all z < ec , Ud (IU (ec , z), IU (e, z)) = Ud (0, z). Thus, we obtain Ud (0, z) = z for all z < ec which is a contradiction. • If ec < we can take an y0 such that 0 < y0 < ec < . But then there will exist a z 0 < such that IU (ec , z 0 ) < IU (y0 , z 0 ) < ed and then IU (Uc (ec , y0 ), z 0 ) = IU (y0 , z 0 ) whereas Ud (IU (ec , z 0 ), IU (y0 , z 0 )) min(IU (ec , z 0 ), IU (y0 , z 0 )) < IU (y0 , z 0 ) obtaining again contradiction. Thus, ed = 0 and finally, for all z we have z = IU (0, z) = IU (Uc (0, 0), z) = Ud (IU (0, z), IU (0, z)) = Ud (z, z).
Lemma 18. Let Uc be a conjunctive uninorm and Ud a t-conorm, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. If they satisfy (18), then ec > and Uc is in Umin with Uc (x, x) = x for all x . Proof. We give the proof in several steps: • First we prove that ec > . Suppose that ec and take z such that < z < 1. Then on one hand IU (Uc (ec , 1), z) = IU (1, z) = 0, and on the other hand Ud (IU (ec , z), IU (1, z)) = Ud (z, 0) = z, which gives a contradiction. • Now we prove that Uc (1, x) = x for all x > . Suppose that there is an x > such that Uc (1, x) > x > , we can take z such that x < z < Uc (1, x) and for this z we have 0 = IU (Uc (1, x), z) = Ud (IU (1, z), IU (x, z)) = Ud (0, IU (x, z)) = IU (x, z) > 0 obtaining contradiction. • From the step above we have by increasingness that Uc is in Umin . • Finally we prove that Uc (x, x) = x for all x . We prove it by distinguishing two cases: ◦ Suppose there is an x such that Uc (x, x) < x and take z 0 such that Uc (x, x) < z 0 < x. Then IU (Uc (x, x), z 0 ) = Ud (IU (x, z 0 ), IU (x, z 0 )) = Ud (0, 0) = 0 which is a contradiction since Uc (x, x) < z 0 . ◦ Suppose now that there is an x such that Uc (x, x) > x. Similarly, taking z 0 such that x < z 0 < Uc (x, x) we have 0 = IU (Uc (x, x), z 0 ) = Ud (IU (x, z 0 ), IU (x, z 0 )) = IU (x, z 0 ) which again is a contradiction because x < z 0 and consequently IU (x, z 0 ) . Lemma 19. Let Uc be a conjunctive uninorm and Ud a t-conorm, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. If they satisfy (18), then U (x, x) = x for all x ec .
D. Ruiz-Aguilera, J. Torrens / Fuzzy Sets and Systems 160 (2009) 832 – 852
851
max
min ec
ec min
max
max
S0
max
min
min
e
T0
e
e
min
e
ec
ec
y ec
y RS
0
e
IR
e
0
ec
Fig. 9. Structure of Uc (top left), Ud (top right), IU (bottom), with U ∈ Ucos,max , that satisfy Eq. (18).
Proof. Suppose U (x, x) > x for some x > ec . There is an interval [a, b] containing x and U (x, x) in which IU is strictly monotone. But then, taking z such that ec < x < z < U (x, x), we have IU (Uc (x, U (x, x)), z) = IU (U (x, x), z) = 0 and also Ud (IU (x, z), IU (U (x, x), z)) = Ud (IU (x, z), 0) = IU (x, z) obtaining a contradiction. Theorem 11. Let Uc = (ec , Tc , Sc ) and Ud be a conjunctive and a disjunctive uninorm, respectively, U ≡ (R, e), , S, 1 cos,max and IU its residual implication. Then, Uc , Ud and IU satisfy (18) if and only there exist a continuous t-norm T0 and a continuous t-conorm S0 such that (i) (ii) (iii) (iv)
Ud is the t-conorm given by the ordinal sum Ud = (0, , S0 ) , Uc is in Umin with ec > , Sc = max and Tc = (0, /ec , T0 ) , U (x, x) = x for all x ec and T0 , S0 and I R satisfy Eq. (18), that is, T0 and S0 are N R -dual and: • T0 = min and S0 = max or, • S0 is strict and if s is its additive generator with s(/ec ) = 1 then s is also a multiplicative generator of R.
Proof. Completely similar to the proof of Theorem 10. The structure of solutions given in the previous theorem can be viewed in Fig. 9.
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5. Conclusion Implication functions are essential not only in fuzzy logic, but also in many applications like approximate reasoning and fuzzy control. Among other approaches S-implications and R-implications are the most widely used in this context. In this paper, these kinds of implications have been derived from uninorms continuous in ]0, 1[2 pointing out their structure and studying some of their properties. One of them, particularly important, is the distributivity of such implications over conjunctive and disjunctive uninorms, that has been successfully used in avoiding combinatorial rule explosion in fuzzy systems (see [1]). In this paper, new solutions of this equation have been found among S- and R-implications derived from uninorms continuous in ]0, 1[2 . Acknowledgments The authors want to thank the referees for their valuable comments. This work has been partially supported by the Spanish Grant MTM2006-05540 and the Government of the Balearic Islands Project PRIB-2004-9250. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
[21] [22]
W.E. Combs, Combinatorial rule explosion eliminated by a fuzzy rule configuration, IEEE Trans. Fuzzy Systems 6 (1998) 1–11. B. De Baets, Idempotent uninorms, European J. Oper. Res. 118 (1999) 631–642. B. De Baets, J.C. Fodor, Residual operators of uninorms, Soft Comput. 3 (1999) 89–100. B. De Baets, R. Mesiar, Residual implicators of continuous t-norms, in: Proc. EUFIT-96 Aachen, Germany, 1996, pp. 27–31. S. Dick, A. Kandel, Comments on “Combinatorial rule explosion eliminated by a fuzzy rule configuration”, IEEE Trans. Fuzzy Systems 7 (1999) 475–477. J. Dombi, Basic concepts for the theory of evaluation: the aggregative operator, European J. Oper. Res. 10 (1982) 282–293. P. Dryga´s, Discussion of the structure of uninorms, Kybernetika 41 (2005) 213–226. P. Dryga´s, Remarks about idempotent uninorms, J. Elec. Eng. 57 (2006) 92–94. J.C. Fodor, M. Roubens, Fuzzy preference modelling and multicriteria decision support, Theory and Decision Library, Serie D: System Theory, Knowledge Engineering and Problem Solving, Kluwer Academic Publishers, Dordrecht, 1994. J.C. Fodor, R.R. Yager, A. Rybalov, Structure of uninorms, Int. J. Uncertainty Fuzziness Knowledge-based Systems 5 (1997) 411–427. S. Hu, Z. Li, The structure of continuous uni-norms, Fuzzy Sets and Systems 124 (2001) 43–52. E.P. Klement, R. Mesiar, E. Pap, Triangular Norms, Kluwer Academic Publishers, Dordrecht, 2000. J. Martín, G. Mayor, J. Torrens, On locally internal monotonic operations, Fuzzy Sets and Systems 137 (2003) 27–42. M. Mas, G. Mayor, J. Torrens, The distributivity condition for uninorms and t-operators, Fuzzy Sets and Systems 128 (2002) 209–225. M. Mas, G. Mayor, J. Torrens, Corrigendum to “The distributivity condition for uninorms and t-operators” [Fuzzy Sets and Systems 128 (2002) 209–225], Fuzzy Sets and Systems 153 (2005) 297–299. J.M. Mendel, Q. Liang, Comments on “Combinatorial rule explosion eliminated by a fuzzy rule configuration”, IEEE Trans. Fuzzy Systems 7 (1999) 369–371. D. Ruiz, J. Torrens, Distributive idempotent uninorms, Internat. J. Uncertainty, Fuzziness Knowledge-Based Systems 11 (2003) 413–428. D. Ruiz, J. Torrens, Residual implications and co-implications from idempotent uninorms, Kybernetika 40 (2004) 21–38. D. Ruiz, J. Torrens, Distributivity and conditional distributivity of a uninorm and a continuous t-conorm, IEEE Trans. Fuzzy Systems 14 (2006) 180–190. D. Ruiz-Aguilera, Contribució a l’estudi de les uninormes en el marc de les equacions funcionals, Aplicacions a la morfologia matemàtica, Ph.D. Thesis (in catalan) University of the Balearic Islands, Palma de Mallorca, Spain, 2007, Available at: http://www.tesisenxarxa.net/ TDX-1121107-124008/ . D. Ruiz-Aguilera, J. Torrens, Distributivity of strong implications over conjunctive and disjunctive uninorms, Kybernetika 42 (2005) 319–336. D. Ruiz-Aguilera, J. Torrens, Distributivity of residual implications over conjunctive and disjunctive uninorms, Fuzzy Sets and Systems 158 (2007) 23–37.