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Automorphisms, negations and implication operators
Fuzzy Sets and Systems 134 (2003) 209 – 229
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Automorphisms, negations and implication operators H. Bustince ∗ , P. Burillo, F. Soria Departamento de Autom atica y Computaci on, Universidad P ublica de Navarra, Campus de Arrosadia, 31006 Pamplona, Spain Received 5 April 2000; received in revised form 20 February 2002; accepted 5 April 2002
Abstract In this paper we study the conditions in which the implication operators satisfy the property I (x; c(x)) = c(x) for all x ∈ [0; 1], c being any strong negation. This study has led us to present di1erent implication operator characterization theorems from automorphisms, obtaining a theorem similar to the one presented by Smets and Magrez (Internat. J. Approx. Reason. 1 (1987) 327), in which the strong negation c used is not generated by the same automorphism that generates c 2002 Elsevier Science B.V. All rights reserved. the implication. Keywords: Fuzzy negation; Strong negation; t-Norm; t-Conorm; Automorphism; Implication operations
Introduction The purpose of this paper is to explore the conditions in which the implication operators satisfy the following property: I (x; c(x)) = c(x)
for all x ∈ [0; 1];
(1)
c being any strong negation. We have divided the introduction to this paper in two parts. In part (A) we present the manner in which the study of property (1) arose in our research. In part (B) we show the strategy carried out in the making of this paper. (A) Among the axioms that Sinha and Dougherty [35,36,37] demand of the inclusion grade indicator , the following can be found: If A is a re$nement of B, that is, if A (x)6B (x) when B (x)6c(e) = e ∈(0; 1), and A (x)¿B (x), when B (x)¿c(e) = e ∈(0; 1), then (A ∨ c(A); A ∧ c(A)) 6 ((B ∨ c(B); B ∧ c(B)): At present the main objective of our investigation is the construction of di1erent expressions of fuzzy entropy from the inclusion grade indicators that satisfy the axiom above. ∗
Corresponding author. Tel.: +34-94816-9254; fax: +34-94816-9565. E-mail address: [email protected] (H. Bustince).
c 2002 Elsevier Science B.V. All rights reserved. 0165-0114/03/$ - see front matter PII: S 0 1 6 5 - 0 1 1 4 ( 0 2 ) 0 0 2 1 4 - 2
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With said objective in mind, reviewing the fuzzy literature for the di1erent expressions presented for the inclusion grade indicator, we End the expression given by De Baets and Kerre [5], that is: for any A and B on the same referential X; (A; B) = Inf (I (A (x); B (x))): x∈X
Studying the conditions in which the expression above satisEes Sinha and Dougherty’s axiom we have proven that when the implication satisEes property (1), we have that De Baets and Kerre’s indicator satisEes Sinha and Dougherty’s axiom. This fact has led us to analyze property (1) in fuzzy implications, leaving the study of the functions that simultaneously generate fuzzy entropies and inclusion grade indicators for future papers. (B) In the Erst part of the paper we present the notations and basic deEnitions that we are going to use. Next, we focus on the di1erent axioms demanded of the implication operators subsequently analyzing some of the interrelations between said axioms. Then we present property (1) and we observe that when (1) holds then the following axiom (frequently demanded by fuzzy implications) does not: I (x; y) = 1 if and only if x 6 y
(2)
and vice versa, when I satisEes (2), it does not satisfy (1). Our interest in Ending the conditions in which I satisEes property (1) has led us to study, in Section 4, di1erent fuzzy implication characterization theorems from automorphisms, analyzing at all times the conditions in which the implications obtained satisfy (1) or (2) or neither. And so, in Section 4.1, where we study the S-implications, we have obtained a theorem similar to the one presented by Smets and Magrez [38], in which the strong negation c used is not generated by the same automorphism that generates the implication. Afterwards, we have obtained another S-implication characterization theorem from automorphisms such that the S-implications obtained satisfy (1). In Sections 4.2 and 4.3 we carry out a study similar to the one above, but for QL-implications and Rimplications, respectively. We end the paper with the conclusions, future research and references. 1. Notations and basic denitions 1.1. Fuzzy negations Let c : [0; 1] → [0; 1]. c is a fuzzy negation, i1: (i) c(0) = 1 and c(1) = 0, (ii) c(x)6c(y), if x¿y (monotonicity). A fuzzy negation is strict, i1, (iii) c(x) is continuous, (iv) c(x)¡c(y), for x¿y for all x; y ∈[0; 1]. A strict fuzzy negation is involutive, i1 (v) c(c(x)) = x, for all x ∈[0; 1]. The strict fuzzy negations that are involutive are called strong negations. In this paper we will only use strong negations. It is important to mention that Fodor and Roubens [12] prove that involutive fuzzy negations are automatically strict.
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Di1erent approaches to the study of fuzzy negations were used by Lowen [26], Esteva et al. [11], and Ovchinnikov [29,30], Higashi and Klir [21,25]. Yager investigated fuzzy negations for the purpose of developing useful measures of fuzziness [45,46]. The representation theorem of strong negations was obtained by Trillas [39,40]. We cite here that result in a slightly modiEed form which is more suitable in the sequel, see [31]. First we need the deEnition of an automorphism of a real interval [a; b] ⊂ R. This notion will be used extensively in this paper, especially in the case of the unit interval [0; 1]. Denition 0. A continuous, strictly increasing function ’ : [a; b] → [a; b] with boundary conditions ’(a) = a; ’(b) = b is called an automorphism of the interval [a; b] ⊂ R. Theorem 0. A function c : [0; 1] → [0; 1] is a strong negation if and only if there exists an automorphism ’ of the unit interval such that c(x) = ’−1 (1 − ’(x)). Many properties of the fuzzy negations have been studied, among which we will mention those relative to equilibrium points. An equilibrium point of a fuzzy negation c is any value e ∈(0; 1) such that c(e) = e. Furthermore, regarding these points we will highlight the following property: if c is a continuous fuzzy negation, then c has a unique equilibrium point. As we have said before, hereinafter c will always stand for strong negations (such that c(e) = e). 1.2. t-norms and t-conorms A function T : [0; 1]2 → [0; 1] satisfying the properties: (i) (ii) (iii) (iv)
Boundary conditions, T (x; 1) = x for all x ∈[0; 1], Monotony, T (x; y)6T (z; u) if x6z and y6u, Commutative, T (x; y) = T (y; x) ∀x; y ∈[0; 1], Associative, T (T (x; y); z) = T (x; T (y; z)) for all x; y; z ∈[0; 1],
is called a t-norm. Many papers on the di1erent properties of t-norms have been published. The deEnitions, properties, and notations that we are going to use in this paper can be found in the book by Fodor and Roubens, Fuzzy Preference Modelling and Multicriteria Decision Support, [15]. For example the deEnitions of: continuous t-norms, Archimedean t-norms, strict t-norms, etc. In Section 4.3 of this paper we will use the concept of left-continuous t-norm [15,23,24]. A t-norm is said to be left-continuous if it is left-continuous as a two-place function. A function S : [0; 1]2 → [0; 1] satisfying the properties: (i) Boundary conditions, S(x; 0) = x for all x ∈[0; 1], (ii) Monotony, S(x; y)6S(z; u) if x6z and y6u,
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(iii) Commutative, S(x; y) = S(y; x) ∀x; y ∈[0; 1], (iv) Associative, S(S(x; y); z) = S(x; S(y; z)) for all x; y; z ∈[0; 1], is called a t-conorm. Like with the t-norms, the deEnitions, properties, and notations of the t-conorms that we are going to use throughout the paper are the ones gathered by Fodor and Roubens in [15]. We say that a t-norm T and a t-conorm S are dual with respect to a fuzzy negation c if and only if c(T (a; b)) = S(c(a); c(b)) and c(S(a; b)) = T (c(a); c(b)). Let the triple T; S; c denote that T and S are dual with respect to c [6]. Please note that in a De Morgan triple the negation is automatically strong. The easy proof is: x = T (x; 1) = c(S(c(x); 0)) = c(c(x)) by duality, whence c should be involutive. Therefore we know that ∧; ∨; c is dual with respect to any strong negation c, being ∨ = max and ∧ = min. Due to their importance, we must point out other works where the most important properties of t-norms and t-conorms are investigated, like for example: [2,4,8,14,17,18,19,20,27,32,42,44]. Now we present three t-norm and t-conorm representation theorems that will be used further on. In these three theorems we will use the same notation Fodor and Roubens used in [15], though we will at all times indicate the origin of these theorems. In 1991 Ovchinnikov and Roubens [31] proved the following two theorems: Theorem 1. A continuous t-conorm S satis$es condition S(x; c(x)) = 1 for all x ∈[0; 1] with a strict negation c if and only if there exists an automorphism ’ of the unit interval such that S(x; y) = ’−1 (∧(’(x) + ’(y); 1)) and c(x) ¿ ’−1 (1 − ’(x)): Theorem 2. A continuous t-norm T is such that T (x; c(x)) = 0 holds for all x ∈[0; 1] with a strict negation c if and only if there exists an automorphism ’ of the unit interval such that T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)) and c(x) 6 ’−1 (1 − ’(x)): Please note that if in the two theorems above, we take c(x) = ’−1 (1−’(x)), this strict negation is involutive, that is, it is a strong negation. In 1949, AczMel [1] proved the following theorem (also see: [34]). Theorem 3. A continuous t-norm T is strict if and only if there exists an automorphism ’ of the unit interval such that T (x; y) = ’−1 (’(x) · ’(y)): 2. Implication operators We know [22,33] that a [0; 1]2 → [0; 1] mapping I is called an implication operator if and only if it satisEes the boundary conditions I (0; 0) = I (0; 1) = I (1; 1) = 1 and I (1; 0) = 0.
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The above conditions are, of course, the least we can demand from an implication operator. Other interesting potential properties of implication operators are listed in [3,12,16,42,47]. All fuzzy implications are obtained by generalizing the implication operator of classical logic. In this sense Fodor and Roubens [15] establish the following deEnition: Denition 1. An implication is a function I : [0; 1]2 → [0; 1] that satisEes the following properties: I1: I2: I3: I4: I5:
x6z implies I (x; y)¿I (z; y) for all y ∈[0; 1]. y6t implies I (x; y)6I (x; t) for all x ∈[0; 1]. I (0; x) = 1 (dominance of falsity) for all x ∈[0; 1]. I (x; 1) = 1 for all x ∈[0; 1]. I (1; 0) = 0.
It is important to point out the two following considerations: • If function I : [0; 1]2 → [0; 1] satisEes I2 and I (0; 0) = 1, then I satisEes I3. • If function I : [0; 1]2 → [0; 1] satisEes I1 and I (1; 1) = 1, then I satisEes I4. Now we recall further axioms, in terms of function I . These properties are required in di1erent papers and they could also be important in some applications. I6: I7: I8: I9: I10: I11: I12: I13:
I (1; x) = x (neutrality of truth). I (x; I (y; z)) = I (y; I (x; z)) (exchange property). I (x; y) = 1 if and only if x6y (boundary condition). I (x; 0) = c(x) is a strong negation. I (x; y)¿y. I (x; x) = 1 (identity). I (x; y) = I (c(y); c(x)) (contraposition) with a strong negation c. I is a continuous function (continuity).
It is important to point out that hereinafter when we speak of functions I : [0; 1]2 → [0; 1] we do not suppose it is an implication in Fodor’s sense (DeEnition 1), but that it is a function that satisEes the properties demanded in each particular case. Proposition 1. Let I : [0; 1]2 → [0; 1]. For all x; y; z ∈[0; 1], the following properties hold: (i) (ii) (iii) (iv)
I I I I
satis$es satis$es satis$es satis$es
I1 I1 I2 I2
if if if if
and and and and
only only only only
if if if if
I (x ∨ y; z) = ∧(I (x; z); I (y; z)). I (x ∧ y; z) = ∨(I (x; z); I (y; z)). I (x; y ∧ z) = ∧(I (x; y); I (x; z)). I (x; y ∨ z) = ∨(I (x; y); I (x; z)).
Proof. (i) ( ⇒) Two things can occur: (a) If x6y, then I (x ∨ y; z) = I (y; z), besides by I1, I (x; z)¿I (y; z), therefore ∧(I (x; z); I (y; z)) = I (y; z). (b) If x¿y, then I (x ∨ y; z) = I (x; z), besides in these conditions by I1, I (y; z)¿I (x; z), therefore ∧(I (y; z); I (x; z)) = I (x; z). (⇐) If I (x ∨ y; z) = ∧(I (x; z); I (y; z)) two things can occur:
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(a) If x6y, then I (x ∨ y; z) = I (y; z) = ∧(I (x; z); I (y; z)), then I (x; z)¿I (y; z). (b) If x¿y, then I (x ∨ y; z) = I (x; z) = ∧(I (x; z); I (y; z)), therefore I (y; z)¿I (x; z). The rest of the items are proven in a similar manner. 2.1. Some interdependencies among the axioms Evidently, the axioms I1–I13 are not independent of one another. There are axioms that are direct consequence of others, for example: • If I satisEes I6, then it satisEes I5, • If I satisEes I10, then it also satisEes I4, • If I satisEes I8, then I satisEes I3 and I11. In the following lemma we present other dependencies. Lemma 1. Let I : [0; 1]2 → [0; 1]. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
If If If If If If If If If If If
I I I I I I I I I I I
satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es
I1 and I12, then I satis$es I2; I2 and I12, then I satis$es I1; I3 and I12, then I satis$es I4; I4 and I12, then I satis$es I3; I6 and I12, then I satis$es I9; I9 and I12, then I satis$es I6; I2 and I9, then I satis$es I3; I1 and I6, then I satis$es I10; I7 and I9, then I satis$es I12; I1, I6 and I12, then I satis$es I2; I3; I4; I5; I9 and I10; I2; I7 and I8, then I satis$es I1; I3; I4; I5; I6; I10 and I11;
hold. Proof. (i) If y6t, then c(y)¿c(t), by I1 we have I (c(y); c(x))6I (c(t); c(x)), since I fulElls I12 then I (x; y)6I (x; t). (ii) If y6t, then c(t)6c(y) by I2 we have I (c(x); c(t))6I (c(x); c(y)) for all c(x) ∈[0; 1], besides I fulElls I12, therefore I (c(x); c(t)) = I (t; x)6I (c(x); c(y)) = I (y; x). (iii) I (x; 1) = I (0; c(x)) = 1 for all x ∈[0; 1]. (iv) I (0; x) = I (c(x); 1) = 1 for all x ∈[0; 1]. (v) I (x; 0) = I (c(0); c(x)) = I (1; c(x)) = c(x). (vi) c(x) = I (x; 0) = I (1; c(x)). (vii) I (0; x) = I (0; ∨(x; 0)) = ∨(I (0; x); I (0; 0)) = ∨(I (0; x); c(0)) = 1 for all x ∈[0; 1]. (viii) Since x61 by I1 and I6, we have I (x; y)¿I (1; y) = y. (ix) I (x; y) = I (x; c(c(y))) = I (x; I (c(y); 0)) = I (c(y); I (x; 0)) = I (c(y); c(x)). (x) Consequence of the above. (xi) See [15]. It is important to point out that item (xi) of the lemma above can be found proven in [15] and item (ix) in [23].
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Table 1
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
I1; I12 ⇒ I2 I2; I12 ⇒ I1 I3; I12 ⇒ I4 I4; I12 ⇒ I3 I6; I12 ⇒ I9 I9; I12 ⇒ I6 I2; I9 ⇒ I3 I1; I6 ⇒ I10 I7; I9 ⇒ I12 I1; I6; I12 ⇒ I2; I3; I4; I5; I9; I10 I2; I7; I8 ⇒ I1; I3; I4; I5; I6; I10; I11
Table 1, represents the results of the lemma above. From items (i) – (vi) of Lemma 1 we deduce the following: If I : [0; 1]2 → [0; 1] satis$es axiom I12, then the following properties hold: (i) I1 holds if and only if I2 holds; (ii) I3 holds if and only if I4 holds; (iii) I6 holds if and only if I9 holds. The results of the lemmas that we demonstrate below will be used in further sections of the paper. Lemma 2. Let I : [0; 1]2 → [0; 1] be any function that satis$es at least one of the following items: (i) (ii) (iii) (iv)
If If If If
I I I I
satis$es satis$es satis$es satis$es
I2; I1; I4, I6,
I6 and I9, or I6 and I9, or I7, I9 and I11, or I7, I9 and I (x; x) = I (0; x) for all x ∈[0; 1], then
I (x; y) = 0 and only if x = 1 and y = 0: Proof. (i) (⇐) If x = 1 and y = 0, then by I9 we have I (1; 0) = c(1) = 0. (⇒) If I (x; y) = 0, then by item (iv) of Proposition 1 we have 0 = I (x; y) = I (x; y ∨ 0) = ∨(I (x; y); I (x; 0)) = ∨(0; c(x)) = c(x), then x = 1. Besides, by I6 we have 0 = I (x; y) = I (1; y) = y. (ii) (⇐) By I9 we have I (1; 0) = c(1) = 0. (⇒) If I (x; y) = 0, then by item (ii) of Proposition 1 and axiom I6 we have 0 = I (x; y) = I (x ∧ 1; y) = ∨(I (x; y); I (1; y)) = ∨(0; y) = y. Besides by I9, 0 = I (x; y) = I (x; 0) = c(x), therefore x = 1. (iii) (⇐) By item (ix) of Lemma 1, we have that I satisEes I12, furthermore, by item (vi) of the same lemma I satisEes I6, and therefore I satisEes I5, thus I (1; 0) = 0. (⇒) Since I (x; y) = 0 we have c(y) = I (y; 0) = I (y; I (x; y)) = I (x; I (y; y)) = I (x; 1) = 1, therefore y = 0, then 0 = I (x; y) = I (x; 0) = c(x), therefore x = 1. (iv) Similar to the above. Lemma 3. For I : [0; 1]2 → [0; 1] the following properties hold: (i) If I satis$es I2 and I9, then I (x; y)¿c(x) for all x; y ∈[0; 1]. (ii) If I satis$es I2, I9 and I10, then I (x; y)¿ ∨(c(x); y) for all x; y ∈[0; 1]. (iii) If I satis$es I2, I9 and I12, then I (x; y)¿ ∨(c(x); y) for all x; y ∈[0; 1].
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Proof. (i) I (x; y) = I (x; y ∨ 0) = ∨(I (x; y); I (x; 0)) = ∨(I (x; y); c(x)), [0; 1]. (ii) Evident. (iii) c(x)6I (x; y) = I (c(y); c(x))¿c(c(y)) = y.
then
I (x; y)¿c(x)
for
all
x; y ∈
3. The property: I (x; c(x)) = c(x) for all x belonging to [0; 1] Let I : [0; 1]2 → [0; 1] be any function such that I (x; c(x)) = c(x) for all x ∈[0; 1]. In these conditions we can easily see: (a) I (1; 0) = 0 and I (0; 1) = 1, (b) I (c(x); x) = x for all x ∈[0; 1] due to the fact that we only consider strong negations. (c) There is a contradiction with axiom I8 as proven by the following theorem. Theorem 4. If I : [0; 1]2 → [0; 1] satis$es I8, then it does not satisfy I (x; c(x)) = c(x) for all x ∈[0; 1]. Proof. Let us suppose it is not true, that is, I satisEes: I (x; c(x)) = c(x)
for all x ∈ [0; 1]; and I8:
Because c is a strong negation, it has a single equilibrium point e ∈(0; 1). Since e = c(e), we have e = c(e) = I (e; c(e)) = 1, which is a contradiction with the fact that e ∈(0; 1). Corollary 1. If I : [0; 1]2 → [0; 1] satis$es I (x; c(x)) = c(x) for all x ∈[0; 1], then it does not satisfy I8. Proof. Similar to the demonstration of the theorem above. It is important to point out that there exist I functions that do not satisfy axiom I8 nor the property I (x; c(x)) = c(x) for all x ∈[0; 1], for example, Reichenbach’s implication operator I (x; y) = 1 − x + x · y, with the strong negation c(x) = 1 − x. 4. Implications by t-norms, t-conorms and negations Three general classes of implication operator, I , have been identiEed in the literature [7,9,10], the Simplications, the QL-implications and the R-implications. An S-implication associated with a t-conorm S and a strong negation c is deEned by I (x; y) = S(c(x); y): A QL-implication associated with a t-conorm S, a t-norm T and a strong negation c is deEned by I (x; y) = S(c(x); T (x; y)): An R-implication associated with a t-norm T is deEned by I (x; y) = sup{z | T (x; z) 6 y}: In the particular case that the t-norm considered be left-continuous, then we will refer to the R-implication deEned above writing: residual implication [23,24].
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Di1erent characterizations of S-implications, QL-implications and R-implications can be found in Trillas and Valverde [41]. It is easy to see that both S-implications and R-implications satisfy properties I1–I5 for any t-norm T , t-conorm S and strong negation c, thus they are implications in Fodor and Roubens’ sense. In general QLimplications violate property I1. The conditions under which I1 is satisEed can be found in Fodor [13]. Proposition 2. If I is an S-implication or a QL-implication, the following items hold: (i) The axioms: I2, I6 and I9 are satis$ed. (ii) I (x; y)¿c(x) for all x; y ∈[0; 1]. (iii) I (x; y) = 0 if and only if x = 1 and y = 0. Proof. (i) If I (x; y) = S(c(x); y), since S is increasing in the second argument, then by the deEnition of S-implication, I is as well. I (1; x) = S(c(1); x) = S(0; x) = x and I (x; 0) = S(c(x); 0) = c(x). If I (x; y) = S(c(x); T (x; y)), since T and S are increasing in the second argument, by the deEnition of QLimplication, I is as well. I (1; x) = S(c(1); T (1; x)) = S(0; T (1; x)) = T (1; x) = x and I (x; 0) = S(c(x); T (x; 0)) = S(c(x); 0) = c(x). (ii) and (iii) are consequence of the Lemmas 2 and 3. 4.1. S-implications and the I (x; c(x)) = c(x) property Theorem 5. Let I be an S-implication with an appropriate t-conorm S and a strong negation c. I (x; c(x)) = c(x) for all x ∈ [0; 1] if and only if S = ∨: Proof. Keeping in mind that: S is idempotent if and only if S = ∨, and that c is a strong negation, i.e. an involutive order reversing bijection of the closed unit interval, we have c(x) = I (x; c(x)) = S(c(x); c(x)) = ∨(c(x); c(x)). The following theorem establishes the axioms that we must demand of the function I : [0; 1]2 → [0; 1] in order for it to be given by the expression I (x; y) = ∨(c(x); y) for all x; y ∈[0; 1]. Theorem 6. A function I : [0; 1]2 → [0; 1] is such that I2, I7, I9 and I (x; c(x)) = c(x) for all x ∈[0; 1] if and only if I is an S-implication and I (x; y) = ∨(c(x); y) for all x; y ∈[0; 1]. Proof. (⇒) Suppose that I is any function which fulEls I2, I7, I9 and I (x; c(x)) = c(x) for all x ∈[0; 1]. By Lemma 1, I satisEes I12, I6 and I1. We deEne a function S by S(x; y) = I (c(x); y). Let us prove that S is a t-conorm. S is nondecreasing in both places by I1 and I2. By I9, S(x; 0) = I (c(x); 0) = c(c(x)) = x for all x ∈[0; 1]. S is commutative by I12: S(x; y) = I (c(x); y) = I (c(y); x) = S(y; x): By I7 and commutativity, we have that S(x; S(y; z)) = S(x; S(z; y)) = I (c(x); I (c(z); y)) = I (c(z); I (c(x); y)) = S(z; S(x; y)) = S(S(x; y); z):
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Therefore, S is a t-conorm. We obviously have that I (x; y) = S(c(x); y), by the deEnition of S. Since furthermore by hypothesis I (x; c(x)) = c(x) for all x ∈[0; 1], and S(x; x) = x if and only if s = ∨, we have c(x) = I (x; c(x)) = S(c(x); c(x)), therefore S = ∨. Then I (x; y) = ∨(c(x); y). (⇐) If y6z, then I (x; y) = ∨(c(x); y)6 ∨(c(x); z) = I (x; z), therefore I2 holds. I (x; 0) = ∨(c(x); 0) = c(x) for all x ∈[0; 1], thus I satisEes I9. I7 holds because I (x; I (y; z)) = ∨(c(x); ∨(c(y); z)) = ∨(c(y); ∨(c(x); z)) = I (y; I (x; z)); by the associativity and commutativity of ∨. Finally, by the idempotency of ∨ we have that I (x; c(x)) = ∨(c(x); c(x)) = c(x)
for all x ∈ [0; 1]:
Corollary 2. The S-implication, I (x; y) = ∨(c(x); y), is such that (i) (ii) (iii) (iv)
It It If If
ful$lls the axioms I1–I7; I9; I10; I12 and I13. is the only S-implication that veri$es I (x; c(x)) = c(x) for all x ∈[0; 1]. e ∈(0; 1) is the equilibrium point of c, then I (e; e) = e. x ∈[e; 1], then I (x; x) = x.
Proof. (i) We only need to keep in mind the theorem above and Lemma 1. (ii) Evident by Theorem 5. (iii) I (e; e) = ∨(c(e); e) = ∨(e; e) = e. (iv) Since x¿e, then c(x)6c(e) = e6x. Please note that if in I (x; y) = ∨(c(x); y) we take c(x) = 1 − x for all x ∈[0; 1], we have Kleene–Dienes’ operator. Taking into account Corollaries 1 and 2, the S-implications studied in the two theorems above do not satisfy axiom I8. The following theorem allows us to build S-implications from automorphisms in the unit interval. Nevertheless, we will see further on that when I is generated by automorphisms: I satisEes I8 if and only if the strong negation is generated by the same automorphism that generates the implication I . Theorem 7. A function I : [0; 1]2 → [0; 1] is such that I2; I7; I9; I11 and I13 are satis$ed if and only if there exists an automorphism ’ of the unit interval such that I (x; y) = ’−1 (∧(’(c(x)) + ’(y); 1)) and c(x) ¿ ’−1 (1 − ’(x)): Proof. (⇒) Since I satisEes I2, I7, I9, I11 and I13 by Lemma 1 we know that I also fulElls I12. We deEne a function S by S(x; y) = I (c(x); y). In the demonstration of Theorem 6 we have proven that in these conditions S is a t-conorm. Moreover, by I11, S(x; c(x)) = I (c(x); c(x)) = I (x; x) = 1. Also, since I is continuous, S is as well. Keeping in mind Theorem 1 we have that I (x; y) = S(c(x); y) = ’−1 (∧(’(c(x)) + ’(y); 1))
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and c(x) ¿ ’−1 (1 − ’(x)): (⇐) By the deEnition of automorphism, if y6z, then I (x; y) = ’−1 (∧(’(c(x)) + ’(y); 1))6’−1 (∧ (’(c(x)) + ’(z); 1)) = I (x; z), therefore I2 holds. I7 holds because I (x; I (y; z)) = ’−1 (∧(’(c(x)) + ∧(’(c(y)) + ’(z); 1); 1) = ’−1 (∧(’(c(y)) + ∧(’(c(x)) + ’(z); 1); 1) = I (y; I (x; z)): Besides I (x; 0) = ’−1 (∧(’(c(x)) + 0; 1)) = ’−1 (’(c(x))) = c(x), therefore I9 holds. Finally, since c(x)¿’−1 (1−’(x)), we have that ’(c(x))+’(x)¿1−’(x)+’(x) = 1, therefore I (x; x) = ’−1 (∧(’(c(x)) + ’(x); 1)) = 1, therefore I fulElls I11. Moreover since ’ is an automorphism and ∧ and c are continuous then I13 also holds. Corollary 3. Let I : [0; 1]2 → [0; 1] be such that I2; I7; I9; I11 and I13 hold. Then I (x; y) = 1 if and only if c(x) ¿ ’−1 (1 − ’(y)): Proof. (⇒) Since I (x; y) = 1 = ’−1 (∧(’(c(x)) + ’(y); 1)), we have 1 = ∧ (’(c(x)) + ’(y); 1), therefore ’(c(x)) + ’(y)¿1, that is, ’(c(x))¿1 − ’(x), then c(x)¿’−1 (1 − ’(y)). (⇐) If c(x)¿’−1 (1 − ’(y)), then ’(c(x))¿1 − ’(y), therefore ’(c(x)) + ’(y)¿1, then I (x; y) = 1. Remark. The corollary above allows us to conclude that the implication operator generated from the automorphism ’ used in Theorem 7 does not always fulEll axiom I8. The following example makes this result clear: Example. Let us take ’(x) = x for all x ∈[0; 1], and the strong negation c(x) = (1−x2 )1=2 , evidently, c(x) = (1− x2 )1=2 ¿1 − x, 1 − x being the strong negation generated by the identity automorphism. In these conditions, I (x; y) = ∧ (c(x) + y; 1) = ∧ ((1 − x2 )1=2 + y; 1). For x = 0:5 and y = 0:4, (obviously x = 0:5¿0:4 = y), we have c(0:5) = (1 − 0:52 )1=2 = 0:87¿1 − 0:4 = 0:6 and I (0:5; 0:4) = 1. In 1987 Smets and Magrez proved the following theorem [15,38]: A function I : [0; 1]2 → [0; 1] is such that I2, I7, I8 and I13 are satis$ed if and only if there exists an automorphism ’ of the unit interval such that I (x; y) = ’−1 (∧(1 − ’(x) + ’(y); 1)): As a consequence of this theorem and of Theorem 7 we have the following result. Corollary 4. Let I : [0; 1]2 → [0; 1] be such that I2; I7; I9; I11 and I13 are satis$ed. In these conditions: c(x) = ’−1 (1 − ’(x)) for all x ∈ [0; 1] if and only if I fulfills I8: Proof. Evident by Theorem 7 and Smets and Magrez’s Theorem. Remark. (a) From Theorem 4 and the corollary above we deduce that the functions I generated from the automorphism ’, according to the Corollary 4, do not satisfy the property I (x; c(x)) = c(x) for all x ∈[0; 1]. (b) Theorem 7 is a generalization of Smets and Magrez’s theorem in the following sense: they always considered that the strong negation c of the axioms I9 and I12 was generated by the automorphism that
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Fig. 1.
generates the implication I , which is something we do not do. That is, the strong negation that we use can be generated by an automorphism % di1erent from the automorphism ’ that generates I , provided that c(x) = %−1 (1 − %(x))¿’−1 (1 − ’(x)). (c) From Corollary 4 we deduce that the only functions I generated from the automorphism ’ according to Theorem 7 that satisfy I8 are the ones we obtain generated by the same automorphism as in Smets and Magrez’s theorem. (d) If in the conditions of Corollary 4 we take the automorphism ’(x) = x for all x ∈[0; 1], the resulting implication operator is Lukasiewicz’s, that is, I (x; y) = ∧(1 − x + y; 1). It is easy to prove, (Lemma 1), that this implication operator satisEes I1–I8, I9 with c(x) = 1 − x, I10–I13. (e) Fig. 1 represents two implication operators obtained from Theorem 7 taking in both cases the identity automorphism. In Fig. 1(a) the strong negation considered is Yager’s: c(x) = (1 − xw )1=w with w = 3, in Fig. 1(b) Sugeno’s: c(x) = (1 − x)=(1 + 'x) with ' = −0:5. In both cases c(x)¿1 − x for all x ∈[0; 1]. We know that a consequence of axiom I8 is axiom I11, since the latter is imposed in Theorem 7, we present below a theorem that allows us to obtain implications from automorphisms ’ without demanding I11. We will see that in these conditions the S-implications I generated do not fulEll I8 nor I (x; c(x)) = c(x) for all x ∈[0; 1]. Denition 2. It is said that I is strictly monotone, if it is strictly decreasing in the Erst argument and strictly increasing in the second, both on (0; 1)2 . Theorem 8. A function I : [0; 1]2 → [0; 1] is such that it is strictly monotone, and I7; I9 and I13 are satis$ed if and only if there exists an automorphism ’ of the unit interval such that I (x; y) = c(’−1 (’(x) · ’(c(y)))): Proof. (⇒) Since I fulElls I7 and I9, by Lemma 1 we know I also fulElls I12. We deEne a function T given by T (x; y) = c(I (x; c(y))).
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Evidently, since c is a strong negation and I is strictly monotone, then T is strictly increasing in both arguments on (0; 1)2 . Moreover, by I12 we have T (x; y) = c(I (x; c(y))) = c(I (y; c(x))) = T (y; x), therefore T satisEes the commutativity. In addition, T (1; x) = c(I (1; c(x))) = c(I (x; c(1))) = c(I (x; 0)) = c(c(x)) = x for all x ∈[0; 1]. By I12 and I7 we have that T (x; T (y; z)) = T (x; c(I (y; c(z)))) = T (x; c(I (z; c(y)))) = c(I (x; c(c(I (z; c(y)))))) = c(I (x; I (z; c(y)))) = c(I (z; I (x; c(y)))) = c(I (z; c(c(I (x; c(y)))))) = c(I (z; c(T (x; y)))) = T (z; T (x; y)) = T (T (x; y); z): Since I fulElls I13 and c is a strong negation, T is continuous. Therefore, T is a continuous, strict t-norm. Using Theorem 3 we have that I (x; y) = c(T (x; c(y))) = c(’−1 (’(x) · ’(c(y)))). (⇐) Since c is a strong negation and ’ is an automorphism, I (x; y) = c(’−1 (’(x) · ’(c(y)))) is strictly monotone and fulElls I13. I (x; 0) = c(’−1 (’(x) · ’(c(0)))) = c(’−1 (’(x) · ’(1))) = c((’−1 (’(x)))) = c(x); thus I satisEes I 9: I also satisEes I7, because: I (x; I (y; z)) = c(’−1 (’(x) · ’(c(I (y; z))))) = c(’−1 (’(x) · ’(c(c(’−1 (’(y) · ’(c(z)))))))) = c(’−1 (’(x) · ’(y) · ’(c(z)))); I (y; I (x; z)) = c(’−1 (’(y) · ’(c(I (x; z))))) = c(’−1 (’(y) · ’(c(c(’−1 (’(x) · ’(c(z)))))))) = c(’−1 (’(y) · ’(x) · ’(c(z)))) therefore I (x; I (y; z)) = I (y; I (x; z)), that is I fulElls I7. Corollary 5. Let I : [0; 1]2 → [0; 1] which is strictly monotone, and satis$es the axioms: I7; I9 and I13. Then (i) I (x; c(x)) = c(’−1 (’(x) · ’(x))); (ii) I (x; y) = 1 if and only if x = 0 or y = 1; (iii) If c(x) = ’−1 (1 − ’(x)) for all x ∈[0; 1], I (x; y) = ’−1 (1 − ’(x) + ’(x) · ’(y)); hold. Proof. (i) Evident. (ii) (⇒) I (x; y) = 1 = c(’−1 (’(x) · ’(c(y)))), then 0 = ’(x)’(c(y)), therefore x = 0 or y = 1. (⇐) Evident. (iii) Since c(x) = ’−1 (1 − ’(x)), we have I (x; y) = c(’−1 (’(x) · ’(’−1 (1 − ’(y))))) = c(’−1 (’(x) · (1−’(y)))) = c(’−1 (’(x)−’(x)·’(y))) = ’−1 (1−’(’−1 (’(x)−’(x)·’(y)))) = ’−1 (1−’(x)+’(x)·’(y))).
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Remark. (a) Keeping in mind the Erst two items of Corollary 5, we have that the implication operators obtained from Theorem 8 do not satisfy axiom I8, nor the property I (x; c(x)) = c(x) for all x ∈[0; 1]. (b) Please note that in the conditions of item (iii) of the corollary above, if ’ is the identity automorphism then we obtain the implication operator of Rechenbach, I (x; y) = 1 − x + x · y, it is easy to prove, (Lemma 1), that this implication operator satisEes the axioms: I1–I7, I9 with c(x) = 1 − x, I10, I12 and I13. 4.2. QL-implications and the I (x; c(x)) = c(x) property Theorem 9. Let I be a QL-implication associated to the t-conorm S = ∨, to any t-norm T and a strong negation c. Then (i) I (x; c(x)) = c(x) for all x ∈[0; 1]; (ii) I (x; y) = 1 if and only if x = 0 or x = y = 1; hold. Proof. (i) We only need to recall that T (x; c(x))6 ∧(x; c(x))6c(x) for all x ∈[0; 1]. (ii) (⇒) If I (x; y) = ∨(c(x); T (x; y)) = 1, then c(x) = 1 or T (x; y) = 1. Evidently, if c(x) = 1 then x = 0. If 1 = T (x; y)6 ∧(x; y), then x = y = 1. (⇐) If x = 0, then I (x; y) = ∨(c(0); T (0; y)) = ∨(1; T (0; y)) = 1. If x = y = 1, then I (1; 1) = 1 because by item (i) of Proposition 2 we know that all QL-implication satisfy the axioms: I2; I6 and I9, therefore, by I6 we have I (1; 1) = 1. Theorem 10. Let I be a QL-implication associated to a t-conorm S, to a strong negation c and a continuous t-norm T such that T (x; c(x)) = 0 for all x ∈[0; 1]. Then there exists an automorphism ’ of the unit interval such that I (x; y) = S(c(x); ’−1 (∨(’(x) + ’(y) − 1; 0))) and c(x)6’−1 (1 − ’(x)): Proof. We only need to recall Theorem 2. Corollary 6. In the conditions of the theorem above, the property: I (x; c(x)) = c(x) for all x ∈ [0; 1] holds. Proof. By item (ii) of Proposition 2 we know that I (x; y)¿c(x) for all x; y ∈[0; 1], therefore I (x; c(x))¿c(x). Moreover I (x; c(x)) = S(c(x); ’−1 (∨(’(x) + ’(c(x)) − 1; 0))) with c(x)6’−1 (1 − ’(x)), by the monotony of S we have I (x; c(x)) = S(c(x); ’−1 (∨(’(x) + ’(c(x)) − 1; 0)))6S(c(x); ’−1 (∨(’(x) + 1 − ’(x) − 1; 0))) = S(c(x); 0) = c(x). Therefore I (x; c(x)) = c(x). Remark. Corollary 6 makes clear that the reciprocal of item (i) of Theorem 9 does not always hold. The conditions in which the reciprocal holds are given in the following theorem.
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Theorem 11. Let T be the continuous t-norm such that T (x; c(x)) = 0 for all x ∈[0; 1], generated by the automorphism ’ according to Theorem 2, and let c be the strong negation generated by that automorphism, that is, c(x) = ’−1 (1 − ’(x)). A function I : [0; 1]2 → [0; 1] is such that: (i) (ii) (iii) (iv)
I (x; c(x)) = c(x) for all x ∈[0; 1], Ful$lls the axioms: I2 and I9, If 2’(c(x))¿’(y), then I (x; y) = c(x), If 2’(c(x))6’(y), then I (x; y) = T (x; y),
if and only if I (x; y) = ∨(c(x); T (x; y)). Proof. (⇒) Since 06y by I2 and I9 we have c(x) = I (x; 0)6I (x; y) for all x; y ∈[0; 1]. Two things can occur: (a) y6c(x), then by I2 and item (i) c(x) = I (x; c(x))¿I (x; y), therefore I (x; y) = c(x). Besides since y6c(x) we have ’(y)6’(c(x))62’(c(x)), then ’(c(x))¿’(y) − ’(c(x)), therefore we have T (x; y) = ’−1 (∨(’(x)+’(y)−1; 0)) = ’−1 (∨(’(y)−(1−’(x)); 0)), since c(x) = ’−1 (1−’(x)), we have ’(c(x)) = 1−’(x), therefore T (x; y) = ’−1 (∨(’(y) − ’(c(x)); 0))6’−1 (’(c(x))) = c(x), then T (x; y)6c(x), therefore I (x; y) = c(x) = ∨(c(x); T (x; y)). (b) y¿c(x), by I2 we have I (x; y)¿I (x; c(x)) = c(x). Since y¿c(x), then ’(y)¿’(c(x)). Two things can occur: (b1) ’(c(x))6’(y)62’(c(x)), therefore ’(y)−’(c(x))6’(c(x)), then T (x; y) = ’−1 (∨(’(y)−’(c(x)); 0)) 6’−1 (∨(’(c(x)); 0)) = ’−1 (’(c(x))) = c(x), then T (x; y)6c(x). By item (iii) we have I (x; y) = c(x) = ∨ (c(x); T (x; y)). (b2) ’(c(x))62’(c(x))6’(y), then ’(c(x))6’(y)−’(c(x)) therefore T (x; y) = ’−1 (∨(’(y)−’(c(x)); 0)) ¿’−1 (∨(’(c(x)); 0)) = c(x). By item (iv) we have I (x; y) = T (x; y) = ∨(c(x); T (x; y)). (⇐) By hypothesis I (x; y) = ∨(c(x); T (x; y)). I (x; 0) = ∨(c(x); T (x; 0)) = c(x), then I9 holds. By Ovchinnikov and Roubens’ Theorem 2 we know that the t-norm generated by the automorphism ’ satisEes T (x; c(x)) = 0, therefore I (x; c(x)) = ∨(c(x); T (x; c(x)) = ∨(c(x); 0) = c(x)
for all x ∈ [0; 1]:
Evidently, I (x; y) = ∨ (c(x); T (x; y)) fulElls I2, we only need to keep in mind the monotony of T and ∨. If 2’(c(x))¿’(y), then ’(c(x))¿’(y) − ’(c(x)) therefore T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)) = −1 ’ (∨(’(y)−(1−’(x)); 0)), since c(x) = ’−1 (1−’(x)), we have ’(c(x)) = 1−’(x), therefore T (x; y) = ’−1 (∨(’(y)−’(c(x)); 0))6’−1 (∨(’(c(x)); 0)) = c(x), that is, c(x)¿T (x; y), then I (x; y) = ∨(c(x); T (x; y)) = c(x). If 2’(c(x))6’(y), then ’(c(x))6’(y) − ’(c(x)), therefore T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)) = ’−1 (∨(’(y) − ’(c(x)); 0))¿’−1 (∨(’(c(x)); 0)) = c(x), that is, c(x)6T (x; y), then I (x; y) = ∨(c(x); T (x; y)) = T (x; y). Remark. (a) Evidently the functions I generated by the theorem above do not fulEll I8, we only need to recall Corollary 1. (b) If in the conditions of the theorem above we take the identity automorphism, (evidently c(x) = 1 − x), we have I (x; y) = ∨(1 − x; x + y − 1), (see Fig. 2(a)). This statement does not fulEll, among other properties, I1 and I4 therefore it is not an implication. If we take ’(x) = x2 for all x ∈[0; 1], then we have I (x; y) = ∨((1 − x2 )1=2 ; (∨(x2 + y2 − 1; 0))1=2 ) and c(x) = (1 − x2 )1=2 , (see Fig. 2(b)). In both cases I2, I3, I5, I6 and I9 stand.
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Fig. 2.
Denition 3 (Dombi [6] and Fodor and Roubens [15]). A De Morgan triple (T; S; c) is called a strong (or Lukasiewicz) De Morgan triple if there exists an automorphism ’ of the unit interval such that T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)); S(x; y) = ’−1 (∧(’(x) + ’(y); 1)); c(x) = ’−1 (1 − ’(x)): Corollary 7. In the conditions of Theorem 10, if (T; S; c) is a strong De Morgan triple, then I (x; y) = ’−1 (∨(1 − ’(x); ’(y))): Proof. I (x; y) = S(’−1 (1 − ’(x)); ’−1 (∨(’(x) + ’(y) − 1; 0))) = ’−1 (∧(1 − ’(x) + ∨(’(x) + ’(y) − 1; 0); 1)) = ’−1 (∧(∨(1 − ’(x); ’(y)); 1)) = ’−1 (∨(1 − ’(x); ’(y))): Remark. If in Corollary 7 we take ’(x) = x for all x ∈[0; 1], we have Kleene–Dienes’ implication operator. Theorem 12. Let I be a QL-implication associated to a t-conorm S, to a strong negation c and a continuous and strict t-norm T. Then there exists an automorphism ’ of the unit interval such that I (x; y) = S(c(x); ’−1 (’(x) · ’(y))): Proof. We only need to recall Theorem 3.
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The QL-implications obtained from the theorem above do not satisfy I8, nor the property I (x; c(x)) = c(x) for all x ∈[0; 1]. The following theorem establishes the conditions in which this property is satisEed. Theorem 13. Let T be the continuous, strict t-norm generated by the automorphism ’ according to Theorem 3 and let c be the strong negation generated by this automorphism, that is, c(x) = ’−1 (1 − ’(x)). A function I : [0; 1]2 → [0; 1] is such that: (i) (ii) (iii) (iv)
I (x; c(x)) = c(x) for all x ∈[0; 1], Ful$lls the axioms I2 and I9, If ’(x) + ’(x)’(y)¿1, then I (x; y) = T (x; y), If ’(x) + ’(x)’(y)61, then I (x; y) = c(x),
if and only if I (x; y) = ∨(c(x); T (x; y)). Proof. (⇒) Since 06y by I2 and I9 we have c(x) = I (x; 0)6I (x; y) for all x; y ∈[0; 1]. Two things can occur: (a) If y6c(x), then by I2 I (x; y)6I (x; c(x)) = c(x), therefore I (x; y) = c(x). Besides T (x; y)6T (x; c(x))6 ∧ (x; c(x))6c(x). Therefore I (x; y) = c(x) = ∨(c(x); T (x; y)). (b) If y¿c(x), then ’(y)¿’(c(x)) = 1 − ’(x), therefore ’(x) + ’(y)¿1. Two things can occur: (b1) ’(x) + ’(y)¿’(x) + ’(x)’(y)¿1, by item (iii) we have I (x; y) = T (x; y). Moreover, ’(x)’(y)¿1 − ’(x), therefore T (x; y) = ’−1 (’(x)’(y))¿’−1 (1 − ’(x)) = c(x), then I (x; y) = T (x; y) = ∨(c(x); T (x; y)). (b2) ’(x) + ’(y)¿1¿’(x) + ’(x)’(y), by item (iv) we have I (x; y) = c(x), besides, ’(x)’(y)61 − ’(x), therefore T (x; y) = ’−1 (’(x)’(y))6’−1 (1 − ’(x)) = c(x), then I (x; y) = c(x) = ∨(c(x); T (x; y)). (⇐) I (x; 0) = ∨(c(x); T (x; 0)) = ∨(c(x); 0) = 0, thus I9 holds. By the monotony of ∨ and of T satisEes I2. Since T (x; c(x))6 ∧ (x; c(x))6c(x), we have I (x; c(x)) = ∨(c(x); T (x; c(x))) = c(x) for all x ∈[0; 1]. If ’(x)+’(x)’(y)¿1, then ’(x)’(y)¿1−’(x), therefore T (x; y) = ’−1 (’(x)’(y))¿’−1 (1−’(x)) = c(x), then I (x; y) = ∨(c(x); T (x; y)) = T (x; y). If ’(x)+’(x)’(y)61, then ’(x)’(y)61−’(x), therefore T (x; y) = ’−1 (’(x)’(y))6’−1 (1−’(x)) = c(x), then I (x; y) = ∨(c(x); T (x; y)) = c(x) therefore item (iv) holds. Remark. If in the conditions of the theorem above we take the identity automorphism, we have I (x; y) = ∨(1− x; xy) and c(x) = 1−x, (see Fig. 3(a)), statement that does not satisfy I1 and I4, among other axioms. Evidently it does not satisfy I8 either. If we take ’(x) = x2 for all x ∈[0; 1] we have I (x; y) = ∨((1 − x2 )1=2 ; xy) and c(x) = (1 − x2 )1=2 , (see Fig. 3(b)). In both cases I2, I3, I5, I6 and I9 stand. Denition 4 (Dombi [6] and Fodor and Roubens [15]). A De Morgan triple (T; S; c) is said to be a strict (or productlike) De Morgan triple if there exists an automorphism ’ of the unit interval such that: T (x; y) = ’−1 (’(x)’(y)); S(x; y) = ’−1 (’(x) + ’(y) − ’(x)’(y)); c(x) = ’−1 (1 − ’(x)): Corollary 8. In the conditions of Theorem 12, if (T; S; c) is a strict De Morgan triple, then the following items hold: (i) I (x; y) = ’−1 (1 − ’(x) + ’(x)’(x)’(y)). (ii) I (x; y) = 1 if and only if x = 0 or x = y = 1.
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Fig. 3.
Proof. (i) I (x; y) = S(c(x); T (x; y)) = S(’−1 (1 − ’(x)); ’−1 (’(x)’(y))) = ’−1 (1 − ’(x) + ’(x)’(y) − (1 − ’(x))(’(x)’(y))) = ’−1 (1 − ’(x) + ’(x)’(y) − ’(x)’(y) + ’(x))’(x)’(y)) = ’−1 (1 − ’(x) + ’(x)’(x)’(y)): (ii) Evident. Remark. (a) Evidently the I obtained in Corollary 8 does not fulEll I (x; c(x)) = c(x) for all x ∈[0; 1], nor axiom I8. (b) If in the corollary above we take ’(x) = x for all x ∈[0; 1], we have Klir and Yuan’s function (1994), I (x; y) = 1 − x + x2 y, that fulElls the axioms: I2, I3, I5, I6 and I9, with c(x) = 1 − x for all x ∈[0; 1]. 4.3. R-implications and the I (x; c(x)) = c(x) property The following result appears frequently in fuzzy literature [10,15,24]. Let I be an R-implication associated to a t-norm T. Then the following property holds. If x 6 y; then I (x; y) = 1: Corollary 9. Let I be an R-implication associated to a t-norm T, then I (x; c(x)) = c(x) for all x ∈[0; 1] does not hold.
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Proof. Since c is a strong negation, then there exists a single equilibrium point e ∈(0; 1) such that c(e) = e, therefore I (e; c(e)) = I (e; e). Keeping in mind the result indicated above we know that if x6y, then I (x; y) = 1, therefore I (e; c(e)) = I (e; e) = 1 = c(e). Remark. Keeping in mind the result indicated above, we know that all R-implications fulEll I11 and in addition they fulEll one of the implications of I8. In this paper we do not analyze the behavior of I8 in the R-implications, as it is a problem that has already been studied in fuzzy literature [10,15,24] (for example in the R-implication characterization theorems based on left-continuous t-norms I8 is always demanded). In this sense we quote Klement and Navara [24]: A reasonable way of constructing connectives in fuzzy logics is to start with left-continuous t-norm T and to use the residual implication de$ned by I (x; y) = sup{z | T (x; z) 6 y} as the interpretation of the implication. It is immediate that we have I8, that is, I (x; y) = 1 if and only if x6y. Therefore said implications do not satisfy the property: I (x; c(x)) = c(x) for all x ∈[0; 1].
5. Conclusions and future research The most important results of this paper can be summarized in four points. (It is necessary to keep in mind that in the entire paper we have only used strong negations.) (1) There are di1erent interdependencies among the axioms demanded of the implication operators. Moreover, the property (1) and axiom I8 cannot be fulElled simultaneously. (2) Theorem 6 has made clear that I (x; y) = ∨(c(x); x) are the only S-implications that fulEll (1). The study of axiom I8 in S-implications has led us to Theorem 7, which is similar to Smets and Magrez’s theorem with the di1erence that in Theorem 7 we can use di1erent automorphisms in order to generate the implications and strong negations. In Corollary 4 we have proven that the only implications generated from Theorem 7 that fulEll I8 are the ones obtained by Smets and Magrez. In addition, since axiom I11 is a consequence of axiom I8, and we impose axiom I11 in Theorem 7, we have also wanted to study the case of implications generated from automorphisms without imposing I11, this has led us to Theorem 8. By means of this theorem we can generate S-implications from automorphisms such that these implications do not fulEll (1) nor I8. Evidently, there are S-implications, which are not generated by automorphisms, that do not fulEll the two properties above either. (3) For the QL-implications we have proven that if the t-conorm used is S = ∨, then (1) holds, however Theorem 10 has made clear that the reciprocal is not true. In the next theorem, we have presented the conditions in which we can generate QL-implications that fulEll (1) from automorphisms. In the theorems that followed we have analyzed the behavior (with respect to (1) and I8), of the QL-implications obtained from automorphisms when we demand that the corresponding t-norm be strict. (4) Regarding the R-implications we have proven that these never fulEll property (1). In the near future, among other things, we plan to analyze property (1) when the negation considered is not a strong negation. As future research we can point out the study of the behavior of De Baets and Kerre’s inclusion grade indicator, (A; B) = Inf x∈X (I (A (x); B (x))), when the implication operators used fulEll (1) or the axioms in the theorems studied in this paper. With respect to this point we can advance some of the results (which are direct consequence of the studies carried out in this paper). For example, (i) If I satisEes I8, then: (A; B) = 1 if and only if A6B.
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(ii) If: I (x; y) = 0 if and only if x = 1 and y = 0 (property fulElled by S-implications and QL-implications, as we have proven in Proposition 2), then: (A; B) = 0 if and only if A = {x; 1 | x ∈X } and B = {x; 0 | x ∈X }, obtaining a stronger result than the one demanded by Sinha and Dougherty. (iii) If I satisEes (1), then if A is a reEnement of B, that is, if A (x)6B (x) when B (x)6c(e) = e ∈(0; 1), and A (x)¿B (x), when B (x)¿c(e) = e ∈(0; 1), then (A ∨ c(A); A ∧ c(A)) 6 ((B ∨ c(B); B ∧ c(B))); that is, if I satisEes (1), then De Baets and Kerre’s inclusion grade indicator satisEes one of the axioms demanded by Sinha and Dougherty, when using the inclusion grade indicator in Approximate Reasoning. In general, our aim for the near future is to carry out a study of the application of the results of this paper to De Baets and Kerre’s inclusion grade indicators, and, in accordance with the selection of implication operators we make, see what Sinha and Dougherty’s axioms these indicators fulEll. Acknowledgements We thank the referees for their helpful comments. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24]
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www.elsevier.com/locate/fss
Automorphisms, negations and implication operators H. Bustince ∗ , P. Burillo, F. Soria Departamento de Autom atica y Computaci on, Universidad P ublica de Navarra, Campus de Arrosadia, 31006 Pamplona, Spain Received 5 April 2000; received in revised form 20 February 2002; accepted 5 April 2002
Abstract In this paper we study the conditions in which the implication operators satisfy the property I (x; c(x)) = c(x) for all x ∈ [0; 1], c being any strong negation. This study has led us to present di1erent implication operator characterization theorems from automorphisms, obtaining a theorem similar to the one presented by Smets and Magrez (Internat. J. Approx. Reason. 1 (1987) 327), in which the strong negation c used is not generated by the same automorphism that generates c 2002 Elsevier Science B.V. All rights reserved. the implication. Keywords: Fuzzy negation; Strong negation; t-Norm; t-Conorm; Automorphism; Implication operations
Introduction The purpose of this paper is to explore the conditions in which the implication operators satisfy the following property: I (x; c(x)) = c(x)
for all x ∈ [0; 1];
(1)
c being any strong negation. We have divided the introduction to this paper in two parts. In part (A) we present the manner in which the study of property (1) arose in our research. In part (B) we show the strategy carried out in the making of this paper. (A) Among the axioms that Sinha and Dougherty [35,36,37] demand of the inclusion grade indicator , the following can be found: If A is a re$nement of B, that is, if A (x)6B (x) when B (x)6c(e) = e ∈(0; 1), and A (x)¿B (x), when B (x)¿c(e) = e ∈(0; 1), then (A ∨ c(A); A ∧ c(A)) 6 ((B ∨ c(B); B ∧ c(B)): At present the main objective of our investigation is the construction of di1erent expressions of fuzzy entropy from the inclusion grade indicators that satisfy the axiom above. ∗
Corresponding author. Tel.: +34-94816-9254; fax: +34-94816-9565. E-mail address: [email protected] (H. Bustince).
c 2002 Elsevier Science B.V. All rights reserved. 0165-0114/03/$ - see front matter PII: S 0 1 6 5 - 0 1 1 4 ( 0 2 ) 0 0 2 1 4 - 2
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With said objective in mind, reviewing the fuzzy literature for the di1erent expressions presented for the inclusion grade indicator, we End the expression given by De Baets and Kerre [5], that is: for any A and B on the same referential X; (A; B) = Inf (I (A (x); B (x))): x∈X
Studying the conditions in which the expression above satisEes Sinha and Dougherty’s axiom we have proven that when the implication satisEes property (1), we have that De Baets and Kerre’s indicator satisEes Sinha and Dougherty’s axiom. This fact has led us to analyze property (1) in fuzzy implications, leaving the study of the functions that simultaneously generate fuzzy entropies and inclusion grade indicators for future papers. (B) In the Erst part of the paper we present the notations and basic deEnitions that we are going to use. Next, we focus on the di1erent axioms demanded of the implication operators subsequently analyzing some of the interrelations between said axioms. Then we present property (1) and we observe that when (1) holds then the following axiom (frequently demanded by fuzzy implications) does not: I (x; y) = 1 if and only if x 6 y
(2)
and vice versa, when I satisEes (2), it does not satisfy (1). Our interest in Ending the conditions in which I satisEes property (1) has led us to study, in Section 4, di1erent fuzzy implication characterization theorems from automorphisms, analyzing at all times the conditions in which the implications obtained satisfy (1) or (2) or neither. And so, in Section 4.1, where we study the S-implications, we have obtained a theorem similar to the one presented by Smets and Magrez [38], in which the strong negation c used is not generated by the same automorphism that generates the implication. Afterwards, we have obtained another S-implication characterization theorem from automorphisms such that the S-implications obtained satisfy (1). In Sections 4.2 and 4.3 we carry out a study similar to the one above, but for QL-implications and Rimplications, respectively. We end the paper with the conclusions, future research and references. 1. Notations and basic denitions 1.1. Fuzzy negations Let c : [0; 1] → [0; 1]. c is a fuzzy negation, i1: (i) c(0) = 1 and c(1) = 0, (ii) c(x)6c(y), if x¿y (monotonicity). A fuzzy negation is strict, i1, (iii) c(x) is continuous, (iv) c(x)¡c(y), for x¿y for all x; y ∈[0; 1]. A strict fuzzy negation is involutive, i1 (v) c(c(x)) = x, for all x ∈[0; 1]. The strict fuzzy negations that are involutive are called strong negations. In this paper we will only use strong negations. It is important to mention that Fodor and Roubens [12] prove that involutive fuzzy negations are automatically strict.
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211
Di1erent approaches to the study of fuzzy negations were used by Lowen [26], Esteva et al. [11], and Ovchinnikov [29,30], Higashi and Klir [21,25]. Yager investigated fuzzy negations for the purpose of developing useful measures of fuzziness [45,46]. The representation theorem of strong negations was obtained by Trillas [39,40]. We cite here that result in a slightly modiEed form which is more suitable in the sequel, see [31]. First we need the deEnition of an automorphism of a real interval [a; b] ⊂ R. This notion will be used extensively in this paper, especially in the case of the unit interval [0; 1]. Denition 0. A continuous, strictly increasing function ’ : [a; b] → [a; b] with boundary conditions ’(a) = a; ’(b) = b is called an automorphism of the interval [a; b] ⊂ R. Theorem 0. A function c : [0; 1] → [0; 1] is a strong negation if and only if there exists an automorphism ’ of the unit interval such that c(x) = ’−1 (1 − ’(x)). Many properties of the fuzzy negations have been studied, among which we will mention those relative to equilibrium points. An equilibrium point of a fuzzy negation c is any value e ∈(0; 1) such that c(e) = e. Furthermore, regarding these points we will highlight the following property: if c is a continuous fuzzy negation, then c has a unique equilibrium point. As we have said before, hereinafter c will always stand for strong negations (such that c(e) = e). 1.2. t-norms and t-conorms A function T : [0; 1]2 → [0; 1] satisfying the properties: (i) (ii) (iii) (iv)
Boundary conditions, T (x; 1) = x for all x ∈[0; 1], Monotony, T (x; y)6T (z; u) if x6z and y6u, Commutative, T (x; y) = T (y; x) ∀x; y ∈[0; 1], Associative, T (T (x; y); z) = T (x; T (y; z)) for all x; y; z ∈[0; 1],
is called a t-norm. Many papers on the di1erent properties of t-norms have been published. The deEnitions, properties, and notations that we are going to use in this paper can be found in the book by Fodor and Roubens, Fuzzy Preference Modelling and Multicriteria Decision Support, [15]. For example the deEnitions of: continuous t-norms, Archimedean t-norms, strict t-norms, etc. In Section 4.3 of this paper we will use the concept of left-continuous t-norm [15,23,24]. A t-norm is said to be left-continuous if it is left-continuous as a two-place function. A function S : [0; 1]2 → [0; 1] satisfying the properties: (i) Boundary conditions, S(x; 0) = x for all x ∈[0; 1], (ii) Monotony, S(x; y)6S(z; u) if x6z and y6u,
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(iii) Commutative, S(x; y) = S(y; x) ∀x; y ∈[0; 1], (iv) Associative, S(S(x; y); z) = S(x; S(y; z)) for all x; y; z ∈[0; 1], is called a t-conorm. Like with the t-norms, the deEnitions, properties, and notations of the t-conorms that we are going to use throughout the paper are the ones gathered by Fodor and Roubens in [15]. We say that a t-norm T and a t-conorm S are dual with respect to a fuzzy negation c if and only if c(T (a; b)) = S(c(a); c(b)) and c(S(a; b)) = T (c(a); c(b)). Let the triple T; S; c denote that T and S are dual with respect to c [6]. Please note that in a De Morgan triple the negation is automatically strong. The easy proof is: x = T (x; 1) = c(S(c(x); 0)) = c(c(x)) by duality, whence c should be involutive. Therefore we know that ∧; ∨; c is dual with respect to any strong negation c, being ∨ = max and ∧ = min. Due to their importance, we must point out other works where the most important properties of t-norms and t-conorms are investigated, like for example: [2,4,8,14,17,18,19,20,27,32,42,44]. Now we present three t-norm and t-conorm representation theorems that will be used further on. In these three theorems we will use the same notation Fodor and Roubens used in [15], though we will at all times indicate the origin of these theorems. In 1991 Ovchinnikov and Roubens [31] proved the following two theorems: Theorem 1. A continuous t-conorm S satis$es condition S(x; c(x)) = 1 for all x ∈[0; 1] with a strict negation c if and only if there exists an automorphism ’ of the unit interval such that S(x; y) = ’−1 (∧(’(x) + ’(y); 1)) and c(x) ¿ ’−1 (1 − ’(x)): Theorem 2. A continuous t-norm T is such that T (x; c(x)) = 0 holds for all x ∈[0; 1] with a strict negation c if and only if there exists an automorphism ’ of the unit interval such that T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)) and c(x) 6 ’−1 (1 − ’(x)): Please note that if in the two theorems above, we take c(x) = ’−1 (1−’(x)), this strict negation is involutive, that is, it is a strong negation. In 1949, AczMel [1] proved the following theorem (also see: [34]). Theorem 3. A continuous t-norm T is strict if and only if there exists an automorphism ’ of the unit interval such that T (x; y) = ’−1 (’(x) · ’(y)): 2. Implication operators We know [22,33] that a [0; 1]2 → [0; 1] mapping I is called an implication operator if and only if it satisEes the boundary conditions I (0; 0) = I (0; 1) = I (1; 1) = 1 and I (1; 0) = 0.
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The above conditions are, of course, the least we can demand from an implication operator. Other interesting potential properties of implication operators are listed in [3,12,16,42,47]. All fuzzy implications are obtained by generalizing the implication operator of classical logic. In this sense Fodor and Roubens [15] establish the following deEnition: Denition 1. An implication is a function I : [0; 1]2 → [0; 1] that satisEes the following properties: I1: I2: I3: I4: I5:
x6z implies I (x; y)¿I (z; y) for all y ∈[0; 1]. y6t implies I (x; y)6I (x; t) for all x ∈[0; 1]. I (0; x) = 1 (dominance of falsity) for all x ∈[0; 1]. I (x; 1) = 1 for all x ∈[0; 1]. I (1; 0) = 0.
It is important to point out the two following considerations: • If function I : [0; 1]2 → [0; 1] satisEes I2 and I (0; 0) = 1, then I satisEes I3. • If function I : [0; 1]2 → [0; 1] satisEes I1 and I (1; 1) = 1, then I satisEes I4. Now we recall further axioms, in terms of function I . These properties are required in di1erent papers and they could also be important in some applications. I6: I7: I8: I9: I10: I11: I12: I13:
I (1; x) = x (neutrality of truth). I (x; I (y; z)) = I (y; I (x; z)) (exchange property). I (x; y) = 1 if and only if x6y (boundary condition). I (x; 0) = c(x) is a strong negation. I (x; y)¿y. I (x; x) = 1 (identity). I (x; y) = I (c(y); c(x)) (contraposition) with a strong negation c. I is a continuous function (continuity).
It is important to point out that hereinafter when we speak of functions I : [0; 1]2 → [0; 1] we do not suppose it is an implication in Fodor’s sense (DeEnition 1), but that it is a function that satisEes the properties demanded in each particular case. Proposition 1. Let I : [0; 1]2 → [0; 1]. For all x; y; z ∈[0; 1], the following properties hold: (i) (ii) (iii) (iv)
I I I I
satis$es satis$es satis$es satis$es
I1 I1 I2 I2
if if if if
and and and and
only only only only
if if if if
I (x ∨ y; z) = ∧(I (x; z); I (y; z)). I (x ∧ y; z) = ∨(I (x; z); I (y; z)). I (x; y ∧ z) = ∧(I (x; y); I (x; z)). I (x; y ∨ z) = ∨(I (x; y); I (x; z)).
Proof. (i) ( ⇒) Two things can occur: (a) If x6y, then I (x ∨ y; z) = I (y; z), besides by I1, I (x; z)¿I (y; z), therefore ∧(I (x; z); I (y; z)) = I (y; z). (b) If x¿y, then I (x ∨ y; z) = I (x; z), besides in these conditions by I1, I (y; z)¿I (x; z), therefore ∧(I (y; z); I (x; z)) = I (x; z). (⇐) If I (x ∨ y; z) = ∧(I (x; z); I (y; z)) two things can occur:
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(a) If x6y, then I (x ∨ y; z) = I (y; z) = ∧(I (x; z); I (y; z)), then I (x; z)¿I (y; z). (b) If x¿y, then I (x ∨ y; z) = I (x; z) = ∧(I (x; z); I (y; z)), therefore I (y; z)¿I (x; z). The rest of the items are proven in a similar manner. 2.1. Some interdependencies among the axioms Evidently, the axioms I1–I13 are not independent of one another. There are axioms that are direct consequence of others, for example: • If I satisEes I6, then it satisEes I5, • If I satisEes I10, then it also satisEes I4, • If I satisEes I8, then I satisEes I3 and I11. In the following lemma we present other dependencies. Lemma 1. Let I : [0; 1]2 → [0; 1]. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
If If If If If If If If If If If
I I I I I I I I I I I
satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es satis$es
I1 and I12, then I satis$es I2; I2 and I12, then I satis$es I1; I3 and I12, then I satis$es I4; I4 and I12, then I satis$es I3; I6 and I12, then I satis$es I9; I9 and I12, then I satis$es I6; I2 and I9, then I satis$es I3; I1 and I6, then I satis$es I10; I7 and I9, then I satis$es I12; I1, I6 and I12, then I satis$es I2; I3; I4; I5; I9 and I10; I2; I7 and I8, then I satis$es I1; I3; I4; I5; I6; I10 and I11;
hold. Proof. (i) If y6t, then c(y)¿c(t), by I1 we have I (c(y); c(x))6I (c(t); c(x)), since I fulElls I12 then I (x; y)6I (x; t). (ii) If y6t, then c(t)6c(y) by I2 we have I (c(x); c(t))6I (c(x); c(y)) for all c(x) ∈[0; 1], besides I fulElls I12, therefore I (c(x); c(t)) = I (t; x)6I (c(x); c(y)) = I (y; x). (iii) I (x; 1) = I (0; c(x)) = 1 for all x ∈[0; 1]. (iv) I (0; x) = I (c(x); 1) = 1 for all x ∈[0; 1]. (v) I (x; 0) = I (c(0); c(x)) = I (1; c(x)) = c(x). (vi) c(x) = I (x; 0) = I (1; c(x)). (vii) I (0; x) = I (0; ∨(x; 0)) = ∨(I (0; x); I (0; 0)) = ∨(I (0; x); c(0)) = 1 for all x ∈[0; 1]. (viii) Since x61 by I1 and I6, we have I (x; y)¿I (1; y) = y. (ix) I (x; y) = I (x; c(c(y))) = I (x; I (c(y); 0)) = I (c(y); I (x; 0)) = I (c(y); c(x)). (x) Consequence of the above. (xi) See [15]. It is important to point out that item (xi) of the lemma above can be found proven in [15] and item (ix) in [23].
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Table 1
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
I1; I12 ⇒ I2 I2; I12 ⇒ I1 I3; I12 ⇒ I4 I4; I12 ⇒ I3 I6; I12 ⇒ I9 I9; I12 ⇒ I6 I2; I9 ⇒ I3 I1; I6 ⇒ I10 I7; I9 ⇒ I12 I1; I6; I12 ⇒ I2; I3; I4; I5; I9; I10 I2; I7; I8 ⇒ I1; I3; I4; I5; I6; I10; I11
Table 1, represents the results of the lemma above. From items (i) – (vi) of Lemma 1 we deduce the following: If I : [0; 1]2 → [0; 1] satis$es axiom I12, then the following properties hold: (i) I1 holds if and only if I2 holds; (ii) I3 holds if and only if I4 holds; (iii) I6 holds if and only if I9 holds. The results of the lemmas that we demonstrate below will be used in further sections of the paper. Lemma 2. Let I : [0; 1]2 → [0; 1] be any function that satis$es at least one of the following items: (i) (ii) (iii) (iv)
If If If If
I I I I
satis$es satis$es satis$es satis$es
I2; I1; I4, I6,
I6 and I9, or I6 and I9, or I7, I9 and I11, or I7, I9 and I (x; x) = I (0; x) for all x ∈[0; 1], then
I (x; y) = 0 and only if x = 1 and y = 0: Proof. (i) (⇐) If x = 1 and y = 0, then by I9 we have I (1; 0) = c(1) = 0. (⇒) If I (x; y) = 0, then by item (iv) of Proposition 1 we have 0 = I (x; y) = I (x; y ∨ 0) = ∨(I (x; y); I (x; 0)) = ∨(0; c(x)) = c(x), then x = 1. Besides, by I6 we have 0 = I (x; y) = I (1; y) = y. (ii) (⇐) By I9 we have I (1; 0) = c(1) = 0. (⇒) If I (x; y) = 0, then by item (ii) of Proposition 1 and axiom I6 we have 0 = I (x; y) = I (x ∧ 1; y) = ∨(I (x; y); I (1; y)) = ∨(0; y) = y. Besides by I9, 0 = I (x; y) = I (x; 0) = c(x), therefore x = 1. (iii) (⇐) By item (ix) of Lemma 1, we have that I satisEes I12, furthermore, by item (vi) of the same lemma I satisEes I6, and therefore I satisEes I5, thus I (1; 0) = 0. (⇒) Since I (x; y) = 0 we have c(y) = I (y; 0) = I (y; I (x; y)) = I (x; I (y; y)) = I (x; 1) = 1, therefore y = 0, then 0 = I (x; y) = I (x; 0) = c(x), therefore x = 1. (iv) Similar to the above. Lemma 3. For I : [0; 1]2 → [0; 1] the following properties hold: (i) If I satis$es I2 and I9, then I (x; y)¿c(x) for all x; y ∈[0; 1]. (ii) If I satis$es I2, I9 and I10, then I (x; y)¿ ∨(c(x); y) for all x; y ∈[0; 1]. (iii) If I satis$es I2, I9 and I12, then I (x; y)¿ ∨(c(x); y) for all x; y ∈[0; 1].
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Proof. (i) I (x; y) = I (x; y ∨ 0) = ∨(I (x; y); I (x; 0)) = ∨(I (x; y); c(x)), [0; 1]. (ii) Evident. (iii) c(x)6I (x; y) = I (c(y); c(x))¿c(c(y)) = y.
then
I (x; y)¿c(x)
for
all
x; y ∈
3. The property: I (x; c(x)) = c(x) for all x belonging to [0; 1] Let I : [0; 1]2 → [0; 1] be any function such that I (x; c(x)) = c(x) for all x ∈[0; 1]. In these conditions we can easily see: (a) I (1; 0) = 0 and I (0; 1) = 1, (b) I (c(x); x) = x for all x ∈[0; 1] due to the fact that we only consider strong negations. (c) There is a contradiction with axiom I8 as proven by the following theorem. Theorem 4. If I : [0; 1]2 → [0; 1] satis$es I8, then it does not satisfy I (x; c(x)) = c(x) for all x ∈[0; 1]. Proof. Let us suppose it is not true, that is, I satisEes: I (x; c(x)) = c(x)
for all x ∈ [0; 1]; and I8:
Because c is a strong negation, it has a single equilibrium point e ∈(0; 1). Since e = c(e), we have e = c(e) = I (e; c(e)) = 1, which is a contradiction with the fact that e ∈(0; 1). Corollary 1. If I : [0; 1]2 → [0; 1] satis$es I (x; c(x)) = c(x) for all x ∈[0; 1], then it does not satisfy I8. Proof. Similar to the demonstration of the theorem above. It is important to point out that there exist I functions that do not satisfy axiom I8 nor the property I (x; c(x)) = c(x) for all x ∈[0; 1], for example, Reichenbach’s implication operator I (x; y) = 1 − x + x · y, with the strong negation c(x) = 1 − x. 4. Implications by t-norms, t-conorms and negations Three general classes of implication operator, I , have been identiEed in the literature [7,9,10], the Simplications, the QL-implications and the R-implications. An S-implication associated with a t-conorm S and a strong negation c is deEned by I (x; y) = S(c(x); y): A QL-implication associated with a t-conorm S, a t-norm T and a strong negation c is deEned by I (x; y) = S(c(x); T (x; y)): An R-implication associated with a t-norm T is deEned by I (x; y) = sup{z | T (x; z) 6 y}: In the particular case that the t-norm considered be left-continuous, then we will refer to the R-implication deEned above writing: residual implication [23,24].
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Di1erent characterizations of S-implications, QL-implications and R-implications can be found in Trillas and Valverde [41]. It is easy to see that both S-implications and R-implications satisfy properties I1–I5 for any t-norm T , t-conorm S and strong negation c, thus they are implications in Fodor and Roubens’ sense. In general QLimplications violate property I1. The conditions under which I1 is satisEed can be found in Fodor [13]. Proposition 2. If I is an S-implication or a QL-implication, the following items hold: (i) The axioms: I2, I6 and I9 are satis$ed. (ii) I (x; y)¿c(x) for all x; y ∈[0; 1]. (iii) I (x; y) = 0 if and only if x = 1 and y = 0. Proof. (i) If I (x; y) = S(c(x); y), since S is increasing in the second argument, then by the deEnition of S-implication, I is as well. I (1; x) = S(c(1); x) = S(0; x) = x and I (x; 0) = S(c(x); 0) = c(x). If I (x; y) = S(c(x); T (x; y)), since T and S are increasing in the second argument, by the deEnition of QLimplication, I is as well. I (1; x) = S(c(1); T (1; x)) = S(0; T (1; x)) = T (1; x) = x and I (x; 0) = S(c(x); T (x; 0)) = S(c(x); 0) = c(x). (ii) and (iii) are consequence of the Lemmas 2 and 3. 4.1. S-implications and the I (x; c(x)) = c(x) property Theorem 5. Let I be an S-implication with an appropriate t-conorm S and a strong negation c. I (x; c(x)) = c(x) for all x ∈ [0; 1] if and only if S = ∨: Proof. Keeping in mind that: S is idempotent if and only if S = ∨, and that c is a strong negation, i.e. an involutive order reversing bijection of the closed unit interval, we have c(x) = I (x; c(x)) = S(c(x); c(x)) = ∨(c(x); c(x)). The following theorem establishes the axioms that we must demand of the function I : [0; 1]2 → [0; 1] in order for it to be given by the expression I (x; y) = ∨(c(x); y) for all x; y ∈[0; 1]. Theorem 6. A function I : [0; 1]2 → [0; 1] is such that I2, I7, I9 and I (x; c(x)) = c(x) for all x ∈[0; 1] if and only if I is an S-implication and I (x; y) = ∨(c(x); y) for all x; y ∈[0; 1]. Proof. (⇒) Suppose that I is any function which fulEls I2, I7, I9 and I (x; c(x)) = c(x) for all x ∈[0; 1]. By Lemma 1, I satisEes I12, I6 and I1. We deEne a function S by S(x; y) = I (c(x); y). Let us prove that S is a t-conorm. S is nondecreasing in both places by I1 and I2. By I9, S(x; 0) = I (c(x); 0) = c(c(x)) = x for all x ∈[0; 1]. S is commutative by I12: S(x; y) = I (c(x); y) = I (c(y); x) = S(y; x): By I7 and commutativity, we have that S(x; S(y; z)) = S(x; S(z; y)) = I (c(x); I (c(z); y)) = I (c(z); I (c(x); y)) = S(z; S(x; y)) = S(S(x; y); z):
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Therefore, S is a t-conorm. We obviously have that I (x; y) = S(c(x); y), by the deEnition of S. Since furthermore by hypothesis I (x; c(x)) = c(x) for all x ∈[0; 1], and S(x; x) = x if and only if s = ∨, we have c(x) = I (x; c(x)) = S(c(x); c(x)), therefore S = ∨. Then I (x; y) = ∨(c(x); y). (⇐) If y6z, then I (x; y) = ∨(c(x); y)6 ∨(c(x); z) = I (x; z), therefore I2 holds. I (x; 0) = ∨(c(x); 0) = c(x) for all x ∈[0; 1], thus I satisEes I9. I7 holds because I (x; I (y; z)) = ∨(c(x); ∨(c(y); z)) = ∨(c(y); ∨(c(x); z)) = I (y; I (x; z)); by the associativity and commutativity of ∨. Finally, by the idempotency of ∨ we have that I (x; c(x)) = ∨(c(x); c(x)) = c(x)
for all x ∈ [0; 1]:
Corollary 2. The S-implication, I (x; y) = ∨(c(x); y), is such that (i) (ii) (iii) (iv)
It It If If
ful$lls the axioms I1–I7; I9; I10; I12 and I13. is the only S-implication that veri$es I (x; c(x)) = c(x) for all x ∈[0; 1]. e ∈(0; 1) is the equilibrium point of c, then I (e; e) = e. x ∈[e; 1], then I (x; x) = x.
Proof. (i) We only need to keep in mind the theorem above and Lemma 1. (ii) Evident by Theorem 5. (iii) I (e; e) = ∨(c(e); e) = ∨(e; e) = e. (iv) Since x¿e, then c(x)6c(e) = e6x. Please note that if in I (x; y) = ∨(c(x); y) we take c(x) = 1 − x for all x ∈[0; 1], we have Kleene–Dienes’ operator. Taking into account Corollaries 1 and 2, the S-implications studied in the two theorems above do not satisfy axiom I8. The following theorem allows us to build S-implications from automorphisms in the unit interval. Nevertheless, we will see further on that when I is generated by automorphisms: I satisEes I8 if and only if the strong negation is generated by the same automorphism that generates the implication I . Theorem 7. A function I : [0; 1]2 → [0; 1] is such that I2; I7; I9; I11 and I13 are satis$ed if and only if there exists an automorphism ’ of the unit interval such that I (x; y) = ’−1 (∧(’(c(x)) + ’(y); 1)) and c(x) ¿ ’−1 (1 − ’(x)): Proof. (⇒) Since I satisEes I2, I7, I9, I11 and I13 by Lemma 1 we know that I also fulElls I12. We deEne a function S by S(x; y) = I (c(x); y). In the demonstration of Theorem 6 we have proven that in these conditions S is a t-conorm. Moreover, by I11, S(x; c(x)) = I (c(x); c(x)) = I (x; x) = 1. Also, since I is continuous, S is as well. Keeping in mind Theorem 1 we have that I (x; y) = S(c(x); y) = ’−1 (∧(’(c(x)) + ’(y); 1))
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and c(x) ¿ ’−1 (1 − ’(x)): (⇐) By the deEnition of automorphism, if y6z, then I (x; y) = ’−1 (∧(’(c(x)) + ’(y); 1))6’−1 (∧ (’(c(x)) + ’(z); 1)) = I (x; z), therefore I2 holds. I7 holds because I (x; I (y; z)) = ’−1 (∧(’(c(x)) + ∧(’(c(y)) + ’(z); 1); 1) = ’−1 (∧(’(c(y)) + ∧(’(c(x)) + ’(z); 1); 1) = I (y; I (x; z)): Besides I (x; 0) = ’−1 (∧(’(c(x)) + 0; 1)) = ’−1 (’(c(x))) = c(x), therefore I9 holds. Finally, since c(x)¿’−1 (1−’(x)), we have that ’(c(x))+’(x)¿1−’(x)+’(x) = 1, therefore I (x; x) = ’−1 (∧(’(c(x)) + ’(x); 1)) = 1, therefore I fulElls I11. Moreover since ’ is an automorphism and ∧ and c are continuous then I13 also holds. Corollary 3. Let I : [0; 1]2 → [0; 1] be such that I2; I7; I9; I11 and I13 hold. Then I (x; y) = 1 if and only if c(x) ¿ ’−1 (1 − ’(y)): Proof. (⇒) Since I (x; y) = 1 = ’−1 (∧(’(c(x)) + ’(y); 1)), we have 1 = ∧ (’(c(x)) + ’(y); 1), therefore ’(c(x)) + ’(y)¿1, that is, ’(c(x))¿1 − ’(x), then c(x)¿’−1 (1 − ’(y)). (⇐) If c(x)¿’−1 (1 − ’(y)), then ’(c(x))¿1 − ’(y), therefore ’(c(x)) + ’(y)¿1, then I (x; y) = 1. Remark. The corollary above allows us to conclude that the implication operator generated from the automorphism ’ used in Theorem 7 does not always fulEll axiom I8. The following example makes this result clear: Example. Let us take ’(x) = x for all x ∈[0; 1], and the strong negation c(x) = (1−x2 )1=2 , evidently, c(x) = (1− x2 )1=2 ¿1 − x, 1 − x being the strong negation generated by the identity automorphism. In these conditions, I (x; y) = ∧ (c(x) + y; 1) = ∧ ((1 − x2 )1=2 + y; 1). For x = 0:5 and y = 0:4, (obviously x = 0:5¿0:4 = y), we have c(0:5) = (1 − 0:52 )1=2 = 0:87¿1 − 0:4 = 0:6 and I (0:5; 0:4) = 1. In 1987 Smets and Magrez proved the following theorem [15,38]: A function I : [0; 1]2 → [0; 1] is such that I2, I7, I8 and I13 are satis$ed if and only if there exists an automorphism ’ of the unit interval such that I (x; y) = ’−1 (∧(1 − ’(x) + ’(y); 1)): As a consequence of this theorem and of Theorem 7 we have the following result. Corollary 4. Let I : [0; 1]2 → [0; 1] be such that I2; I7; I9; I11 and I13 are satis$ed. In these conditions: c(x) = ’−1 (1 − ’(x)) for all x ∈ [0; 1] if and only if I fulfills I8: Proof. Evident by Theorem 7 and Smets and Magrez’s Theorem. Remark. (a) From Theorem 4 and the corollary above we deduce that the functions I generated from the automorphism ’, according to the Corollary 4, do not satisfy the property I (x; c(x)) = c(x) for all x ∈[0; 1]. (b) Theorem 7 is a generalization of Smets and Magrez’s theorem in the following sense: they always considered that the strong negation c of the axioms I9 and I12 was generated by the automorphism that
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Fig. 1.
generates the implication I , which is something we do not do. That is, the strong negation that we use can be generated by an automorphism % di1erent from the automorphism ’ that generates I , provided that c(x) = %−1 (1 − %(x))¿’−1 (1 − ’(x)). (c) From Corollary 4 we deduce that the only functions I generated from the automorphism ’ according to Theorem 7 that satisfy I8 are the ones we obtain generated by the same automorphism as in Smets and Magrez’s theorem. (d) If in the conditions of Corollary 4 we take the automorphism ’(x) = x for all x ∈[0; 1], the resulting implication operator is Lukasiewicz’s, that is, I (x; y) = ∧(1 − x + y; 1). It is easy to prove, (Lemma 1), that this implication operator satisEes I1–I8, I9 with c(x) = 1 − x, I10–I13. (e) Fig. 1 represents two implication operators obtained from Theorem 7 taking in both cases the identity automorphism. In Fig. 1(a) the strong negation considered is Yager’s: c(x) = (1 − xw )1=w with w = 3, in Fig. 1(b) Sugeno’s: c(x) = (1 − x)=(1 + 'x) with ' = −0:5. In both cases c(x)¿1 − x for all x ∈[0; 1]. We know that a consequence of axiom I8 is axiom I11, since the latter is imposed in Theorem 7, we present below a theorem that allows us to obtain implications from automorphisms ’ without demanding I11. We will see that in these conditions the S-implications I generated do not fulEll I8 nor I (x; c(x)) = c(x) for all x ∈[0; 1]. Denition 2. It is said that I is strictly monotone, if it is strictly decreasing in the Erst argument and strictly increasing in the second, both on (0; 1)2 . Theorem 8. A function I : [0; 1]2 → [0; 1] is such that it is strictly monotone, and I7; I9 and I13 are satis$ed if and only if there exists an automorphism ’ of the unit interval such that I (x; y) = c(’−1 (’(x) · ’(c(y)))): Proof. (⇒) Since I fulElls I7 and I9, by Lemma 1 we know I also fulElls I12. We deEne a function T given by T (x; y) = c(I (x; c(y))).
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Evidently, since c is a strong negation and I is strictly monotone, then T is strictly increasing in both arguments on (0; 1)2 . Moreover, by I12 we have T (x; y) = c(I (x; c(y))) = c(I (y; c(x))) = T (y; x), therefore T satisEes the commutativity. In addition, T (1; x) = c(I (1; c(x))) = c(I (x; c(1))) = c(I (x; 0)) = c(c(x)) = x for all x ∈[0; 1]. By I12 and I7 we have that T (x; T (y; z)) = T (x; c(I (y; c(z)))) = T (x; c(I (z; c(y)))) = c(I (x; c(c(I (z; c(y)))))) = c(I (x; I (z; c(y)))) = c(I (z; I (x; c(y)))) = c(I (z; c(c(I (x; c(y)))))) = c(I (z; c(T (x; y)))) = T (z; T (x; y)) = T (T (x; y); z): Since I fulElls I13 and c is a strong negation, T is continuous. Therefore, T is a continuous, strict t-norm. Using Theorem 3 we have that I (x; y) = c(T (x; c(y))) = c(’−1 (’(x) · ’(c(y)))). (⇐) Since c is a strong negation and ’ is an automorphism, I (x; y) = c(’−1 (’(x) · ’(c(y)))) is strictly monotone and fulElls I13. I (x; 0) = c(’−1 (’(x) · ’(c(0)))) = c(’−1 (’(x) · ’(1))) = c((’−1 (’(x)))) = c(x); thus I satisEes I 9: I also satisEes I7, because: I (x; I (y; z)) = c(’−1 (’(x) · ’(c(I (y; z))))) = c(’−1 (’(x) · ’(c(c(’−1 (’(y) · ’(c(z)))))))) = c(’−1 (’(x) · ’(y) · ’(c(z)))); I (y; I (x; z)) = c(’−1 (’(y) · ’(c(I (x; z))))) = c(’−1 (’(y) · ’(c(c(’−1 (’(x) · ’(c(z)))))))) = c(’−1 (’(y) · ’(x) · ’(c(z)))) therefore I (x; I (y; z)) = I (y; I (x; z)), that is I fulElls I7. Corollary 5. Let I : [0; 1]2 → [0; 1] which is strictly monotone, and satis$es the axioms: I7; I9 and I13. Then (i) I (x; c(x)) = c(’−1 (’(x) · ’(x))); (ii) I (x; y) = 1 if and only if x = 0 or y = 1; (iii) If c(x) = ’−1 (1 − ’(x)) for all x ∈[0; 1], I (x; y) = ’−1 (1 − ’(x) + ’(x) · ’(y)); hold. Proof. (i) Evident. (ii) (⇒) I (x; y) = 1 = c(’−1 (’(x) · ’(c(y)))), then 0 = ’(x)’(c(y)), therefore x = 0 or y = 1. (⇐) Evident. (iii) Since c(x) = ’−1 (1 − ’(x)), we have I (x; y) = c(’−1 (’(x) · ’(’−1 (1 − ’(y))))) = c(’−1 (’(x) · (1−’(y)))) = c(’−1 (’(x)−’(x)·’(y))) = ’−1 (1−’(’−1 (’(x)−’(x)·’(y)))) = ’−1 (1−’(x)+’(x)·’(y))).
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Remark. (a) Keeping in mind the Erst two items of Corollary 5, we have that the implication operators obtained from Theorem 8 do not satisfy axiom I8, nor the property I (x; c(x)) = c(x) for all x ∈[0; 1]. (b) Please note that in the conditions of item (iii) of the corollary above, if ’ is the identity automorphism then we obtain the implication operator of Rechenbach, I (x; y) = 1 − x + x · y, it is easy to prove, (Lemma 1), that this implication operator satisEes the axioms: I1–I7, I9 with c(x) = 1 − x, I10, I12 and I13. 4.2. QL-implications and the I (x; c(x)) = c(x) property Theorem 9. Let I be a QL-implication associated to the t-conorm S = ∨, to any t-norm T and a strong negation c. Then (i) I (x; c(x)) = c(x) for all x ∈[0; 1]; (ii) I (x; y) = 1 if and only if x = 0 or x = y = 1; hold. Proof. (i) We only need to recall that T (x; c(x))6 ∧(x; c(x))6c(x) for all x ∈[0; 1]. (ii) (⇒) If I (x; y) = ∨(c(x); T (x; y)) = 1, then c(x) = 1 or T (x; y) = 1. Evidently, if c(x) = 1 then x = 0. If 1 = T (x; y)6 ∧(x; y), then x = y = 1. (⇐) If x = 0, then I (x; y) = ∨(c(0); T (0; y)) = ∨(1; T (0; y)) = 1. If x = y = 1, then I (1; 1) = 1 because by item (i) of Proposition 2 we know that all QL-implication satisfy the axioms: I2; I6 and I9, therefore, by I6 we have I (1; 1) = 1. Theorem 10. Let I be a QL-implication associated to a t-conorm S, to a strong negation c and a continuous t-norm T such that T (x; c(x)) = 0 for all x ∈[0; 1]. Then there exists an automorphism ’ of the unit interval such that I (x; y) = S(c(x); ’−1 (∨(’(x) + ’(y) − 1; 0))) and c(x)6’−1 (1 − ’(x)): Proof. We only need to recall Theorem 2. Corollary 6. In the conditions of the theorem above, the property: I (x; c(x)) = c(x) for all x ∈ [0; 1] holds. Proof. By item (ii) of Proposition 2 we know that I (x; y)¿c(x) for all x; y ∈[0; 1], therefore I (x; c(x))¿c(x). Moreover I (x; c(x)) = S(c(x); ’−1 (∨(’(x) + ’(c(x)) − 1; 0))) with c(x)6’−1 (1 − ’(x)), by the monotony of S we have I (x; c(x)) = S(c(x); ’−1 (∨(’(x) + ’(c(x)) − 1; 0)))6S(c(x); ’−1 (∨(’(x) + 1 − ’(x) − 1; 0))) = S(c(x); 0) = c(x). Therefore I (x; c(x)) = c(x). Remark. Corollary 6 makes clear that the reciprocal of item (i) of Theorem 9 does not always hold. The conditions in which the reciprocal holds are given in the following theorem.
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Theorem 11. Let T be the continuous t-norm such that T (x; c(x)) = 0 for all x ∈[0; 1], generated by the automorphism ’ according to Theorem 2, and let c be the strong negation generated by that automorphism, that is, c(x) = ’−1 (1 − ’(x)). A function I : [0; 1]2 → [0; 1] is such that: (i) (ii) (iii) (iv)
I (x; c(x)) = c(x) for all x ∈[0; 1], Ful$lls the axioms: I2 and I9, If 2’(c(x))¿’(y), then I (x; y) = c(x), If 2’(c(x))6’(y), then I (x; y) = T (x; y),
if and only if I (x; y) = ∨(c(x); T (x; y)). Proof. (⇒) Since 06y by I2 and I9 we have c(x) = I (x; 0)6I (x; y) for all x; y ∈[0; 1]. Two things can occur: (a) y6c(x), then by I2 and item (i) c(x) = I (x; c(x))¿I (x; y), therefore I (x; y) = c(x). Besides since y6c(x) we have ’(y)6’(c(x))62’(c(x)), then ’(c(x))¿’(y) − ’(c(x)), therefore we have T (x; y) = ’−1 (∨(’(x)+’(y)−1; 0)) = ’−1 (∨(’(y)−(1−’(x)); 0)), since c(x) = ’−1 (1−’(x)), we have ’(c(x)) = 1−’(x), therefore T (x; y) = ’−1 (∨(’(y) − ’(c(x)); 0))6’−1 (’(c(x))) = c(x), then T (x; y)6c(x), therefore I (x; y) = c(x) = ∨(c(x); T (x; y)). (b) y¿c(x), by I2 we have I (x; y)¿I (x; c(x)) = c(x). Since y¿c(x), then ’(y)¿’(c(x)). Two things can occur: (b1) ’(c(x))6’(y)62’(c(x)), therefore ’(y)−’(c(x))6’(c(x)), then T (x; y) = ’−1 (∨(’(y)−’(c(x)); 0)) 6’−1 (∨(’(c(x)); 0)) = ’−1 (’(c(x))) = c(x), then T (x; y)6c(x). By item (iii) we have I (x; y) = c(x) = ∨ (c(x); T (x; y)). (b2) ’(c(x))62’(c(x))6’(y), then ’(c(x))6’(y)−’(c(x)) therefore T (x; y) = ’−1 (∨(’(y)−’(c(x)); 0)) ¿’−1 (∨(’(c(x)); 0)) = c(x). By item (iv) we have I (x; y) = T (x; y) = ∨(c(x); T (x; y)). (⇐) By hypothesis I (x; y) = ∨(c(x); T (x; y)). I (x; 0) = ∨(c(x); T (x; 0)) = c(x), then I9 holds. By Ovchinnikov and Roubens’ Theorem 2 we know that the t-norm generated by the automorphism ’ satisEes T (x; c(x)) = 0, therefore I (x; c(x)) = ∨(c(x); T (x; c(x)) = ∨(c(x); 0) = c(x)
for all x ∈ [0; 1]:
Evidently, I (x; y) = ∨ (c(x); T (x; y)) fulElls I2, we only need to keep in mind the monotony of T and ∨. If 2’(c(x))¿’(y), then ’(c(x))¿’(y) − ’(c(x)) therefore T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)) = −1 ’ (∨(’(y)−(1−’(x)); 0)), since c(x) = ’−1 (1−’(x)), we have ’(c(x)) = 1−’(x), therefore T (x; y) = ’−1 (∨(’(y)−’(c(x)); 0))6’−1 (∨(’(c(x)); 0)) = c(x), that is, c(x)¿T (x; y), then I (x; y) = ∨(c(x); T (x; y)) = c(x). If 2’(c(x))6’(y), then ’(c(x))6’(y) − ’(c(x)), therefore T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)) = ’−1 (∨(’(y) − ’(c(x)); 0))¿’−1 (∨(’(c(x)); 0)) = c(x), that is, c(x)6T (x; y), then I (x; y) = ∨(c(x); T (x; y)) = T (x; y). Remark. (a) Evidently the functions I generated by the theorem above do not fulEll I8, we only need to recall Corollary 1. (b) If in the conditions of the theorem above we take the identity automorphism, (evidently c(x) = 1 − x), we have I (x; y) = ∨(1 − x; x + y − 1), (see Fig. 2(a)). This statement does not fulEll, among other properties, I1 and I4 therefore it is not an implication. If we take ’(x) = x2 for all x ∈[0; 1], then we have I (x; y) = ∨((1 − x2 )1=2 ; (∨(x2 + y2 − 1; 0))1=2 ) and c(x) = (1 − x2 )1=2 , (see Fig. 2(b)). In both cases I2, I3, I5, I6 and I9 stand.
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Fig. 2.
Denition 3 (Dombi [6] and Fodor and Roubens [15]). A De Morgan triple (T; S; c) is called a strong (or Lukasiewicz) De Morgan triple if there exists an automorphism ’ of the unit interval such that T (x; y) = ’−1 (∨(’(x) + ’(y) − 1; 0)); S(x; y) = ’−1 (∧(’(x) + ’(y); 1)); c(x) = ’−1 (1 − ’(x)): Corollary 7. In the conditions of Theorem 10, if (T; S; c) is a strong De Morgan triple, then I (x; y) = ’−1 (∨(1 − ’(x); ’(y))): Proof. I (x; y) = S(’−1 (1 − ’(x)); ’−1 (∨(’(x) + ’(y) − 1; 0))) = ’−1 (∧(1 − ’(x) + ∨(’(x) + ’(y) − 1; 0); 1)) = ’−1 (∧(∨(1 − ’(x); ’(y)); 1)) = ’−1 (∨(1 − ’(x); ’(y))): Remark. If in Corollary 7 we take ’(x) = x for all x ∈[0; 1], we have Kleene–Dienes’ implication operator. Theorem 12. Let I be a QL-implication associated to a t-conorm S, to a strong negation c and a continuous and strict t-norm T. Then there exists an automorphism ’ of the unit interval such that I (x; y) = S(c(x); ’−1 (’(x) · ’(y))): Proof. We only need to recall Theorem 3.
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The QL-implications obtained from the theorem above do not satisfy I8, nor the property I (x; c(x)) = c(x) for all x ∈[0; 1]. The following theorem establishes the conditions in which this property is satisEed. Theorem 13. Let T be the continuous, strict t-norm generated by the automorphism ’ according to Theorem 3 and let c be the strong negation generated by this automorphism, that is, c(x) = ’−1 (1 − ’(x)). A function I : [0; 1]2 → [0; 1] is such that: (i) (ii) (iii) (iv)
I (x; c(x)) = c(x) for all x ∈[0; 1], Ful$lls the axioms I2 and I9, If ’(x) + ’(x)’(y)¿1, then I (x; y) = T (x; y), If ’(x) + ’(x)’(y)61, then I (x; y) = c(x),
if and only if I (x; y) = ∨(c(x); T (x; y)). Proof. (⇒) Since 06y by I2 and I9 we have c(x) = I (x; 0)6I (x; y) for all x; y ∈[0; 1]. Two things can occur: (a) If y6c(x), then by I2 I (x; y)6I (x; c(x)) = c(x), therefore I (x; y) = c(x). Besides T (x; y)6T (x; c(x))6 ∧ (x; c(x))6c(x). Therefore I (x; y) = c(x) = ∨(c(x); T (x; y)). (b) If y¿c(x), then ’(y)¿’(c(x)) = 1 − ’(x), therefore ’(x) + ’(y)¿1. Two things can occur: (b1) ’(x) + ’(y)¿’(x) + ’(x)’(y)¿1, by item (iii) we have I (x; y) = T (x; y). Moreover, ’(x)’(y)¿1 − ’(x), therefore T (x; y) = ’−1 (’(x)’(y))¿’−1 (1 − ’(x)) = c(x), then I (x; y) = T (x; y) = ∨(c(x); T (x; y)). (b2) ’(x) + ’(y)¿1¿’(x) + ’(x)’(y), by item (iv) we have I (x; y) = c(x), besides, ’(x)’(y)61 − ’(x), therefore T (x; y) = ’−1 (’(x)’(y))6’−1 (1 − ’(x)) = c(x), then I (x; y) = c(x) = ∨(c(x); T (x; y)). (⇐) I (x; 0) = ∨(c(x); T (x; 0)) = ∨(c(x); 0) = 0, thus I9 holds. By the monotony of ∨ and of T satisEes I2. Since T (x; c(x))6 ∧ (x; c(x))6c(x), we have I (x; c(x)) = ∨(c(x); T (x; c(x))) = c(x) for all x ∈[0; 1]. If ’(x)+’(x)’(y)¿1, then ’(x)’(y)¿1−’(x), therefore T (x; y) = ’−1 (’(x)’(y))¿’−1 (1−’(x)) = c(x), then I (x; y) = ∨(c(x); T (x; y)) = T (x; y). If ’(x)+’(x)’(y)61, then ’(x)’(y)61−’(x), therefore T (x; y) = ’−1 (’(x)’(y))6’−1 (1−’(x)) = c(x), then I (x; y) = ∨(c(x); T (x; y)) = c(x) therefore item (iv) holds. Remark. If in the conditions of the theorem above we take the identity automorphism, we have I (x; y) = ∨(1− x; xy) and c(x) = 1−x, (see Fig. 3(a)), statement that does not satisfy I1 and I4, among other axioms. Evidently it does not satisfy I8 either. If we take ’(x) = x2 for all x ∈[0; 1] we have I (x; y) = ∨((1 − x2 )1=2 ; xy) and c(x) = (1 − x2 )1=2 , (see Fig. 3(b)). In both cases I2, I3, I5, I6 and I9 stand. Denition 4 (Dombi [6] and Fodor and Roubens [15]). A De Morgan triple (T; S; c) is said to be a strict (or productlike) De Morgan triple if there exists an automorphism ’ of the unit interval such that: T (x; y) = ’−1 (’(x)’(y)); S(x; y) = ’−1 (’(x) + ’(y) − ’(x)’(y)); c(x) = ’−1 (1 − ’(x)): Corollary 8. In the conditions of Theorem 12, if (T; S; c) is a strict De Morgan triple, then the following items hold: (i) I (x; y) = ’−1 (1 − ’(x) + ’(x)’(x)’(y)). (ii) I (x; y) = 1 if and only if x = 0 or x = y = 1.
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Fig. 3.
Proof. (i) I (x; y) = S(c(x); T (x; y)) = S(’−1 (1 − ’(x)); ’−1 (’(x)’(y))) = ’−1 (1 − ’(x) + ’(x)’(y) − (1 − ’(x))(’(x)’(y))) = ’−1 (1 − ’(x) + ’(x)’(y) − ’(x)’(y) + ’(x))’(x)’(y)) = ’−1 (1 − ’(x) + ’(x)’(x)’(y)): (ii) Evident. Remark. (a) Evidently the I obtained in Corollary 8 does not fulEll I (x; c(x)) = c(x) for all x ∈[0; 1], nor axiom I8. (b) If in the corollary above we take ’(x) = x for all x ∈[0; 1], we have Klir and Yuan’s function (1994), I (x; y) = 1 − x + x2 y, that fulElls the axioms: I2, I3, I5, I6 and I9, with c(x) = 1 − x for all x ∈[0; 1]. 4.3. R-implications and the I (x; c(x)) = c(x) property The following result appears frequently in fuzzy literature [10,15,24]. Let I be an R-implication associated to a t-norm T. Then the following property holds. If x 6 y; then I (x; y) = 1: Corollary 9. Let I be an R-implication associated to a t-norm T, then I (x; c(x)) = c(x) for all x ∈[0; 1] does not hold.
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Proof. Since c is a strong negation, then there exists a single equilibrium point e ∈(0; 1) such that c(e) = e, therefore I (e; c(e)) = I (e; e). Keeping in mind the result indicated above we know that if x6y, then I (x; y) = 1, therefore I (e; c(e)) = I (e; e) = 1 = c(e). Remark. Keeping in mind the result indicated above, we know that all R-implications fulEll I11 and in addition they fulEll one of the implications of I8. In this paper we do not analyze the behavior of I8 in the R-implications, as it is a problem that has already been studied in fuzzy literature [10,15,24] (for example in the R-implication characterization theorems based on left-continuous t-norms I8 is always demanded). In this sense we quote Klement and Navara [24]: A reasonable way of constructing connectives in fuzzy logics is to start with left-continuous t-norm T and to use the residual implication de$ned by I (x; y) = sup{z | T (x; z) 6 y} as the interpretation of the implication. It is immediate that we have I8, that is, I (x; y) = 1 if and only if x6y. Therefore said implications do not satisfy the property: I (x; c(x)) = c(x) for all x ∈[0; 1].
5. Conclusions and future research The most important results of this paper can be summarized in four points. (It is necessary to keep in mind that in the entire paper we have only used strong negations.) (1) There are di1erent interdependencies among the axioms demanded of the implication operators. Moreover, the property (1) and axiom I8 cannot be fulElled simultaneously. (2) Theorem 6 has made clear that I (x; y) = ∨(c(x); x) are the only S-implications that fulEll (1). The study of axiom I8 in S-implications has led us to Theorem 7, which is similar to Smets and Magrez’s theorem with the di1erence that in Theorem 7 we can use di1erent automorphisms in order to generate the implications and strong negations. In Corollary 4 we have proven that the only implications generated from Theorem 7 that fulEll I8 are the ones obtained by Smets and Magrez. In addition, since axiom I11 is a consequence of axiom I8, and we impose axiom I11 in Theorem 7, we have also wanted to study the case of implications generated from automorphisms without imposing I11, this has led us to Theorem 8. By means of this theorem we can generate S-implications from automorphisms such that these implications do not fulEll (1) nor I8. Evidently, there are S-implications, which are not generated by automorphisms, that do not fulEll the two properties above either. (3) For the QL-implications we have proven that if the t-conorm used is S = ∨, then (1) holds, however Theorem 10 has made clear that the reciprocal is not true. In the next theorem, we have presented the conditions in which we can generate QL-implications that fulEll (1) from automorphisms. In the theorems that followed we have analyzed the behavior (with respect to (1) and I8), of the QL-implications obtained from automorphisms when we demand that the corresponding t-norm be strict. (4) Regarding the R-implications we have proven that these never fulEll property (1). In the near future, among other things, we plan to analyze property (1) when the negation considered is not a strong negation. As future research we can point out the study of the behavior of De Baets and Kerre’s inclusion grade indicator, (A; B) = Inf x∈X (I (A (x); B (x))), when the implication operators used fulEll (1) or the axioms in the theorems studied in this paper. With respect to this point we can advance some of the results (which are direct consequence of the studies carried out in this paper). For example, (i) If I satisEes I8, then: (A; B) = 1 if and only if A6B.
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(ii) If: I (x; y) = 0 if and only if x = 1 and y = 0 (property fulElled by S-implications and QL-implications, as we have proven in Proposition 2), then: (A; B) = 0 if and only if A = {x; 1 | x ∈X } and B = {x; 0 | x ∈X }, obtaining a stronger result than the one demanded by Sinha and Dougherty. (iii) If I satisEes (1), then if A is a reEnement of B, that is, if A (x)6B (x) when B (x)6c(e) = e ∈(0; 1), and A (x)¿B (x), when B (x)¿c(e) = e ∈(0; 1), then (A ∨ c(A); A ∧ c(A)) 6 ((B ∨ c(B); B ∧ c(B))); that is, if I satisEes (1), then De Baets and Kerre’s inclusion grade indicator satisEes one of the axioms demanded by Sinha and Dougherty, when using the inclusion grade indicator in Approximate Reasoning. In general, our aim for the near future is to carry out a study of the application of the results of this paper to De Baets and Kerre’s inclusion grade indicators, and, in accordance with the selection of implication operators we make, see what Sinha and Dougherty’s axioms these indicators fulEll. Acknowledgements We thank the referees for their helpful comments. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24]
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