De Morgan Systems on the Unit Interval

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1 De Morgan Systems on the Unit Interval Mai Gehrke y, Carol Walker, and Elbert Walker Department of Mathematical Sciences New Mexico State University Las Cruces, NM mgehrke, hardy, Abstract Logical connectives on fuzzy sets arise from those on the unit interval. The basic theory of these connectives is cast in an algebraic spirit with an emphasis on equivalence between the various systems that arise. Special attention is given to De Morgan systems with strict Archimedean t-norms and strong negations. A typical result is that any De Morgan system with strict t-norm and strong negation is isomorphic to one whose t-norm is multiplication. 1 Introduction A fuzzy subset A of a set S is a mapping A : S! [0; 1]. Operations on the set F(S) of all such fuzzy subsets of S come from operations on [0; 1]. Standard ones are ^; _; and 0 given by (A ^ B)(s) = minfa(s); B(s)g (A _ B)(s) = maxfa(s); B(s)g A 0 (s) = 1 A(s) These (logical) connectives are usually referred to as and, or, and not, or intersection, union, and complement. The original theory of fuzzy sets was formulated using these operations, and they still play a fundamental role in both theory and applications. Viewing subsets of S as mappings S! f0; 1g, these operations clearly generalize the usual notions of intersection, union, and complement. But there are many other such generalizations and a huge literature dealing with them. International Journal of Intelligent Systems 11(1996) y Partially supported by NSF grant DMS and an NMSU Summer Research Award 1

2 There are extensive bibliographies in [4, 5, 7], for example. Our main concern will be with special classes of these connectives, particularly strict Archimedean t-norms and t-conorms, and strong negations. Such operators have generators. We give the notion of generators a new emphasis, and investigate systematically the isomorphisms between De Morgan systems on the unit interval. The unit interval Operations on fuzzy sets come from those on the unit interval [0; 1]. This interval is endowed with an order, and has many other mathematical features. But the system I = ([0; 1]; ) is the basic building block of the theory. Our concern will be with putting additional operations on it, namely t-norms and the like. But rst we need some discussion of automorphisms and anti-automorphisms of I. De nition 1 An automorphism of I is a one-to-one mapping f of I onto I such that f(a) f(b) if and only if a b. An anti-automorphism of I is a one-to-one mapping g of I onto I such that g(a) g(b) if and only if a b. Thus automorphisms preserve and anti-automorphisms reverse. It should be noted that f(0) = 0 and f(1) = 1 for automorphisms f and g(0) = 1 and g(1) = 0 for anti-automorphisms g. Since discontinuities of monotone functions are jumps, these automorphisms and anti-automorphisms are continuous. Of course, automorphisms are strictly increasing, and anti-automorphisms are strictly decreasing. These mappings are plentiful. Any continuous strictly increasing map connecting (0; 0) and (1; 1) in the plane is an automorphism of I. Figure 1 shows a couple of pictures. Figure 1. Automorphisms on the left, anti-automorphisms on the right Let M ap(i) be the set consisting of all automorphisms and all anti-automorphisms of I, and let Aut(I) be the set of all automorphisms of I. The elements of Map(I) are functions, and may be composed. That is, if f and g are in Map(I), fg is the element of Map(I) given by (fg)(x) = f(g(x)). With this operation, Map(I) is a

3 group. That is, composition of functions is a binary operation on M ap(i) that is associative, has an identity, and every element has an inverse. This means that f(gh) = (fg)h. There is an element 1 in Map(I) such that 1f = f1 = f for all f. (The function 1 is the function given by 1(x) = x for all x. It is called the identity of the group.) For each f Map(I), there is an element f 1 Map(I) such that ff 1 = f 1 f = 1. (The element f 1 is simply the inverse of f as a function on [0; 1].) Aut(I) is a subgroup of Map(I). This means that the restriction of the operation on Map(I) to Aut(I) makes Aut(I) into a group. This subgroup happens to be normal, that is, for every element f of Aut(I) and g of Map(I), the element g 1 fg belongs to the subgroup Aut(I). Elements of the form g 1 fg are conjugates of f. Normal subgroups allow one to form the quotient group M ap(i)=aut(i) whose elements are cosets f Aut(I) = ff g : g Aut(I)g and with the operation (f Aut(I)) (gaut(i)) = f gaut(i). The normality of Aut(I) makes this all work. The group Map(I)=Aut(I) has only two elements, Aut(I) and faut(i) for any f = Aut(I). There are a couple of other important subgroups. Each positive real number r gives an automorphism by r(x) = x r. Identifying r with this automorphism, the set R + of positive real numbers is a subgroup of Aut(I). For any subset S of Map(I) the set fx Map(I) : xs = sx for all s Sg is the centralizer Z(S) of S in Map(I) and is a subgroup of M ap(i). A particularly important anti-automorphism is, given by (x) = 1 x. Its centralizer consists of those f Map(I) such that f = f. In this particular case, we will only be interested in those f which are in Aut(I), that is, in the centralizer of in Aut(I), which is the group Z(fg) \ Aut(I) = ff Aut(I) : f = fg For ease of notation, we are going to denote this group by Z(), and more generally, for any g Map(I); Z(g) = ff Aut(I) : fg = gfg The group Z() consists exactly of those elements of Aut(I) which commute with, which is equivalent to or that f(x) = f(1 x) = f(x) = 1 f(x) f(x) + f(1 x) = 1: Thus automorphisms that are in Z() look like those in Figure. 3

4 Figure. Some elements of Z(). Elements of Z() are easy to construct. For any f Aut(I), the element f+f Z(). In fact, Theorem Z() = f f+f : f Aut(I)g. Proof. If f Z(), then f = f+f. For f Aut(I), f + f 1 f(1 x) + f(x) (x) = f(1 x) f(x) = (1 f(x)) + f(1 x) = f + f = (x) so f+f Z(). The map : Aut(I)! Z() : f! f+f xes Z() elementwise, but its group theoretical signi cance is not clear. For example, it is not a homomorphism: (f)(g) = (f(g)) 6= (fg). We will need the following later on. First, notice that for any f Map(I) and any subgroup G of Map(I), is a subgroup of Map(I). f 1 Gf = ff 1 gf : g Gg Proposition 3 For any f and g Map(I), f 1 R + f \ g 1 Z()g = f1g Proof. If f 1 rf = g 1 zg, then gf 1 r = zgf 1. There is x [0; 1] such that gf 1 (x) = 1. For this x, gf 1 r(x) = zgf 1 (x) = z 1 = 1, and so gf 1 (x r ) = 1. But gf 1 (x) = 1, and since gf 1 is one-to-one, r = 1 and the proposition follows. 4

5 3 t-norms We will put additional structure on the system I, and rst we consider t-norms. They are generalizations of intersection. De nition 4 A binary operation on [0; 1] is a t-norm if for all x; y; z [0; 1]; 1. 1 x = x. x y = y x 3. (x y) z = x (y z) 4. The operation is increasing in each variable. That is, x x 1 and y y 1 imply that x y x 1 y 1. Thus a binary operation on [0; 1] is a t-norm if 1 is an identity, it is commutative, associative, and increasing in each variable. A t-norm on [0; 1] gives a corresponding operation on fuzzy sets: (A B) (s) = A(s) B(s). Of course, the associative property 3 gives unambiguous meaning to x 1 x ::: x n, and in particular to x x ::: x, which we write as x n, where n is the number of x s. We have to be a little careful with this: x is a real number so for any real number r, x r has meaning as the r-th power of x. The context will make clear the meaning of x n : A t-norm has the following additional properties. 0 x = 0. This follows since 0 x 0 1 = 0. x y = 1 if and only if x = y = 1. Clearly 1 = 1 1. If 1 = x y, then and similarly 1 = x. 1 = x y 1 y = y Two often used t-norms are min and ordinary multiplication. Many examples are given in the references. Let be a t-norm and consider the system (I; ). This system is simply I with an additional structure on it, namely the operation. Let be another t-norm on I. We need to make precise the notion of the systems (I; ) and (I; ) being structurally the same. De nition 5 Let and be t-norms. The systems (I; ) and (I; ) are isomorphic if there is an element h Aut(I) such that h(x y) = h(x) h(y). We write (I; ) t (I; ). The mapping h is an isomorphism. The t-norms and are equivalent if (I; ) t (I; ). 5

6 This means that the systems ([0; 1]; ; ) and ([0; 1]; ; ) are isomorphic in the sense of universal algebra: there is a one-to-one map from [0; 1] onto [0; 1] that preserves the operations and relations involved. Equivalence between t-norms is an equivalence relation and partitions t-norms into equivalence classes. The t-norm min is rather special. A t-norm is idempotent if aa = a for all a [0; 1]. If is idempotent, then for a b, a = aa ab a1 = a, so = min. Thus min is the only idempotent t-norm. It is in an equivalence class all by itself. In this article, we will restrict our attention to those t-norms such that a a < a for a (0; 1). An isomorphism of a system with itself is called an automorphism. It is easy to show that the set of automorphisms of (I; ) is a subgroup of Aut(I). Thus, with each t-norm, there is a group associated with it, namely its automorphism group Aut(I; ) = ff Aut(I) : f(x y) = f(x) f(y)g For the t-norm a ^ b = minfa; bg, it is clear that Aut(I; ^) = Aut(I). For the t-norm multiplication, Aut(I; ) = ff Aut(I) : f(x) = x c ; c R + g ' R +. This is a wellknown, classical result related to one of the four basic functional equations of Cauchy (see, for example, Aczél [3]). If H is a subgroup of a group G, and g G, then we have already noted that g 1 Hg = fg 1 hg : h Hg is a subgroup of G. This subgroup is said to be conjugate to H, or a conjugate of H: The map h! g 1 hg is an isomorphism from H to its conjugate g 1 Hg. By an isomorphism from a group G to a group H we mean a one-to-one onto map ' : G! H such that '(xy) = '(x)'(y). Theorem 6 If two t-norms are equivalent then their automorphism groups are conjugate. Proof. Suppose that and are equivalent. Then there is an isomorphism f : (I; )! (I; ). The map g! f 1 gf is an isomorphism from Aut(I; ) to Aut(I; ), so f 1 Aut(I; )f = Aut(I; ). De nition 7 A t-norm is convex if whenever x y c x 1 y 1, then there is an r between x and x 1 and an s between y and y 1 such that c = r s. A t-norm is Archimedean if for each a; b (0; 1), there is a positive integer n such that n times a n z } { = a a a < b. For t-norms, the condition of convexity is equivalent to continuity. In this article, we refer to the condition as convex. This formulation has the advantage of being strictly order theoretic, allowing us to remain within the algebraic context of I as a lattice. For convex t-norms, the condition for Archimedean simpli es, as the following well-known proposition attests. 6

7 Proposition 8 The following are equivalent for a convex t-norm. 1. isarchimedean.. a a < a for all a (0; 1). The theorem below is fundamental in determining equivalences of convex Archimedean t-norms. It has usually been thought of as a theorem about representing t-norms by generators. A principle reference is [8]. The theorem in essence goes back at least to Abel [1]. There is a proof for the strict case in [], some discussion in [10], and a proof in [11]. We will give only a very brief outline of a proof here. Theorem 9 If is a convex Archimedean t-norm then there is an a [0; 1) and an isomorphism f : I! ([a; 1] ; ) such that f(x y) = max ff(x)f(y); ag for all x; y [0; 1]. Also if g : I! ([b; 1] ; ) is another such isomorphism, then g(x y) = max fg(x)g(y); bg if and only if f = rg for some r > 0. Proof. The proof consists of a construction of a map f satisfying f(xy) = f(x)f(y) for x y 6= 0. An increasing sequence fx n g 1 n= 1 in [0; 1) is de ned inductively by the condition x n x n = x n 1. The set of all points of the form x i1 x i x in is dense in the unit interval. De ne a function f on the sequence fx n : x n 6= 0g by f(x n ) = x n 0 if x n 6= 0 The function f can be extended to nite nonzero products under of the elements of this sequence by f (x i1 x i x in ) = f(x i1 )f(x i ) f(x in ): and then extended to an isomorphism from I to ([f(0); 1] ; ) by convexity. Then f(x y) = f(x)f(y) whenever x y 6= 0, so that x y = f 1 (f(x)f(y)) if f(x)f(y) f(0) and x y = 0 otherwise. Suppose that an isomorphism g : I! ([g(0); 1] ; ) gives the same t-norm as the f just constructed. Take r = (ln x 0 ) = (ln g (x 0 )). Then rg(x 0 ) = (g(x 0 )) r = x 0 = f(x 0 ), and it can be seen from the construction of f that rg must agree with f on the points x n and hence everywhere. Conversely, it is easy to show that if f = rg, then f and g give the same t-norm. A function f such that x y = f 1 (maxff(x)f(y); f(0)g) is called a generator of the t-norm. Two functions f and g are generators of the same t-norm if and only if f = rg for some r > 0, that is, f(x) = (g(x)) r for all x [0; 1]. 7

8 A t-norm is nilpotent if for a 6= 1, a n = 0 for some positive integer n, the n depending on a. It is clear from the theorem that is nilpotent if and only if f(0) > 0. So Archimedean t-norms fall naturally into two classes: nilpotent ones and those not nilpotent. Those not nilpotent are called strict. In the next section, we start by examining the strict ones. Historically, Archimedean t-norms have been represented by maps g : I! [0; 1], where g is a strictly decreasing function with 0 < g(0) 1 and g(1) = 0. In this case the binary operation satis es g(x g + y) = minfg(x) + g(y); g(0)g and since this minimum is in the range of g, x g + y = g 1 (minfg(x) + g(y); g(0)g) These two types of representations give the same t-norms. In particular, if g : I! [0; 1] is a continuous, strictly decreasing function, with 0 < g(0) 1 and g(1) = 0, let f(x) = e g(x), x f y = f 1 (maxff(x)f(y); f(0)g) and x g y = g 1 (minfg(x) + g(y); g(0)g). Then f : I! [f (0) ; 1] is an isomorphism and g + = f. (See [11], for example. ) In this paper, we use the multiplicative representation, since this allows us to remain within the context of the unit interval and to have a natural group structure on the set of generators. We restate the previous theorem for the strict t-norm case. Theorem 10 The Archimedean t-norm is strict if and only if there is an element f Aut(I) such that f(x y) = f(x)f(y). Another element g Aut(I) satis es this condition if and only if f = rg for some r > 0. So a generator of a strict t-norm is just an isomorphism from Aut(I; ) to Aut(I; ). Corollary 11 For any strict t-norm, Aut(I; ) t Aut(I; ). Corollary 1 For any two strict t-norms and, Aut(I; ) t Aut(I; ). Additional properties of strict t-norms are these: On (0; 1) the operation is strictly increasing in each variable. In fact, if is strict, then f(x) = y x is a one-to-one map of [0; 1] onto [0; y]. This follows from the convexity and strict monotonicity of. If is strict, then f(x) = x x is an automorphism and g(x) = (1 x) (1 x) is an anti-automorphism of I. Again, this follows from the convexity and strict monotonicity of. 8

9 Multiplication is a strict t-norm. Call two automorphisms f and g equivalent if they give the same strict t-norm, and write f s g. Then s is an equivalence relation and so induces a partition of the group Aut(I). The members of this partition are the right cosets fr + f : f Aut(I)g of R +. So the set of strict t-norms of I is in natural one-to-one correspondence with the right cosets in Aut(I) of the subgroup R +. Rephrasing, we have Corollary 13 For an automorphism f of I, let f be given by x f y = f 1 (f (x) f(y)). Then f! R + f is a one-to-one correspondence between the strict t-norms on [0; 1] and the right cosets of the subgroup R + in Aut(I). We know that for strict t-norms and ; the systems (I; ) and (I; ) are isomorphic. We spell out exactly what those isomorphisms are. Theorem 14 Let and be strict t-norms with generators f and g, respectively. Then h : (I; )! (I; ) is an isomorphism if and only if g 1 rf = h for some r > 0. That is, the set of isomorphisms from (I; ) to (I; ) is the set g 1 R + f = fg 1 rf : r R + g: Proof. An isomorphism h : (I; )! (I; ) gets an isomorphism (I; )! (I; ) which must be rf for some r R +. So h = g 1 rf. For any r, g 1 rf is an isomorphism. (I; ) h - (I; f@ g (I; ) Corollary 15 Let f be a generator of the strict t-norm. Then Aut(I; ) = f 1 R + f t R + : Proof. The set of automorphisms of (I; ) is f 1 R + f. It is a subgroup of Aut(I), and is isomorphic to R + via the mapping f 1 rf! r. Corollary 16 Aut(I; ) = R +. There are two basic facts about nilpotent (convex Archimedean) t-norms: any two are equivalent, and each has a trivial automorphism group. 9

10 Theorem 17 Let and be nilpotent t-norms with generators f and g respectively. Let r R + with g(0) = (f(0)) r : Then g 1 rf is the unique isomorphism from (I; ) to (I; ). Proof. We may take r = 1, so that f (0) = g(0): First note that g 1 f Aut(I). We need to show that g 1 f(a b) = g 1 f(a) g 1 f(b), that is, that g 1 ff 1 (maxff(a)f(b); f(0)g) = g 1 (maxfgg 1 f(a)gg 1 f(b); g(0)g) which is clear since f(0) = g(0). Suppose that ' : (I; )! (I; ) is an isomorphism. Then 'f 1 (maxff(a)f(b); f(0)g) = g 1 (maxf(g')(a)(g')(b); g'(0)g) Thus f 1 (maxff(a)f(b); f(0)g) = ' 1 g 1 (maxf(g')(a)(g')(b); g'(0)g) Since f and g' generate the same nilpotent t-norm and agree on 0; f = g' or ' = g 1 f as asserted. Corollary 18 If is a nilpotent t-norm, then Aut(I; ) = f1g. In the case of strict (convex Archimedean) t-norms we have that Aut(I; ) = f 1 R + f Aut(I). It turns out that these are the only convex t-norms with such automorphism groups. Proposition 19 Let be a convex t-norm. f Aut(I) if and only if is a strict t-norm. Then Aut(I; ) = f 1 R + f for some Proof. Suppose that Aut(I; ) = f 1 R + f and suppose that for some a (0; 1), a a = a. Then for any b (0; 1), there is an element g Aut(I; ) such that g(a) = b, namely g = f 1 rf where r = ln f (b) = ln f (a). Thus b b = g(a) g(a) = g(a a) = g(a) = b; so that is idempotent. But the only idempotent t-norm is min, and Aut(I; min) = Aut(I) 6= f 1 R + f. Thus a a < a for all a (0; 1), and is Archimedean. The t-norm is not nilpotent, by the preceding corollary, and thus it is strict. We cannot conclude that the function f in the proposition is a generator of. Comment 3 at the end of this article refers to example of convex t-norms di erent from multiplication for which Aut(I; ) = R +. In that case, f = 1 but 1 is clearly not a generator for. 10

11 4 Negations In this section, we will prove results analogous to those for t-norms in the previous sections. An element f Map(I) has order n if f n = 1, and n is the smallest such positive integer. If no such integer exists, the element has in nite order. All the elements of Aut(I) have in nite order except 1 which has order 1. All antiautomorphisms are either of order two or of in nite order. Anti-automorphisms of order are called involutions. We will only concern ourselves with what are usually called strong negations. We will call them simply negations. De nition 0 A negation (or involution) on I is an anti-automorphism of I of order two. Negations will generally be denoted by small Greek letters. Thus a negation is an order-reversing, one-to-one mapping of I onto I such that ((x)) = x, or equivalently such that = 1. We reserve the notation to denote the negation given by (x) = 1 x. It is a trivial fact that conjugates of negations are negations, but it is a little less trivial and a bit surprising that every negation is a conjugate of by an automorphism. Theorem 1 Let be a negation in Map(I). Let f(x) = 1 (x) + x Then f Aut(I), and = f 1 f. Furthermore, g 1 g = if and only if gf 1 Z(). Proof. Since f is the average of two automorphisms, it is an automorphism. and 1 ((x) + (x) f(x) = = 1 x + (x) 1 (x) + x f(x) = 1 = 1 + (x) x so we have = f 1 f. Now, g 1 g = f 1 f if and only if gf 1 fg 1 = if and only if gf 1 = gf 1 if and only if gf 1 Z(). 11

12 We note that since Z(f 1 f) = f 1 Z()f, the centralizer Z() of a negation = f 1 f is the group g + g f 1 Z()f = f 1 : g Aut(I) f An automorphism f such that = f 1 f is a generator of. So every involution has a generator, and we know when two elements of Aut(I) give the same involution. This theorem seems to be due to Trillas [1] who takes as generators functions from [0; 1] to [0; 1] [6]. A thrust of this paper is to use elements of Map(I) as generators of t-norms, t-conorms, and negations. This enables us to use the language of group theory and to involve only functions on [0; 1]. Also, we are automatically provided with natural operations between generators, being elements of the group M ap(i). Theorem Let be a negation and let f be a generator of. The map! Z()f is a one-to-one correspondence between the negations of I and the set of right cosets of the centralizer Z() of. Consider two systems (I; ) and (I; ) where and are negations. They are isomorphic if there is a map h Aut(I) with h((x)) = h(x), that is if h = h, or equivalently if = h 1 h. Let f and g be generators of and, respectively. If h is an isomorphism, then hf 1 f = g 1 gh which means that f 1 f = h 1 g 1 gh = (gh) 1 gh Therefore, f and gh generate the same negation, and so zf = gh for some z Z(). Thus h g 1 Z()f. It is easy to check that elements of g 1 Z()f are isomorphisms (I; )! (I; ). We have the following theorem. Theorem 3 Let and be negations with generators f and g, respectively. Then the set of isomorphisms from (I; ) to (I; ) is g 1 Z()f. In particular, g 1 f is an isomorphism from (I; ) to (I; ). Note that Z() plays a role for negations analogous to that of R + for strict t- norms. See Corollary 13. If we call two negations and equivalent if (I; ) t (I; ); then the previous theorem says in particular that any two negations are equivalent. We have the following special cases. Corollary 4 The set of isomorphisms from (I; ) to (I; ) is the right coset Z()f of Z(). In particular, the generator f of is an isomorphism from (I; ) to (I; ). Noting that f 1 Z()f = Z(), we have Corollary 5 Aut(I; ) = f 1 Z()f = Z(). In particular, Aut(I; ) = Z(). 1

13 Since z! f 1 zf is an isomorphism from Z() = Aut(I; ) to f 1 Z()f = Aut(I; ), we get Corollary 6 For any two negations and, Aut(I; ) t Aut(I; ). Of course this last corollary follows also because the two systems (I; ) and (I; ) are isomorphic. The upshot of all this is that furnishing I with any negation yields a system isomorphic to that gotten by furnishing I with the negation : x! 1 x. 5 De Morgan systems Let be a t-norm and a negation. Then de ned by x y = ((x) (y)) de nes a binary operation on [0; 1] called a t-conorm. It has the following characterizing properties. 0 x = x. x y = y x. (x y) z = x (y z). is increasing in each variable. If is convex and for x (0; 1), xx > x0 = x, the t-conorm is Archimedean. All binary operations satisfying these properties come from t-norms. If satis es these properties, then de ned by x y = ((x) (y)) using any negation is a t-norm, and x y = ((x) (y)). If a t-norm and t-conorm are related in this way by the negation, then (I; ; ; ) is a De Morgan system, and the t-norm and the t-conorm are said to be dual to one another via the negation. From now on, we are going to restrict ourselves to strict t-norms. Their duals are called strict t-conorms. Suppose that f is a generator of the strict t-norm, is a negation, and x y = ((x) (y)) = f 1 (f(x))(f(y)) = (f) 1 (f(x))(f(y)) Thus the anti-automorphism f and multiplication determine the strict t-conorm. In general, an anti-automorphism g of I is a cogenerator of a strict t-conorm if x y = g 1 (g(x)(g(y)). It should be clear that every t-conorm has a cogenerator, and it is easy to check that g and h are cogenerators of the same t-conorm if and only if g = rh for some r R +. 13

14 For notational reasons, we are going to adorn our operators with their generators. Thus a strict t-norm will be written f, meaning that f is a generator of. Ordinary multiplication is r for any r R +, but that will be denoted as usual by. A conorm with generator h will be denoted h. Finally, f denotes the negation with generator f. We write 1 simply as. So a De Morgan system looks like (I; f ; g ; h ). Being a De Morgan system implies however that h = fg. Now suppose that q : (I; f ; g ; h )! (I; u ; v ; w ) is an isomorphism. Then q Aut(I) and the following hold. q(x f y) = q(x) u q(y) q( g (x)) = v q(x) q(x h y) = q(x) w q(y) But since x h y = g ( g (x) f g (y)) and x w y = v ( v (x) u v (y)), if the rst two equations hold, then q(x h y) = q ( g ( g (x) f g (y))) = v q(( g (x) f g (y)) = v (q( g (x)) u q ( g (y))) = v ( v (q(x)) u ( v q(y))) = q(x) w q(y) Therefore to be an isomorphism, q need only be required to satisfy the rst two conditions. That is, isomorphisms from (I; f ; g ; h ) to (I; u ; v ; w ) are the same as isomorphisms from (I; f ; g ) to (I; u ; v ). We will also call these systems De Morgan systems. To determine the isomorphisms q from (I; f ; g ) to (I; u ; v ), we just note that such a q must be an isomorphism from (I; f ) to (I; u ) and from (I; g ) to (I; v ). Therefore, from Theorems 14 and 3, we get the following theorem. Theorem 7 The set of isomorphisms from (I; f ; g ) to (I; u ; v ) is the set u 1 R + f \ v 1 Z()g This intersection may be empty, of course. That is the case when the equation u 1 rf = v 1 zg has no solution for r > 0 and z Z(). A particular example of this is the case where f = g = u = 1, v = Z(), and v 1 = 1. Then r = v 1 z with r > 0 and z Z(). But then r 1 = v 1 z 1 = r 1 = 1 Thus r = 1, and so v = z. But v = Z(). So there are De Morgan systems (I; f ; g ) and (I; u ; v ) which are not isomorphic. When two De Morgan systems are isomorphic, the isomorphism is unique and the situation is this. 14

15 Theorem 8 (I; f ; g ) t (I; u ; v ) if and only if (I; u ; v ) = (I; fh ; gh ) for some h Aut(I), in which case h 1 is the only such isomorphism. In particular, (I; f ; g ) t (I; ; gf 1). Proof. It is easy to check that h 1 is an isomorphism from (I; f ; g ) to (I; fh ; gh ). If k is such an isomorphism, then k = u 1 rf = v 1 zg for some r R + and z Z(). Thus u = rfk 1 and v = zgk 1 and so (I; u ; v ) = (I; fk 1; gk 1). If k were distinct from h 1, then kh would be a non-trivial automorphism of (I; f ; g ). But by Proposition 3, and Theorem 7, this is impossible. One implication of this theorem, taking f = g, is that the theory of the De Morgan system (I; f ; f ) is the same as that of (I; ; ). More generally this holds for (I; f ; g ) and (I; ; gf 1). This suggests that in applications of De Morgan systems, one may as well take the strict t-norm to be ordinary multiplication. Corollary 9 Aut((I; f ; g )) = f1g. Corollary 30 (I; ; ) t (I; ; ) if and only if = r 1 r for some r R +. Taking = in this last Corollary, we see that (I; ; ) t (I; ; ) if and only if = r 1 r for some r R +. So De Morgan systems isomorphic to (I; ; ) are exactly those of the form (I; ; r ) with r R +. Negations of the form r 1 r are Yager negations [13]. Thus we can state Corollary 31 De Morgan systems (I; ; ) which are isomorphic to (I; ; ) are precisely those with a Yager negation. We close this section with the following remark. The system ([0; 1]; ^; _; 0 ), where ^, _, and 0 are max, min, and x 0 = 1 x forms a De Morgan algebra in the usual lattice theoretic sense. If we replace 0, which we have been denoting by, by any other involution, then the systems ([0; 1]; ^; _; 0 ) and ([0; 1]; ^; _; ) are isomorphic. Isomorphisms between these algebras are exactly the isomorphisms between (I; ) and (I; ). There are many and these are spelled out in Theorem 3. This suggests that in applications of De Morgan systems where ^ and _ are taken for the t-norm and t-conorm, respectively, the negation may as well be (x) = 1 x. 6 The non-uniqueness of negations in De Morgan systems We have noted that a De Morgan system (I; ; ; ) is determined by the system (I; ; ). Of course, it is also determined by the system (I; ; ). Is it determined by (I; ; )? Another way to put it is this. How unique is the negation in a De Morgan system? Suppose that and are two involutions, is a strict t-norm, and ((x) (y)) = ((x) (y)) 15

16 That is, they both give the same strict t-conorm. Then ((x) (y)) = (x) (y) = ((x)) ((y)) so that is an automorphism of (I; ). Let f be a generator of. Then automorphisms of (I; ) are of the form f 1 rf for r > 0. Thus = f 1 rf and = f 1 rf. Since is of order, so is f 1 f. Thus and = f 1 rf = f 1 r 1 f ff 1 = f f 1 r 1 f f 1 = r 1 ff 1 = ff 1 r = rff 1 r and so rff 1 r = ff 1. Let = ff 1. Then is an involution and rr =. On the other hand, if is an involution such that for some r > 0, rr =, then it is routine to check that for any t-norm f ; f 1 f and f 1 rf are negations which give the same t-conorm. Are there such involutions? Yes, of course, with r = 1. But when r = 1, = f 1 rf = 1 and =. Are there such involutions with r 6= 1? a For a positive real number a, let a (x) = e ln x. Then a is an involution satisfying r a r = a for all r > 0. So f 1 a f and f 1 a rf are negations which give the same t-conorm. But it is easy to see that a r = a. So f 1 r a rf = f 1 a f. We get the r following theorem. Theorem 3 Let be a strict t-norm with generator f, and let a and b be positive real numbers. Then the negations f 1 a f and f 1 b f give the same t-conorm. We look a moment at the case when the t-norm is multiplication. Let a (x) = a e ln s and consider the De Morgan system (I; ; a ). Suppose it is isomorphic to (I; ; ). By the previous theorem, = a r for some r R +. But the isomorphism must be r R + since it is an automorphism of (I; ). It is easy to check that a r = a and r that r a = ar. We sum up. Corollary 33 The De Morgan systems (I; ; a ) are all isomorphic: If (I; ; a ) t (I; ; ), then = r for some r R +. 7 Some comments 1. The negations a satisfy r a r = a for all r R +. There are no other negations with this property. [See (3) below.] However, Professor Richard Bagby at New Mexico State University has constructed a large family of negations such that 16

17 rr = for a xed r 6= 1. Thus for each one of these negations, the negations = f 1 f and = f 1 rf give the same t-conorm from the t-norm given by f. Constructing and somehow classifying all such seems not to have been done.. Let and be a strict t-norm and strict t-conorm with generator f and cogenerator g, respectively. When does there exist a negation such that and are dual with respect to? This means nding an involution such that (f 1 (f((x))f((y))) = g 1 (g(x)g(y)) This in turns means that f = rg for some r > 0. The existence of such a then is the same as (f 1 R + g) \ Inv(I) 6= ;, where Inv(I) is the set of involution of I. Consider the equation f = rg. If f and are given, then the dual conorm is determined, or if g and are given, then the dual norm is determined. However, if f and g are given, then the equation must be solved for r and, and there may be many or there may be no solutions. Of course, f and g are determined only up to left multiples of elements of R +. In the special case f = 1, it means for the anti-automorphism g, one must nd an r such that rg is an involution, as the statement (f 1 R + g) \ Inv(I) 6= ; above says. a ln x 3. The involutions a (x) = e all have the property that r a is an involution. In fact, as we have observed, r a = ra : Also a r = a, and a r b = a. This b means that the a and the elements of R + form a subgroup of Map(I) and R + is normal in that subgroup. The elements a are in the normalizer N(R + ) = fg Map(I) : g 1 rg R + for all r R + g of R +. Professor Fred Richman at Florida Atlantic University has determined N(R + ): There are automorphisms in N(R + ) besides those in R + ; but no negations besides the a that satisfy r = r 1. Associating elements f of N(R + ) \ Aut(I) with the t-norm f puts a group structure isomorphic to (N(R + ) \ Aut(I)) =R + on those t-norms coming from N(R + ) \ Aut(I). So this is a non-trivial group. These considerations will be the subject of a later article. 4. Viewing the a as cogenerators, they all give the same conorm since r a = ra. They give the conorm x y = x ln y ln xy which doesn t look commutative but is, and x x = p x. 5. It s easy to calculate a generator for a given negation. One generator for is +1. But a generator for the negation f 1 f is f and this formula gives the more complicated f 1 f+1. Of course they di er by a composition with an element of Z(). Getting a nice expression for the negation generated by an automorphism f may be impossible: it involves calculating f 1. 17

18 8 Bibliography References [1] Abel, N., Untersuchungen der Funktionen Zweier unabhängig veränderlichen Grössen x und y, welche Eigenschaft haben, dass f(z; f(x; y)) eine symmetrische Funktion von x; y und z ist, J. Reine Angew. Math. 1 (186), [] Aczél, J., Sur les opérations dé nies pour nombres réels, Bull. Soc. Math. France 76 (1949), [3] Aczél, J., Lectures on Functional Equations and Applications, Academic Press, New York, [4] Alsina, C., E. Trillas, and L. Valverde, On some logical connectives for fuzzy sets, 71-76, in Fuzzy Sets for Intelligent Systems, D. Dubois, H. Prade, and R. Yager, eds., Morgan Kaufmann, San Mateo, [5] Dubois, D., H. Prade, and R. Yager, eds., Fuzzy Sets for Intelligent Systems, Morgan Kaufmann, San Mateo, [6] Fodor, J., and M. Roubens, Fuzzy Preference Modelling and Multicriteria Decision Support, Kluwer, Dordrecht, [7] Klir, G. T. and T. A. Folger, Fuzzy Sets, Uncertainty, and Information, Prentice Hall, Englewood Cli s, [8] Ling, C. H., Representation of associative functions, Publ. Math. Debrecen 1 (1965), [9] Ovchinnikov, S., and M. Roubens, On strict preference relations, Fuzzy Sets and Systems, 43 (1991), [10] Schweizer, B. and Sklar, A., Associative functions and statistical triangle inequalities, Publ. Math. Debrecen 8 (1961), [11] Schweizer, B. and Sklar, A., Probabilistic Metric Spaces, North-Holland, Amsterdam, [1] Trillas, E., Sobre funciones de negación en la teoria de conjuntos difusos, Stochastica III-1 (1979), [13] Yager, R., On a general class of fuzzy connectives, Fuzzy Sets and Systems 4 (1980),

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