The constructions of non-continuous additive generators of t-conorms based on discrete additive generators of discrete t-conorms

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The constructions of non-continuous additive generators of t-conorms based on discrete additive generators of discrete t-conorms Peter Viceník Institute of Information Engineering, Automation, and Mathematics, Faculty of Chemical and Food Technology, Slovak University of Technology, Radlinského 9, 812 37 Bratislava, Slovakia Received 20 March 2018; received in revised form 12 November 2018; accepted 13 November 2018

Abstract The relation between discrete additive generators of discrete t-conorms (t-norms) and one class of non-continuous additive generators of t-conorms (t-norms) is analysed. The constructions of non-continuous additive generators of t-conorms (t-norms) based on discrete additive generators of discrete t-conorms (t-norms) are introduced. © 2018 Elsevier B.V. All rights reserved. Keywords: Additive generator; Aggregation function; Associativity; Discrete additive generator; Discrete t-conorm; Discrete t-norm; T-conorm; T-norm

1. Introduction The results we introduce in the paper build a bridge between the class of all discrete additive generators of discrete t-conorms (t-norms) and one class of non-continuous additive generators of t-conorms (t-norms). We show how to construct non-continuous additive generators of t-conorms (t-norms) starting from discrete additive generators of discrete t-conorms (t-norms) and vice versa. As we will see the constructions introduced in the paper are based on constructing of ranges of non-continuous additive generators of associative operations starting from ranges of discrete additive generators of discrete associative operations and vice versa. The main problem we deal with in the paper is the following one: Given a discrete strictly increasing (strictly decreasing) additive generator g : X → [0, ∞] of a discrete associative operation G : X 2 → X, where a set X ⊆ [0, 1] be finite and 0, 1 ∈ X, is it possible to construct a strictly increasing (strictly decreasing) additive generator f : [0, 1] → [0, ∞] of an associative operation F : [0, 1]2 → [0, 1] such that F /X 2 = G where F /X 2 : X 2 → X is the restriction of F to X? Although we have already known that for every discrete additively generated t-conorm (t-norm) G : X 2 → X where a set X ⊆ [0, 1] be finite and 0, 1 ∈ X, there exists an extension F : [0, 1]2 → [0, 1] of G E-mail address: [email protected]. https://doi.org/10.1016/j.fss.2018.11.009 0165-0114/© 2018 Elsevier B.V. All rights reserved.

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which is a t-conorm (t-norm) [2, Proposition 7.30 formulated for t-norms] we show in the paper that there exists an extension F of G which is even an additively generated t-conorm (t-norm). Works [3–5] are devoted to the study of discrete additive generators of discrete associative operations. Noncontinuous additive generators acting on the closed unit interval [0, 1] of associative operations are studied in works [7–10]. Note that different aspects of t-norms, t-conorms and related operations are studied in monographs [1] and [2]. Work [6] offers a treatment of discrete t-norms and t-conorms. 2. Additively generated operations Let us start by recalling some basic definitions and introducing some notations. The set of all non-negative integers is denoted by N ∪ {0}. The symbol ≤ denotes the standard linear order on R ∪ {−∞, ∞}. The same symbol also denotes the restriction of ≤ to X ⊆ R ∪ {−∞, ∞}. Let X, Y ⊆ R ∪ {−∞, ∞} be non-empty sets and Y ⊆ X. Suppose that f : X → R ∪ {−∞, ∞} is a function. The range of f is denoted by R(f ). The restriction of f from X to Y is denoted by f/Y , that is, f/Y : Y → R ∪ {−∞, ∞}, f/Y (x) = f (x) for all x ∈ Y . The standard inverse of f (if it exists) is denoted by f −1 . The symbol f −1 (Z) where Z ⊆ R(f ) denotes the set {x ∈ X | f (x) ∈ Z}. Suppose that F : X 2 → X is a binary operation on X. If F (x, y) ∈ Y for all x, y ∈ Y , then F can be restricted from X to Y and this restriction is denoted by F /Y 2 , that is, F /Y 2 : Y 2 → Y , F /Y 2 (x, y) = F (x, y) for all x, y ∈ Y . Definition 1. Let X, Y ⊆ R ∪ {−∞, ∞} be non-empty linearly ordered sets with the usual linear order ≤. (i) A binary operation  : X 2 → X is non-decreasing if x  y ≤ u  v for all x, y, u, v ∈ X, x ≤ u, y ≤ v. (ii) Non-decreasing binary operations  : X 2 → X and ⊗ : Y 2 → Y are isomorphic if there exists a strictly increasing bijection f : X → Y such that x  y = f −1 (f (x) ⊗ f (y)) for all x, y ∈ X where f −1 is the standard inverse of f . The function f is an isomorphism of  and ⊗. (iii) Suppose that there exists a minimum min X of X. A binary operation  : X 2 → X is a t-conorm if it is nondecreasing, commutative, associative and min X is its neutral element. The next definition covers the definition of a strictly increasing additive generator acting on [0, 1] [7, Definition 2] and the definition of a discrete strictly increasing additive generator acting on a non-empty finite set X ⊆ R ∪ {−∞, ∞}, in particular, acting on {0, 1, . . . , n}, n ∈ N [3, Definition 3]. Definition 2. Let X = [0, 1] or X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. A binary operation F : X 2 → X is additively generated if there exists a strictly increasing function f : X → [0, ∞] such that F (x, y) = f (−1) (f (x) + f (y)) where f (−1) : [0, ∞] → X, f (−1) (t) = sup{x ∈ X | f (x) ≤ t}, sup ∅ = min X. The function f (−1) is the pseudoinverse of f . We say that F is additively generated by f and that f is a strictly increasing additive generator (briefly, additive generator) of F . If X is finite we say that f and F are discrete. Let us denote by F the family of all strictly increasing functions f : X → [0, ∞] where either X = [0, 1] or X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. Remark 1. A binary operation F additively generated by a strictly increasing additive generator is always nondecreasing, commutative and max{x, y} ≤ F (x, y) for all x, y ∈ X. In general, F need not be associative and min X need not be a neutral element of F . Obviously, F is a t-conorm if and only if F is associative and min X is its neutral element. Example 1. (i) The function f : [0, 1] → [0, ∞], f (x) = x is an additive generator of the Łukasiewicz t-conorm F : [0, 1]2 → [0, 1], F (x, y) = min{x + y, 1}. (ii) Let X = { nl | l ∈ {0, 1, . . . , n}}, n ∈ N . The function f : X → [0, ∞], f ( nl ) = nl is a discrete additive generator of a discrete t-conorm F : X2 → X, F ( ni , jn ) = min{ i+j n , 1}.

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Remark 2. Let F be a discrete binary operation acting on a non-empty finite set X ⊆ R ∪ {−∞, ∞} additively generated by a discrete strictly increasing additive generator. (i) Then there exists a discrete strictly increasing additive generator f : X → [0, ∞] of F such that f (max X) < ∞. (ii) Suppose that min X is a neutral element of F . Then there exists a discrete strictly increasing additive generator f : X → [0, ∞] of F such that f (min X) = 0 and f (max X) < ∞. Let us recall that a binary operation F : [0, 1]2 → [0, 1] is border-continuous if it is continuous at each point of the border of the unit square [0, 1]2 . The next result gives a necessary and sufficient condition for an additively generated associative operation F : [0, 1]2 → [0, 1] to be a border-continuous t-conorm, and it follows from [8, Proposition 1]. Lemma 1. Let f : [0, 1] → [0, ∞] be a strictly increasing additive generator of an associative operation F : [0, 1]2 → [0, 1]. Then F is a border-continuous t-conorm if and only if limx→0+ f (x) = 0. Note that strictly decreasing additive generators of border-continuous t-norms were studied in details by Viceník [8]. 3. Lower additions We start by defining so called lower additions. As we will see in Section 5 additively generated operations are closely related to lower additions. Let us denote by L the family of all sets X ⊆ [0, ∞] such that there exists a minimum min X of X and sup{x ∈ X | x ≤ t} ∈ X for all t ∈ [min X, ∞]. Definition 3. Let X ∈ L. A binary operation ⊕ : X 2 → X is a lower addition on X if x ⊕ y = S(x + y) where S : [0, ∞] → X, S(t) = sup{x ∈ X | x ≤ t}, sup ∅ = min X. The function S is the lower function determined by X. Remark 3. The lower function S determined by X satisfies S(t) = max{x ∈ X | x ≤ t} for all t ∈ [min X, ∞]. It is non-decreasing on [0, ∞], S(x) ≤ x for all x ∈ [min X, ∞], and S(x) = x if and only if x ∈ X. The lower addition ⊕ on X satisfies x ⊕ y = max{z ∈ X | z ≤ x + y} for all x, y ∈ X. It is non-decreasing, commutative and max{x, y} ≤ x ⊕ y ≤ min{x + y, max X} ≤ x + y for all x, y ∈ X. Moreover, x ⊕ y = x + y if and only if x + y ∈ X. In general, ⊕ need not be associative and min X need not be a neutral element of ⊕. Both the lower function determined by X and the lower addition on X are uniquely determined by X. Example 2. The lower addition ⊕ on [0, ∞] coincides with the usual addition + on [0, ∞], that is, x ⊕ y = x + y for all x, y ∈ [0, ∞]. The lower addition ⊕ on [0, 1] is given by x ⊕ y = min{x + y, 1} for all x, y ∈ [0, 1]. The lower addition ⊕ on X = {0, 1, . . . , n} is given by x ⊕ y = min{x + y, n}. The next remark explains the relation between lower additions introduced in this paper and additions introduced by Viceník [7]. Remark 4. Recall that additions [7, Definition 6] were defined on sets X ∈ M where M denotes the family of all sets X ⊆ [0, ∞] such that there exists a strictly increasing function f : [0, 1] → [0, ∞] with the range R(f ) = X. We can state that the lower addition and the addition coincide on every set X ∈ M ∩ L. We conclude this section by two assertions which will be used in the paper. Proposition 1. Let X, Y ∈ L and X ⊆ Y . If the lower addition on Y can be restricted to X then this restriction coincides with the lower addition on X.

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Proof. Let ⊕X and ⊕Y be the lower additions on X and Y , respectively. Suppose that the lower addition on Y can be restricted to X and denote this restriction by ⊕Y /X 2 . Fix x, y ∈ X. Recall that x ⊕Y y = x(⊕Y /X 2 )y. On the one hand, because X ⊆ Y , it is x ⊕X y = max{z ∈ X | z ≤ x + y} ≤ max{z ∈ Y | z ≤ x + y} = x ⊕Y y. Thus, x ⊕X y ≤ x(⊕Y /X 2 )y. On the other hand, because x ⊕Y y ∈ X, and since x ⊕Y y ≤ x + y, it is x ⊕Y y ∈ {z ∈ X | z ≤ x + y} implying x ⊕Y y ≤ x ⊕X y by definition of ⊕X . Thus, x(⊕Y /X 2 )y ≤ x ⊕X y. 2 Corollary 1. Let X, Y, Z ∈ L and X ⊆ Y ⊆ Z. (i) If the lower addition on Z can be restricted to Y and to X, then the lower addition on Y can be restricted to X. (ii) If the lower addition on Z can be restricted to Y and the lower addition on Y can be restricted to X, then the lower addition on Z can be restricted to X. Proof. (i) If the lower addition ⊕ on Z can be restricted to Y and to X, then, obviously, ⊕/Y 2 can also be restricted to X and (⊕/Y 2 )/X 2 = ⊕/X 2 . The rest follows from the fact that ⊕/Y 2 is the lower addition on Y by Proposition 1. (ii) If the lower addition ⊕ on Z can be restricted to Y , and with respect to Proposition 1 this restriction ⊕/Y 2 is just the lower addition on Y , and if the lower addition ⊕/Y 2 on Y can be restricted to X, then, obviously, the lower addition on Z can also be restricted to X and (⊕/Y 2 )/X 2 = ⊕/X 2 . 2 Proposition 2. (i) Let X ∈ L and a ∈ X. Then X ∩ [a, ∞] ∈ L and the lower addition on X can be restricted to X ∩ [a, ∞] and this restriction coincides with the lower addition on X ∩ [a, ∞]. (ii) Let X ∈ L and c ∈ ]0, ∞[. Then cX = {cx | x ∈ X} ∈ L and the lower additions on X and cX are isomorphic. Proof. (i) First, we will prove that X ∩ [a, ∞] ∈ L. Since a ∈ X, it is obvious that there exists a minimum of X ∩ [a, ∞] and min(X ∩ [a, ∞]) = a. It remains to prove that sup{x ∈ X ∩ [a, ∞] | x ≤ t} ∈ X ∩ [a, ∞] for all t ∈ [a, ∞]. Fix t ∈ [a, ∞]. Since X ∈ L, it is sup{x ∈ X | x ≤ t} ∈ X, and since sup{x ∈ X | x ≤ t} = sup{x ∈ X ∩ [a, ∞] | x ≤ t}, it is sup{x ∈ X ∩ [a, ∞] | x ≤ t} ∈ X. Further, since a ∈ {x ∈ X ∩ [a, ∞] | x ≤ t}, it is sup{x ∈ X ∩ [a, ∞] | x ≤ t} ≥ a. Hence sup{x ∈ X ∩ [a, ∞] | x ≤ t} ∈ X ∩ [a, ∞]. Finally, we will prove that the lower addition ⊕ on X can be restricted to X ∩ [a, ∞] and this restriction coincides with the lower addition on X ∩ [a, ∞]. With respect to Proposition 1 it is sufficient to show that x ⊕ y ∈ X ∩ [a, ∞] for all x, y ∈ X ∩ [a, ∞]. Fix x, y ∈ X ∩ [a, ∞]. Since X ∈ L, it is x ⊕ y = sup{z ∈ X | z ≤ x + y} ∈ X. Further, since a ∈ {z ∈ X | z ≤ x + y}, it is x ⊕ y = sup{z ∈ X | z ≤ x + y} ≥ a. Hence, x ⊕ y ∈ X ∩ [a, ∞]. (ii) First, we will prove that cX = {cx | x ∈ X} ∈ L. Since there exists a minimum of X, it is obvious that there exists a minimum of cX and that min(cX) = c min X. It remains to prove that sup{y ∈ cX | y ≤ s} ∈ cX for all s ∈ [min(cX), ∞], which can be equivalently written as follows: sup{cx ∈ cX | cx ≤ ct} ∈ cX for all t ∈ [min X, ∞]. Fix t ∈ [min X, ∞]. Since X ∈ L, it is sup{x ∈ X | x ≤ t} ∈ X, and since sup{cx ∈ cX | cx ≤ ct} = c sup{x ∈ X | x ≤ t}, it is sup{cx ∈ cX | cx ≤ ct} ∈ cX. Finally, we will prove that the lower additions ⊕ on X and ⊕c on cX are isomorphic. The strictly increasing bijection f : X → cX, f (x) = cx is an isomorphism of ⊕ and ⊕c since c(x ⊕ y) = c sup{z ∈ X | z ≤ x + y} = sup{cz ∈ cX | cz ≤ cx + cy} = (cx) ⊕c (cy). 2 Remark 5. Let X ∈ L. Put Y = (X \ {min X}) ∪ {0}. (i) If the lower addition on X is associative then so is the lower addition on Y . (ii) If min X is a neutral element of the lower addition on X then the lower additions on X and Y are isomorphic. 4. Lower additions on intervals’ sets of the first kind Suppose that a set M ⊆ [0, ∞] is in the form of  M = {0} ∪ ( ]al , bl ]), L = {0, 1, . . . , n}, n ∈ N ∪ {0} l∈L

(1)

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or M = {0} ∪ (



]al , bl ]) ∪ {∞}, L = N ∪ {0}, lim al = ∞,

l∈L

l→∞

5

(2)

where al , bl ∈ [0, ∞], al < bl for all l ∈ L, and bl < al+1 for all l, l + 1 ∈ L. Suppose that for all i, j, k ∈ L, if (]ai , bi ] + ]aj , bj ]) ∩ ]ak , bk ] = ∅, then ak ≤ ai + aj and bk ≤ min{ai + bj , bi + aj }.

(3)

Definition 4. (i) A set M ⊆ [0, ∞] is given by a sequence of intervals {]al , bl ]}l∈L if it is in the form of (1) or (2). The set B = {bl | l ∈ L} is a gap-point set of M. (ii) A set M ⊆ [0, ∞] is an intervals’ set of the first kind if it is in the form of (1) and satisfies (3). (iii) A strictly increasing function f : [0, 1] → [0, ∞] is a strictly increasing additive generator of the first kind (briefly, additive generator of the first kind) if its range is an intervals’ set of the first kind. Every intervals’ set of the first kind is an element of L and every strictly increasing additive generator of the first kind is an element of F . The next remark explains the relation between a set given by a sequence of intervals and a set determined by a sequence of triples introduced by Viceník [8]. Remark 6. Let L = {0, 1, . . . , n}, n ∈ N ∪ {0} and let M ⊆ [0, ∞]. Write L + 1 = {l + 1 | l ∈ L}. Then M is given by a sequence of intervals {]al , bl ]}l∈L , a0 = 0 satisfying (3) if and only if M is determined by a sequence of triples {(ul , vl , vl )}l∈(L+1) [8, Definition 2] satisfying for all i, j, k ∈ L + 1, it is either uk ≤ ui + uj and vk ≤ min{ui + vj , vi + uj } or vi + vj ≤ uk [8, (C7) on page 1637]. In fact, this equivalence follows from definitions if we realize that the relation between al , bl and ul+1 , vl+1 (l ∈ L) is ul+1 = al and vl+1 = bl for all l ∈ L. The next lemma describes some specific properties of lower additions acting on sets given by sequences of intervals satisfying (3). Let M ⊆ [0, ∞] be given by a sequence of intervals {]al , bl ]}l∈L and satisfy (3). Define  al+1 if l + 1 ∈ L (4) dl = ∞ if l + 1 ∈ / L. Clearly, [bl , dl ] ∩ M = {bl } for all l ∈ L. Lemma 2. Let M be given by a sequence of intervals {]al , bl ]}l∈L and satisfy (3). Let B = {bl | l ∈ L}, and dl , l ∈ L be given by (4), and let ⊕ be the lower addition on M. Suppose that i, j ∈ L, x ∈]ai , bi ] and y ∈]aj , bj ]. Write bk = max{bl ∈ B | bl ≤ bi + bj }. Then (i) (ii) (iii) (iv)

ak ≤ ai + aj < x + y ≤ bi + bj ≤ dk and bk ≤ min{ai + bj , bi + aj }. x ⊕ y ∈ ]a k , bk ]. if x + y ∈ / l∈L ]al , bl ] then x ⊕ y = bi ⊕ bj = bk . if x = bi or y = bj then x ⊕ y = bi ⊕ bj = bk .

  Proof. Let i, j ∈ L, x ∈ ]ai , bi ], y ∈ aj , bj and bk = max{bl ∈ B | bl ≤ bi + bj }. (i) Clearly, ai + aj < x + y ≤ bi + bj . Recall that bk ≤ bi + bj . It is ak ≤ ai + aj and bk ≤ min{ai + bj , bi + aj }, since if it would be ak > ai + aj or bk > min{ai + bj , bi + aj }, then we would have (]ai , bi ] + ]aj , bj ]) ∩ ]ak , bk ] = ∅, and by (3), ak ≤ ai + aj and bk ≤ min{ai + bj , bi + aj }, which would be a contradiction. Recall that bi + bj < bk+1 by definition of bk whenever k + 1 ∈ L. It is bi + bj ≤ dk since if it would be bi + bj > dk then dk = ak+1 , k + 1 ∈ L, and so, (]ai , bi ] + ]aj , bj ]) ∩ ak+1 , bk+1 = ∅. By (3), it would be bk+1 ≤ min{ai + bj , bi + aj } ≤ bi + bj , which would be a contradiction.

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(ii) By (i), x + y ∈ ]ak , dk ], that is, x + y ∈ ]ak , bk ] or x + y ∈ [bk , dk ]. If x + y ∈ ]ak , bk ] then x ⊕ y = x + y. If x + y ∈ [bk , dk ]  then x ⊕ y = bk because of [bk , dk ] ∩ M = {bk }. Hence x ⊕ y ∈ ]ak , bk ]. (iii) Let x + y ∈ / l∈L ]al , bl ]. By (i), x + y ∈ ]ak , dk ], and since x + y ∈ / ]ak , bk ] by assumption, it is x + y ∈ ]bk , dk ]. By (i), bi + bj ∈ [bk , dk ]. Hence, x ⊕ y = bi ⊕ bj = bk because of [bk , dk ] ∩ M = {bk }. (vi) Suppose that x = bi or y = bj . Then x + y ∈ [min{ai + bj , bi + aj }, bi + bj ], and by (i), x + y ∈ [bk , dk ]. By (i), bi + bj ∈ [bk , dk ]. Hence, x ⊕ y = bi ⊕ bj = bk because of [bk , dk ] ∩ M = {bk }. 2 We will often work with the following three binary operations which are closely related to a set given by a sequence of intervals satisfying (3), and which are, as stated in the next proposition, pairwise isomorphic. Proposition 3. Let M be given by a sequence of intervals {]al , bl ]}l∈L and satisfy (3). Write A = {al | l ∈ L}, B = {bl | l ∈ L} and define binary operations A : A2 → A, ai A aj = max{al ∈ A | al ≤ ai + aj } and B : B 2 → B, bi B bj = max{bl ∈ B | bl ≤ bi + bj } and ⊗ : L2 → L, i ⊗ j = max{l ∈ L | bl ≤ bi + bj }. Then, the binary operations A , B and ⊗ are pairwise isomorphic. Proof. First we show that for all i, j, k ∈ L, ak ≤ ai + aj if and only if bk ≤ bi + bj . Suppose ak ≤ ai + aj for some i, j, k ∈ L. Then bk ≤ bi + bj since if it would be bi + bj < bk then (]ai , bi ] + ]aj , bj ])∩]ak , bk ] = ∅, and by (3), bk ≤ min{ai + bj , bi + aj } ≤ bi + bj , which would be a contradiction. Suppose bk ≤ bi + bj for some i, j, k ∈ L. Then ak ≤ ai + aj since if it would be ai + aj < ak then (]ai , bi ] + ]aj , bj ]) ∩ ]ak , bk ] = ∅, and by (3), ak ≤ ai + aj , which would be a contradiction. Now, we will prove that the strictly increasing bijection h : A → B, h(al ) = bl for all l ∈ L, is an isomorphism of A and B . From what has already been shown we obtain for all i, j, k ∈ L, ai A aj = ak if and only if bi B bj = bk by definition of A and B , or equivalently, ai A aj = h−1 (h(ai ) B h(aj )), which proves that h is an isomorphism of A and B . Finally, we will prove that the strictly increasing bijection f : L → B, f (l) = bl for all l ∈ L, is an isomorphism of ⊗ and B . Obviously, for all i, j, k ∈ L, i ⊗ j = k if and only if k = f −1 (f (i) B f (j )) by definitions of ⊗ and B , or equivalently, i ⊗ j = f −1 (f (i) B f (j )), which proves that f is an isomorphism of ⊗ and B . 2 Corollary 2. Let M be an intervals’ set of the first kind given by a sequence of intervals {]al , bl ]}l∈L . Then the lower additions on A = {al | l ∈ L} and B = {bl | l ∈ L} are isomorphic. Definition 5. Let M be given by a sequence of intervals {]al , bl ]}l∈L and satisfy (3). A binary operation ⊗ : L2 → L, i ⊗ j = max{l ∈ L | bl ≤ bi + bj } is the index operation corresponding to M. Remark 7. An index k of bk = max{bl | bl ≤ bi + bj } in Lemma 2 is just i ⊗ j where ⊗ is the index operation corresponding to M, that is, k = i ⊗ j . As we will see in Section 8 index operations play an important role in formal expressions of binary operations additively generated by additive generators of the first kind. 5. The relation between lower additions and additively generated operations Before we describe the relation between lower additions and additively generated operations we look at the relation between L and {R(f ) | f ∈ F}. Observe that [0, 1] ∪ [2, 3] ∈ L but [0, 1] ∪ [2, 3] ∈ / {R(f ) | f ∈ F}, and that [0, 1[ ∪ {2} ∈ / L but [0, 1[ ∪ {2} ∈ {R(f ) | f ∈ F}. It is clear that L contains ranges of all discrete additive generators but it does not contain ranges of all additive generators acting on the unit interval [0, 1], for instance [0, 1[ ∪ {2} ∈ / L. Lemma 3. Let [0, 1] be the domain of f ∈ F . Then, R(f ) ∈ L if and only if f is left-continuous. Proof. (⇒) The proof is by contradiction. Suppose that R(f ) ∈ L and that there exists p ∈ ]0, 1] such that limx→p− f (x) = l < f (p). Obviously, l ∈ / R(f ). Choose t ∈ [l, f (p)[. Then, t ∈ [min R(f ), ∞] and sup{x ∈ R(f ) | x ≤ t} = l ∈ / R(f ), which contradicts to R(f ) ∈ L.

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(⇐) The proof is by contradiction. Suppose that f is left-continuous and that there exists t ∈ [min R(f ), ∞] such that sup{x ∈ R(f ) | x ≤ t} = s ∈ / R(f ). Write p = sup{x ∈ [0, 1] | f (x) < s}. Obviously, p ∈]0, 1] and limx→p− f (x) = s < f (p), which contradicts to the left-continuity of f . 2 Theorem 1. Let X be the domain of f ∈ F and R(f ) ∈ L. Let F : X 2 → X be additively generated by f and let ⊕ : R(f )2 → R(f ) be the lower addition on R(f ). Then the strictly increasing bijection f : X → R(f ) is an isomorphism of F and ⊕, that is, F (x, y) = f −1 (f (x) ⊕ f (y)) for all x, y ∈ X. Proof. First, we will prove that f −1 (S(t)) = f (−1) (t) for all t ∈ [0, ∞] where S is the lower function determined by R(f ). If t ∈ [0, min R(f )[ then f −1 (S(t)) = min X = f (−1) (t) by definitions. Fix t ∈ [min R(f ), ∞]. Since R(f ) ∈ L it is S(t) = sup{x ∈ R(f ) | x ≤ t} ∈ R(f ), that is, there exists one and only one p ∈ X such that S(t) = f (p), and f −1 (S(t)) = p. It remains to prove that f (−1) (t) = p. On the one hand, S(t) ≤ t , and so f (p) ≤ t implying p ∈ {x ∈ X | f (x) ≤ t}, and so, p ≤ sup{x ∈ X | f (x) ≤ t} = f (−1) (t). On the other hand, for all x ∈ X, x > p it is f (x) > t since if it would be f (x) ≤ t for some x ∈ X, x > p we would have S(t) ≥ f (x) > f (p) which would contradict to S(t) = f (p). Thus, {x ∈ X | f (x) ≤ t} ⊆ [−∞, p], and so, f (−1) (t) = sup{x ∈ X | f (x) ≤ t} ≤ p. We have already proved f (−1) (t) = f −1 (S(t)). For all x, y ∈ X, it is F (x, y) = f (−1) (f (x) + f (y)) = f −1 (S(f (x) + f (y))) = f −1 (f (x) ⊕ f (y)) by definition of F and ⊕. 2 Corollary 3. Let X be the domain of f ∈ F with R(f ) ∈ L. If a binary operation G : X 2 → X satisfies G(x, y) = f −1 (f (x) ⊕ f (y)) for all x, y ∈ X where ⊕ is the lower addition on R(f ) then G is additively generated by f . Remark 8. Under the assumption X = [0, 1] Theorem 1 can also be derived from the result proved by Viceník [7, Proposition 3]. Example 3. Let X = [1, 2] ∪ ]4, 5]. The lower addition ⊕ : X 2 → X is given by  2 if x, y ∈ [1, 2] x ⊕y = 5 otherwise. It is easy to show that the lower addition ⊕ on X is associative. Therefore, every strictly increasing bijection f : [0, 1] → X is an additive generator of an associative operation F : [0, 1]2 → [0, 1], because the operations ⊕ and F are isomorphic by Theorem 1. For instance, f : [0, 1] → X, f (x) = 2x + 1 if x ∈ [0, 12 ], and f (x) = 2(x − 12 ) + 4 if   x ∈ 12 , 1 , is an additive generator of an associative operation F : [0, 1]2 → [0, 1], F (x, y) = 12 if x, y ∈ [0, 12 ], and F (x, y) = 1 otherwise. We conclude this section by two assertions which will be used in the paper. Proposition 4. Let a finite set X be the domain of functions f, g ∈ F . If the lower additions on R(f ) and R(g) are isomorphic then the strictly increasing bijections f : X → R(f ) and g : X → R(g) additively generate the same operation. Proof. By Theorem 1, for all x, y ∈ X, F : X2 → X additively generated by f satisfies F (x, y) = f −1 (f (x) ⊕R(f ) f (y)) where ⊕R(f ) is the lower addition on R(f ), and G : X 2 → X additively generated by g satisfies G(x, y) = g −1 (g(x) ⊕R(g) (y)) where ⊕R(g) is the lower addition on R(g). If the lower additions ⊕R(f ) and ⊕R(g) are isomorphic, then, because of the finiteness of R(f ) and R(g) which follows from the finiteness of X, there exists one and only one isomorphism h : R(f ) → R(g) of ⊕R(f ) and ⊕R(g) given by h(f (x)) = g(x) for all x ∈ X such that f (x) ⊕R(f ) f (y) = h−1 (h(f (x)) ⊕R(g) h(f (y))) for all x, y ∈ X. Hence, for all x, y ∈ X, F (x, y) = f −1 (f (x) ⊕R(f ) f (y))= f −1 (h−1 (h(f (x)) ⊕R(g) h(f (y))) = g −1 (g(x) ⊕R(g) g(y)) = G(x, y). 2 Proposition 5. Let X be the domain of f ∈ F and let R(f ) ∈ L. If the lower addition on R(f ) can be restricted to a non-empty finite set Z ⊆ R(f ) then the operation F additively generated by f can be restricted to Y = f −1 (Z) and this restriction F /Y 2 : Y 2 → Y is additively generated by f/Y .

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Proof. By Theorem 1, the strictly increasing bijection f : X → R(f ) is an isomorphism of the operation F : X 2 → X additively generated by f and the lower addition ⊕ : R(f )2 → R(f ) on R(f ) such that F (x, y) = f −1 (f (x) ⊕ f (y)) for all x, y ∈ X. If the lower addition on R(f ) can be restricted to a non-empty finite set Z ⊆ R(f ) then F can be restricted to Y = f −1 (Z) and the strictly increasing bijection f/Y : Y → Z is an isomorphism of F /Y 2 : Y 2 → Y and ⊕/Z 2 : Z 2 → Z such that F /Y 2 (x, y) = (f/Y )−1 (f/Y (x)(⊕/Z 2 )f/Y (y)) for all x, y ∈ Y . Clearly, Z ∈ L because Z ⊆ [0, ∞] is a non-empty finite set. By Proposition 1, ⊕/Z 2 coincides with the lower addition ⊕Z : Z 2 → Z on Z, that is, f/Y (x)(⊕/Z 2 )f/Y (y) = f/Y (x) ⊕Z f/Y (y) for all x, y ∈ Y . Hence, F /Y 2 (x, y) = (f/Y )−1 (f/Y (x) ⊕Z f/Y (y)) for all x, y ∈ Y . Clearly, Y is the domain of f/Y ∈ F and R(f/Y ) = Z ∈ L. Thus, by Corollary 3, F /Y 2 is additively generated by f/Y . 2 6. Restrictions of additive generators of the first kind to discrete additive generators We will show that every strictly increasing additive generator of the first kind of a t-conorm can be restricted to a discrete strictly increasing additive generator of a discrete t-conorm. Lemma 4. Let M be given by a sequence of intervals {]al , bl ]}l∈L and satisfy (3). Write B = {bl | l ∈ L} and define the binary operation B : B 2 → B, bi B bj = max{bl ∈ B | bl ≤ bi + bj }. Then, the lower addition on M can be restricted to B and this restriction coincides with the binary operation B . Moreover, the lower addition on M is associative if and only if the binary operation B on B is associative. Proof. From Lemma 2 (iv) it follows that the lower addition on M can be restricted to B and its restriction coincides with the binary operation B . Obviously, if the lower addition on M is associative then so is the operation B . Suppose that the operation B is associative. Since B coincides with the restriction of the lower addition ⊕ from M to B we can write (bi ⊕ bj ) ⊕ bk = bi ⊕ (bj ⊕ bk ) for all i, j, k ∈ L. We will prove that the lower addition ⊕ on M is associative, that is, (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) for all x, y, z ∈ M. The last equation is obviously true if x or y or z is 0 or ∞. Suppose that x ∈]ai , bi ], y ∈]aj , bj ], z ∈]ak , bk ] for some i, j, k ∈ L. Let S be the lower function determined by M. We will consider the following  three cases. Suppose that x + y, y + z ∈ l∈L ]al , bl ]. Obviously, x ⊕ y = x + y, and so (x ⊕ y) ⊕ z = S((x ⊕ y) + z) = S(x + y + z). Similarly, x ⊕ (y ⊕ z) = S(x + y + z). Suppose that x + y, y + z ∈ / l∈L ]al , bl ]. By Lemma 2 (iii), x ⊕ y = bi ⊕ bj = bm where bm = max{bl ∈ B | bl ≤ bi + bj }, and by Lemma 2 (iv), bm ⊕ z = bm ⊕ bk . Hence (x ⊕ y) ⊕ z = (bi ⊕ bj ) ⊕ bk . Similarly, x ⊕ (y ⊕ z) = bi ⊕ (bj ⊕ bk ).   Suppose that x + y ∈ l∈L ]al , bl ] and y + z ∈ / l∈L ]al , bl ]. As in the first case (x ⊕ y) ⊕ z = S(x + y + z), and as in the second case x ⊕ (y ⊕ z) = bi ⊕ (bj ⊕ bk ). Recall that (bi ⊕ bj ) ⊕ bk = bi ⊕ (bj ⊕ bk ). On the one hand x ≤ bi , y ≤ bj and z ≤ bk , and because of the monotonicity of ⊕, (x ⊕ y) ⊕ z ≤ (bi ⊕ bj ) ⊕ bk = x ⊕ (y ⊕ z). On the other hand, y ⊕ z = max{w ∈ M | w ≤ y + z} ≤ y + z, and because of the monotonicity of S, x ⊕ (y ⊕ z) = S(x + (y ⊕ z)) ≤ S(x + y + z) = (x ⊕ y) ⊕ z.

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Remark 9. Under the assumption a0 = 0 Lemma 4 can also be derived from the result proved by Viceník [8, Theorem 2].    Example 4. Let M = {0} ∪ ( l∈N∪{0} 3l − 1, 3l ) ∪ {∞} and B = {3l | l ∈ N ∪ {0}}. It is a matter of straightforward verification to show that M is given by a sequence of intervals {]3l − 1, 3l ]}l∈N ∪{0} and satisfies (3). The binary operation  : B 2 → B, 3i  3j = max{3l | 3l ≤ 3i + 3j } = max{3i , 3j } for all i, j ∈ N ∪ {0}, is associative, and by Lemma 4, so is the lower addition on M. Corollary 4. Let M be an intervals’ set of the first kind with the gap-point set B. Then, (i) the lower addition on M can be restricted to B and this restriction coincides with the lower addition on B. Moreover, the lower addition on M is associative if and only if the lower addition on B is associative.

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(ii) the lower addition on M can be restricted to B ∪ {0} and this restriction coincides with the lower addition on B ∪ {0}. Moreover, the lower addition on M is associative if and only if the lower addition on B ∪ {0} is associative. Proof. (i) Because B is non-empty and finite, B ∈ L, and B is the lower addition on B. The rest follows immediately from Lemma 4. (ii) Since the lower addition on M can be restricted to B by (i), and since 0 is a neutral element of the lower addition on M, the lower addition on M can even be restricted to B ∪ {0}. Because B ∪ {0} is non-empty and finite, B ∪ {0} ∈ L, and by Proposition 1, this restriction coincides with the lower addition on B ∪ {0}. Further, it is a matter of straightforward verification to show that the lower addition on B ∪ {0} is associative if and only if the lower addition on B is associative. The rest follows from the second part of (i). 2 Theorem 2. Let f : [0, 1] → [0, ∞] be a strictly increasing additive generator of the first kind of a t-conorm F and let B be a gap-point set of R(f ). Then (i) F can be restricted to X = f −1 (B), and this restriction F /X 2 : X 2 → X is a discrete associative operation additively generated by f/X. (ii) F can be restricted to X = f −1 (B ∪ {0}), and this restriction F /X 2 : X 2 → X is a discrete t-conorm additively generated by f/X. Proof. Assertion (i) follows from Corollary 4 (i) and Proposition 5. Assertion (ii) follows from Corollary 4 (ii) and Proposition 5. 2 Example 5. Let X = { nl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N . The strictly increasing function f : [0, 1] → [0, ∞],   l−1 − 1) for all x ∈ l−1 , l and l ∈ L \ {0} is an additive generator of the first kind f (0) = 0, f (x) = n(x − l−1 n ) + (3 n n of a t-conorm F . Further, f/X : X → [0, ∞], f/X(0) = 0, f/X( nl ) = 3l−1 for all l ∈ L \ {0}, is a discrete additive generator of a discrete t-conorm F /X 2 ( ni , jn ) = max{ ni , jn } for all i, j ∈ L by Theorem 2 (ii). 7. The constructions of non-continuous additive generators based on discrete additive generators We introduce several constructions of non-continuous strictly increasing additive generators of associative operations starting from discrete strictly increasing additive generators of discrete associative operations. All these constructions are based on constructing of an intervals’ set of the first kind starting from a non-empty finite set X ⊆ [0, ∞[ of certain properties. Suppose that a set X ⊆ [0, ∞[ is in the form of X = {xl ∈ [0, ∞[ | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where xl < xl+1 for all l, l + 1 ∈ L. Write  min P if P = ∅, δ(X) = ∞ if P = ∅, where P = {xl − xl−1 | l ∈ L \ {0}}, and  min Q if Q = ∅, ε(X) = ∞ if Q = ∅,

(5)

(6)

where Q = {|xi +xj −xk | | xi +xj −xk = 0, i, j, k ∈ L}. Clearly, δ(X) > 0 and ε(X) > 0. It is obvious that δ(X) = ∞ if and only if X = {x0 }, and ε(X) = ∞ if and only if X = {0}. Definition 6. A non-empty set X ⊆ [0, ∞] is anti-additive if x + y ∈ / X for all x, y ∈ X. Observe that if a non-empty set X ⊆ [0, ∞] is anti-additive then 0, ∞ ∈ / X, that is X ⊆ ]0, ∞[.

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Lemma 5. Let M be given by a sequence of intervals {]al , bl ]}l∈L , satisfy (3), and let B = {bl | l ∈ L} be its gap-point set. If B ⊆ ]0, ∞[ then B is anti-additive. Proof. The proof is by contradiction. Suppose that B ⊆ ]0, ∞[ and that bi + bj = bk for some i, j, k ∈ L. Then (]ai , bi ] + ]aj , bj ]) ∩ ]ak , bk ] = ∅, and by (3), bk ≤ min{ai + bj , bi + aj }. Since bi , bj < ∞, min{ai + bj , bi + aj } < bi + bj , and so, bk < bi + bj , which is a contradiction. 2 7.1. Extensions We describe an extension of discrete strictly increasing additive generators of discrete associative operations (t-conorms) to non-continuous strictly increasing additive generators acting on [0, 1] of associative operations (t-conorms). This extension is based on constructing of an intervals’ set of the first kind starting from an arbitrary non-empty finite set B ⊆ ]0, ∞[ which is anti-additive. Lemma 6. Let B = {bl ∈ ]0, ∞[ | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where bl < bl+1 for all l, l + 1 ∈ L. Let δ = δ(B) and ε = ε(B) be given by (5) and (6), respectively. Suppose that B is anti-additive. Then the set  M = {0} ∪ ( ]bl − dl , bl ]), l∈L

where {dl }l∈L is a non-decreasing sequence of positive real numbers such that dn < δ and dn ≤ ε, is an intervals’ set of the first kind with the gap-point set B. Proof. In order to show that M is an intervals’ set of the first kind with the gap-point set B we should prove the following: 0 ≤ b0 − d0 < b0 , bi−1 < bi − di < bi for all i ∈ L \ {0}, and that for all i, j, k ∈ L, if (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅, then bk − dk ≤ bi − di + bj − dj and bk ≤ min{bi − di + bj , bi + bj − dj }. From 0 < d0 ≤ ε ≤ |b0 + b0 − b0 | = b0 we obtain 0 ≤ b0 − d0 < b0 . For all i ∈ L \ {0}, from 0 < di < δ ≤ bi − bi−1 we obtain bi−1 < bi − di < bi . Suppose that (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅ for some i, j, k ∈ L for the rest of the proof. First, we will prove that bk < bi + bj and that i ≤ k and j ≤ k. Clearly, bi + bj = bk because B is anti-additive. If it would be bi + bj < bk , from dk ≤ ε ≤ |bi + bj − bk | = bk − bi − bj we would have bi + bj ≤ bk − dk , and consequently (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅, which would be a contradiction. If it would be k < i or k < j and without loss of generality suppose that it would be k < i, we would have bk ≤ bi−1 < bi − di ≤ bi − di + bj − dj , and consequently (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅, which would be a contradiction. Thus, bk < bi + bj and i ≤ k and j ≤ k. From ε ≤ |bi − bj − bk | = bi + bj − bk and di ≤ dk ≤ ε, dj ≤ dk ≤ ε we obtain bk ≤ bi + bj − ε ≤ min{bi − di + bj , bi + bj − dj }, and consequently bk − dk ≤ bi − di + bj − dj . 2  Corollary 5. Under the hypotheses of Lemma 6, for all d ∈ ]0, ε], d < δ, the set M = {0} ∪ ( l∈L ]bl − d, bl ]) is an intervals’ set of the first kind. Proof. Our assertion follows from Lemma 6 where d0 = · · · = dn = d, d ∈ ]0, ε], d < δ.

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Corollary 6. Let B ⊆ ]0, ∞[ be a non-empty finite set. Then the following assertions are equivalent: (i) There exists an intervals’ set M of the first kind with the gap-point set B such that the lower addition on M is associative. (ii) The set B is anti-additive and the lower addition on B is associative. Proof. (⇒) Suppose that (i) holds. Then, B is anti-additive by Lemma 5, and the lower addition on B is associative by Corollary 4 (i). (⇐) Suppose that (ii) holds. Since B is anti-additive, there exists an intervals’ set M of the first kind with the gap-point set B by Lemma 6, and since the lower addition on B is associative, the lower addition on M is also associative by Corollary 4 (i). 2

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Theorem 3. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. (i) If g : X → [0, ∞] is a discrete strictly increasing additive generator of a discrete associative operation G with an anti-additive set R(g) then there exists a strictly increasing additive generator f : [0, 1] → [0, ∞] of an associative operation F such that f/X = g and F /X2 = G. (ii) If g : X → [0, ∞], g(0) = 0 is a discrete strictly increasing additive generator of a discrete t-conorm G with an anti-additive set R(g) \ {0} then there exists a strictly increasing additive generator f : [0, 1] → [0, ∞] of the first kind of a t-conorm F such that f/X = g and F /X 2 = G. Proof. (i) Suppose that g : X → [0, ∞] is a discrete additive generator of a discrete associative operation G with an anti-additive set R(g). Write X = {xl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N where xl < xl+1 for all l, l + 1 ∈ L. First, we will describe the construction of f . As in Lemma 6 we construct an intervals’ set M = {0} ∪ ( l∈L ]bl − dl , bl ]) of the first kind with the gap-point set B = R(g) = {bl | bl = g(xl ), l ∈ L} where 0 < d0 ≤ · · · ≤ dn < δ(B) and dn ≤ ε(B), and then, we choose an arbitrary strictly increasing bijection f : [0, 1] → M ∩ [b0 , ∞]with dl (x − xl−1 ) + (bl − dl ) for all x ∈ xl−1 , xl and f (xl ) = bl for all l ∈ L, for instance, f (0) = b0 and f (x) = (xl −x l−1 ) l ∈ L \ {0}. Now, we will prove that F is an associative operation. Since the strictly increasing bijection g : X → B additively generates an associative operation G, by Theorem 1, the lower addition on B is associative, and by Corollary 4 (i), so is the lower addition on M. Further, the lower addition on M ∩ [b0 , ∞] is associative since, by Proposition 2 (i), it is a restriction of the lower addition from M to M ∩ [b0 , ∞]. Thus, by Theorem 1, the strictly increasing bijection f : [0, 1] → M ∩ [b0 , ∞] additively generates an associative operation F . Finally, we will show that f/X = g and F /X 2 = G. By construction of f it is obvious that f/X = g. From B ⊆ (M ∩ [b0 , ∞]) ⊆ M, and since the lower addition on M can be restricted to M ∩ [b0 , ∞] by Proposition 2 (i) and to B by Corollary 4 (i), it follows that the lower addition on M ∩ [b0 , ∞] can also be restricted to B by Corollary 1 (i). Thus, by Proposition 5, the operation F can be restricted to X = f −1 (B), and this restriction F /X 2 is additively generated by f/X, that is, F /X 2 = G. (ii) Suppose that g : X → [0, ∞], g(0) = 0 is a discrete additive generator of a t-conorm G with an anti-additive set R(g) \ {0}. Write X \ {0} = {xl | l ∈ L}, L = {0, . . . , n}, n ∈ N ∪ {0} where xl < xl+1 for all l, l + 1 ∈ L. First, we will describe the construction of f . As in Lemma 6 we construct an intervals’ set M = {0} ∪ ( l∈L ]bl − dl , bl ]) of the first kind with the gap point set B = R(g) \ {0} = {bl | bl = g(xl ), l ∈ L} where 0 < d0 ≤ · · · ≤ dn < δ(B) and dn ≤ ε(B), and then, we choose an arbitrary strictly increasing bijection f : [0, 1] → M dl (x − xl−1 ) + (bl − dl ) for all with f (0) = 0 and f (xl ) = bl for all l ∈ L, for instance, f (0) = 0 and f (x) = (xl −x l−1 )   x ∈ xl−1 , xl and l ∈ L where x−1 = 0. Now, we will prove that F is a t-conorm. Since the strictly increasing bijection g : X → B ∪ {0} additively generates an associative operation G, by Theorem 1, the lower addition on B ∪ {0} is associative, and by Corollary 4 (ii), so is the lower addition on M. Thus, by Theorem 1, the strictly increasing bijection f : [0, 1] → M additively generates an associative operation F . Moreover, F is a t-conorm since f (0) = 0. Finally, we will show that f/X = g and F /X 2 = G. By construction of f it is obvious that f/X = g. By Corollary 4 (ii), the lower addition on M can be restricted to B ∪ {0}. Thus, by Proposition 5, the operation F can be restricted to X = f −1 (B ∪ {0}), and this restriction F /X 2 is additively generated by f/X, that is, F /X 2 = G. 2 Example 6. Let X = { nl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N . The function g : X → [0, ∞], g( nl ) = 10l for all l ∈ L is a discrete additive generator of a discrete t-conorm G( ni , jn ) = max{ ni , jn } for all i, j ∈ L, with the anti-additive    set B = R(g) = {10l | l ∈ L}. Since δ(B) = 9 and ε(B) = 1, the set M = {0} ∪ ( l∈L 10l − 1, 10l ) is an intervals’ set The strictly increasing  the gap-point set B by Corollary 5.  bijection f : [0, 1] → {1} ∪   of the first kind with l − 1) for all x ∈ l−1 , l and l ∈ L \ {0} is an additive ) + (10 ( l∈L\{0} 10l − 1, 10l ), f (0) = 1, f (x) = n(x − l−1 n n n generator of an associative function F such that f/X = g and F /X 2 = G by the proof of Theorem 3 (i). Observe that the associative operation F constructed in the proof of Theorem 3 (i) is never a t-conorm because f (0) > 0 whereas the operation F constructed in the proof of Theorem 3 (ii) is always a t-conorm because f (0) = 0. In

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general, the resulting operation F need not be border-continuous. The construction which leads to a border-continuous t-conorm will be introduced in the next subsection. 7.2. Extensions and border-continuity We describe an extension of discrete strictly increasing additive generators of discrete t-conorms to strictly increasing additive generators of the first kind of border-continuous t-conorms. This extension is based on constructing of an intervals’ set of the first kind which contains a right-hand neighborhood of 0 starting from an arbitrary non-empty finite set B ⊆ ]0, ∞[ which is anti-additive and min B < δ(B). Lemma 7. Let B = {bl ∈ ]0, ∞[ | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where bl < bl+1 for all l, l + 1 ∈ L. Let δ = δ(B) and ε = ε(B) be given by (5) and (6), respectively. Suppose that B is anti-additive and b0 < δ. Then the set  M = [0, b0 ] ∪ ( ]bl − dl , bl ]), l∈L\{0}

where {dl }l∈L\{0} is a non-decreasing sequence of positive real numbers such that dl ≤ ε for all l ∈ L \ {0}, is an intervals’ set of the first kind with the gap-point set B and with the first interval [0, b0 ].  Proof. Put d0 = b0 for formal reasons. Then M = {0} ∪ ( l∈L ]bl − dl , bl ]). Clearly, the first interval of M is [0, b0 ]. In order to show that M is an intervals’ set of the first kind with the gap-point set B we should prove the following: bi−1 < bi − di < bi for all i ∈ L \ {0}, and that for all i, j, k ∈ L, if (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅, then bk − dk ≤ bi − di + bj − dj and bk ≤ min{bi − di + bj , bi + bj − dj }. For all i ∈ L \ {0}, from 0 < di ≤ ε ≤ |b0 + b0 − b0 | = b0 < δ ≤ bi − bi−1 we obtain bi−1 < bi − di < bi . Suppose that (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅ for some i, j, k ∈ L for the rest of the proof. First, we will prove that bk < bi + bj and that i ≤ k and j ≤ k. Clearly, bi + bj = bk because B is anti-additive. If it would be bi + bj < bk , we would have k = 0, and from dk ≤ ε ≤ |bi + bj − bk | = bk − bi − bj we would have bi + bj ≤ bk − dk , and consequently (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅, which would be a contradiction. If it would be k < i or k < j and without loss of generality suppose that it would be k < i, we would have bk ≤ bi−1 < bi − di ≤ bi − di + bj − dj , and consequently (]bi − di , bi ] + ]bj − dj , bj ]) ∩ ]bk − dk , bk ] = ∅, which would be a contradiction. Thus, bk < bi + bj and i ≤ k and j ≤ k. We will consider the following two cases. Suppose that i = 0 and j = 0. From ε ≤ |bi + bj − bk | = bi + bj − bk and di ≤ dk ≤ ε, dj ≤ dk ≤ ε we obtain bk ≤ bi + bj − ε ≤ min{bi − di + bj , bi + bj − dj }, and consequently, bk − dk ≤ bi − di + bj − dj . Suppose that i = 0 or j = 0. Without loss of generality suppose that j = 0. Recall that bk < bi + b0 and i ≤ k. First, we will show that i = k. If it would be i < k, from b0 < δ ≤ bi+1 − bi we would have bi + b0 < bi+1 ≤ bk , which would be a contradiction to bk < bi + b0 . We have already proved that i = k, and so, our assumption can be written in the form of (]bi − di , bi ] + ]0, b0 ]) ∩ ]bi − di , bi ] = ∅. Clearly, bi − di ≤ bi − di + 0. Because of dl ≤ ε ≤ |b0 + b0 − b0 | = b0 for all l ∈ L \ {0} and d0 = b0 we obtain dl ≤ b0 for all l ∈ L, in particular di ≤ b0 . Hence bi ≤ min{bi − di + b0 , bi + 0}. 2 Corollary 7. Let B = {bl ∈ ]0, ∞[ | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where bl < bl+1 for all l, l + 1 ∈ L. Let δ = δ(B) be given by (5). Then the following assertions are equivalent: (i) There exists an intervals’ set M of the first kind with the gap-point set B and with the first interval [0, b0 ] such that the lower addition on M is associative. (ii) The set B is anti-additive and the lower addition on B is associative and b0 < δ. Proof. (⇒) Suppose that (i) holds. Then B is anti-additive and the lower addition on B is associative by Corollary 6. It remains to prove that b0 < δ. The proof is by contradiction. Let M be given by a sequence of intervals {]al , bl ]}l∈L with a0 = 0. Suppose that b0 ≥ δ, that is, b0 ≥ bi − bi−1 for some i ∈ L \ {0}. Then (]0, b0 ] + ]ai−1 , bi−1 ]) ∩ ]ai , bi ] = ∅, and by (3), ai ≤ 0 + ai−1 = ai−1 , which is a contradiction. (⇐) Suppose that (ii) holds. Since B is anti-additive and b0 < δ, there exists an intervals’ set M of the first kind with the gap-point set B and with the first interval [0, b0 ] by Lemma 7, and since the lower addition on B is associative, the lower addition on M is also associative by Corollary 4 (i). 2

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Theorem 4. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. If g : X → [0, ∞], g(0) = 0 is a discrete strictly increasing additive generator of a discrete t-conorm G with an anti-additive set R(g) \ {0} = {b0 , . . . , bn }, 0 < b0 < · · · < bn < ∞ such that b0 < δ where δ = δ(R(g) \ {0}) be given by (5), then there exists a strictly increasing additive generator f : [0, 1] → [0, ∞] of the first kind of a border-continuous t-conorm F such that f/X = g and F /X 2 = G. Proof. Suppose that g : X → [0, ∞], g(0) = 0 is a discrete additive generator of a discrete t-conorm G with an anti-additive set R(g) \ {0} = {b0 , . . . , bn }, 0 < b0 < · · · < bn < ∞ such that b0 < δ where δ = δ(R(g) \ {0}) be given by (5). Write X \ {0} = {xl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where xl < xl+1 for all l, l + 1 ∈ L. First, we will describe the construction of f . As in Lemma 7 we construct an intervals’ set M = [0, b0 ] ∪ ( l∈L\{0} ]bl − dl , bl ]) of the first kind with the gap point set B = R(g) \ {0} = {bl | bl = g(xl ), l ∈ L} where {dl }l∈L\{0} is a non-decreasing sequence of positive real numbers such that dl ≤ ε for all l ∈ L \ {0} where ε = ε(B) be given by (6), and then, we choose an arbitrary strictly increasing bijection f : [0, 1] → M with f  (0) = 0 and dl f (xl ) = bl for all l ∈ L, for instance, f (0) = 0 and f (x) = (xl −x (x − x ) + (b − d ) for all x ∈ xl−1 , xl and l−1 l l l−1 ) l ∈ L where x−1 = 0. Now, we will prove that F is a border-continuous t-conorm. Since the strictly increasing bijection g : X → B ∪ {0} additively generates an associative operation G, by Theorem 1, the lower addition on B ∪ {0} is associative, and by Corollary 4 (ii), so is the lower addition on M. By Theorem 1, the strictly increasing bijection f : [0, 1] → M additively generates an associative operation F . Since [0, b0 ] ⊆ M, we have limx→0+ f (x) = 0, and by Lemma 1, the associative operation F is a border-continuous t-conorm. Finally, we will prove that f/X = g and F /X 2 = G. By construction of f it is obvious that f/X = g. By Corollary 4 (ii), the lower addition on M can be restricted to B ∪ {0}, and by Proposition 5, the operation F can be restricted to X = f −1 (B ∪ {0}), and this restriction F /X 2 is additively generated by f/X, that is, F /X 2 = G. 2 Example 7. Let X = { nl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N . The function g : X → [0, ∞], g(0) = 0, g( nl ) = 10l−1 for all l ∈ L \ {0} is a discrete additive generator of a discrete t-conorm G( ni , jn ) = max{ ni , jn } for all i, j ∈ L, with l−1 | l ∈ L \ {0}}. Since b = min B = 1 < 9 ≤ δ(B) and ε(B) = 1, the set the anti-additive R(g) \ {0} = {10 0   set B = l−1 l−1 M = {0} ∪ ( l∈L\{0} 10 − 1, 10 ) is an intervals’ set of the first kind with the gap-point set B and with the first interval [0, 1] by Lemma 7. The strictly increasing bijection f : [0, 1] → M, f (0) = 0, f (x) = n(x − l−1 n )+  l−1 l  l−1 (10 − 1) for all x ∈ n , n and l ∈ L \ {0} is an additive generator of a border-continuous t-conorm F such that f/X = g and F /X 2 = G by the proof of Theorem 4. 7.3. The general construction We describe the construction of non-continuous strictly increasing additive generators acting on [0, 1] of associative operations starting from discrete strictly increasing additive generators of discrete associative operations. This construction is based on constructing of an intervals’ set of the first kind starting from an arbitrary non-empty finite set A ⊆ [0, ∞[. Definition 7. A set M ⊆ [0, ∞] given by a sequence of intervals {]al , bl ]}l∈L is additive if for all i, j, k ∈ L, if (]ai , bi ] + ]aj , bj ]) ∩ ]ak , bk ] = ∅, then ak = ai + aj and bk ≤ min{ai + bj , bi + aj }.

(7)

Remark 10. Let M ⊆ [0, ∞]. (i) If M is given by a sequence of intervals {]al , bl ]}l∈L , then (7) implies (3). (ii) M is an additive intervals’ set of the first kind if and only if M is in the form of (1) and satisfies (7). (iii) Let M be given by a sequence of intervals {]al , bl ]}l∈L . We say that the set A = {al | l ∈ L} is the set of all left endpoints of intervals of M.

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Lemma 8. Let A = {al ∈ [0, ∞[ | l ∈ L}, L = {0, 1 . . . , n}, n ∈ N ∪ {0} where al < al+1 for all l, l + 1 ∈ L. Let δ = δ(A) and ε = ε(A) be given by (5) and (6), respectively. Then the set  M = {0} ∪ ( ]al , al + dl ]), l∈L

where {dl }l∈L is a non-increasing sequence of positive real numbers such that δ > d0 and 12 ε ≥ d0 , is an additive intervals’ set of the first kind with the set A of all left endpoints of intervals of M. Proof. In order to show that M is an additive intervals’ set of the first kind we should prove that ai−1 < ai−1 + di−1 < ai for all i ∈ L \ {0}, and that for all i, j, k ∈ L, if (]ai , ai + di ] + ]aj , aj + dj ]) ∩ ]ak , ak + dk ] = ∅, then ak = ai + aj and ak + dk ≤ min{ai + aj + dj , ai + di + aj }. For all i ∈ L \ {0} we have 0 < di−1 < δ ≤ ai − ai−1 , and so ai−1 < ai−1 + di−1 < ai . Suppose that (]ai , ai + di ] + ]aj , aj + dj ]) ∩ ]ak , ak + dk ] = ∅ for some i, j, k ∈ L for the rest of the proof. First, we will prove that ai + aj = ak and that i ≤ k and j ≤ k. If it would be ai + aj < ak , from ε ≤ |ai + aj − ak | = ak −ai −aj , and because of di ≤ 12 ε and dj ≤ 12 ε we would have ai +di +aj +dj ≤ ai +aj +ε ≤ ak , and consequently (]ai , ai + di ] + ]aj , aj + dj ]) ∩ ]ak , ak + dk ] = ∅, which would be a contradiction. If it would be ai + aj > ak , from ε ≤ |ai + aj − ak | = ai + aj − ak , and because of dk ≤ 2ε we would have ak + dk ≤ ak + ε ≤ ai + aj , and consequently (]ai , ai + di ] + ]aj , aj + dj ]) ∩ ]ak , ak + dk ] = ∅, which would be a contradiction. We have already proved that ai + aj = ak . If it would be k < i or k < j , and without loss of generality suppose that k < i, it would be ak < ai , and so ak < ai + aj , which would be a contradiction to ai + aj = ak . Thus, i ≤ k and j ≤ k. Finally, we will prove that ak + dk ≤ min{ai + aj + dj , ai + di + aj }. Since ak = ai + aj , dk ≤ di and dk ≤ dj we have ak + dk ≤ min{ai + aj + dj , ai + di + aj }. 2 Corollary 8. Under the hypotheses of Lemma 8, for all d ∈ ]0, η] where η =  ( l∈L ]al , al + d]) is an additive intervals’ set of the first kind.

1 2

min{δ, ε}, the set M = {0} ∪

Proof. If η < ∞ then η < δ, and so, for all d ∈ ]0, η], it is d < δ and d ≤ 12 ε. Thus, our assertion follows from Lemma 8 where d0 = · · · = dn = d. If η = ∞ then A = {0}, and so, M = [0, d] where d ∈ ]0, ∞]. Obviously, our assertion is true. 2 Corollary 9. Let A = {al ∈ [0, ∞[ | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where al < al+1 for all l, l + 1 ∈ L. Then the following assertions are equivalent: (i) There exists an additive intervals’ set M of the first kind with the set A of all left endpoints of intervals of M such that the lower addition on M is associative. (ii) The lower addition on A is associative. Proof. (⇒) Suppose that (i) holds. Then, by Corollary 4 (i), the lower addition on a gap-point set B of M is associative, and by Corollary 2, so is the lower addition on A. (⇐) Suppose that (ii) holds. By Lemma 8, there exists an additive intervals’ set M of the first kind with the set A of all left endpoints of intervals of M. By Corollary 2, the lower addition on a gap-point set B of M is associative, and by Corollary 4 (i), so is the lower addition on M. 2 With respect to Remark 2 (i) we can assume without loss of generality that a discrete strictly increasing additive generator g : X → [0, ∞] satisfies g(max X) < ∞. Theorem 5. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. If g : X → [0, ∞[ is a discrete strictly increasing additive generator of a discrete associative operation G then there exists a strictly increasing additive generator f : [0, 1] → [0, ∞] of an associative operation F such that f/X is a discrete strictly increasing additive generator of G and F /X 2 = G.

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Proof. Suppose that g : X → [0, ∞[ is a discrete additive generator of a discrete associative operation G. Write X = {xl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N where xl < xl+1 for all l, l + 1 ∈ L. First, we will describe the construction of f . As in Lemma 8 we construct an additive intervals’ set M = {0} ∪ ( l∈L ]ai , ai + di ]) of the first kind where A = R(g) = {al | al = g(xl ), l ∈ L} and δ(A) > d0 ≥ · · · ≥ dn > 0 and 1 )= 2 ε(A) ≥ d0 , and then, we choose an arbitrary strictly increasing bijection f : [0, 1] → M ∩ [a0 + d0 , ∞]with f (x l dl and (x − x ) + a for all x ∈ x , x al + dl for all l ∈ L, for instance, f (0) = a0 + d0 and f (x) = (xl −x l−1 l l−1 l l−1 ) l ∈ L \ {0}. Now, we will prove that F is an associative operation. Since the strictly increasing bijection g : X → A additively generates an associative operation G, by Theorem 1, the lower addition on A is associative. By Corollary 2, the lower addition on B = {al + dl | l ∈ L} is associative, and by Corollary 4 (i), so is the lower addition on M. Further, the lower addition on M ∩ [a0 + d0 , ∞] is also associative since, by Proposition 2 (i), it is a restriction of the lower addition from M to M ∩ [a0 + d0 , ∞]. Thus, by Theorem 1, the strictly increasing bijection f : [0, 1] → M ∩ [a0 + d0 , ∞] additively generates an associative operation F . We will prove that f/X is a discrete additive generator of G. Since the lower additions on A = R(g) and B = R(f/X) are isomorphic by Corollary 2, both strictly increasing bijections g : X → A and f/X : X → B additively generate the same operation G by Proposition 4. Finally, we will prove that F /X 2 = G. From B ⊆ (M ∩ [a0 + d0 , ∞]) ⊆ M, and since the lower addition on M can be restricted to M ∩ [a0 + d0 , ∞] by Proposition 2 (i) and to B by Corollary 4 (i), it follows that the lower addition on M ∩ [a0 + d0 , ∞] can also be restricted to B by Corollary 1 (i). Thus, by Proposition 5, the operation F can be restricted to X = f −1 (B), and this restriction F /X 2 is additively generated by f/X, that is, F /X 2 = G. 2 Corollary 10. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. If F : X 2 → X is a discrete associative operation additively generated by a strictly increasing additive generator then there exists a discrete strictly increasing additive generator f of F with an anti-additive set R(f ). Example 8. Let X = { nl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N . The function g : X → [0, ∞], g( nl ) = 2l + 2 for all l ∈ L, is a discrete additive generator of a discrete associative operation G( ni , jn) = min{ i+jn+1 , 1} for all i, j ∈ L. Since δ(A) = ε(A) = 2 where A = R(g) = {2l + 2 | l ∈ L}, the set M = {0} ∪ ( l∈L ]2l + 2, 2l + 3]) is an additive intervals’ set of the first kind by Corollary 8. The  strictly increasing bijection f : [0, 1] → M ∩ [3, ∞], f (0) = 3,  l−1 l f (x) = n(x − l−1 ) + (2l + 2) for all x ∈ , n n n and l ∈ L \ {0} is an additive generator of an associative operation F such that the function f/X is a discrete additive generator of G and F /X 2 = G by the proof of Theorem 5. Observe that the associative operation F constructed in the proof of Theorem 5 is never a t-conorm because f (0) > 0. The construction which leads to a border-continuous t-conorm will be introduced in the next subsection. 7.4. The general construction and border-continuity We describe the construction of strictly increasing additive generators of the first kind of border-continuous t-conorms starting from discrete strictly increasing additive generators of discrete t-conorms. This construction is based on constructing of an intervals’ set of the first kind which contains a right-hand neighborhood of 0 starting from an arbitrary non-empty finite set A ⊆ [0, ∞[, 0 ∈ A. Lemma 9. Let A = {al ∈ [0, ∞[ | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N ∪ {0} where al < al+1 for all l, l + 1 ∈ L and a0 = 0. Let ε = ε(A) be given by (6). Then the set  M = {0} ∪ ( ]al , al + dl ]), l∈L

where {dl }l∈L is a non-increasing sequence of positive real numbers such that 12 ε ≥ d0 , is an additive intervals’ set of the first kind with the set A of all left endpoints of intervals of M and with the first interval [0, d0 ]. Proof. For all l ∈ L \ {0}, it is al − al−1 = a0 + al − al−1 because a0 = 0. Thus, {al − al−1 | l ∈ L \ {0}} ⊆ {|ai + aj − ak | | ai + aj − ak = 0, i, j, k ∈ L}, and by definition of ε and δ where δ = δ(A) be given by (5), it is

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ε ≤ δ. If ε < ∞ then d0 < δ. Thus, our assertion follows from Lemma 8. If ε = ∞ then A = {0}, and so, M = [0, d0 ], d0 ∈ ]0, ∞]. Obviously, our assertion is true. 2    Corollary 11. Under the hypotheses of Lemma 9, for all d ∈ 0, 12 ε , the set M = {0} ∪ ( l∈L ]al , al + d]) is an additive intervals’ set of the first kind with the first interval [0, d].   Proof. Our assertion follows from Lemma 9 where d0 = · · · = dn = d, d ∈ 0, 12 ε .

2

With respect to Remark 2 (ii) we can assume without loss of generality that a discrete additive generator g : X → [0, ∞] of a discrete t-conorm satisfies g(min X) = 0 and g(max X) < ∞. Theorem 6. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. If g : X → [0, ∞[, g(0) = 0 is a discrete strictly increasing additive generator of a discrete t-conorm G then there exists a strictly increasing additive generator f : [0, 1] → [0, ∞] of the first kind of a border-continuous t-conorm F such that f/X is a discrete strictly increasing additive generator of G and F /X 2 = G. Proof. Suppose that g : X → [0, ∞[, g(0) = 0 is a discrete additive generator of a discrete t-conorm G. Write X = {xl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N where xl < xl+1 for all l, l + 1 ∈ L. First, we will describe the construction of f . As in Lemma 9 we construct an additive intervals’ set M = {0} ∪ ( l∈L ]al , al + dl ]) of the first kind with the first interval [0, d0 ] where A = R(g) = {al | al = g(xl ), l ∈ L} and 1 2 ε(A) ≥ d0 ≥ · · · ≥ dn > 0, and then, we choose an arbitrary number x 21 ∈ ]x0 , x1 [ and an arbitrary strictly increasing bijection f : [0, 1] → M with f (x0 ) = a0 , (x0 = a0 = 0), f (x 1 ) = a0 + d0 and f (xl ) = al + dl and for all l ∈ L \ {0}, 2 for instance,

f (x) =

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

d0 x1

x

2

d1 (x1 −x 1 ) (x 2

− x 1 ) + a1

dl (xl −xl−1 ) (x

2

− xl−1 ) + al

 for all x ∈ 0, x 1 2   for all x ∈ x 1 , x1 2   for all x ∈ xl−1 , xl and l ∈ L \ {0, 1}.

Now, we will prove that F is a border-continuous t-conorm. Since the strictly increasing bijection g : X → A additively generates an associative operation, by Theorem 1, the lower addition on A is associative. By Corollary 2, the lower addition on B = {al + dl | l ∈ L} is associative too, and by Corollary 4 (i), so is the lower addition on M. By Theorem 1, the strictly increasing bijection f : [0, 1] → M additively generates an associative operation F . Since [0, d0 ] ⊆ M we have limx→0+ f (x) = 0, and by Lemma 1, the associative operation F is a border-continuous tconorm. We will prove that f/X is a discrete additive generator of G. Since the lower additions on A and B are isomorphic by Corollary 2, and since a0 is a neutral element of the lower addition on A, the number a0 + d0 has to be a neutral element of the lower addition on B. Thus, by Remark 5 (ii), the lower additions on B and B0 = {0} ∪ (B \ {a0 + d0 }) are isomorphic. Hence, the lower additions on A and B0 are isomorphic too, and so, both strictly increasing bijections g : X → A and f/X : X → B0 additively generate the same operation G by Proposition 4. Finally, we will prove that F /X 2 = G. First, we will prove that the lower addition on M can be restricted to B0 . If B = {a0 + d0 } then B0 = {0}, and our assertion is obviously true. If B = {a0 + d0 } then a1 + d1 ∈ B. From (B ∩ [a1 + d1 , ∞]) ⊆ B ⊆ M, and since the lower addition on M can be restricted to B by Corollary 4 (i) and the lower addition on B can be restricted to B ∩ [a1 + d1 , ∞] by Proposition 2 (i), it follows that the lower addition on M can be also be restricted to B ∩ [a1 + d1 , ∞] by Corollary 1 (ii). It is clear that the lower addition on M can even be restricted to {0} ∪ (B ∩ [a1 + d1 , ∞]) = B0 . Thus, by Proposition 5, the operation F can be restricted to X = f −1 (B0 ), and this restriction F /X 2 is additively generated by f/X, that is, F /X 2 = G. 2 With respect to Remark 2 (ii) we immediately obtain the following consequence of Theorem 6.

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Corollary 12. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. For every discrete additively generated t-conorm G : X 2 → X there exists an additively generated border-continuous t-conorm F : [0, 1]2 → [0, 1] such that F /X 2 = G. Remark 11. It follows from the result proved by Viceník [9, Theorem 1] that the construction method described in the proof of Theorem 6 enables us to constructs from one discrete strictly increasing additive generator of a discrete t-conorm infinitely many additively generated border-continuous t-conorms which are pairwise non-isomorphic. Example 9. Let X = { nl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N . The function g : X → [0, ∞], g( nl ) = 2l for all l ∈ L, is a discrete additive generator of a discrete t-conorm G( ni , jn ) = min{ i+j n , 1} for all i, j ∈ L. Since ε(A) = 2 where A = {2l | l ∈ L}, the set M = {0} ∪ ( l∈L ]2l, 2l + 1]) is an intervals’ set of the first kind with the first interval [0, 1] by 1 1 Corollary 11. The strictly increasing bijection f : [0, 1] → M, f (x) = 2nx for all x ∈ [0, 2n ], f (x) = 2n(x − 2n )+2  1 1   l−1 l−1 l for all x ∈ 2n , n , and f (x) = n(x − n ) + 2l for all x ∈ n , n and l ∈ L \ {0, 1} is an additive generator of a border-continuous t-conorm F such that the function f/X is a discrete additive generator of G and F /X 2 = G by the proof of Theorem 6. 8. T-conorms additively generated by strictly increasing additive generators of the first kind Let f be a strictly increasing additive generator of the first kind of F and let the range R(f ) of f be given by a sequence of intervals {]al , bl ]}l∈L where L = {0, 1, . . . , n − 1}, n ∈ N . Let ⊗ be the index operation corresponding −1 to R(f ) and ⊕ be the  lower addition on R(f ). Put x0 = 0 and xl+1  all l ∈L. Write fl (x) = f (x)  = f  (bl ) for for all x ∈ xl , xl+1 , l ∈ L, and Fi,j (x, y) = F (x, y) for all x ∈ xi , xi+1 , y ∈ xj , xj +1 and i, j ∈ L. Observe that fl : ]xl , xl+1 ] → ]al , bl ] and fl−1 :]al , bl ] →]x bijections   l , xl+1 ] are strictly   increasing  for all l ∈ L.  its inverse Then, for all i, j ∈ L and x ∈ xi , xi+1 and y ∈ xj , xj +1 we have fi (x) ∈ ai , bj and fj (y) ∈ aj , bj , and so, ai⊗j < fi (x) + fj (y) ≤ di⊗j by Lemma 2 (i) and Remark 7. Thus, fi (xi ) ⊕ fj (y) ∈ ai⊗j , bi⊗j and  fi (x) + fj (y) if fi (x) + fj (y) ≤ bi⊗j fi (xi ) ⊕ fj (y) = bi⊗j if fi (x) + fj (y) > bi⊗j .     −1 −1 Since Fi,j (x, y) = fi⊗j (fi (x) ⊕ fj (y)) where fi⊗j : ai⊗j , bi⊗j → xi⊗j , x(i⊗j )+1 , it is Fi,j (x, y) ∈   xi⊗j , x(i⊗j )+1 and  Fi,j (x, y) =

−1 fi⊗j (fi (x) + fj (y)) x(i⊗j )+1

if fi (x) + fj (y) ≤ fi⊗j (x(i⊗j )+1 ) if fi (x) + fj (y) > fi⊗j (x(i⊗j )+1 ),

or equivalently −1 Fi,j (x, y) = fi⊗j (min{fi (x) + fj (y), fi⊗j (x(i⊗j )+1 )}).

Moreover, if (f (]xi , xi+1 ]) + f (]xj , xj +1 ])) ∩ f (]xi⊗j , x(i⊗j )+1 ]) = ∅ then Fi,j is constant and Fi,j (x, y) = x(i⊗j )+1 . We give examples of two t-conorms additively generated by additive generators of the first kind. The first one is finitely-valued on ]0, 1[2 and the second one is infinitely-valued on ]0, 1[2 . Example 10. (i) Let L = {0, 1, . . . , n − 1}, n ∈ N . The strictly increasing function f : [0, 1] → [0, ∞], f (0) = 0 and and l ∈ L is an additive generator of the first kind of a t-conorm f (x) = fl (x) = n(x − nl ) + 10l for all x ∈ nl , l+1 n  F (x, y) = Fi,j (x, y) = max

i +1 j +1 , n n



   j j +1 i+1 , y ∈ , , and i, j ∈ L (F (x, 0) = F (0, x) = x for all x ∈ [0, 1]). (The index operation n n n n    l corresponding to R(f ) = {0} ∪ ( l∈L 10 , 10l + 1 ) is given by i ⊗ j = max{i, j } for all i, j ∈ L.) for all x ∈

i

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(ii) Let L = {0, 1, . . . , n − 1}, n ∈ N . The  strictly increasing function f : [0, 1] → [0, ∞], f (0) = 0 and f (x) = fl (x) = n(x − nl ) + 2l for all x ∈ nl , l+1 and l ∈ L, is an additive generator of the first kind of a border-continuous n t-conorm  min{x + y, i+jn+1 } if i + j ≤ n − 1 F (x, y) = Fi,j (x, y) = 1 if i + j > n − 1    i i+1  for all x ∈ n , n , y ∈ jn , j +1 and i, j ∈ L (F (x, 0) = F (0, x) = x for all x ∈ [0, 1]). (The index operation n  corresponding to R(f ) = {0} ∪ ( l∈L ]2l, 2l + 1]) is given by i ⊗ j = min{i + j, n − 1} for all i, j ∈ L.) 9. Remarks on constructing of strictly decreasing additive generators of t-norms We have dealt with strictly increasing additive generators so far. Now we define strictly decreasing ones. The next two definitions are analogous to Definitions 2 and 4 (iii). Definition 8. Let X = [0, 1] or X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. A binary operation F : X 2 → X is additively generated if there exists a strictly decreasing function f : X → [0, ∞] such that F (x, y) = f (−1) (f (x) + f (y)) where f (−1) : [0, ∞] → X, f (−1) (t) = inf{x ∈ X | f (x) ≤ t}, inf ∅ = max X. The function f (−1) is the pseudo-inverse of f . We say that F is additively generated by f and that f is a strictly decreasing additive generator of F . If X is finite we say that f and F are discrete. Let us denote by G the family of all strictly decreasing functions f : X → [0, ∞] where either X = [0, 1] or X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. The next remark concerns the pseudo-inverse of a function defined on a one-element set. Remark 12. Let f : X → [0, ∞] be a function and X = {a}, a ∈ R ∪ {−∞, ∞}. On the one hand, since f ∈ F , the pseudo-inverse f (−1) : [0, ∞] → X of f is given by the formula f (−1) (t) = sup{x ∈ X | f (x) ≤ t} (sup ∅ = min X) for all t ∈ [0, ∞]. On the other hand, since f ∈ G, the pseudo-inverse f (−1) : [0, ∞] → X of f is given by the formula f (−1) (t) = inf{x ∈ X | f (x) ≤ t} (inf ∅ = max X) for all t ∈ [0, ∞]. Observe that both formulae lead to the same function f (−1) (t) = a for all t ∈ [0, ∞]. Definition 9. A strictly decreasing function f : [0, 1] → [0, ∞] is a strictly decreasing additive generator of the first kind if its range is an intervals’ set of the first kind. Let us recall the definitions of a t-norm and a v-dual binary operation [2, Definition 2.33]. Definition 10. Let X, Y ⊆ R ∪ {−∞, ∞} be non-empty linearly ordered sets with the usual linear order ≤. (i) Suppose that there exists a maximum max X of X. A binary operation  : X 2 → X is a t-norm if it is nondecreasing, commutative, associative and max X is its neutral element. (ii) Suppose that f : X → Y be a strictly decreasing bijection. A non-decreasing binary operation  : X2 → X is the f -dual operation to a non-decreasing binary operation ⊗ : Y 2 → Y if x  y = f −1 (f (x) ⊗ f (y)) for all x, y ∈ X. Observe that if f : X → Y is a strictly decreasing bijection then so is its inverse f −1 : Y → X, and  is the f -dual operation to ⊗ if and only if ⊗ is the f −1 -dual operation to . The composition of functions f and g (if it exists) will be denoted by g ◦ f where g ◦ f (x) = g(f (x)) for all x in the domain of g ◦ f . Lemma 10. Let X = [0, 1] or X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. Let f : X → [0, ∞] be a strictly monotone function and v : X → X be a strictly decreasing bijection. Then (f ◦ v)(−1) (t) = v −1 ◦ f (−1) (t) for all t ∈ [0, ∞]. Proof. If X = {a} then (f ◦ v)(−1) (t) = a and v −1 ◦ f (−1) (t) = a for all t ∈ [0, ∞] by Remark 12. Suppose that X is not a one-element set for the rest of the proof. The function f is either strictly increasing or strictly decreasing.

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Suppose that f is strictly increasing. Then, the composition f ◦ v is strictly decreasing and its pseudo-inverse (f ◦ v)(−1) : [0, ∞] → X is given by (f ◦ v)(−1) (t) = inf{x ∈ X | f (v(x)) ≤ t} where inf ∅ = max X. The composition v −1 ◦ f (−1) : [0, ∞] → X is given by v −1 ◦ f (−1) (t) = v −1 (sup{x ∈ X | f (x) ≤ t}) where sup ∅ = min X. Fix t ∈ [0, ∞] and write i = inf{x ∈ X | f (v(x)) ≤ t} and s = sup{x ∈ X | f (x) ≤ t}. We should prove that i = v −1 (s). By definition of s we have (i) for all x ∈ X, if x < s then f (x) ≤ t , and (ii) for all x ∈ X, if x > s then f (x) > t. On the one hand, for all x ∈ X, if x < v −1 (s) then v(x) > s, and by (ii), f (v(x)) > t. Thus, {x ∈ X | f (v(x)) ≤ t} ⊆ [v −1 (s), max X] ∩ X, and so, i ≥ v −1 (s). On the other hand, for all x ∈ X, if x > v −1 (s) then v(x) < s, and by (i), f (v(x)) ≤ t . Thus, ]v −1 (s), max X] ∩ X ⊆ {x ∈ X | f (v(x)) ≤ t}, and so, i ≤ inf(]v −1 (s), max X] ∩ X).

(8)

It remains to show that i ≤ v −1 (s). We will consider the following four cases. Suppose that X = [0, 1] and v −1 (s) < 1. Then, by (8), i ≤ inf(]v −1 (s), 1]) = v −1 (s). Suppose that X = [0, 1] and v −1 (s) = 1. Then, i ≤ 1 = v −1 (s). Suppose that X is finite and {x ∈ X | f (x) ≤ t} = ∅. Then s = max{x ∈ X | f (x) ≤ t}, and so s ∈ {x ∈ X | f (x) ≤ t}, and consequently v −1 (s) ∈ {x ∈ X | f (v(x)) ≤ t}. Thus, i ≤ v −1 (s). Suppose that X is finite and {x ∈ X | f (x) ≤ t} = ∅. Then s = min X and v −1 (s) = max X. Thus, i ≤ max X = v −1 (s). Suppose that f is strictly decreasing. Then, the composition f ◦ v is strictly increasing and its pseudo-inverse (f ◦ v)(−1) : [0, ∞] → X is given by (f ◦ v)(−1) (t) = sup{x ∈ X | f (v(x)) ≤ t} where sup ∅ = min X. The composition v −1 ◦ f (−1) : [0, ∞] → X is given by v −1 ◦ f (−1) (t) = v −1 (inf{x ∈ X | f (x) ≤ t}) where inf ∅ = max X. Fix t ∈ [0, ∞] and write s = sup{x ∈ X | f (v(x)) ≤ t} and i = inf{x ∈ X | f (x) ≤ t}. We should prove that s = v −1 (i). By definition of i we have (iii) for all x ∈ X, if x < i then f (x) > t, and (iv) for all x ∈ X, if x > i then f (x) ≤ t. On the one hand, for all x ∈ X, if x > v −1 (i) then v(x) < i, and by (iii), f (v(x)) > t. Thus, −1 {x ∈ X | f (v(x)) ≤ t} ⊆ [min X, v −1 (i)] ∩ X, and so, s ≤ v −1

(i). On the other hand, for all x ∈ X, if x < v (i) then v(x) > i, and by (iv), f (v(x)) ≤ t . Thus, min X, v −1 (i) ∩ X ⊆ {x ∈ X | f (v(x)) ≤ t}, and so, s ≥ sup([min X, v −1 (i)[ ∩ X).

(9)

It remains to show that s ≥ v −1 (i). We will consider the following four cases. Suppose that X = [0, 1] and v −1 (i) > 0. Then, by (9), s ≥ sup([0, v −1 (i)[) = v −1 (i). Suppose that X = [0, 1] and v −1 (i) = 0. Then, s ≥ 0 = v −1 (i). Suppose that X is finite and {x ∈ X | f (x) ≤ t} = ∅. Then i = min{x ∈ X | f (x) ≤ t}, and so i ∈ {x ∈ X | f (x) ≤ t}, and consequently v −1 (i) ∈ {x ∈ X | f (v(x)) ≤ t}. Thus, s ≥ v −1 (i). Suppose that X is finite and {x ∈ X | f (x) ≤ t} = ∅. Then i = max X and v −1 (i) = min X. Thus, s ≥ min X = v −1 (i). 2 The next two propositions enable us to use any of the constructions of strictly increasing additive generators introduced in the paper for constructing strictly decreasing additive generators. Assertions (ii) and (iii) of Proposition 6 where X = [0, 1] can be found in the monograph by Klement et al. [2, Proposition 2.34]. Proposition 6. Let X = [0, 1] or X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. Let v : X → X be a strictly decreasing  : X 2 → X be binary operations such that F  be the v-dual operation to F . bijection and F, F  is additively (i) If F is additively generated by a strictly increasing (strictly decreasing) additive generator f then F generated by a strictly decreasing (strictly increasing) additive generator f ◦ v. Moreover, assuming X = [0, 1], if f is of the first kind then so is f ◦ v.  is associative. (ii) F is associative if and only if F  is a t-norm (t-conorm). (iii) F is a t-conorm (t-norm) if and only if F  is border-continuous. (iv) Suppose that X = [0, 1]. Then F is border-continuous if and only if F Proof. (i) Suppose that f : X → [0, ∞] is a strictly increasing (strictly decreasing) additive generator of F , that is, F (x, y) = f (−1) (f (x) + f (y)) for all x, y ∈ X. Since v : X → X is a strictly decreasing bijection, f ◦ v : X →  is the v-dual operation to F , that is, F (x, y) = [0, ∞] is a strictly decreasing (strictly increasing) function. Since F −1 −1 (−1)  v (F (v(x), v(y)) for all x, y ∈ X, we obtain F (x, y) = v ◦ f ((f ◦ v)(x) + (f ◦ v)(y)) for all x, y ∈ X. By Lemma 10, the pseudo-inverse (f ◦ v)(−1) : [0, ∞] → X satisfies (f ◦ v)(−1) (t) = v −1 (f (−1) (t)) for all t ∈ [0, ∞].

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(x, y) = (f ◦ v)(−1) ((f ◦ v)(x) + (f ◦ v)(y)) for all x, y ∈ X, that is, F  is additively generated by a It follows that F strictly decreasing (strictly increasing) additive generator f ◦ v. Let X = [0, 1]. Suppose that f : [0, 1] → [0, ∞] is of the first kind, that is, the range of f is an interval’s set of the first kind. Since R(f ) = R(f ◦ v), the range of f ◦ v : [0, 1] → [0, ∞] is an interval’s set of the first kind, that is, f ◦ v is of the first kind. (ii) and (iii) The assertions are obviously true. (iv) Denote by B the border of [0, 1]2 . Since v(0) = 1 and v(1) = 0, we have for all (x, y) ∈ [0, 1]2 , (x, y) ∈ B if and only if (v(x), v(y)) ∈ B. . Suppose that F is border-continuous. Fix First, we will prove that if F is border-continuous then so is F (x, y) ∈ B and choose an arbitrary sequence {(xn , yn )} of points (xn , yn ) ∈ [0, 1]2 satisfying limn→∞ xn = x and (x, y). Since v is continuous, the sequence (xn , yn ) = F limn→∞ yn = y. It is sufficient to show that limn→∞ F 2 {(v(xn ), v(yn ))} of points (v(xn ), v(yn )) ∈ [0, 1] satisfies limn→∞ v(xn ) = v(x) and limn→∞ v(yn ) = v(y). Since (v(x), v(y)) ∈ B and F is border-continuous, limn→∞ F (v(xn ), v(yn )) = F (v(x), v(y)). Finally, since v −1 is contin(x, y). (xn , yn ) = F uous, it is limn→∞ v −1 (F (v(xn ), v(yn ))) = v −1 (F (v(x), v(y))) which is equivalent to limn→∞ F −1 −1  Since v : X → X is a strictly decreasing bijection and since F is the v -dual operation to F , from what we  is border-continuous then so is F . 2 have already proved it follows that if F Proposition 7. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. Let v : X → X and w : [0, 1] → [0, 1] be strictly decreasing  : [0, 1]2 → [0, 1] be binary operations such that G   : X 2 → X and F, F bijections such that w/X = v, and let G, G  be the v-dual operation to G and F be the w-dual operation to F . (i) If f is a strictly increasing (strictly decreasing) additive generator of F and f/X is a strictly increasing (strictly decreasing) additive generator of G, then f ◦ w is a strictly decreasing (strictly increasing) additive generator of  and (f ◦ w)/X is a strictly decreasing (strictly increasing) additive generator of G.  F 2 2   (ii) F /X = G if and only if F /X = G. Proof. (i) Suppose that f : [0, 1] → [0, ∞] is a strictly increasing (strictly decreasing) additive generator of F and f/X : X → [0, ∞] is a strictly increasing (strictly decreasing) additive generator of G. By Proposition 6 (i), f ◦ w :  and (f/X) ◦ v : X → [0, ∞] [0, 1] → [0, ∞] is a strictly decreasing (strictly increasing) additive generator of F  is a strictly decreasing (strictly increasing) additive generator of G. Since w/X = v and R(w/X) = X, we have (f/X) ◦ v = (f/X) ◦ (w/X) = f ◦ (w/X) = (f ◦ w)/X. /X 2 = G.  Suppose that F /X 2 = G. Since w/X = v and R(w/X) = (ii) First, we will prove that if F /X 2 = G then F −1 −1 (x, y) = w −1 (F (w(x), w(y))) = v −1 (G(v(x), v(y))) = G(x,  y), X and w /X = v , we obtain for all x, y ∈ X, F 2  /X = G. that is, F Since v −1 : X → X and w −1 : [0, 1] → [0, 1] are strictly decreasing bijections such that w −1 /X = v −1 and since  and F is the w −1 -dual operation to F , from what we have already proved it follows G is the v −1 -dual operation to G 2 2  then F /X = G. 2 /X = G that if F Remark 13. Let X ⊆ R ∪ {−∞, ∞} be a non-empty finite set. Then, there exists uniquely determined strictly decreasing bijection v : X → X. In fact, writing X = {x0 , . . . , xn }, (n ∈ N ∪ {0}), x0 < · · · < xn , we have v(xl ) = xn−l for all l ∈ {0, . . . , n}. Observe that, v −1 (x) = v(x) and v(v(x)) = x for all x ∈ X. Moreover, assuming X = {x0 , . . . , xn }, (n ∈ N ), 0 = x0 < · · · < xn = 1, there exists a strictly decreasing bijection w : [0, 1] → [0, 1] such that w/X = v. For −xn−l−1 ) l instance, w(x) = xn−l − (xn−l (xl+1 −xl ) (x −xl ) for all x ∈ [xl , xl+1 ] and l ∈ {0, . . . , n −1}. Especially, if X = { n | l ∈ L}, L = {0, . . . , n}, (n ∈ N ) then v( nl ) = 1 −

l n

for all l ∈ L, and w can be defined as w(x) = 1 − x for all x ∈ [0, 1].

The next theorem formulated for strictly decreasing additive generators is an analogue of Theorem 1. Theorem 7. Let X be the domain of f ∈ G and R(f ) ∈ L. Let F : X 2 → X be additively generated by f and let ⊕ : R(f )2 → R(f ) be the lower addition on R(f ). Then F is the f -dual operation to ⊕ where f : X → R(f ) is a strictly decreasing bijection, that is, F (x, y) = f −1 (f (x) ⊕ f (y)) for all x, y ∈ X.

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 as the v-dual operation to F . Write Proof. Choose a strictly decreasing bijection v : X → X and define F . f = f ◦ v. By Proposition 6 (i), the function f : X → [0, ∞] is a strictly increasing additive generator of F  Clearly, R(f) = R(f ). By Theorem 1, the strictly increasing bijection f : X → R(f ) is an isomorphism of F , (x, y) = f−1 (f(x) ⊕ f(y)) for all x, y ∈ X. Obviously, F is the v −1 -dual operation to F and ⊕, that is, F (v −1 (x), v −1 (y))) for all x, y ∈ X where v −1 : X → X is a strictly decreasing bijection. that is, F (x, y) = v(F Clearly, R(f◦ v −1 ) = R(f ). It follows that F (x, y) = (f◦ v −1 )−1 (f◦ v −1 (x) ⊕ f◦ v −1 (y)) for all x, y ∈ X where f◦ v −1 : X → R(f ) is a strictly decreasing bijection, that is, F is the f◦ v −1 -dual operation to ⊕. Finally, observe that f◦ v −1 = f . 2 The next theorem formulated for strictly decreasing additive generators is an analogue of Theorem 6. Theorem 8. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. If g : X → [0, ∞[, g(1) = 0 is a discrete strictly decreasing additive generator of a discrete t-norm G then there exists a strictly decreasing additive generator f : [0, 1] → [0, ∞] of the first kind of a border-continuous t-norm F such that f/X is a discrete strictly decreasing additive generator of G and F /X 2 = G. Proof. Suppose that g : X → [0, ∞[, g(1) = 0 is a discrete strictly decreasing additive generator of a discrete  as the v-dual operation to G. Write  t-norm G. Choose a strictly decreasing bijection v : X → X. Define G g = g ◦ v. Clearly, R( g ) = R(g). By Proposition 6 (i) and (iii), the function  g : X → [0, ∞[,  g (0) = 0 is a discrete strictly in By Theorem 6, there exists a strictly increasing additive generator creasing additive generator of a discrete t-conorm G.  such that f/X is a discrete strictly increasf: [0, 1] → [0, ∞] of the first kind of a border-continuous t-conorm F 2    ing additive generator of G and F /X = G. Choose an arbitrary strictly decreasing bijection w : [0, 1] → [0, 1] such that w/X = v. Observe that v −1 : X → X and w −1 : [0, 1] → [0, 1] are strictly decreasing bijections such  Define F as the w −1 -dual operation to F . Write that w −1 /X = v −1 . Recall that G is the v −1 -dual operation to G. −1  f = f ◦ w . Hence, by Proposition 6 (i) and (iii)–(iv), the function f : [0, 1] → [0, ∞] is a strictly decreasing additive generator of the first kind of a border-continuous t-norm F , by Proposition 7 (i), the function f/X is a discrete strictly decreasing additive generator of G, and finally, by Proposition 7 (ii), F /X 2 = G. 2 The next remark is analogous to Remark 2. Remark 14. Let F be a discrete binary operation acting on a non-empty finite set X ⊆ R ∪ {−∞, ∞} additively generated by a discrete strictly decreasing additive generator. (i) Then there exists a discrete strictly decreasing additive generator f : X → [0, ∞] of F such that f (min X) < ∞. (ii) Suppose that max X is a neutral element of F . Then there exists a discrete strictly decreasing additive generator f : X → [0, ∞] of F such that f (min X) < ∞ and f (max X) = 0. With respect to Remark 14 (ii) we immediately obtain the following consequence of Theorem 8. Corollary 13. Let X ⊆ [0, 1] be a finite set and 0, 1 ∈ X. For every discrete additively generated t-norm G : X 2 → X there exists an additively generated border-continuous t-norm F : [0, 1]2 → [0, 1] such that F /X 2 = G. The next example is based on Example 9 and Propositions 6 and 7. Example 11. Let X = { nl | l ∈ L}, L = {0, 1, . . . , n}, n ∈ N . The function g : X → [0, ∞], g( nl ) = 2(n − l) for all l ∈ L is a discrete strictly decreasing additive generator of the discrete Łukasiewicz t-norm G where G( ni , jn ) = max{ i+j n − 1, 0} for all i, j ∈ L.  as the v-dual operation to G where v : X → X, v( l ) = 1 − l is a strictly decreasing bijection. The Define G n n  is the discrete Łukasiewicz t-conorm, that is, G(  i , j ) = min{ i+j , 1} for all i, j ∈ L. Writing  operation G g = g ◦ v, n n n l  the function  g : X → [0, ∞],  g ( n ) = 2l for all l ∈ L, is a strictly increasing additive generator of G.

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As in Example 9 we construct the strictly increasing function f: [0, 1] → [0, ∞],

1 ⎧ , for all x ∈ 0, 2n ⎪ ⎨ 2nx   1 1 1 f(x) = 2n x − 2n + 2 for all x ∈ 2n , n , ⎪    ⎩  l−1 l n x − n + 2l for all x ∈ l−1 n , n and l ∈ L \ {0, 1},  such that f/X is a discrete strictly which is a strictly increasing additive generator of a border-continuous t-conorm F 2    increasing additive generator of G and F /X = G. The function w : [0, 1] → [0, 1], w(x) = 1 − x is a strictly decreasing bijection and w/X = v. Define F as the . Writing f = f◦ w −1 , the function f : [0, 1] → [0, ∞], w −1 -dual operation to F 

⎧  l−1 l l−1 ⎪ ⎨ n 1 − n − x  + 2l for all x ∈ 1 − n , 1 − n and l ∈ L \ {0, 1}, 1 1 f (x) = 2n 1 − 2n −x +2 for all x ∈ 1 − n1 , 1 − 2n , ⎪ 

⎩ 1 2n(1 − x) for all x ∈ 1 − 2n , 1 , is a strictly decreasing additive generator of a border-continuous t-norm F such that f/X is a discrete strictly decreasing additive generator of G and F /X 2 = G. Note that all results of the paper formulated for strictly increasing additive generators can be reformulated for strictly decreasing additive generators in an analogous way or by using the principles of duality. 10. Conclusions Although Theorems 3–6 claim an existence of strictly increasing additive generators of associative operations of certain properties their proofs are constructive and offer methods of constructing strictly increasing additive generators of associative operations starting from discrete strictly increasing additive generators of discrete associative operations. Theorem 2 offers a method of constructing discrete strictly increasing additive generators of discrete associative operations starting from strictly increasing additive generators of the first kind of associative operations. We can conclude this article by saying that the results of the paper enable us to use any method for constructing discrete strictly increasing (strictly decreasing) additive generators of a discrete t-conorms (t-norms) for constructing strictly increasing (strictly decreasing) additive generators of the first kind of t-conorms (t-norms), and vice versa, to use any method for constructing strictly increasing (strictly decreasing) additive generators of the first kind of t-conorms (t-conorms) for constructing discrete strictly increasing (strictly decreasing) additive generators of discrete t-conorms (t-norms). Acknowledgements This work was supported by the Grant VEGA 1/0614/18. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

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