Ranks of additive generators

Fuzzy Sets and Systems 160 (2009) 2032 – 2048 www.elsevier.com/locate/fss

Ranks of additive generators A. Mesiarová-Zemánkováa, b,∗ a Department of Computer Science, Trinity College, Dublin, Ireland b Mathematical Institute of Slovak Academy of Sciences, Bratislava, Slovakia

Received 19 August 2008; received in revised form 23 December 2008; accepted 23 December 2008 Available online 3 January 2009

Abstract For additive generators of t-norms and t-conorms, the concept of a rank is introduced. It is shown that cancellative and conditionally cancellative t-norms have only additive generators with an infinite rank. The conditions for additive generators with finite ranks are introduced and several examples are given. It is shown that the pseudo-inverse to the Cantor function is an additive generator with rank 2 of a t-conorm C and that for all m ∈ N, m  2 this t-conorm has an additive generator with rank m. © 2009 Elsevier B.V. All rights reserved. Keywords: t-Norm; t-Conorm; Additive generator; Strong additive generator

1. Introduction The class of triangular norms is an important class of aggregation operators [3], which was introduced in the framework of probabilistic metric spaces [11,14]. A triangular norm is a binary operation T : [0, 1]2 →[0, 1] which is commutative, associative, non-decreasing in both variables and 1 is its neutral element. More details on t-norms can be found in monographs [1,8]. An important approach which allows to define t-norms (as two-place functions) by means of a one-place function is the definition by means of an additive generator. Definition 1. Let t: [0, 1]→[0, ∞] with t(1) = 0 be a strictly decreasing function such that an operation T : [0, 1]2 → [0, 1] given by T (x, y) = t (−1) (t(x) + t(y)) is a t-norm. Here t (−1) : [0, ∞]→[0, 1] is the pseudo-inverse of t given by t (−1) (u) = sup{x ∈ [0, 1] | t(x) > u}, see [7], and t is called an additive generator of the t-norm T . This approach is important mainly because it reduces the computational complexity. The class of all continuous additive generators has already been characterized and t-norms generated by continuous additive generators are exactly the continuous Archimedean t-norms [10,13]. The t-norms generated by non-continuous additive generators are not yet fully characterized. The main problem in the case of non-continuous additive generators is to ensure the associativity of ∗ Corresponding author at: Mathematical Institute of Slovak Academy of Sciences, Bratislava, Slovakia. Tel.: +421 2 57 510 501.

E-mail address: [email protected]. 0165-0114/$ - see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2008.12.015

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the generated operation. Many important results concerning the additive generators of t-norms are due to Viceník [15] and Jenei [5,6]. Due to the associativity any t-norm can be uniquely extended to an n-ary aggregation operator and thus can process n inputs. Therefore we can write T (x1 , . . . , xn ) which will denote the aggregation of x1 , . . . , xn ∈ [0, 1] by the t-norm T . Although any n-ary form of an additively generated t-norm can be again expressed by means of its additive generator this need not save computational time at all. The problem is that in general T (x1 , . . . , xn ) = t (−1) (t(x1 ) + · · · + t(xn )) need not hold and the computation of t (−1) (t(x1 ) + t(t (−1) (t(x2 ) + t(t (−1) (t(x3 ) + · · · + t(xn ))) · · ·))) is then similarly complex as the computation of T (x1 , . . . , xn ). For any n ∈ N let us assume a function Tn : [0, 1]n →[0, 1] given by Tn (x1 , . . . , xn ) = t (−1) (t(x1 ) + · · · + t(xn )) for (x1 , . . . , xn ) ∈ [0, 1]n . It is evident that T2 = T . In this paper we will investigate additive generators t such that for a given m ∈ N we have T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) for all n  m, n ∈ N and all (x 1 , . . . , xn ) ∈ [0, 1]n . It is evident that if xi = 0 for some i ∈ {1, . . . , n} then T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) = 0. Similarly, if xi = 1 for some i ∈ {1, . . . , n} then T (x1 , . . . , xn ) = T (x1 , . . . , xi−1 , xi+1 , . . . , xn ) and Tn (x1 , . . . , xn ) = t (−1) (t(x1 ) + · · · + t(xn )) = t (−1) (t(x1 ) + · · · + t(xi−1 ) + t(xi+1 ) + · · · t(xn )) = Tn−1 (x1 , . . . , xi−1 , xi+1 , . . . , xn ). Therefore we can always assume (x1 , . . . , xn ) ∈ ]0, 1[n . In [16] it was shown that in order to determine the associativity of an operation generated by some strictly decreasing fixed function t with t(1) = 0 only the range of t is important. Let us denote the range of an additive generator t ˆ for bˆ ∈ [0, ∞] and ⊕: (Range(t))2 −→Range(t) by ˆ = t(t (−1) (b)) by Range(t). Define F: [0, ∞]→Range(t) by F(b) ˆ ˆ ˆ ˆ cˆ ⊕ d = F(cˆ + d) for all c, ˆ d ∈ Range(t), i.e., cˆ ⊕ d = cˆ + dˆ if cˆ + dˆ ∈ Range(t) (for more precise definition of ⊕, where only Range(t) is needed, see [16]). Then the operation T given by T (x, y) = t (−1) (t(x) + t(y)) for x, y ∈ [0, 1] is associative if and only if (Range(t), ⊕) is a semigroup. The same principle can also be used for the investigation of the equality T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ). This equality holds for all (x1 , . . . , xn ) ∈ [0, 1]n if and only if for all cˆ1 , . . . , cˆn ∈ Range(t) we have F(cˆ1 + · · · + cˆn ) = cˆ1 ⊕ · · · ⊕ cˆn . This is due to the fact that the strict monotonicity of t implies F(cˆ1 + · · · + cˆn ) = cˆ1 ⊕ · · · ⊕ cˆn , i.e., t(T (x1 , . . . , xn )) = t(Tn (x1 , . . . , xn )) if and only if T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ), where t(xi ) = cˆi for i = 1, . . . , n. 2. Rank of an additive generator First we introduce the notion of a rank of an additive generator. Definition 2. Let T be a t-norm and let t be its additive generator. For every n ∈ N assume a function Tn : [0, 1]n →[0, 1] given by Tn (x1 , . . . , xn ) = t (−1) (t(x1 ) + · · · + t(xn )) for (x1 , . . . , xn ) ∈ [0, 1]n . (i) If T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) for all n ∈ N and all (x1 , . . . , xn ) ∈ [0, 1]n then we say that the rank of the additive generator t is infinity, Rank(t) = ∞. Such an additive generator is also called a strong additive generator in [12]. (ii) Let m ∈ N and T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) for all n  m and T (x1 , . . . , xm+1 )  Tm+1 (x1 , . . . , xm+1 ) for some (x1 , . . . , xm+1 ) ∈ [0, 1]m+1 . Then the rank of the additive generator t is m, i.e., Rank(t) = m.

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Let us first note that since the t-norm T is given by T (x, y) = t (−1) (t(x) + t(y)) = T2 (x, y) for all (x, y) ∈ [0, 1]2 we know that the rank of any additive generator is greater than or equal to 2. Remark 1. (i) Since T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) for some (x1 , . . . , xn ) ∈ [0, 1]n if and only if F(cˆ1 +· · ·+cˆn ) = cˆ1 ⊕· · ·⊕cˆn , for t(xi ) = cˆi , i = 1, . . . , n and the operation ⊕ can be derived from Range(t) only, we immediately see that Range(t1 ) = Range(t2 ) implies Rank(t1 ) = Rank(t2 ). (ii) The t-conorms are dual operations to t-norms, i.e., a binary operation S: [0, 1]2 →[0, 1] is a t-conorm if and only if the binary operation T : [0, 1]2 →[0, 1] given by T (x, y) = 1 − S(1 − x, 1 − y) is a t-norm. The t-conorms are associative, commutative, non-decreasing and 0 is their neutral element. Evidently, a t-conorm S is generated by an additive generator s: [0, 1]→[0, ∞] if and only if the dual t-norm T is generated by an additive generator t: [0, 1]→[0, ∞] such that t(x) = s(1−x) for x ∈ [0, 1]. For more details see [8]. This observation allows to define the rank also for additive generators of t-conorms, and evidently Rank(s) = Rank(t) whenever t(x) = s(1 − x) for all x ∈ [0, 1]. Due to the duality of t-norms and t-conorms, all results for the rank of additive generators of t-norms can be straightforwardly reformulated for the rank of additive generators of t-conorms, and vice versa. (iii) Not going into details, recall that uninorms [4] can also be generated by means of additive generators. However, each such additive generator is a continuous strictly monotone function and thus it always has an infinite rank. At this point we should also note that the value t(0) is not important and the generated t-norm is still the same if we put t(0) = t(0+ ). Therefore in the rest of this paper we will assume that t(0) = t(0+ ). Inspired by Proposition 2 in [16] we introduce the following proposition. Proposition 1. Let t 1 and t 2 be two additive generators of t-norms T 1 and T 2 , respectively. Denote Range∗ (t 1 ) = Range(t 1 ) \ {0} and Range∗ (t 2 ) = Range(t 2 ) \ {0}: (i) If Range∗ (t 1 ) ∩ [u, v] = Range∗ (t 2 ) for some u < v then Rank(t 1 )  Rank(t 2 ). (ii) If Range∗ (t 1 ) = cˆ · Range∗ (t 2 ) for some positive constant cˆ then Rank(t 1 ) = Rank(t 2 ). ˆ where 2t 1 (0)  cˆ < ∞ then Rank(t 2 ) = min(Rank(t 1 ), (iii) If Range∗ (t 2 ) = Range∗ (t 1 ) ∪ (Range∗ (t 1 ) + c), 1 − 1 ˆ (0) ). (t (1 ) + c)/t Proof. Validity of (ii) is obvious. For (i) let us assume that Rank(t 1 ) > Rank(t 2 ) = n−1. Then there exist (x1 , . . . , xn ) ∈ ]0, 1[n such that T 2 (x1 , . . . , xn )  Tn2 (x1 , . . . , xn ). Since Range(t 2 ) ⊆ Range(t 1 ) there exist (y1 , . . . , yn ) ∈ ]0, 1[n such that t 2 (xi ) = t 1 (yi ) for i = 1, . . . , n. Also t 2 (0) = t 1 (z) and t 2 (0+ ) = t 1 (z + ) for some z ∈ [0, 1]. If T 2 (x1 , . . . , xn ) = 0 then T 1 (y1 , . . . , yn )  z and T 2 (x1 , . . . , xn )  Tn2 (x1 , . . . , xn ) implies t 2 (x1 ) + · · · + t 2 (xn ) < t 2 (0+ ) and thus t 1 (y1 ) + · · · + t 1 (yn ) < t 1 (z + ). Then (t 1 )(−1) (t 1 (y1 ) + · · · + t 1 (yn )) > z which is a contradiction. Thus T 2 (x1 , . . . , xn ) = q > 0. Let cˆi = t 2 (xi ) = t 1 (yi ) for i = 1, . . . , n. Then t 2 (T 2 (x1 , . . . , xn )) = cˆ1 ⊕· · ·⊕ cˆn = t 1 (T 1 (y1 , . . . , yn )) = t 1 (Tn1 (y1 , . . . , yn ))  t 2 (Tn2 (x1 , . . . , xn )). Therefore for cˆ1 + · · · + cˆn = cˆ we have ˆ  t 2 ((t 2 )(−1) (c)), ˆ t 1 ((t 1 )(−1) (c)) ˆ > v and thus which means cˆ > v. However, then t 1 ((t 1 )(−1) (c)) t 2 (T 2 (x1 , . . . , xn )) > v, which is a contradiction. In (iii) it is obvious that Rank(t 1 ) Rank(t 2 ). Assume y ∈ [0, 1] such that t 1 (0) = t 2 (y). We have T 2 (y, y) = y since 2 ˆ For all  > 0 there exists a y > y such that cˆ + t 1 (1 − ) = t 2 (y ). If Rank(t 2 ) = n t (y − ) = cˆ + t 1 (1− ) and 2t 2 (y) c. 2 2 then we have t (y ) n · t (y) for all  > 0. Since t 2 (y ) = cˆ + t 1 (1 − ) and t 2 (y) = t 1 (0) we get cˆ + t 1 (1 − )  n · t 1 (0) ˆ 1 (0) ) > Rank(t 2 ) = n − 1. and for  → 0 we get n  (cˆ + t 1 (1− ))/t 1 (0) . Assume min(Rank(t 1 ), (t 1 (1− ) + c)/t

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In such a case there exist (x1 , . . . , xn ) ∈ [0, 1]n such that T 2 (x1 , . . . , xn )  Tn2 (x1 , . . . , xn ). We know that at most one of t 2 (xi ) ∈ / Range(t 1 ) since otherwise there would be T 2 (x1 , . . . , xn ) = Tn2 (x1 , . . . , xn ) = 0. If for all i = 1, . . . , n we have t 2 (xi ) ∈ Range(t 1 ) then let t 2 (xi ) = t 1 (yi ) for all i = 1, . . . , n and some (y1 , . . . , yn ) ∈ [0, 1]n . Further if T 1 (y1 , . . . , yn ) = 0 we have T 2 (x1 , . . . , xn ) = y and t 2 (xi )  t 2 (y) = t 1 (0) and thus t 2 (x1 ) + · · · + t 2 (xn )  n · t 2 (y). If t 2 (x1 ) + · · · + t 2 (xn ) > t 2 (y − ) then n · t 2 (y) > t 2 (y − ) and thus n · t 1 (0) > cˆ + t 1 (1− ) and   cˆ + t 1 (1− ) , n> t 1 (0) which is a contradiction. This means that t 2 (x1 ) + · · · + t 2 (xn ) < t 2 (y) and t 1 (y1 ) + · · · + t 1 (yn ) < t 1 (0) and thus the rank of t 1 is smaller than n which is a contradiction. Therefore T 1 (y1 , . . . , yn ) = z > 0 and we have t 1 (y1 ) + · · · + t 1 (yn ) ∈ [t 1 (z + ), t 1 (z − )], where t 1 (z + ) = t 2 (q + ), t 1 (z − ) = t 2 (q − ), t 1 (z) = t 2 (q) for some q ∈ [0, 1]. Then t 2 (x1 ) + · · · + t 2 (xn ) = t 1 (y1 ) + · · · + t 1 (yn ) ∈ [t 2 (q + ), t 2 (q − )] and Tn2 (x1 , . . . , xn ) = q. Further we have t 1 (T 1 (y1 , . . . , yn )) = t 2 (T 2 (x1 . . . , xn )) and thus t 2 (T 2 (x1 , . . . , xn )) = t 1 (z) = t 2 (q) and since t 2 is strictly decreasing we also have T 2 (x1 . . . , xn ) = q. Therefore the only possibility is that there is exactly one i ∈ {1, . . . , n} such that t 2 (xi ) ∈ / Range(t 1 ). Without any loss of generality we can assume that i = 1. n Then there exist (y1 , . . . yn ) ∈ [0, 1] such that t 1 (y1 ) + cˆ = t 2 (x1 ) and t 2 (xi ) = t 1 (yi ) for all i = 2, . . . , n. Now if T 1 (y1 , . . . , yn ) = z we have t 1 (y1 ) + · · · + t 1 (yn ) ∈ [t 1 (z + ), t 1 (z − )] and thus t 2 (x1 ) + · · · + t 2 (xn ) = cˆ + t 1 (y1 ) + · · · + t 1 (yn ) ∈ [cˆ + t 1 (z + ), cˆ + t 1 (z − )]. Moreover, there is a q ∈ [0, 1] such that t 1 (z + ) + cˆ = t 2 (q + ), t 1 (z − ) + cˆ = t 2 (q − ) and t 1 (z) + cˆ = t 2 (q). Therefore Tn2 (x1 , . . . , xn ) = q. If T 1 (y2 , . . . , yn ) = b we have t 1 (y2 ) + · · · + t 1 (yn ) ∈ [t 1 (b+ ), t 1 (b− )] and thus t 2 (x2 ) + · · · + t 2 (xn ) = t 1 (y2 ) + · · · + t 1 (yn ) ∈ [t 2 (w + ), t 2 (w − )], 2 (x , . . . , x ) = w = T 2 where t 1 (b+ ) = t 2 (w + ), t 1 (b− ) = t 2 (w − ), t 1 (b) = t 2 (w). Similarly as before Tn−1 2 n (x2 , . . . , xn ). Now

T 2 (x1 , . . . , xn ) = (t 2 )(−1) (t 2 (x1 ) + t 2 (w)) and t 2 (x1 ) + t 2 (w) = t 1 (y1 ) + cˆ + t 1 (b) ∈ [cˆ + t 1 (z + ), cˆ + t 1 (z − )]. Therefore T 2 (x1 , . . . , xn ) = q.  2.1. Strong additive generators We will first mention a few results about strong generators. Note that some of the following statements can also be derived from [15]. Since any additive generator is a monotone function, we know that it has countably many points of discontinuity. Let us denote the set of these points of discontinuity by Q = {qa }a∈A , where A is an index set which is either finite or countably infinite. Let us further denote Ia = [t(qa+ ), t(qa− )] for all a ∈ A.

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Proposition 2. Let t be an additive generator of a t-norm T such that the range of t is relatively closed, i.e., for all x, y ∈ [0, 1] we have t(x) + t(y) ∈ Range(t) ∪ [t(0), ∞]. Then the rank of t is equal to infinity. Proof. If the range of t is closed then for t(T (x, y)) we have either t(T (x, y)) = t(x) + t(y) or t(T (x, y)) = t(0) in the case when t(x) + t(y) > t(0). Thus for any n ∈ N and any (x1 , . . . , xn ) ∈ [0, 1]n we have T (x1 , . . . , xn ) = t (−1) (t(T (x1 , . . . , xn−1 )) + t(xn )) and either t(T (x1 , . . . , xn−1 )) = t(x1 ) + · · · + t(xn−1 ) or t(T (x1 , . . . , xn−1 )) = t(0) in the case when t(x1 ) + · · · + t(xn−1 ) > t(0). Therefore T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) in the case when t(x1 ) + · · · + t(xn−1 )  t(0). If t(x1 ) + · · · + t(xn−1 ) > t(0) then T (x1 , . . . , xn ) = 0 = Tn (x1 , . . . , xn ).  Now we will show the connection between existence of non-trivial idempotent elements and the Archimedean property of a generated t-norm. Just note that in general the Archimedean property implies that there are no non-trivial idempotent elements, but not vice versa. For example the left-continuous ordinal sum of the zero t-subnorm, where Z (x, y) = 0 for all (x, y) ∈ [0, 1]2 , and the product t-norm ( 0, 21 , Z , 21 , 1, TP ) has no non-trivial idempotent elements, but it is not Archimedean. Proposition 3. If T is a generated t-norm which has no idempotent elements then it is Archimedean. (n)

Proof. Let us suppose the converse. Then for some x ∈ ]0, 1[ we have limn→∞ x T = q > 0, but T (q, q) < q. Therefore T (q + , q + ) > q and t(q + ) > 0 for all 1 − q >  > 0. Then there exists an 1 such that t(q + 1 ) + t(q + 1 ) > t(q + ) and thus T (q + 1 , q + 1 ) q which is a contradiction.  Further we will show that a strong additive generator cannot generate a t-norm with a non-trivial idempotent element. Note that any idempotent point of a t-norm is necessarily a point of discontinuity of its additive generator. This is due to the fact that if an idempotent point b > 0 would be a point of continuity of an additive generator t then it would be t(b) + t(b) = t(b) and thus t(b) = 0 which implies b = 1. Lemma 1. Let t be an additive generator of a t-norm T such that the rank of t is infinity. Then T is Archimedean, which (n) means that it has no non-trivial idempotent elements and limn→∞ x T = 0 for all x ∈ [0, 1[. (n)

Proof. If t is a strong additive generator we have x T = t (−1) (n · t(x)). Since for any x < 1 there is t(x) > 0 we have (n) limn→∞ n · t(x) = ∞ and thus limn→∞ x T = 0.  Lemma 1 shows that an additive generator with an infinite rank generates an Archimedean t-norm. However, the converse is not true. In the following example we introduce a t-norm which is Archimedean, but its additive generator has a finite rank. Example 1. Let a ∈ ]0, 41 [ and t : [0, 1] → [0, ∞] be given by ⎧ 3 − 4ax if x ∈ [0, 41 ], ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 3−x if x ∈ ] 41 , 21 [, ⎪ ⎪ ⎨ t(x) = (4a − 1)x + 49 − 2a if x ∈ [ 21 43 ], ⎪ ⎪ ⎪ ⎪ ⎪ (4a − 1)x + 2 − 4a if x ∈ ] 43 , 1[, ⎪ ⎪ ⎪ ⎩ 0 if x = 1. Then t is strictly decreasing with t(1) = 0. For all x, y, z ∈ ] 43 , 1[ we have T (x, y) = 21 , T ( 21 , z) = 41 , i.e., T (T (x, y), z) = 41 . However, T3 (x, y, z) = 0, since t(x) + t(y) + t(z)3. Therefore the rank of t is Rank(t) = 2. On the other hand, t is an additive generator of the t-norm T : [0, 1]2 → [0, 1] which is Archimedean and it

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3.5

1 0.8

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0.2 1.5 0.5 0.25

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0.75

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Fig. 1. The additive generator t from Example 1, the t-norm T generated by t and the strong additive generator t 2 of T for a = 18 .

is given by

T (x, y) =

⎧ min(x, y) ⎪ ⎪ ⎪ ⎪ ⎪ ⎨1 2

if max(x, y) = 1, if (x, y) ∈ ] 43 , 1[2 ,

⎪ ⎪ max((1 − 4a)(x + y) + 6a − 45 , 41 ) if (x, y) ∈ [ 21 , 43 ]×] 43 , 1[∪ ] 43 , 1[×[ 21 , 43 ], ⎪ ⎪ ⎪ ⎩ 0 otherwise.

Note that if we assume this additive generator for a = 18 then the generated t-norm will be given by ⎧ min(x, y) if max(x, y) = 1, ⎪ ⎪ ⎪ ⎪  2 ⎪ ⎪ 3 ⎪1 ⎪ ⎨ if (x, y) ∈ ,1 , 4 T[ 1 ] (x, y) = 2

     8 ⎪ ⎪ 1 3 3 3 1 3 x +y−1 1 ⎪ ⎪ , if (x, y) ∈ , × , 1 ∪ , 1 × , , max ⎪ ⎪ 2 4 2 4 4 4 2 4 ⎪ ⎪ ⎩ 0 otherwise, see Fig. 1. Let t 2 : [0, 1]→[0, ∞] be defined by ⎧7 x if x ∈ [0, 41 ], ⎪ 2 − 2 ⎪ ⎪ ⎪ ⎪ 13 ⎪ − x if x ∈ ] 41 , 21 [, ⎪ ⎪ ⎨ 4 t 2 (x) = 49 − x2 if x ∈ [ 21 , 43 ], ⎪ ⎪ ⎪ 3 x ⎪ ⎪ if x ∈ ] 43 , 1[, ⎪ 2 − 2 ⎪ ⎪ ⎩ 0 if x = 1. Then t 2 is a strong additive generator of the t-norm T[ 1 ] . 8

Remark 2. In the above results we have shown that every non-trivial idempotent point of a t-norm T is necessarily a point of discontinuity of every additive generator t of T . Moreover, we know that there are only countably many points of discontinuity of a monotone function and therefore every generated t-norm can only have countably many idempotent points. In [2,9,10] it was shown that the minimum t-norm cannot be generated by a continuous function. In [8] the same result was shown for functions with a range relatively closed under addition. From our results we

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immediately see that the minimum t-norm cannot be generated by a non-continuous function with a range not relatively closed under addition either. 2.2. Cancellative and conditionally cancellative t-norms The cancellativity of a t-norm is an important property and cancellative t-norms have several special features. We say that a t-norm T is cancellative if T (x, y) = T (x, z) implies y = z for all x, y, z ∈ [0, 1], x  0. A t-norm T is conditionally cancellative if T (x, y) = T (x, z) > 0 implies y = z for all x, y, z ∈ [0, 1]. From [12] we know that cancellative t-norms can be generated only by additive generators that have a closed range, i.e., by strong additive generators. There exist conditionally cancellative t-norms which have no additive generator with a relatively closed range. For example if a t-norm T has a piece-wise linear additive generator t where Range(t) = {0} ∪ ]80, 84[ ∪ {86} ∪ ]92, 96[ ∪ {100} ∪ ]124, 128[ ∪ {132} ∪ ]140, 141[ ∪ {142} ∪ ]160, 228[ ∪ {230} ∪ ]232, 240] then T has no additive generator with a relatively closed range. This t-norm is Archimedean and the additive generator t is strong. Theorem 1. If t is an additive generator of a conditionally cancellative t-norm T then t is a strong generator. Proof. If T is a conditionally cancellative t-norm then for all points b of continuity of t the point T (x, b) is again a point of continuity of T for all x ∈ [0, 1] (when we take also 0 as a point of continuity) and thus t(x) + t(b) = t(T (x, b)). This is due to the fact that if T (x, b) = q > 0, where q is a point of discontinuity of T, then t(x) + t(b) ∈ [t(q + ), t(q − )] and there exists a point b  b such that t(x) + t(b ) ∈ [t(q + ), t(q − )]. This means that T (x, b) = T (x, b ) = q, which is a contradiction with the conditional cancellativity of T . Further assume that the rank of t is n − 1. Then there exist (x1 , . . . , xn ) ∈ ]0, 1[n with n > 2 such that T (x1 , . . . , xn )  Tn (x1 , . . . , xn ). If t(xi ) + t(x j ) ∈ Range(t) for some i, j ∈ {1, . . . , n}, i  j, then t(xi ) + t(x j ) = t(z) for some z ∈ [0, 1] and we can substitute T (xi , x j ) by z and t(xi ) + t(x j ) by t(z). Therefore the rank of t is smaller than n − 1 which is a contradiction. This means that t(xi ) + t(x j ) ∈ / Range(t) for all i, j ∈ {1, . . . , n}, i  j. Also t(xi ) + t(x j ) < t(0) for all i, j ∈ {1, . . . , n}, i  j, since otherwise T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) = 0. / Range(t). Let T (x1 , x2 ) = z > 0. If there exists a y < 1 such that Thus for i = 1, j = 2 we have t(x1 ) + t(x2 ) ∈ T (x1 , x2 , y) > 0 then there exists also a point of continuity b such that T (x1 , x2 , b) > 0 (because there are only countably many points of discontinuity). Then t(T (x1 , x2 , b)) = t(x1 ) + t(x2 ) + t(b) and also t(T (z, b)) = t(z)+t(b) and thus t(x1 )+t(x2 ) = t(z) which is a contradiction. Therefore T (x1 , x2 , y) = 0 for all y < 1 and we have T (x1 , x2 , x3 ) = 0. That is why T3 (x1 , x2 , x3 ) > 0 and for a point of continuity b1 with x3 < b1 < 1 we have t (−1) (t(x1 ) + t(x2 ) + t(x3 ))t (−1) (t(x1 ) + t(x2 ) + t(b)) = T (x1 , x2 , b) = 0 which is a contradiction.  2.3. Generators with finite rank If an additive generator t has no point of discontinuity then its range is relatively closed and we know that its rank is infinity. Therefore for additive generators with finite ranks the set of points of discontinuity Q = {qa }a∈A is not empty. Evidently, the lengths of intervals Ia = [t(qa− ), t(qa+ )] play the most important role in the determination of the rank of an additive generator. We will now introduce some useful lemmas.

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Lemma 2. Let t be an additive generator of a t-norm T . Then for all a ∈ A such that there exists an xa ∈ [0, 1[ with t(qa ) + t(xa ) ∈ Ia we have Rank(t)  (sup Ia − t(qa ))/t(xa ) + 1. Proof. Let t(qa ) + t(xa ) ∈ Ia for some a ∈ A and xa ∈ [0, 1[. Then T (qa , xa ) = t (−1) (t(qa ) + t(xa )) = t (−1) (t(qa )) = qa . Moreover, then T (qa , T (xa , xa )) = T (T (qa , xa ), xa ) = T (qa , xa ) = qa (n)

and by induction we can show for all n ∈ N that T ((xa )T , qa ) = qa . Assume n > (sup Ia − t(qa ))/t(xa ). We get n · t(xa ) + t(qa ) > sup Ia and thus t (−1) (n · t(xa ) + t(qa )) < qa , which means that the rank of t is smaller than n + 1.  Corollary 1. Let t be an additive generator of a t-norm T . Let X be the set of all pairs (x, a), x ∈ [0, 1[, a ∈ A, such that t(qa ) + t(x) ∈ Ia . Denote r x,a = (sup Ia − t(qa ))/t(x) + 1 for all (x, a) ∈ X . Then for r = inf (x,a)∈X r x,a we have Rank(t) r . Corollary 2. Consider the set X from the previous corollary. For a given a ∈ A let Sa = {x ∈ [0, 1[ | (x, a) ∈ X }. Define X 1 = {(x, a) | a ∈ A, x = inf Sa }. Denote ra∗ = (sup Ia − t(qa ))/t(x) + 1 for all a ∈ A, (x, a) ∈ X 1 . Then for r ∗ = inf a∈A ra∗ we have r ∗ = r = inf (x,a)∈X r x,a and thus Rank(t) r ∗ . Assume sets X, X 1 from Corollary 2. It is clear that for additive generators with infinite ranks we have X = X 1 = ∅. In Example 1 we have also seen an Archimedean t-norm with an additive generator t with a finite rank, where the set X related to t was empty. It is evident that if the set X related to an additive generator t is empty then the t-norm T generated by t has no idempotent elements and Proposition 3 implies that T is Archimedean. 3. Ranks of additive generators continuous at point 1 The class of additive generators continuous at point 1 is very important. These additive generators have special properties which are then translated to similar properties of t-norms generated by these generators. First we introduce the following lemma which can be derived from results in [15]. Lemma 3 (Viceník [15]). If an additive generator of a t-norm T is continuous at point 1 and if there exists a point of discontinuity of t then there exists a point of discontinuity qa ∈ ]0, 1[ with T (qa , qa ) = qa , i.e., qa is an idempotent element of T . This also means that for Ia we have t(qa ) + t(qa ) ∈ Ia . Lemma 3 shows us that an additive generator which is continuous at point 1 generates an Archimedean t-norm if and only if it is continuous everywhere. The following example shows that if an additive generator t is continuous at point 1 not all of its points of discontinuity need to be idempotent elements of the t-norm T generated by t. Example 2. Let t be given by ⎧ 1−x if x ∈ [ 21 , 1], ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 5 − 2x if x ∈ ] 1 , 1 [, 2 4 2 t(x) = 21 1 ⎪ ⎪ if x = ⎪ 10 4, ⎪ ⎪ ⎩7 2 − 2x else, see Fig. 2. Then t is continuous at point 1 and there are two points of discontinuity 21 , 41 where T ( 21 , 21 ) = 21 and T ( 41 , 41 ) = 0. 21 31 = 10 This additive generator has rank 2 since T ( 41 , 21 , 21 ) = 41 because of T ( 41 , 21 ) = 41 , but 2t( 21 ) + t( 41 ) = 2 21 + 10 31 and t (−1) ( 10 ) = 15 .

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A. Mesiarová-Zemánková / Fuzzy Sets and Systems 160 (2009) 2032 – 2048 3.5 1 3 0.8 2.5 0.6 2 0.4 1.5 0.2 1 0.5

0.5

0.25

0.5

0.75

0

1

0.2

0.4

0.6

0.8

1

Fig. 2. The additive generator t from Example 2 and the t-norm T generated by t.

The t-norm T generated by t is given by ⎧ 

⎪ 1 ⎪ ⎪ , x + y − 1 max ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ 

⎪ ⎪ 3 ⎪ ⎪ ⎪ max 0, x + y − ⎪ ⎪ 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

if (x, y) ∈ 

2

1 ,1 2

,

1 1 2 , , 4 2  1 2 if (x, y) ∈ 0, , 4  1 1 1 × , , if (x, y) ∈ 0, 4 4 2  1 1 1 T (x, y) = 0 if (x, y) ∈ × 0, , , ⎪ ⎪ 4 2 4 ⎪ ⎪ ⎪

  ⎪ ⎪ y 1 1 1 ⎪ ⎪ max 0, x + − if (x, y) ∈ 0, × ,1 , ⎪ ⎪ ⎪ 2 2 4 2 ⎪ ⎪ ⎪ 

 ⎪ ⎪ 1 x 1 1 ⎪ ⎪ if (x, y) ∈ , max 0, y + − , 1 × 0, ⎪ ⎪ ⎪ 2 2 2 4 ⎪ ⎪ ⎪

  ⎪ ⎪ 1 1 1 1 y 1 ⎪ ⎪ max if (x, y) ∈ × ,x + − , ,1 , ⎪ ⎪ ⎪ 4 2 2 4 2 2 ⎪ ⎪ ⎪ 

 ⎪ ⎪ 1 1 x 1 1 1 ⎪ ⎪ if (x, y) ∈ . ,y+ − ,1 × , ⎩ max 4 2 2 2 4 2 if (x, y) ∈

If we assume that all points of discontinuity of t are idempotent points of the t-norm T generated by t then we get the following result. Theorem 2. Let t be an additive generator continuous at point 1 such that for all a ∈ A we have T (qa , qa ) = qa . Then for each a ∈ A we have inf Sa = qa and therefore X 1 = {(qa , a)}a∈A . Thus ra∗ = sup Ia /t(qa ), a ∈ A. Moreover, Rank t = inf a∈A ra∗ .

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Proof. Let the rank of t be n − 1. If inf a∈A ra∗ < n − 1 then inf a∈A ra∗ < n − 1 which means that there exists an (n−1) a ∈ A such that n − 1 > ra∗ = sup Ia /t(qa ). Thus (n − 1)t(qa ) > sup Ia , i.e., t (−1) ((n − 1)t(qa )) < qa = (qa )T which means that the rank of t is smaller than n − 1. Assume inf a∈A ra∗ > n − 1. Since t has rank n − 1, we have T (x 1 , . . . , xn )  Tn (x1 , . . . , xn ) for some (x1 , . . . , xn ) ∈ ]0, 1[n . Because of the commutativity of T we can assume x1  · · ·  xn . For every two xi , x j ∈ {x1 , . . . , xn } we have t(xi ) + t(x j ) < t(0) since otherwise we get T (x1 , . . . , xn ) = Tn (x1 , . . . , xn ) = 0. If there exist xi and x j , i < j, such that t(xi ) + t(x j ) ∈ Range(t) then t(xi ) + t(x j ) = t(v) for some v ∈ ]0, 1[ and we get T (v, x1 , . . . , xi−1 , xi+1 , . . . , x j−1 , x j+1 , . . . , xn ) = T (x1 , . . . , xn )  Tn (x1 , . . . , xn ) = Tn−1 (v, x1 , . . . , xi−1 , xi+1 , . . . , x j−1 , x j+1 , . . . , xn ). Thus in such a case the rank of t would be smaller than n −1 which is a contradiction. Therefore we have t(x1 )+t(x2 ) < / Range(t) and then surely t(x1 )+t(x2 ) ∈ Ia for some a ∈ A. Here T (qa , qa ) = qa = T (x1 , x2 ) = t(0) and t(x1 )+t(x2 ) ∈ T (qa , 1) and we have qa  x1 , x2 . Since x1  · · ·  xn we get T (x1 , x2 ) = qa = T (qa , 1) = T (qa , x3 ) = T (x1 , x2 , x3 ) = · · · = T (x1 , . . . , xn ). Then since T (x1 , . . . , xn )  Tn (x1 , . . . , xn ) we have Tn (x1 , . . . , xn ) < qa because of T (x1 , . . . , xn )  T (x1 , x2 ) = qa . Therefore t(x1 ) + · · · + t(xn ) n · t(qa ) implies qa > Tn (x1 , . . . , xn )  t (−1) (n · t(qa )) which means that ra∗ < n since (n) (qa )T = qa . Then inf a∈A ra∗ < n and thus inf a∈A ra∗  n − 1 which is a contradiction.  However, to determine the rank of an additive generator in general it is not enough to know the length |Ia | of the interval Ia for all a ∈ A such that qa is an idempotent point. If we have for example two additive generators t 1 and t 2 with Range(t 1 ) = [0, 21 ] ∪ ] 23 , 2] ∪ ] 25 , 3] and Range(t 2 ) = [0, 21 ] ∪ ] 23 , 2] ∪ ]3, 27 ] then both T 1 and T 2 have just one idempotent point t 1 (x) = 0.5 = t 2 (y), but the first additive generator has rank 2 and the second has rank 3. Remark 3. If the rank of t is n − 1 then there exist (x 1 , . . . , xn ) ∈ ]0, 1[n such that T (x1 , . . . , xn )  Tn (x1 , . . . , xn ) but for any (n − 1)-tuple the equality is satisfied. Thus for any proper subset Y ⊂ {x1 , . . . , xn } with Y = {y1 , . . . , ym }, / Range(t), t(y1 ) + · · · + t(ym ) < t(0). Denote Z = X \ Y, Z = {z 1 , . . . , z n−m }. m  2, we have t(y1 ) + · · · + t(ym ) ∈ Then for T (y1 , . . . , ym ) = b we have Tn (x1 , . . . , xn )  Tm−n+1 (z 1 , . . . , z m−n , b). Proposition 4. Let t be an additive generator of a t-norm T continuous at point 1. Then if qa is a point of discontinuity of t we have t(qa− ) > t(qa ). Proof. Suppose that t(qa− ) = t(qa ) and let d = t(qa− ) − t(qa+ ). Since t is continuous at point 1 there exists an , 0 <  < d, such that t(x ) =  for some x ∈ [0, 1]. The properties of limits ensure that there exists an a > qa such that t(a ) +  > t(qa+ ). Then we have T (a , x ) = qa . There is T (qa , x ) = t (−1) (t(qa ) + t(x )) and since t(qa− ) = t(qa ) we have t(qa ) + t(x ) = t(qa− ) + t(x ) > t(qa− ) and we get T (qa , x ) = z < qa . Assume a point of continuity v of the additive generator t with z < v < qa . Set t(v) − t(qa ) = d2 . Now again there exists an x with t(x ) = , 0 <  < d2 . Further t(x ) + t(qa ) < d2 + t(qa ) = t(v) and thus qa > T (qa , x ) > v. Therefore x  < x and t(x ) > t(x ). Now similarly as before let a be such that t(a ) +  > t(qa+ ). We have T (x , a ) = qa and since t(x ) =  <  < d = t(qa− )−t(qa+ ) we have also T (a , x ) = qa . Thus z = T (qa , x ) = T (a , x , x ) = T (qa , x ) > v which is a contradiction. That is why t(qa− ) > t(qa ).  The previous proposition implies the following result from [15]. Lemma 4 (Viceník [15]). A left-continuous additive generator of a t-norm T is necessarily continuous.

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Proposition 4 shows us that if t is discontinuous, but continuous at point 1 then the set X (defined in Corollary 1) is surely not empty. Therefore the only additive generators which are continuous at point 1 with X = ∅ are those continuous. Corollary 3. If qa is a point of discontinuity of an additive generator t which is continuous at point 1 then there exists an x ∈ ]0, 1[ such that T (qa , x) = qa . Proposition 5. Let t be an additive generator of a t-norm T continuous at point 1. For all qa , a ∈ A we denote by xa the value inf{x ∈ [0, 1] | T (x, qa ) = qa }. Then T (xa , xa ) = xa and for the set X 1 defined in Corollary 2 we have X 1 = {(xa , a)}a∈A . Proof. From Proposition 4 we see that for every qa we have t(qa ) < t(qa− ) and t(xa ) > 0. If T (qa , qa ) = qa then xa = qa . Otherwise T (qa , qa ) < qa . For  > 0 such that t(xa+ ) > t(xa + ) > t(xa+ )/2 we have T (xa + , xa + )  xa and also T (xa + , qa ) = qa . Thus qa = T (xa + , xa + , qa )  T (xa , qa ) qa and we have T (xa , qa ) = qa . Therefore if T (xa , xa ) = v < xa we get qa = T (xa , xa , qa ) = T (v, qa ) < qa , which is a contradiction. Thus T (xa , xa ) = xa . Further if there is an x < xa such that t(x) + t(qa ) ∈ Ia then T (x, qa ) = qa which is a contradiction and therefore X 1 = {(xa , a)}a∈A .  Remark 4. Let t be an additive generator continuous at point 1 and let s ∈ ]0, 1[ be a point of discontinuity of t such that t is continuous on ]s, 1]. Then T (s, s) = s and for the set S = [0, t(0)] \ a∈A Ia we have S = {1} ∪ b∈B Ib ∪ {t(0)}, where B is an index set and for all b ∈ B the set Ib is an open interval such that the additive generator t is continuous on the interval Ib . Then: (i) We have |Ib |  t(s + ) for all b ∈ B. This holds since for any x, y > s such that t(x) + t(y) > t(s + ) and t(x) + t(y)  t(s) and any z we have T (x, y, z) = T (s, z) = t (−1) (t(s) + t(z)). If for some interval of continuity Ib we have |Ib | > t(s + ) then there is a z, t(z) ∈ Ib such that sup Ib − t(s + ) > t(z) and thus t(T (z, s)) ∈ Ib . Therefore t(t (−1) (t(s)+t(z))) = t(s)+t(z). On the other hand for x, y > s such that t(s + ) < t(x)+t(y) < sup Ib −t(z) there is also t(T (x, z)) = t(x)+t(z) and t(T (x, y, z)) = t(x)+t(y)+t(z). Thus we obtain t(s)+t(z) = t(x)+t(y)+t(z) which is a contradiction. It is now evident that if there is a finite number of points of discontinuity then t(0) must be finite. (ii) For all a ∈ A there is t(qa− ) − t(qa ) t(s). Indeed, if qa is any point of discontinuity then t(qa− )  t(qa ) and thus (n) there is an x such that t(qa ) + t(x) t(qa− ). Then T (qa , x T ) = qa for all n ∈ N. Let n ∈ N be such that nt(x) > (n) + − t(s ). If t(qa ) − t(qa ) < t(s) then T (qa , x T ) = T (qa , s) < qa which is a contradiction. (iii) If qa is a point of discontinuity of t then for all x ∈ ]0, 1[ also y = T (qa , x) is a point of discontinuity of t. This is due to the fact that if T (qa , x ) = qa for some x ∈ ]0, 1[ then also T (y, x ) = T (qa , x, x ) = T (qa , x) = y. (iv) If T (qa , x) = y we have also T (qa+ , x) = y. This is due to the fact that otherwise if T (qa+ , x) > y we have t(qa+ ) + t(x) < t(y + ) and there exists a point of continuity p ∈ [0, 1] such that t(qa+ ) + t(x) < t( p) < t(y + ). Denote d = t(y + ) − t( p). Then there exists an x p with t(x p ) < min(d, t(qa− ) − t(qa+ )). Further T (qa+ , x) > p and T ( p, x p ) > y. However, T (qa+ , x p , x) = T (qa , x) = y which is a contradiction. (v) If T (x1 , y1 ) = qa for some a ∈ A and t(x1 ) ∈ Ib1 , t(y1 ) ∈ Ib2 with b1 , b2 ∈ B then T (x, y) = qa for all x, y with t(x) ∈ Ib1 , t(y) ∈ Ib2 , x  x 1 and y  y1 . This is due to the fact that |Ib1 |  t(s + ) and |Ib2 |  t(s + ) imply that there exist x and y such that t(x1 ) + t(x ) = t(x) and t(y1 ) + t(y ) = t(y) and since |Ia |  t(s) we have T (x, y) = T (x1 , y1 , x , y ) = qa . Thus we have Ib1 + Ib2 ⊆ Ib ∪ [t(qa+ ), t(qa− )], for all b1 , b2 ∈ B where b ∈ B and sup Ib = t(qa+ ). We have seen that for all a ∈ A and for all x ∈ ]0, 1[ the point y = T (qa , x) = T (qa+ , x) is a point of discontinuity of t. However, the point T (qa− , x), x ∈ ]0, 1[ or even the point T (qa− , qa− ) can be a point of continuity of t. For example for 27 16 , 5 ], for a point qa with t(qa− ) = 23 we have 2t(qa− ) = 3 = t( p), an additive generator with range [0, 21 ] ∪ ] 23 , 2] ∪ ] 10 where p is a point of continuity of t. 4. Rank of additive generators which generate the same t-norm Remark 5. Let T be a t-norm which is generated and continuous at point (1, 1). Then each of its additive generators is continuous at point 1 [15]. Assume the set W = {x ∈ ]0, 1[ | there exists a y ∈ ]0, 1[, T (x, y) = x}. Then for any additive generator t of T we have Q = W .

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Further, if there exists an s such that [s, 1]∩ W = {s} then T (s, s) = s due to Proposition 5 and all additive generators of T must coincide (up to a positive multiplicative constant) on [s, 1]. This is due to the fact that T |[s,1] is a linear transformation of some continuous Archimedean t-norm. However, this does not mean that every additive generator of T has the same rank. Assume additive generator

1 − x for x ∈ [ 21 , 1], t 1 (x) = 2 − x else (see [15]) and additive generator

1 − x for x ∈ [ 21 , 1], 2 t (x) = 4 − x else. Both additive generators generate the same t-norm, they coincide on [ 21 , 1], but the first has rank 3 and the second has rank 7. Note that if cˆ  23 and

cˆ

t (x) =

1 − x for x ∈ [ 21 , 1], cˆ − x else,

then t cˆ generates the same t-norm as t 1 and t 2 and rank of t cˆ is (cˆ − 21 )/ 21 = 2cˆ − 1 . Thus for any n ∈ N, n 2, we can find a cˆn such that t cˆn has rank n. As we have mentioned, for any t-norm T continuous at point (1, 1) with [s, 1] ∩ W = {s} we can suppose that all its additive generators coincide on [s, 1]. Let t 1 and t 2 be two additive generators of T . Let B 1 (B 2 ) be a set (defined in Remark 4) related to t 1 (t 2 ). Then for all b1 ∈ B 1 (b2 ∈ B 2 ) there exists a b2 ∈ B 2 (b1 ∈ B 1 ) such that (t 1 )(−1) (Ib1 ) = (t 2 )(−1) (Ib2 ). Therefore there is a one-to-one mapping between B 1 and B 2 . Moreover, for each b1 ∈ B 1 there exists a cˆb1 ∈ R such that t 1 (x) = t 2 (x) + cˆb1 for all x ∈ (t 1 )(−1) (Ib1 ). This is due to the fact that for any y, z ∈ (t 1 )(−1) (Ib1 ) there exists x > s such that t 1 (z) = t 1 (y) +t 1 (x) and since t 1 (x) = t 2 (x) also t 2 (z) = t 2 (y) +t 1 (x). Therefore t 1 (z) − t 2 (z) = t 1 (y) − t 2 (y) for all y, z ∈ (t 1 )(−1) (Ib1 ) and we can just take t 1 (z) − t 2 (z) = cˆb1 . Remark 6. (i) It is clear that additive generators with the same range which do not coincide necessarily generate different t-norms. In the other case there is (t 1 )−1 (u ⊕ v) = (t 2 )−1 (u ⊕ v) for all u, v from the given range. Thus if v = 0 there is (t 1 )−1 (u) = (t 2 )−1 (u). Now if u = t 1 (x)  t 2 (x) = v then t 2 (y) = u for some x  y. Further (t 1 )−1 (u) = x  y = (t 2 )−1 (u) which is a contradiction. (ii) Let T (qa , qa ) = qa for some a ∈ A and let t 1 (qa− ) + cˆ  2t 1 (qa ) and 2t 1 (qa− ) + min(0, c) ˆ t 1 (0) for some positive 2 1 2 1 constant c. ˆ Then if t (x) = t (x) for x qa and t (x) = cˆ + t (x) for x < qa the additive generator t 2 generates the same t-norm as the additive generator t 1 . 4.1. Cantor t-conorm In this section we study the t-conorm C generated by the pseudo-inverse of the Cantor function, which was introduced in [16]. Because of the duality between t-norms and t-conorms we can derive the same results for the t-norm T dual to C. We choose to work with a t-conorm in this case since in the area of functions related to the Cantor function it is more transparent to work with increasing functions. The t-conorm C is continuous at point (1, 1) but there is no such s that [0, s] ∩ W = s. This means that 0 is the accumulation point of points of discontinuity for all additive generators of C. Moreover, any additive generator of C has (countably) infinitely many points of discontinuity.

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The Cantor function c∗ : [0, 1]→[0, 1] is given by ⎧ 0 if x = 0, ⎪ ⎪ ⎪ ⎨ m n x  zi  i c∗ (x) = if x ∈ ]0, 1[, x = , ⎪ i i 2 3 ⎪ i=1 i=1 ⎪ ⎩ 1 if x = 1, where n is either finite or infinite, xi ∈ {0, 1, 2},

min{i ∈ N | xi = 1} if {i ∈ N | xi = 1}  ∅, m= n if {i ∈ N | xi = 1} = ∅ and z i = xi /2 for i < m (for all i ∈ N if m = ∞) and z m = 1. Note that when we assume the dyadic expansion of a number, or expansions with other bases, there are always numbers which allow both finite and infinite expansions. In such a case we assume the finite expansion always. The pseudo-inverse to the Cantor function is a strictly decreasing, left-continuous function c: [0, 1]→[0, 1] given by ⎧ 0 if x = 0, ⎪ ⎪ ⎪ ⎨ n n x  zi  i c(x) = if x ∈ ]0, 1[, x = , i i ⎪ 3 2 ⎪ i=1 i=1 ⎪ ⎩ 1 if x = 1, where n is either finite or infinite, xi ∈ {0, 1}, and z i = 2xi for i < n (for all i ∈ N if n = ∞) and z n = xn . The t-conorm C is given by C(x, y) = c∗ (c(x) + c(y)) for all x, y ∈ [0, 1]. The additive generator c has rank 2 since for example C( 21 , 21 ) = 21 and thus C( 21 , 21 , 21 ) = 21 but 3c( 21 ) = 3 · 13 = 1 and c∗ (1) = 1  21 . The question is whether there can exist an additive generator of C with rank higher than 2. The answer is yes, indeed for any n ∈ N, n 2 there exists an additive generator of C with rank n. We will demonstrate the construction of such an additive generator for n = 3. For higher ranks the construction will by analogical. Assume a Cantor function which is not based on bases 3 and 2, but on bases 4 and 2. This means that such a Cantor function c4∗ is given by ⎧ 0 if x = 0, ⎪ ⎪ ⎪ ⎨ m n x  zi  i c4∗ (x) = if x ∈ ]0, 1[, x = , ⎪ i i ⎪ i=1 2 i=1 4 ⎪ ⎩ 1 if x = 1, where n is either finite or infinite, xi ∈ {0, 1, 2, 3},

min{i ∈ N | 0 < xi < 3} if {i ∈ N | 0 < xi < 3}  ∅, m= n if {i ∈ N | 0 < xi < 3} = ∅ and z i = xi /3 for i < m (for all i ∈ N if m = ∞) and z m = 1. The pseudo-inverse to the Cantor function c4∗ is a strictly decreasing left-continuous function c4 : [0, 1]→[0, 1] given by ⎧ 0 if x = 0, ⎪ ⎪ ⎪ ⎨ n n x  zi  i c4 (x) = if x ∈ ]0, 1[, x = , ⎪ i i 4 2 ⎪ i=1 ⎪ ⎩ i=1 1 if x = 1, where n is either finite or infinite, xi ∈ {0, 1}, and z i = 3xi for i < n (for all i ∈ N if n = ∞) and z n = xn . Let C 4 (x, y) = c4∗ (c4 (x) + c4 (y)) for all (x, y) ∈ [0, 1]2 .

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Summarizing, the process how C 4 (x, y) for x, y ∈ ]0, 1[ with x = {0, 1} and m, n ∈ N ∪ ∞ is computed is as follows:

n i=1

xi /2i and y =

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m i=1

yi /2i with xi , yi ∈

1. First x and y are mapped to the system with base 4 to c4 (x) and c4 (y). 2. c4 (x) and c4 (y) are summed up together in base 4. The result of the summation will be denoted by u. 3. u is truncated, so that everything after the first occurrence of number 1 or 2 in sequence {u i } is set to 0, i.e., if the result of the truncation is z then z i = 0 for all i > i 0 , where i 0 is the smallest index (if it exists) where u j ∈ {1, 2} (if u i0 = 2 then z i0 = 1). If such an index does not exist no truncation is made. 4. The result of the truncation is mapped back to the system with base 2 by the inverse to c4 . Thus it is clear that if c4 (C 4 (x, y))  c4 (x) + c4 (y) we have c4 (C 4 (x, y))c4 (x) + c4 (y) and the question is how big can the difference between them be. The following cases can occur: Case 1: If for all i  max(n, m) we have either xi = yi = 0 or xi  yi then i 0 = min(n, m) and after truncation z k = 3xk + 3yk for all k < i 0 and z i0 = 1. In this case the difference between c4 (x) + c4 (y) and c4 (C 4 (x, y)) is smaller than 1/2i0 . Case 2: There exists an i such that xi = yi > 0. In this case we take the smallest such index i 1 . If i 1 = n = m we have u k = 3xk + 3yk for all k < i and u i1 = 2 and u i = 0 for all i > i 1 . Here i 0 = i 1 and thus it will be truncated to z i1 = 1. In such a case the difference between c4 (x) + c4 (y) and c4 (C 4 (x, y)) is 1/2i0 . Case 3: If i 1 = n < m (the case i 1 = m < n is analogous) then u i1 = 0. If p is the biggest index, p < i 1 such that x p + y p = 0 we have i 0 = p, u p = 1 and u k = 3xk + 3yk for all k < i 0 . The difference between c4 (x) + c4 (y) and c4 (C 4 (x, y)) is smaller than 1/2i1 . If there is no such index p (which means that the result before truncation is greater than 1) the result is 1 and the truncation difference is again smaller than 1/2i1 . Case 4: If i 1 < min(n, m) in this case it can happen that xi = yi also for i > i 1 . Since 6 = 1 · 4 + 2 we know that maximally 1 can be added to the lower index (the only case when 2 could be added to the lower index is when xi = yi = 1 for all i > n for some n ∈ N; however, in such a case both x and y admit finite dyadic expansions). Thus indices bigger than i 1 are irrelevant. Now again let p be the biggest index, p < i 1 such that x p + y p = 0. Then we have i 0 = p and u p = 1 and u k = 3xk + 3yk for all k < p and the difference between c4 (x) + c4 (y) and c4 (C 4 (x, y)) is smaller than 1/2i1 −1 . In all mentioned cases the difference between c4 (x)+c4 (y) and c4 (C 4 (x, y)) is either smaller than 1/2i0 or C 4 (x, y) = 1. In all cases we also have i 0  min(n, m). We want to show that the function C 4 generated by the additive generator c4 is the same as the t-conorm C generated by c. Here the same principle holds as described before for c4 , thus we need only to check all four cases mentioned above. Case 1 yields clearly the same result. In Case 2 we have i 1 = n = m with u k = 2xk + 2yk for all k < i and u i1 = 2 and u i = 0 for all i > i 1 . Here also in the system with base 3 we have i 0 = i 1 and thus it will be truncated to u i1 = 1. In Case 3 we have i 1 = n < m (or analogously i 1 = m < n). In the system with base 3 u i = 0 and if p is the biggest index, p < i 1 such that x p + y p = 0 we have i 0 = p and u p = 1. If there is no such index p (which means that the result before truncation is greater than 1) the result is 1. We see that this is the same as in the system with base 4. In Case 4 we have i 1 < min(n, m). In this case xi = yi can happen also for i > i 1 . Since 4 = 1 · 3 + 1 we know that maximally 1 can be added to the lower index (the only case when 2 is added to the lower index is when xi = yi = 1 for all i > n for some n ∈ N; however, in such a case both x and y admit finite dyadic expansions). Thus the indices bigger than i 1 are irrelevant. Now again let p be the biggest index, p < i 1 such that x p + y p = 0. We have i 0 = p and u p = 1 which is again the same as in the case of the system with base 4. Therefore we know that C 4 is the same t-conorm as C. Finally we prove that the additive generator c4 is of rank 3. Since C 4 ( 21 , 21 ) = 21 also C 4 ( 21 , 21 , 21 , 21 ) = 21 . However, c4 ( 21 ) = 41 and 4 · c( 21 ) = 1 and since c4∗ (1) = 1  21 for the rank of c4 we have Rank(c4 )  3. Assume that Rank(c4 ) < 3. Then there exist (x, y, z) ∈ ]0, 1[3 such that (−1)

c4

(c4 (x) + c4 (y) + c4 (z))  C 4 (x, y, z) = a.

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Thus c4 (x) + c4 (y) + c4 (z) > c4 (a + ) and c4 (v) + c4 (z) ∈ [c4 (a − ), c4 (a + )] where v = C 4 (x, y). If C 4 (x, y) = 1 then (−1)

C 4 (x, y, z) = c4

(c4 (x) + c4 (y) + c4 (z)) = 1.

Assume C 4 (x, y), C 4 (x, z), C 4 (y, z) < 1. We know that the difference between c4 (v) and c4 (x) + c4 (y) is smaller than 1/2i0 , where i 0 is connected with C 4 (x, y) (i.e., it is the last index in its dyadic expansion). Now for i 2 connected with C 4 (v, z) we know that c4 (a + ) − c4 (a − ) = 2/2i2 and also that i 2 i 0 . Then we have c4 (v) c4 (x) + c4 (y) c4 (v) +

1 2i 0

and also c4 (v) + c4 (z) c4 (a − ) +

1 2i 2

thus together c4 (x) + c4 (y) + c4 (z) c4 (v) + c4 (z) +

1 1 1 2 c4 (a − ) + i + i  c4 (a − ) + i = c4 (a + ). i 0 0 2 2 2 2 22

Therefore (−1)

c4

(c4 (x) + c4 (y) + c4 (z)) = a,

which is a contradiction. Thus Rank(c4 ) = 3. 5. Conclusion and open problems In this paper we have studied the ranks of additive generators. The basic idea of additive generators, where the two-place function is expressed by means of a one-place function in order to decrease the computational complexity, often cannot be translated to the aggregation of more inputs in the case of discontinuous generators. In our work we study additive generators of triangular norms which allow simple computation of n-ary extensions of t-norms for all n  m for some m ∈ N, m 2. We have shown that cancellative and conditionally cancellative t-norms possess only additive generators with infinite ranks. We have shown several examples and conditions for computation of ranks of additive generators. We have also shown that the pseudo-inverse to the Cantor function c is an additive generator of rank 2 and that the t-conorm C generated by c has an additive generator with rank m for any natural number m 2. However, there are still some unsolved problems in this investigation. We have the following open problems. Open problem 1. Let T be an additively generated t-norm with an additive generator t which has a finite rank. Assume any m ∈ N, m  2. Does there exist an additive generator t m of the t-norm T such that Rank(t m ) = m? Open problem 2. Characterize Archimedean t-norms possessing only strong additive generators. We have seen that the class of t-norms mentioned in Open problem 2 contains cancellative and conditionally cancellative t-norms and it is a proper subclass of the class of all Archimedean t-norms. We also know that if an additive generator is continuous at point 1 then it generates an Archimedean t-norm only if it is continuous. Therefore the question is which Archimedean t-norms discontinuous at point (1, 1) possess only strong additive generators. Note that if for an additive generator t the set X (defined in Corollary 1) is not empty then we have T (x, qa ) = qa > 0 (n) for some x ∈ ]0, 1[ and some a ∈ A. Then also T (x T , qa ) = qa and thus   (n) (n) qa = T lim x T , qa  lim x T , n→∞

n→∞

which means that T is not Archimedean. Therefore we know that Archimedean t-norms are generated by exactly those additive generators for which X = ∅. Further if T is a t-norm discontinuous at point (1, 1) which is Archimedean, generated and has no zero divisors, then T has a strong additive generator if and only if it is cancellative. In the other case there is T (x, y) = T (x, z) > 0

A. Mesiarová-Zemánková / Fuzzy Sets and Systems 160 (2009) 2032 – 2048 (n)

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(n)

for some x, y, z ∈ [0, 1], x0, y < z. Then also T (x, yT ) = T (x, z T ) and if t is a strong additive generator of the t-norm T which has no zero divisors we have t(0) = ∞. Therefore for all n ∈ N there must be a cˆn ∈ [0, 1] such that t(x) + n · t(z), t(x) + n · t(y) ∈ [t(cˆn+ ), t(cˆn− )]. However, for n > t(1− )/(t(y) − t(z)) we obtain t(cˆn− ) − t(cˆn+ ) n(t(y) − t(z)) > t(1− ). This means that the set X is not empty and thus T is not Archimedean. Example 3. Let t: [0, 1]→[0, ∞] be given by ⎧ 0 if x = 1, ⎪ ⎪ ⎪ ⎨ 1 1 n−1 t(x) = n + 1 − x · 2 if x ∈ n , n−1 , ⎪ 2 2 ⎪ ⎪ ⎩ ∞ if x = 0. Then we have t(1/2n ) = n + 21 and t((1/2n−1 )− ) = n. The function t is an additive generator of the t-norm T given by 

1 x y 1 T (x, y) = max m+n , m + n − m+n−1 2 2 2 2 if x ∈ [1/2n , 1/2n−1 [, y ∈ [1/2m , 1/2m−1 [ and T (x, y) = min(x, y) if min(x, y) = 0 or max(x, y) = 1. This t-norm (k) (k) is Archimedean, since for every x ∈ [1/2n , 1/2n−1 [ we have x T ∈ [1/2k·n , 1/2k·n−1 [ and thus limk→∞ x T = 0. The p additive generator t has rank 2, but for every p ∈ N, p  2, there is an additive generator t which generates the t-norm T and has rank p. For example we can choose t p (x) = n + 2/ p − x · 2n / p for x ∈ [1/2n , 1/2n−1 [, with t p (1) = 0 and t p (0) = ∞. Thus we have t p (1/2n ) = n + 1/ p and t p ((1/2n−1 )− ) = n. However, according to the note above the t-norm T cannot possess a strong additive generator. For Archimedean t-norms with zero divisors we know the following facts: Proposition 6. Let T be a generated Archimedean t-norm which has zero divisors and is discontinuous at point (1, 1). (m) Then there exists an m ∈ N such that lim x→1− (x)T = 0. Further for any additive generator t of T we have either Rank(t) < m or Rank(t) = ∞. (n)

Proof. If T is Archimedean and has zero divisors then for each x ∈ ]0, 1[ there exists an n ∈ N such that x T = 0. If T is discontinuous at point (1, 1) then we have t(1− ) > 0 and there exists an  > 0 such that t(1−) > t(1− ) > t(1 − )/2 (n) and the point 1− is a point of continuity of t. Then T (1−, 1−)  T (1− , 1− )  1−. Therefore if (1−)T = 0 we have (2n) lim x→1− (x)T = 0. Thus m = 2n. Then if Rank(t)  m for every (x1 , . . . , xm ) ∈ ]0, 1[m we have T (x1 , . . . , xm ) = Tm (x1 , . . . , xm ) = 0. Thus for any k  m and any (x 1 , . . . , xk ) ∈ ]0, 1[k we have T (x1 , . . . , xk )  T (x1 , . . . , xm ) = 0 and Tk (x1 , . . . , xk ) Tm (x1 , . . . , xm ) = 0,, i.e., T (x1 , . . . , xk ) = Tk (x1 , . . . , xk ).  (3)

Proposition 7. Let T be a generated t-norm. If lim x→1− (x)T = 0 then there always exists a strong additive generator t of T . Proof. Let t be an additive generator of T . Suppose that it is not strong. If T (x, y, z) = t (−1) (t(x) + t(y) + t(z)) for all x, y, z ∈ ]0, 1[ then since T (x, y, z) = 0 the rank of t is infinity. Therefore there must exist x, y, z ∈ ]0, 1[ such that 0 = T (x, y, z) < t (−1) (t(x) + t(y) + t(z)) and thus t(x) + t(y) + t(z) < t(0). Then also 3t(1− ) < t(0) and thus / Rank(t). Let T (1− , 1− ) = qa for some point of discontinuity qa . Consider d = t(0) − 3t(1− ). Now if we 2t(1− ) ∈ 1 define t (x) = t(x) + d for x > qa and t 1 (x) = t(x) + 2d for x qa the additive generator t 1 will generate the same t-norm T . Indeed, for 1 > x, y > qa and T (x, y) = z qa we have t(x) + t(y) ∈ [t(z + ), t(z − )] and t 1 (x) + t 1 (y) = t(x) + t(y) + 2d ∈ [2d + t(z + ), 2d + t(z − )] = [t 1 (z + ), t 1 (z − )] and thus (t 1 )(−1) (t 1 (x) + t 1 (y)) = z. For 1 > x > qa  y we have T (x, y)  T (x, qa ) = T (x, 1− , 1− ) = 0 and thus t(x) + t(y)  t(0). Here we have t 1 (x) + t 1 (y) = t(x) + t(y) + 3d  t(0) + 3d = t 1 (0) + d  t 1 (0).

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This means that (t 1 )(−1) (t 1 (x) + t 1 (y)) = 0. In the case when qa  x, y we have t(x) + t(y)  t(0) and T (x, y) = 0. Then t 1 (x) + t 1 (y) = t(x) + t(y) + 4d  t(0) + 4d = t 1 (0) + 2d  t 1 (0) and (t 1 )(−1) (t 1 (x) + t 1 (y)) = 0. Finally we will show that t 1 is a strong generator. Indeed we have 3t 1 (1− ) = 3t(1− )+3d = t(0)+2d = t 1 (0) and thus 1 (t )(−1) (3t 1 (1− )) = 0. For all x, y, z ∈ ]0, 1[ we have t 1 (x)+t 1 (y)+t 1 (z)  3t 1 (1− ) and thus t 1 (x)+t 1 (y)+t 1 (z)  t 1 (0) and we have T (x, y, z) = t (−1) (t(x) + t(y) + t(z)) = 0 for all x, y, z ∈ ]0, 1[.  Acknowledgements The work on this paper was supported by Grants VEGA 2/71 42/27 and APVV-0071-06. The author is grateful to Professor Khurshid Ahmad, and the support of Trinity College Dublin is also kindly acknowledged. References [1] C. Alsina, M.J. Frank, B. Schweizer, Associative Functions: Triangular Norms and Copulas, World Scientific, Hackensack, London, Singapore, 2006. [2] V.I. Arnold, Concerning the representability of functions of two variables in the form X [(x) + (y)], Uspekhi Mat. Nauk 12 (1957) 119–121. [3] T. Calvo, A. Kolesárová, M. Komorníková, R. Mesiar, Aggregation operators: properties, classes and construction methods, in: T. Calvo, G. Mayor, R. Mesiar (Eds.), Aggregation Operators, Physica Verlag, Heidelberg, 2002, pp. 3–107. [4] J. Fodor, R. Yager, A. Rybalov, Structure of uninorms, Internat. J. Uncertainty Fuzziness Knowledge-Based Systems 5 (1997) 411–427. [5] S. Jenei, On discontinuous triangular norms, BUSEFAL 66 (1996) 38–42. [6] S. Jenei, On Archimedean triangular norms, Fuzzy Sets and Systems 99 (1998) 179–186. [7] E.P. Klement, R. Mesiar, E. Pap, Quasi- and pseudo-inverses of monotone functions, and the construction of t-norms, Fuzzy Sets and Systems 104 (1999) 3–13. [8] E.P. Klement, R. Mesiar, E. Pap, Triangular Norms, Trends in Logic, Studia Logica Library, Vol. 8, Kluwer Academic Publishers, Dordrecht, 2000. [9] G.M. Krause, Interior idempotents and non-representability of groupoids, Stochastica 7 (1983) 5–10. [10] C.M. Ling, Representation of associative functions, Publ. Math. Debrecen 12 (1965) 189–212. [11] K. Menger, Statistical metrics, Proc. Nat. Acad. Sci. U.S.A. 8 (1942) 535–537. [12] A. Mesiarová, Special classes of triangular norms, Ph.D. Thesis, Mathematical Institute of Slovak Academy of Sciences, Bratislava, 2005. [13] P.S. Mostert, A.L. Shields, On the structure of semigroups on a compact manifold with boundary, Ann. of Math. 65 (1957) 117–143. [14] B. Schweizer, A. Sklar, Statistical metric spaces, Pacific J. Math. 10 (1960) 313–334. [15] P. Viceník, Non-continuous additive generators of triangular norms, Ph.D. Thesis, Slovak University of Technology, Bratislava, 2002. [16] P. Viceník, Additive generators of border-continuous triangular norms, Fuzzy Sets and Systems 13 (2008) 1631–1645.