New methods to construct uninorms on bounded lattices

International Journal of Approximate Reasoning 115 (2019) 254–264

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International Journal of Approximate Reasoning www.elsevier.com/locate/ijar

New methods to construct uninorms on bounded lattices Gül Deniz Çaylı Department of Mathematics, Faculty of Science, Karadeniz Technical University, 61080 Trabzon, Turkey

a r t i c l e

i n f o

Article history: Received 14 May 2019 Received in revised form 6 October 2019 Accepted 7 October 2019 Available online 11 October 2019 Keywords: Bounded lattice Uninorm Idempotent uninorm Neutral element t-Norm t-Conorm

a b s t r a c t In this paper, we study idempotent uninorms on bounded lattices. We demonstrate that there need not always exist an idempotent uninorm on a bounded lattice L having the neutral element e ∈ L \{0, 1} which differs from the least and the greatest idempotent uninorms. In addition, we propose two new methods to obtain uninorms on bounded lattices with a neutral element, where some sufficient and necessary conditions on the underlying t-norms and t-conorms are required. In particular, two new types of idempotent uninorms on bounded lattices are obtained. We also give an illustrative example for clarity. © 2019 Elsevier Inc. All rights reserved.

1. Introduction Aggregation functions are nondecreasingly monotone, expressing the idea that an increase of any of the input values cannot decrease the output value. They are mainly used to obtain a single output value from several input values. For this reason, a lot of studies about aggregation functions have appeared in last decades [3]. They play a fundamental role in many application fields, such as classification systems, fuzzy rule-based systems, pattern recognition, decision making and image processing. Uninorms on the real unit interval were introduced by Yager and Rybalov [39] and studied by Fodor et al. [25]. They are a special kind of aggregation functions that are placed in the class of mixed aggregations. Thus, uninorms are extensively used in numerous applications, such as decision making, neural networks, expert systems, approximate reasoning, fuzzy system modeling, fuzzy sets and fuzzy logic [16,23,24,30,31,34,36,38,40,41]. Uninorms as an important generalization and unification of t-norms and t-conorms have a neutral element e lying anywhere in the real unit interval rather than at one or zero as in the case of t-norms and t-conorms, respectively. In particular, a uninorm U is a t-norm T (resp. t-conorm S) when the case e = 1 (resp. e = 0). For more details on the concepts t-norms, t-conorms and uninorms on the real unit interval [0, 1], we refer to [17–22,26,29,32,33]. Uninorms related to algebraic structures on bounded lattices have recently attracted much attention from researchers. Uninorms were defined on a bounded lattice L by Karaçal and Mesiar [28]. They showed that it is possible to choose an arbitrary element e ∈ L \{0, 1} as the neutral element and to construct a uninorm based on a t-norm T on L and a t-conorm S on L. In addition, the least and the greatest uninorms on bounded lattices with a neutral element were constructed. After then, Çaylı, Karaçal and Mesiar [8] introduced two new construction methods to obtain uninorms by using the knowledge of that t-norms and t-conorms on bounded lattices exist. As a consequence of these methods, it was demonstrated the presence of idempotent uninorms on a bounded lattice L for any element e ∈ L \{0, 1} playing the role of a neutral element.

E-mail address: [email protected]. https://doi.org/10.1016/j.ijar.2019.10.006 0888-613X/© 2019 Elsevier Inc. All rights reserved.

G.D. Çaylı / International Journal of Approximate Reasoning 115 (2019) 254–264

255

In particular, the least and the greatest idempotent uninorms on bounded lattices were obtained. Uninorms on lattices were also studied in the papers [5–7,9,11–15,35,37]. In this paper, we study uninorms (especially idempotent uninorms) with a neutral element on bounded lattices. Considering a bounded lattice L, we prove that an idempotent uninorm on L with the neutral element e ∈ L \{0, 1} different from the least and the greatest idempotent uninorms does not need to exist. Moreover, we introduce two new methods for constructing uninorms on L based on a t-norm T e on [0, e]2 and a t-conorm S e on [e , 1]2 under some additional constraints. These methods yield a new approach to the construction of idempotent uninorms on bounded lattices when considering an idempotent t-norm T e on [0, e]2 and an idempotent t-conorm S e on [e , 1]2 . The paper is organized as follows. First, we give some preliminaries about uninorms on bounded lattices in Section 2. In Section 3, we discuss the existence of idempotent uninorms on a bounded lattice L with the fixed neutral element e ∈ L \{0, 1}. We also show that an idempotent uninorm different from two extremal idempotent uninorms on L does not need to exist. In the meantime, we propose two new methods for building uninorms on bounded lattices with a neutral element, where some sufficient and necessary conditions on the underlying t-norm T e on [0, e]2 and t-conorm S e on [e , 1]2 are required. Our methods can be applied to construct new types of idempotent uninorms on some bounded lattices. Finally, some concluding remarks are added. 2. Preliminaries In this section, some preliminaries concerning bounded lattices and uninorms (t-norms, t-conorms) on them are recalled. A bounded lattice ( L , ≤) is a lattice which has the top and bottom elements, which are written as 1 and 0, respectively, i.e., two elements 1, 0 ∈ L exist such that 0 ≤ x ≤ 1, for all x ∈ L. Given x, y ∈ L, we use the notation x  y denote that x and y are incomparable. I e denotes the family of all incomparable elements with e ∈ L \{0, 1}, i.e., I e = {x ∈ L | x  e }. Denote A (e ) = [0, e] × [e , 1] ∪ [e , 1] × [0, e] for e ∈ L. Definition 1 ([4]). Let ( L , ≤, 0, 1) be a bounded lattice and a, b ∈ L such that a ≤ b. A subinterval [a, b] of L is defined as

[a, b] = {x ∈ L | a ≤ x ≤ b}. Similarly, we define ]a, b] = {x ∈ L | a < x ≤ b}, [a, b[ = {x ∈ L | a ≤ x < b} and ]a, b[ = {x ∈ L | a < x < b}. Definition 2 ([8,12,13,28]). Let ( L , ≤, 0, 1) be a bounded lattice. An operation U : L 2 → L is called a uninorm on L (shortly a uninorm, if L is fixed) if it is commutative, associative, increasing with respect to both variables and there is an element e ∈ L, called the neutral element, such that U (e , x) = x, for all x ∈ L. In particular, the uninorm U defined in Definition 2 is a t-norm T (resp. t-conorm S) when the case e = 1 (resp. e = 0) [1,2,10,27]. In the following, we give an example for some t-norms and t-conorms on a bounded lattice L, which will be used in the sequel. Example 1. Let ( L , ≤, 0, 1) be a bounded lattice. The least t-norm T W and the greatest t-norm T ∧ on L are given as

⎧ ⎨ b if a = 1, T W (a, b) = a if b = 1, ⎩

0 otherwise

and

T ∧ (a, b) = a ∧ b. The least t-conorm S ∨ and the greatest t-conorm S W on L are given as

S ∨ (a, b) = a ∨ b and

⎧ ⎨ b if a = 0, S W (a, b) = a if b = 0, ⎩

1 otherwise.

Proposition 1 ([28]). Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and U be a uninorm on L with the neutral element e. Then i) T e = U |[0, e ]2 : [0, e ]2 → [0, e ] is a t-norm on [0, e ]2 . ii) S e = U |[e , 1]2 : [e , 1]2 → [e , 1] is a t-conorm on [e , 1]2 .

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Proposition 2 ([28]). Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and U be a uninorm on L with the neutral element e. Then the following statements hold: i) a ∧ b ≤ U (a, b) ≤ a ∨ b for (a, b) ∈ A (e ). ii) U (a, b) ≤ a for (a, b) ∈ L × [0, e ]. iii) U (a, b) ≤ b for (a, b) ∈ [0, e ] × L. iv) a ≤ U (a, b) for (a, b) ∈ L × [e , 1]. v) b ≤ U (a, b) for (a, b) ∈ [e , 1] × L. Definition 3 ([8]). Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and U be a uninorm on L with the neutral element e. U is called an idempotent uninorm if U (a, a) = a for all a ∈ L. Proposition 3 ([8]). Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and U be an idempotent uninorm on L with the neutral element e. Then the following statements hold: i) U (a, b) = a ∧ b for all (a, b) ∈ [0, e ]2 . ii) U (a, b) = a ∨ b for all (a, b) ∈ [e , 1]2 . iii) a ∧ b ≤ U (a, b) ≤ a ∨ b for all (a, b) ∈ L 2 \([0, e ]2 ∪ [e , 1]2 ). Theorem 1 ([8]). Let ( L , ≤, 0, 1) be a bounded lattice and e ∈ L \{0, 1}. If T e : [0, e]2 → [0, e] is a t-norm on [0, e]2 and S e : [e , 1]2 → [e , 1] is a t-conorm on [e , 1]2 , then the functions U t : L 2 → L and U s : L 2 → L are uninorms on L with the neutral element e, where

⎧ T (a, b) if (a, b) ∈ [0, e ]2 , ⎪ ⎪ ⎨ e b if (a, b) ∈ [0, e ] × I e , U t (a, b) = a if (a, b) ∈ I e × [0, e ], ⎪ ⎪ ⎩ a∨b otherwise

and

⎧ S (a, b) if (a, b) ∈ [e , 1]2 , ⎪ ⎪ ⎨ e b if (a, b) ∈ [e , 1] × I e , U s (a, b) = a if (a, b) ∈ I e × [e , 1], ⎪ ⎪ ⎩ a∧b otherwise.

(1)

(2)

Corollary 1 ([8]). Let e ∈ L \{0, 1}. If we put T e = T ∧ on [0, e]2 and S e = S ∨ on [e , 1]2 in Theorem 1, the following uninorms, are, respectively, the greatest and the least idempotent uninorms on L with the neutral element e, respectively.

⎧ a ∧ b if (a, b) ∈ [0, e ]2 , ⎪ ⎪ ⎨ b if (a, b) ∈ [0, e ] × I e , U t∧ (a, b) = if (a, b) ∈ I e × [0, e ], ⎪a ⎪ ⎩ a ∨ b otherwise

and

⎧ a ∨ b if (a, b) ∈ [e , 1]2 , ⎪ ⎪ ⎨ b if (a, b) ∈ [e , 1] × I e , U s∨ (a, b) = a if (a, b) ∈ I e × [e , 1], ⎪ ⎪ ⎩ a ∧ b otherwise.

(3)

(4)

3. Idempotent uninorms After uninorms were introduced on the unit interval, it has recently been shown that the related and complex definitions on bounded lattices are challenging to obtain. Karaçal and Mesiar [28] extended the concept of uninorms to bounded lattices and showed the existence of these uninorms. It should be noted that the classes of uninorms on bounded lattices described in [28] are not idempotent, in general. Afterwards, Çaylı et al. [8] introduced two new classes of uninorms on a bounded lattice L with the neutral element e ∈ L \{0, 1} based on a t-norm on [0, e]2 and a t-conorm on [e , 1]2 . When considering that the only idempotent t-norm on [0, e]2 is infimum t-norm and the only idempotent t-conorm on [e , 1]2 is supremum t-conorm, they obtained the least and greatest idempotent uninorms on a general bounded lattice L. In this section, we demonstrate that an idempotent uninorm on a bounded lattice L having the neutral element e ∈ L \{0, 1} which differs from two extremal idempotent uninorms does not need to exist. By using the existence of a t-norm T e on [0, e]2 and a t-conorm S e on [e , 1]2 , we propose two constructions for uninorms on a bounded lattice L with a neutral element e ∈ L \{0, 1} under some additional assumptions on L. If we take the t-norm T e as T ∧ and the t-conorm S e as S ∨ , we obtain two new classes of idempotent uninorms. At the same time, an illustrative example is provided in order to show that these classes of idempotent uninorms are different from the least and the greatest idempotent uninorms.

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257

Fig. 1. Lattice L.

Proposition 4. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and U be an idempotent uninorm on L with the neutral element e. In this case, we have that U (a, b) ∈ L \{e }, for all a, b ∈ L \{e }. Proof. Suppose that U (a, b) = e for some elements a, b ∈ L \ {e }. Then we have that U (a, U (a, b)) = U (a, e ) = a and U (U (a, a) , b) = U (a, b) = e. This is a contradiction with the associativity of U . So, it cannot be U (a, b) = e for a, b ∈ L \ {e }. Therefore, we have that U (a, b) ∈ L \{e }, for all a, b ∈ L \{e }. 2 Theorem 2. Consider a bounded lattice L which is isomorphic to the lattice characterized in Hasse diagram in Fig. 1. In that case, there is no idempotent uninorm U on L with the neutral element e such that it differs from the least and the greatest idempotent uninorms. Proof. Let U be an arbitrary idempotent uninorm on the bounded lattice L, which is characterized in Hasse diagram in Fig. 1, for the indicated neutral element e. Due to the fact that e is the neutral element, U (e , x) = U (x, e ) = x for all x ∈ L. Since U is idempotent, U (x, x) = x for all x ∈ L. By Proposition 3, U (x, y ) = x ∧ y for all x, y ∈ [0, e] and U (x, y ) = x ∨ y for all x, y ∈ [e , 1]. By Proposition 3 and Proposition 4, we have that b ≤ U (b, c ) ≤ c, i.e., U (b, c ) ∈ {b, c }. The proof is split into all possible cases. Let U (b, c ) = b. 1. By Proposition 3 and Proposition 4, we have that b ≤ U (t , b) ≤ t, i.e., U (t , b) ∈ {b, n, t }. Suppose that U (t , b) = t. Then we have that U (t , U (b, c )) = U (t , b) = t and U (U (t , b) , c ) = U (t , c ) = 1. This is a contradiction with the associativity of U . So, it cannot be U (t , b) = t. Suppose that U (t , b) = n. Then we have that U (t , U (b, c )) = U (t , b) = n and U (U (t , b) , c ) = U (n, c ) = 1. This is a contradiction with the associativity of U . So, it cannot be U (t , b) = n. Hence, U (t , b) = b. 2. We obtain that b ≤ U (n, b) ≤ U (t , b) = b from the monotonicity of U and Proposition 3. So, U (n, b) = b. 3. By Proposition 3 and Proposition 4, we have that m ≤ U (m, n) ≤ n, i.e., U (m, n) ∈ {m, n}. Suppose that U (m, n) = n. Then we have that U (m, U (n, b)) = U (m, b) = 0 and U (U (m, n) , b) = U (n, b) = b. This is a contradiction with the associativity of U . So, it cannot be U (m, n) = n. So, U (m, n) = m. 4. We have that s = U (s, m) = U (s, U (m, n)) = U (U (s, m) , n) = U (s, n) from the associativity of U . So, U (s, n) = s. 5. By Proposition 3 and Proposition 4, we have that b ≤ U (1, b) ≤ 1, i.e., U (1, b) ∈ {b, n, t , c , 1}. Suppose that U (1, b) = n. Then, we obtain that U (1, U (b, c )) = U (1, b) = n and U (U (1, b) , c ) = U (n, c ) = 1. This is a contradiction with the associativity of U . So, it cannot be U (1, b) = n. Suppose that U (1, b) = t. Then, we obtain that U (1, U (b, c )) = U (1, b) = t and U (U (1, b) , c ) = U (t , c ) = 1. This is a contradiction with the associativity of U . So, it cannot be U (1, b) = t. Suppose that U (1, b) = c. Then, we obtain that U (1, U (b, t )) = U (1, b) = c and U (U (1, b) , t ) = U (c , t ) = 1. This is a contradiction with the associativity of U . So, it cannot be U (1, b) = c. Suppose that U (1, b) = 1. Then, we get that U (b, U (c , t )) = U (b, 1) = 1 and U (U (b, c ) , t ) = U (b, t ) = b. This is a contradiction with the associativity of U . So, it cannot be U (1, b) = 1. Thus, U (1, b) = b. 6. We have that 0 = U (s, b) = U (s, U (b, t )) = U (U (s, b) , t ) = U (0, t ) from the associativity of U . So, U (0, t ) = 0. 7. We obtain that U (0, k) ≤ U (0, t ) = 0 from the monotonicity of U . So, U (0, k) = 0. 8. We obtain that U (0, n) ≤ U (0, t ) = 0 from the monotonicity of U . So, U (0, n) = 0. 9. By Proposition 3 and Proposition 4, we have that m ≤ U (m, t ) ≤ t, i.e., U (m, t ) ∈ {m, n, t }. Suppose that U (m, t ) = n. Then we have that U (m, U (t , b)) = U (m, b) = 0 and U (U (m, t ) , b) = U (n, b) = b. This is a contradiction with the associativity of U . So, it cannot be U (m, t ) = n. Suppose that U (m, t ) = t. Then we have that U (m, U (t , b)) = U (m, b) = 0 and U (U (m, t ) , b) = U (t , b) = b. This is a contradiction with the associativity of U . So, it cannot be U (m, t ) = t. Hence, U (m, t ) = m. 10. We have that s = U (s, m) = U (s, U (m, t )) = U (U (s, m) , t ) = U (s, t ) from the associativity of U . Therefore, U (s, t ) = s. 11. By Proposition 3 and Proposition 4, we have that s ≤ U (c , s) ≤ c, i.e., U (c , s) ∈ {s, m, c }. Suppose that U (c , s) = c. Then we get that U (b, U (c , s)) = U (b, c ) = b and U (U (b, c ) , s) = U (b, s) = 0. This is a contradiction with the associativity of U . So, it cannot be U (c , s) = c. Suppose that U (c , s) = m. Then we obtain that U (c , U (s, s)) = U (c , s) = m and U (U (c , s) , s) = U (m, s) = s. This is a contradiction with the associativity of U . So, it cannot be U (c , s) = m. Thus, U (c , s) = s.

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Table 1 The least idempotent uninorm U on L. U

0

s

m

b

e

k

n

t

c

1

0 s m b e k n t c 1

0 0 0 0 0 0 0 0 0 0

0 s s 0 s s s s s s

0 s m 0 m s m m m m

0 0 0 b b 0 b b b b

0 s m b e k n t c 1

0 s s 0 k k k k k k

0 s m b n k n t 1 1

0 s m b t k t t 1 1

0 s m b c k 1 1 c 1

0 s m b 1 k 1 1 1 1

12. We have that s = U (t , s) = U (t , U (c , s)) = U (U (t , c ) , s) = U (1, s) from the associativity of U . So, U (1, s) = s. 13. We obtain that s ≤ U (k, s) ≤ U (1, s) = s by Proposition 3 and the monotonicity of U . Therefore, U (k, s) = s. 14. We obtain that U (b, k) ≤ U (b, t ) = b and U (b, k) ≤ U (e , k) = k from the monotonicity of U . That is, U (b, k) ≤ b ∧ k = 0. Hence, U (b, k) = 0. 15. We get that U (m, k) ≤ U (m, t ) = m and U (m, k) ≤ U (e , k) = k from the monotonicity of U . That is, U (m, k) ≤ m ∧ k = s. Moreover, by using Proposition 3, we have that s ≤ U (m, k). Therefore, U (m, k) = s. 16. By Proposition 3 and Proposition 4, we have that m ≤ U (m, c ) ≤ c, i.e., U (m, c ) ∈ {m, c }. Suppose that U (m, c ) = c. Then we get that U (m, U (c , b)) = U (m, b) = 0 and U (U (m, c ) , b) = U (c , b) = b. This is a contradiction with the associativity of U . So, it cannot be U (m, c ) = c. Thus, U (m, c ) = m. 17. We have that U (0, c ) ≤ U (s, c ) = s and U (0, c ) ≤ U (b, c ) = b from the monotonicity of U . That is, U (0, c ) ≤ s ∧ b = 0. So, U (0, c ) = 0. 18. We have that U (0, 1) ≤ U (s, 1) = s and U (0, 1) ≤ U (b, 1) = b from the monotonicity of U . That is, U (0, 1) ≤ s ∧ b = 0. So, U (0, 1) = 0. 19. By Proposition 3 and Proposition 4, we have that m ≤ U (m, 1) ≤ 1, i.e., U (m, 1) ∈ {m, n, c , t , 1}. Suppose that U (m, 1) = n. Then we get that U (t , U (m, 1)) = U (t , n) = t and U (U (t , m) , 1) = U (m, 1) = n. This is a contradiction with the associativity of U . So, it cannot be U (m, 1) = n. Suppose that U (m, 1) = c. Then we get that U (t , U (m, 1)) = U (t , c ) = 1 and U (U (t , m) , 1) = U (m, 1) = c. This is a contradiction with the associativity of U . So, it cannot be U (m, 1) = c. Suppose that U (m, 1) = t. Then we obtain that U (c , U (m, 1)) = U (c , t ) = 1 and U (U (c , m) , 1) = U (m, 1) = t. This is a contradiction with the associativity of U . So, it cannot be U (m, 1) = t. Suppose that U (m, 1) = 1. Then we obtain that U (b, U (m, 1)) = U (b, 1) = b and U (U (b, m) , 1) = U (0, 1) = 0. This is a contradiction with the associativity of U . So, it cannot be U (m, 1) = 1. Hence, U (m, 1) = m. 20. By Proposition 3 and Proposition 4, we have that k ≤ U (k, t ) ≤ t, i.e., U (k, t ) ∈ {k, t }. Suppose that U (k, t ) = t. Then we obtain that U (k, U (t , b)) = U (k, b) = 0 and U (U (k, t ) , b) = U (t , b) = b. This is a contradiction with the associativity of U . So, it cannot be U (k, t ) = t. Therefore, U (k, t ) = k. 21. By Proposition 3 and the monotonicity of U , k = U (k, e ) ≤ U (k, c ) ≤ 1, i.e., U (k, c ) ∈ {k, t , 1}. Suppose that U (k, c ) = t. Then we obtain that U (k, U (c , b)) = U (k, b) = 0 and U (U (k, c ) , b) = U (t , b) = b. This is a contradiction with the associativity of U . So, it cannot be U (k, c ) = t. Suppose that U (k, c ) = 1. Then we obtain that U (k, U (c , b)) = U (k, b) = 0 and U (U (k, c ) , b) = U (1, b) = b. This is a contradiction with the associativity of U . So, it cannot be U (k, c ) = 1. Therefore, U (k, c ) = k. 22. By Proposition 3, k ≤ U (k, 1) ≤ 1, i.e., U (k, 1) ∈ {k, t , 1}. Suppose that U (k, 1) = t. Then we obtain that U (k, U (1, b)) = U (k, b) = 0 and U (U (k, 1) , b) = U (t , b) = b. This is a contradiction with the associativity of U . So, it cannot be U (k, 1) = t. Suppose that U (k, 1) = 1. Then we obtain that U (k, U (1, b)) = U (k, b) = 0 and U (U (k, 1) , b) = U (1, b) = b. This is a contradiction with the associativity of U . So, it cannot be U (k, 1) = 1. Hence, U (k, 1) = k. 23. By Proposition 3 and the monotonicity of U , we have that k = U (k, e ) ≤ U (k, n) ≤ t, i.e., U (k, n) ∈ {k, t }. Suppose that U (k, n) = t. Then we obtain that U (k, U (n, b)) = U (k, b) = 0 and U (U (k, n) , b) = U (t , b) = b. This is a contradiction with the associativity of U . So, it cannot be U (k, n) = t. Therefore, U (k, n) = k. By the commutativity of U , if U (b, c ) = b, then we obtain that there is only one idempotent uninorm U on the bounded lattice L, which is characterized in Hasse diagram in Fig. 1, for the neutral element e. This idempotent uninorm on L is given by Table 1 and it is the least idempotent uninorm on L. Similarly, if U (b, c ) = c, then there is only one idempotent uninorm on the bounded lattice L, which is characterized in Hasse diagram in Fig. 1, for the neutral element e. This idempotent uninorm on L is given by Table 2 and it is the greatest idempotent uninorm on L. 2 Now, we investigate the existence of an idempotent uninorm on a bounded lattice such that it differs from the least and the greatest idempotent uninorms. In Example 2, considering the bounded lattice L 1 depicted in Hasse diagram in Fig. 2, we define an idempotent uninorm U having the indicated neutral element e which differs from two extremal idempotent uninorms on L 1 .

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259

Table 2 The greatest idempotent uninorm U on L. U

0

s

m

b

e

k

n

t

c

1

0 s m b e k n t c 1

0 0 0 0 0 k n t c 1

0 s s 0 s k n t c 1

0 s m 0 m k n t c 1

0 0 0 b b k n t c 1

0 s m b e k n t c 1

k k k k k k t t 1 1

n n n n n t n t 1 1

t t t t t t t t 1 1

c c c c c 1 1 1 c 1

1 1 1 1 1 1 1 1 1 1

Fig. 2. Lattice L 1 . Table 3 Idempotent uninorm U on L 1 . U

0

m

x

e

t

1

0 m x e t 1

0 m m 0 m 1

m m m m m 1

m m x x x 1

0 m x e t 1

m m x t t 1

1 1 1 1 1 1

Example 2. Consider a bounded lattice L 1 = {0, m, e , x, t , 1} with the given order in Fig. 2 and define a mapping U : L 1 × L 1 → L 1 by Table 3. Then U is an idempotent uninorm on L 1 with the neutral element e. In addition, it differs from the least and the greatest idempotent uninorms since U (x, t ) = x = t = U t∧ (x, t ) and U (0, x) = m = 0 = U s∨ (0, x). In the following Theorem 3 and Theorem 4, we propose two methods for generating uninorms on a bounded lattice L with the neutral element e ∈ L \{0, 1} based on the t-norm T e on [0, e]2 and the t-conorm S e on [e , 1]2 under some additional constraints. When considering T e = T ∧ and S e = S ∨ , two different idempotent uninorms on bounded lattices are obtained. Theorem 3. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and S e be a t-conorm on [e , 1]2 . S e (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[ if and only if the function U 1 : L 2 → L is a uninorm on L with the neutral element e, where

⎧ S (a, b) if (a, b) ∈ [e , 1]2 , ⎪ ⎪ ⎪ e ⎪ if (a, b) ∈ I e × [e , 1[ , ⎨a U 1 (a, b) = b if (a, b) ∈ [e , 1[ × I e , ⎪ ⎪ if (a, b) ∈ [0, e] × {1} ∪ {1} × [0, e] ∪ I e × {1} ∪ {1} × I e , ⎪a∨b ⎪ ⎩ a∧b otherwise.

(5)

Proof. Let S e (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[. We prove that U 1 is a uninorm on L with the neutral element e. In order to show that U 1 is a uninorm, we prove the monotonicity, associativity, commutativity and the fact that e is a neutral element of U 1 . The commutativity and the fact that e is a neutral element of U 1 can be easily seen. Note that U 1 differs from the uninorm U s given by the formula (2) in Theorem 1, only at points where a = 1 or b = 1. So, to prove the monotonicity and associativity of U 1 , we only consider the cases that at least one of the variables of U 1 is 1. Monotonicity: We demonstrate that if a ≤ b, then U 1 (a, c ) ≤ U 1 (b, c ) for all c ∈ L.

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Fig. 3. Lattice L 2 . Table 4 T-norm T e on [0, e]2 . Te

0

s

r

p

q

e

0 s r p q e

0 0 0 0 0 0

0 s s s s s

0 s s s s r

0 s s r r p

0 s s r r q

0 s r p q e

If a = 1, then b = 1. So, we have that for all c ∈ L, U 1 (a, c ) = U 1 (1, c ) = 1 = U 1 (1, c ) = U 1 (b, c ). If b = 1, then we have that for all a, c ∈ L, U 1 (a, c ) ≤ 1 = U 1 (1, c ) = U 1 (b, c ). If c = 1, then we have that for all a, b ∈ L, U 1 (a, c ) = U 1 (a, 1) = 1 = U 1 (b, 1) = U 1 (b, c ). Associativity: We prove that U 1 (a, U 1 (b, c )) = U 1 (U 1 (a, b), c ) for all a, b, c ∈ L. If a = 1 or c = 1, then we have that U 1 (a, U 1 (b, c )) = 1 = U 1 (U 1 (a, b), c ). If U 1 (a, b) = 1 or U 1 (b, c ) = 1, then it holds a = 1 or b = 1 or c = 1. In this case, we have that U 1 (a, U 1 (b, c )) = 1 = U 1 (U 1 (a, b), c ). Therefore, we obtain that U 1 is a uninorm on L with the neutral element e. Conversely, let U 1 be a uninorm on L with the neutral element e. Then we prove that S e (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[. Suppose that S e (x, y ) = 1 for x, y ∈ [e , 1[. In this case, we obtain that U 1 (0, U 1 (x, y )) = U 1 (0, S e (x, y )) = U 1 (0, 1) = 1 and U 1 (U 1 (0, x), y ) = U 1 (0 ∧ x, y ) = U 1 (0, y ) = 0 ∧ y = 0. This is a contradiction with the associativity of U 1 . So, S e (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[. 2 Remark 1. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and S e be a t-conorm on [e , 1]2 such that S e (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[. In Theorem 3, the uninorm U 1 defined by the formula (5) does not need to coincide with any prescribed t-norm T e except T ∧ on [0, e ]2 . Considering the lattice L 2 with the given order in Fig. 3, force U 1 | [0, e ]2 to be the t-norm T e : [0, e ]2 → [0, e ] defined by Table 4. By applying the construction approach in Theorem 3, we have U 1 ( p , U 1 (q, z)) = U 1 ( p , q ∧ z) = U 1 ( p , r ) = T e ( p , r ) = s and U 1 (U 1 ( p , q), z) = U 1 ( T e ( p , q), z) = U 1 (r , z) = r ∧ z = r for the elements p , q, z ∈ L 2 . Therefore, the associativity of U 1 is violated. So, the underlying t-norm of U 1 cannot be arbitrarily chosen except T ∧ . Corollary 2. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and S e be a t-conorm on [e , 1]2 . S e (x, y ) = S ∨ (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[ if and only if the function U 1∨ : L 2 → L is an idempotent uninorm on L with the neutral element e, where

⎧ a ∨ b if (a, b) ∈ [e , 1]2 ∪ [0, e] × {1} ∪ {1} × [0, e] ∪ I e × {1} ∪ {1} × I e , ⎪ ⎪ ⎨ a if (a, b) ∈ I e × [e , 1[ , U 1∨ (a, b) = if (a, b) ∈ [e , 1[ × I e , ⎪b ⎪ ⎩ a ∧ b otherwise.

(6)

Theorem 4. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and T e be a t-norm on [0, e ]2 . T e (x, y ) ∈ ]0, e] for all x, y ∈ ]0, e] if and only if the function U 2 : L 2 → L is a uninorm on L with the neutral element e, where

⎧ T e (a, b) if (a, b) ∈ [0, e ]2 , ⎪ ⎪ ⎪ ⎪ if (a, b) ∈ I e × ]0, e] , ⎨a U 2 (a, b) = b if (a, b) ∈ ]0, e] × I e , ⎪ ⎪ if (a, b) ∈ [e , 1] × {0} ∪ {0} × [e , 1] ∪ I e × {0} ∪ {0} × I e , ⎪a∧b ⎪ ⎩ a∨b otherwise.

(7)

G.D. Çaylı / International Journal of Approximate Reasoning 115 (2019) 254–264

261

Table 5 T-conorm S e on [e , 1]2 . Se

e

m

n

k

t

1

e m n k t 1

e m n k t 1

m k k t t 1

n k k t t 1

k t t t t 1

t t t t t 1

1 1 1 1 1 1

The result can be proved in a manner similar to the proof of Theorem 3, thus the proof is omitted. Remark 2. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and T e be a t-norm on [0, e ]2 such that T e (x, y ) ∈ ]0, e] for all x, y ∈ ]0, e]. In Theorem 4, the uninorm U 2 defined by the formula (7) does not need to coincide with any prescribed t-conorm S e except S ∨ on [e , 1]2 . Considering the lattice L 2 with the given order in Fig. 3, force U 2 | [e , 1]2 to be the t-conorm S e : [e , 1]2 → [e , 1] defined by Table 5. By applying the construction approach in Theorem 4, we have U 2 (m, U 2 (n, z)) = U 2 (m, n ∨ z) = U 2 (m, k) = S e (m, k) = t and U 2 (U 2 (m, n), z) = U 2 ( S e (m, n), z) = U 2 (k, z) = k ∨ z = k for the elements m, n, z ∈ L 2 . Therefore, the associativity of U 2 is violated. So, the underlying t-conorm of U 2 cannot be arbitrarily chosen except S ∨ . Corollary 3. Let ( L , ≤, 0, 1) be a bounded lattice, e ∈ L \{0, 1} and T e be a t-norm on [0, e ]2 . T e (x, y ) = T ∧ (x, y ) ∈ ]0, e] for all x, y ∈ ]0, e] if and only if the function U 2∧ : L 2 → L is an idempotent uninorm on L with the neutral element e, where

⎧ a ∧ b if (a, b) ∈ [0, e ]2 ∪ [e , 1] × {0} ∪ {0} × [e , 1] ∪ I e × {0} ∪ {0} × I e , ⎪ ⎪ ⎨ a if (a, b) ∈ I e × ]0, e] , U 2∧ (a, b) = b if (a, b) ∈ ]0, e] × I e , ⎪ ⎪ ⎩ a ∨ b otherwise.

(8)

Remark 3. Let e ∈ L \{0, 1}. (i) Observe that the existence of idempotent uninorms on an arbitrary bounded lattice was first demonstrated in [8]. The idempotent uninorm U s∨ introduced in Corollary 1 differs from the idempotent uninorm U 1∨ introduced in Corollary 2 on the domain [0, e] × {1} ∪ {1} × [0, e] ∪ I e × {1} ∪ {1} × I e . While U 1∨ has the value 1 on the domain [0, e] × {1} ∪ {1} × [0, e], U s∨ has the value x ∧ y. On the domain I e × {1} (resp. {1} × I e ), U s∨ has the value x (resp. y) instead of the value 1 in U 1∨ . Similarly, the idempotent uninorm U t∧ introduced in Corollary 1 differs from the idempotent uninorm U 2∧ introduced in Corollary 3 on the domain [e , 1] × {0} ∪ {0} × [e , 1] ∪ I e × {0} ∪ {0} × I e . While U 2∧ has the value 0 on the domain [e , 1] × {0} ∪ {0} × [e , 1], U t∧ has the value x ∨ y. On the domain I e × {0} (resp. {0} × I e ), U t∧ has the value x (resp. y) instead of the value 0 in U 2∧ . (ii) Observe that the uninorms U 1∨ and U 2∧ introduced in Corollaries 2 and 3, respectively, can be equivalently defined by

U 1∨ (a, b) = and

⎧ ⎨a∨b ⎩

1

ϕ (a) ∧ ϕ (b)

if (a, b) ∈ [e , 1]2 , if a = 1 or b = 1, otherwise

⎧ ⎨a∧b

if (a, b) ∈ [0, e ]2 , U 2∧ (a, b) = 0 if a = 0 or b = 0, ⎩ φ (a) ∨ φ (b) otherwise, where



ϕ (a) = and

 φ (a) =

1 if a ∈ [e , 1], a otherwise 0 a

if a ∈ [0, e ], otherwise.

(iii) Observe that U 1∨ = U s∨ and U 2∧ = U t∧ since U 1∨ (0, 1) = 1 = 0 = U s∨ (0, 1) and U 2∧ (0, 1) = 0 = 1 = U t∧ (0, 1). When considering that there is no element in I e and ]e , 1[, then we have U 1∨ = U t∧ . Similarly, when there is no element in I e and ]0, e[, then we have that U 2∧ = U s∨ . It should be pointed out that the idempotent uninorms U 1∨ and U 2∧ do not have to coincide with the idempotent uninorms U t∧ and U s∨ on L, respectively. Let us demonstrate this argument by the following example:

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Fig. 4. Lattice L 3 . Table 6 Idempotent uninorm U 1∨ on L 3 . U 1∨

0

k

e

m

t

s

1

0 k e m t s 1

0 0 0 0 0 0 1

0 k k k k k 1

0 k e m t s 1

0 k m m k m 1

0 k t k t t 1

0 k s m t s 1

1 1 1 1 1 1 1

Table 7 Idempotent uninorm U 2∧ on L 3 . U 2∧

0

k

e

m

t

s

1

0 k e m t s 1

0 0 0 0 0 0 0

0 k k m t s 1

0 k e m t s 1

0 m m m s s 1

0 t t s t s 1

0 s s s s s 1

0 1 1 1 1 1 1

Table 8 The greatest idempotent uninorm U t ∧ on L 3 . U t∧

0

k

e

m

t

s

1

0 k e m t s 1

0 0 0 m t s 1

0 k k m t s 1

0 k e m t s 1

m m m m s s 1

t t t s t s 1

s s s s s s 1

1 1 1 1 1 1 1

Example 3. Given the bounded lattice L 3 = {0, k, e , m, t , s, 1} with the given order in Fig. 4. Note that T ∧ (x, y ) ∈ ]0, e] for all x, y ∈ ]0, e] and S ∨ (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[. By using the formulas (6) in Corollary 2 and (8) in Corollary 3, respectively, the idempotent uninorms U 1∨ : L 3 × L 3 → L 3 and U 2∧ : L 3 × L 3 → L 3 are defined by Table 6 and Table 7, respectively. The greatest idempotent uninorm U t∧ and the least idempotent uninorm U s∨ on L 3 with the neutral element e are defined in Tables 8 and 9. It is easy to see that the idempotent uninorms U 1∨ and U 2∧ are different from not only the greatest idempotent uninorm U t∧ but also the least idempotent uninorm U s∨ on L 3 . Furthermore, the idempotent uninorms U 1∨ and U 2∧ differ from each other. Remark 4. Consider a bounded lattice ( L , ≤, 0, 1) and the set U of all uninorms on L with the following order: For any uninorms U , V ∈ U ,

U ≤ V ⇔ U (a, b) ≤ V (a, b) for all a, b ∈ L .

G.D. Çaylı / International Journal of Approximate Reasoning 115 (2019) 254–264

263

Table 9 The least idempotent uninorm U s∨ on L 3 . U s∨

0

k

e

m

t

s

1

0 k e m t s 1

0 0 0 0 0 0 0

0 k k k k k k

0 k e m t s 1

0 k m m k m m

0 k t k t t t

0 k s m t s 1

0 k 1 m t 1 1

It can be easily shown that U is a partially ordered set with the top element S W : L 2 → L and the bottom element T W : L 2 → L. Similarly, each U (e ) is a partially ordered set such that U (e ) denotes the set of all uninorms on L with the neutral element e ∈ L. It is clear that U s∨ ≤ U 1∨ ≤ U t∧ and U s∨ ≤ U 2∧ ≤ U t∧ . Further, we have U s∨ ≤ U s ≤ U 1 , U 2 ≤ U t ≤ U t∧ , U 1∨ ≤ U 1 and U 2 ≤ U 2∧ . U 1 and U s are not necessarily comparable with U t∧ and U 1∨ , respectively. Considering the bounded lattice L 2 in Fig. 3 and the t-conorm S e : [e , 1]2 → [e , 1] defined by Table 5, it holds U 1  U t∧ and U s  U 1∨ . Because, U 1 ( p , m) = p < m = U t∧ ( p , m), U 1 (k, m) = U s (k, m) = S e (k, m) = t > k = U 1∨ (k, m) = U t∧ (k, m) and U s (0, 1) = 0 < 1 = U 1∨ (0, 1). U s∨ and U 2∧ are not necessarily comparable with U 2 and U t , respectively. Considering the bounded lattice L 2 in Fig. 3 and the t-norm T e : [0, e]2 → [0, e] defined by Table 4, it holds U s∨  U 2 and U 2∧  U t . Because, U 2 ( p , m) = m > p = U s∨ ( p , m), U 2 ( p , r ) = U t ( p , r ) = T e ( p , r ) = s < r = U 2∧ ( p , r ) = U s∨ ( p , r ) and U t (0, 1) = 1 > 0 = U 2∧ (0, 1). Similarly, it can be seen that U 1 , U 2 , U s and U t are not necessarily comparable with U t , U s , U t∧ and U s∨ , respectively. Note that the uninorm U 1∨ cannot be comparable with the uninorm U 2∧ , in general. Considering the bounded lattice L 3 characterized in Hasse diagram in Fig. 4, the uninorms U 1∨ and U 2∧ defined by Table 6 and Table 7, respectively, are incomparable with each other on L 3 . Because, we have that U 1∨ (0, 1) = 1 > 0 = U 2∧ (0, 1) and U 1∨ (k, m) = k < m = U 2∧ (k, m). Similarly, it can be seen that the uninorm U 1 cannot be comparable with the uninorm U 2 , in general. 4. Concluding remarks After the definition of uninorms on bounded lattices and the demonstration of the existence of these uninorms in [28], two new classes of uninorms on bounded lattices were defined by using the knowledge of that a t-norm and a t-conorm exist on bounded lattices in [8]. In the meantime, the least and the greatest idempotent uninorms on bounded lattices L with the neutral element e ∈ L \{0, 1} were obtained based on the fact that the only idempotent t-norm on L is infimum t-norm and the only idempotent t-conorm on L is supremum t-conorm. In this paper, we investigated the presence of different classes of idempotent uninorms on bounded lattices. We also proposed that an idempotent uninorm on a bounded lattice L with a neutral element e ∈ L \{0, 1} which differs from two extremal idempotent uninorms does not need to exist. Furthermore, we introduced two new methods for building uninorms from U (e ) based on the t-norm T e on [0, e]2 such that T e (x, y ) ∈ ]0, e] for all x, y ∈ ]0, e] and the t-conorm S e on [e , 1]2 such that S e (x, y ) ∈ [e , 1[ for all x, y ∈ [e , 1[. In particular, when taking T e = T ∧ and S e = S ∨ , our methods yield two new types of idempotent uninorms U 1∨ and U 2∧ on bounded lattices. In addition, we added an illustrative example in order to show that the these idempotent uninorms are different from the least and the greatest idempotent uninorms. It should be noted that the class of idempotent uninorm U 1∨ (resp. U 2∧ ) differs from the class of idempotent uninorm U s∨ (resp. U t∧ ) introduced in [8] only in the value when a variable of uninorm is 1 (resp. 0). The different classes of idempotent uninorms defined on some special bounded lattices which are not necessarily coincide with U 1∨ and U 2∧ can be found in the papers [12, Corollaries 3 and 5] and [13, Corollaries 1 and 2]. Some new open problems on idempotent uninorms occur. For example, we open new problems of the description of bounded lattices on which we can define idempotent uninorms different from two extremal idempotent uninorms and the introduction of more general structures than the known structures of idempotent uninorms on bounded lattices. In our future research, we aim to study dealing with properties of idempotent uninorms on bounded lattices, and especially the above mentioned problems. We believe that our work will open an intensive and deep research of the challenging subject of idempotent uninorms on bounded lattices. Declaration of competing interest I wish to confirm that there are no known conflicts of interest associated with this publication. Acknowledgements The author expresses her sincere thanks to the editors and reviewers for their most valuable comments and suggestions in improving this paper greatly.

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