Constructing uninorms via closure operators on a bounded lattice

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Constructing uninorms via closure operators on a bounded lattice Yao Ouyang a , Hua-Peng Zhang b,∗ a Faculty of Science, Huzhou University, Huzhou, Zhejiang 313000, China b School of Science, Nanjing University of Posts and Telecommunications, Nanjing 210023, China

Received 30 December 2018; received in revised form 25 March 2019; accepted 12 May 2019

Abstract In this paper, via closure (resp. interior) operators on a bounded lattice L, we propose a rather effective method to construct uninorms on L with a given triangular norm T (resp. triangular conorm S) on the subinterval [0, e] (resp. [e, 1]) of L. Our method encompasses as special cases the construction methods introduced by F. Karaçal and R. Mesiar (2015) [16] as well as by G.D. Çaylı et al. (2016) [5]. We also exploit a pair of novel closure and interior operators on complete lattices and utilize this pair of operators to construct uninorms. Several interesting examples are included to illustrate our method. © 2019 Elsevier B.V. All rights reserved. Keywords: Lattice; Closure operator; Interior operator; Triangular norm; Triangular conorm; Uninorm

1. Introduction Triangular norms (t-norms) and triangular conorms (t-conorms) on the unit interval [0, 1] are indispensable tools in many fields such as probabilistic metric spaces [19], fuzzy logic [11] and generalized theory of measures and integrals [20]. Mathematically speaking, a function T (S) : [0, 1]2 → [0, 1] is a t-norm (t-conorm) if and only if the triple ([0, 1], T , ≤) (([0, 1], S, ≤)) is a linearly ordered commutative semigroup with neutral element 1 (0) and annihilator 0 (1). Uninorms on the unit interval [0, 1] [21], as a unification of t-norms and t-conorms, allow a neutral element e to be an arbitrary number in [0, 1]. Uninorms are also useful in the fields where t-norms and t-conorms play an important role. T-norms and t-conorms on more general structures (e.g., posets and bounded lattices) have been proposed and extensively investigated [2,6,7,9,17,18,22]. In 2015, Karaçal and Mesiar [16] initiated the study of uninorms on a bounded lattice. Although the existence of t-norms and t-conorms on a general bounded lattice L is immediate, the construction of uninorms on L with an arbitrarily given neutral element e ∈ L \ {0, 1} is a challenging problem due to the poor structures of L compared with [0, 1]. To the best of our knowledge, up till now, there are only a few construction * Corresponding author.

E-mail addresses: [email protected] (Y. Ouyang), [email protected] (H.-P. Zhang). https://doi.org/10.1016/j.fss.2019.05.006 0165-0114/© 2019 Elsevier B.V. All rights reserved.

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methods for uninorms on a general bounded lattice in the literature [5,14,16]. In addition, under some constraints, there are several construction methods for uninorms on a bounded lattice [3,4]. In general topology [10], closure (interior) operators on the powerset P(X) of a nonempty set X are common tools to construct topologies on X. Actually, there is a one-to-one correspondence between the set of all closure (interior) operators on P(X) and that of all topologies on X. Note that closure (interior) operators on P(X) are essentially defined on the inherent lattice structure on P(X) with set inclusion, set intersection and set union as the partial order, the meet and the join on P(X), respectively. From this perspective, closure (interior) operators on the powerset P(X) of a nonempty set X can be easily extended to a general lattice [12] by neglecting the axiom that the closure (interior) of ∅ (X) is ∅ (X). The main aim of this paper is to present a rather effective method to construct uninorms via closure (interior) operators on a bounded lattice. The remainder of this paper is organized as follows. After recalling some necessary notions on lattices and uninorms on a bounded lattice in Section 2, we give several examples of closure and interior operators on a lattice in Section 3. Interestingly, we can construct a pair of novel closure and interior operators on a complete lattice. Section 4 is devoted to the construction of uninorms via closure operators on a bounded lattice, while Section 5 includes the dual results of Section 4 and construction methods for uninorms via the pair of closure and interior operators on a complete lattice defined in Section 3. We end with some conclusions and problems for future investigation in Section 6. 2. Preliminaries A lattice is a nonempty set L equipped with a partial order ≤ such that any two elements x and y have a greatest lower bound (called meet or infimum), denoted by x ∧ y, as well as a smallest upper bound (called join or supremum), denoted by x ∨ y. For a, b ∈ L, we also write b ≥ a if a ≤ b holds. The symbol a < b means that a ≤ b and a = b. If neither a ≤ b nor a ≥ b, then we say that a is incomparable with b and write a  b. A lattice (L, ≤, ∧, ∨) is called bounded if it has atop element 1 and a bottom element  0, while it is said to be complete if for any A ⊂ L, the greatest lower bound A and the smallest upper bound A of A exist. Obviously, any finite lattice is necessarily complete and any complete lattice is necessarily bounded. Let (L, ≤, ∧, ∨) be a lattice and a, b ∈ L with a ≤ b. The subinterval [a, b] is a sublattice of L defined as [a, b] = {x ∈ L | a ≤ x ≤ b}. Other subintervals such as [a, b) and (a, b) can be defined similarly. For more information about lattices, we refer to [1]. In the following, aggregation functions such as t-norms, t-conorms and uninorms are defined on a bounded lattice. Definition 2.1. [2] Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and [a, b] ⊂ L be a subinterval. A binary operation T : [a, b] × [a, b] → [a, b] is said to be a t-norm on [a, b] if, for any x, y, z ∈ [a, b], the following conditions are fulfilled: (i) (ii) (iii) (iv)

T (x, y) = T (y, x) (commutativity); If x ≤ y, then T (x, z) ≤ T (y, z) (monotonicity); T (T (x, y), z) = T (x, T (y, z)) (associativity); T (x, b) = x (neutrality).

If [a, b] = [0, 1], then we define t-norms on the lattice L. The strongest t-norm on L is T∧ defined by T∧ (x, y) = x ∧ y for any x, y ∈ L, while the weakest t-norm on L is the drastic product TD which takes value x ∧ y if 1 ∈ {x, y} and 0 otherwise. That is to say, for any t-norm T on L, we have TD ≤ T ≤ T∧ . A t-conorm on [a, b] is a binary operation S : [a, b] ×[a, b] → [a, b], which is commutative, monotonic, associative and has neutral element a, i.e., S(x, a) = x for any x ∈ [a, b]. Definition 2.2. [16] Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and e ∈ L. A binary operation U : L × L → L is called a uninorm on L with neutral element e if, for any x, y, z ∈ L, the following conditions are fulfilled:

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(i) (ii) (iii) (iv)

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U (x, y) = U (y, x) (commutativity); If x ≤ y, then U (x, z) ≤ U (y, z) (monotonicity); U (U (x, y), z) = U (x, U (y, z)) (associativity); U (x, e) = x (neutrality).

Let (L, ≤, ∧, ∨, 0, 1) be element e is given by ⎧ x ∧y ⎪ ⎪ ⎪ ⎨y U e (x, y) = ⎪ x ⎪ ⎪ ⎩ 1

a bounded lattice and e ∈ L be given, then the strongest uninorm U e on L with neutral if x, y ∈ [0, e] if x ∈ [0, e] and y ∈ / [0, e] if y ∈ [0, e] and x ∈ / [0, e] otherwise.

The weakest uninorm Ue on L with neutral element e is given by ⎧ ⎪ ⎪ x ∨ y if x, y ∈ [e, 1] ⎪ ⎨y if x ∈ [e, 1] and y ∈ / [e, 1] Ue (x, y) = ⎪x if y ∈ [e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎩ 0 otherwise. 3. Closure and interior operators on a lattice In this section, we provide some interesting examples of closure and interior operators on a lattice. 3.1. Closure operators on a lattice Definition 3.1. [12] Let (L, ≤, ∧, ∨) be a lattice. A mapping cl : L → L is said to be a closure operator if, for any x, y ∈ L, it satisfies the following three conditions: (i) x ≤ cl(x) (expansion); (ii) cl(x ∨ y) = cl(x) ∨ cl(y) (preservation of join); (iii) cl(cl(x)) = cl(x) (idempotence). By (i), the condition (iii) is equivalent to cl(cl(x)) ≤ cl(x). In addition, (ii) implies (ii)’ x ≤ y =⇒ cl(x) ≤ cl(y). Note that Birkhoff [1] defines a closure operator by (i), (ii)’ and (iii). Any lattice can naturally induce a family of closure operators on itself. Example 3.2. [9] Let (L, ≤, ∧, ∨) be a lattice and a ∈ L be given. Then the mapping cla : L → L defined as cla (x) = x ∨ a (∀ x ∈ L) is a closure operator. Any topology can induce a closure operator on the underlying powerset. Example 3.3. [10] Let P(X) be the powerset of a nonempty set X. It is well known that (P(X), ⊂, ∩, ∪) is a lattice. For any topology T ⊂ P(X) on X, the mapping clT : P(X) → P(X) defined as  clT (A) = {X \ O | O ∈ T , A ⊂ X \ O} (∀ A ∈ P(X)) is a closure operator satisfying clT (∅) = ∅. In the remainder of this subsection we will construct a novel closure operator on a complete lattice.

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1  a

@ @ b

@ @ c  @ d  e  @ f @ g@ @ @ k h @ @

0 Fig. 1. Hasse diagram of the lattice L in Example 3.8.

Let (L, ≤, ∧, ∨) be a lattice. Denote by U C(L) the set of all universally comparable elements in L, i.e., U C(L) = {a ∈ L | ∀ b ∈ L, either a ≤ b or b ≤ a}. Obviously, if L is a chain, i.e., any two elements in L are comparable, then U C(L) = L. The following two propositions list some basic properties of U C(L) when L is bounded or complete. Proposition 3.4. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice. Then (U C(L), ≤, ∧, ∨) is a nonempty chain and a sublattice of L. Proof. It is obvious that 0, 1 ∈ U C(L) and that U C(L) is a chain. For x, y ∈ U C(L), without loss of generality, we suppose that x ≤ y, which implies x ∧ y = x ∈ U C(L) and x ∨ y = y ∈ U C(L). Hence, U C(L) is a sublattice of L. 2 Proposition 3.5. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice. Then (U C(L), ≤, ∧, ∨) is also complete. Proof. Suppose  that (L, ≤, ∧, ∨, 0, 1) is complete. By Theorem 6 in Chapter 1 of [1], it suffices to prove that A ⊂  U C(L) implies A ∈ U C(L). Denote a = A, we need to prove that a can be comparable with any x ∈ L. If A is empty, then a = 1 ∈ U C(L). So we can suppose that A is nonempty. For any x ∈ L, ifthere is some b ∈ A such that x ≥ b, then x ≥ a. Otherwise, we have that x ≤ b for any b ∈ A, which implies x ≤ A = a. Hence, a can be comparable with any x ∈ L, whence a ∈ U C(L). 2 Now we define the mapping ⇑ on a complete lattice. Definition 3.6. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice. The mapping ⇑ : L → L is defined as, for any x ∈ L, ⇑ (x) = {a ∈ U C(L) | a ≥ x}. Since 1 ∈ U C(L), ⇑ is well defined by Proposition 3.5. We give some examples to illustrate the mapping ⇑. Example 3.7. (1) If L is a chain, then ⇑ (x) = x for any x ∈ L. (2) If L is a lattice such that for any a ∈ L \ {0, 1} there is b ∈ L satisfying b  a, then U C(L) = {0, 1}. Hence, ⇑ (0) = 0 and for any c > 0, ⇑ (c) = 1. Example 3.8. Let L be the lattice given by Fig. 1. Obviously, U C(L) = {0, g, c, 1}. Hence, ⇑ (0) = 0, ⇑ (h) =⇑ (k) =⇑ (g) = g, ⇑ (d) =⇑ (e) =⇑ (f ) =⇑ (c) = c and ⇑ (a) =⇑ (b) =⇑ (1) = 1.

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The following example shows that we can even define the mapping ⇑ on some special non-complete lattices. Example 3.9. Let L = [0, 1) ∪{u +iv | u, v ∈ (0, 1]}. In the set [0, 1) we use the usual order. We say that a +ib ≤ c +id if and only if a ≤ c and b ≤ d. We also suppose that x < a + ib for any x ∈ [0, 1) and any a + ib ∈ {u + iv | u, v ∈ (0, 1]}. Then ≤ is a partial order on L and (L, ≤) is a lattice. Since 0L = 01 and 1L = 1 + i, L is bounded. It is easy to see that U C(L) = [0, 1) ∪ {1 + i}. Note that [0, 1) has no smallest upper bound and {u + iv | u, v ∈ (0, 1]} has no greatest lower bound. Hence, L is not complete, but we can define the mapping ⇑ on it. In fact, ⇑ (x) = x for any x ∈ [0, 1) and ⇑ (x) = 1 + i for any x ∈ {u + iv | u, v ∈ (0, 1]}. The following proposition presents some properties of the mapping ⇑ defined in Definition 3.6. Proposition 3.10. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice and ⇑ : L → L be defined in Definition 3.6. Then the following results hold: (i) (ii) (iii) (iv) (v)

x ≤⇑ (x) for any x ∈ L; ⇑ (x) ∈ U C(L) for any x ∈ L; for any x ∈ L, x ∈ U C(L) implies ⇑ (x) = x; ⇑ (⇑ (x)) =⇑ (x) for any x ∈ L; ⇑ is monotonic.

Proof. (i) Follows from the definition of ⇑ (x). (ii) Follows from Proposition 3.5. (iii) Follows from the definition of ⇑ (x). (iv) Follows from (ii) and (iii). (v) For any x, y ∈ L with x ≤ y, it follows from {a ∈ U C(L) | a ≥ x} ⊃ {a ∈ U C(L) | a ≥ y} that ⇑ (x) ≤⇑ (y).

2

The mapping ⇑ defined in Definition 3.6 is a closure operator. Theorem 3.11. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice. The mapping ⇑ : L → L defined in Definition 3.6 is a closure operator. Proof. Due to Proposition 3.10(i) and (iv), we only need to prove that ⇑ preserves join. For any x, y ∈ L, it follows from the monotonicity of ⇑ that ⇑ (x)∨ ⇑ (y) ≤⇑ (x ∨ y). On the other hand, since ⇑ (x), ⇑ (y) ∈ U C(L) and (U C(L), ≤, ∧, ∨) is a lattice, we have that ⇑ (x)∨ ⇑ (y) ∈ U C(L), together with x ∨ y ≤⇑ (x)∨ ⇑ (y) implying that ⇑ (x ∨ y) ≤⇑ (x)∨ ⇑ (y). Hence, ⇑ (x ∨ y) =⇑ (x)∨ ⇑ (y). Therefore, ⇑ is a closure operator.

2

3.2. Interior operators on a lattice Dually, we can define interior operators on a lattice. Definition 3.12. Let (L, ≤, ∧, ∨) be a lattice. A mapping int : L → L is said to be an interior operator if, for any x, y ∈ L, it satisfies the following three conditions: 1 Here we use 0 (1 ) to distinguish the smallest (greatest) element of L from the number 0 (1). L L

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(i) int (x) ≤ x (contraction); (ii) int (x ∧ y) = int (x) ∧ int (y) (preservation of meet); (iii) int (int (x)) = int (x) (idempotence). Any lattice can naturally induce a family of interior operators on itself. Example 3.13. [9] Let (L, ≤, ∧, ∨) be a lattice and a ∈ L be given. Then the mapping inta : L → L defined as inta (x) = x ∧ a (∀ x ∈ L) is an interior operator. Any topology can induce an interior operator on the underlying powerset. Example 3.14. [10] Let P(X) be the powerset of a nonempty set X. For any topology T ⊂ P(X) on X, the mapping intT : P(X) → P(X) defined as intT (A) =



{O | O ∈ T , O ⊂ A} (∀ A ∈ P(X))

is an interior operator on the lattice (P(X), ⊂, ∩, ∪) satisfying intT (X) = X. In a dual way, we can define the mapping ⇓ on a complete lattice. Definition 3.15. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice. The mapping ⇓ : L → L is defined as, for any x ∈ L, ⇓ (x) =



{a ∈ U C(L) | a ≤ x}.

Since 0 ∈ U C(L), ⇓ is well defined by Proposition 3.5. We give some examples to illustrate the mapping ⇓. Example 3.16. (1) If L is a chain, then ⇓ (x) = x for any x ∈ L. (2) If L is a lattice such that for any a ∈ L \ {0, 1} there is b ∈ L satisfying b  a, then ⇓ (1) = 1 and for any c < 1, ⇓ (c) = 0. (3) For the lattice in Example 3.8, we have ⇓ (h) =⇓ (k) =⇓ (0) = 0, ⇓ (d) =⇓ (e) =⇓ (f ) =⇓ (g) = g, ⇓ (a) =⇓ (b) =⇓ (c) = c and ⇓ (1) = 1. The mapping ⇓ defined in Definition 3.15 is an interior operator. Theorem 3.17. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice. The mapping ⇓ : L → L defined in Definition 3.15 is an interior operator. Proof. Similar to Theorem 3.11.

2

4. Uninorms based on a closure operator We can construct uninorms via closure operators on a bounded lattice, as shown in the following theorem. Theorem 4.1. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice, cl : L → L be a closure operator, e ∈ L be given and T : [0, e] × [0, e] → [0, e] be a t-norm. Then the binary operation U : L × L → L defined by

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U (x, y) =

⎧ T (x, y) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y ⎪ ⎪ x ⎪ ⎪ ⎪ ⎪ ⎩ cl(x) ∨ cl(y)

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if x, y ∈ [0, e] if x ∈ [0, e] and y ∈ / [0, e] if y ∈ [0, e] and x ∈ / [0, e] otherwise

is a uninorm on L with neutral element e. Proof. Obviously, U is commutative and e is the neutral element of U . Monotonicity: Let x, y, z ∈ L with x < y. We need to prove that U (x, z) ≤ U (y, z). We distinguish the following four cases: (i) x ∈ [0, e] and y ∈ [0, e]. If z ∈ [0, e], then U (x, z) = T (x, z) ≤ T (y, z) = U (y, z). If z ∈ / [0, e], then U (x, z) = z = U (y, z). (ii) x ∈ [0, e] and y ∈ / [0, e]. If z ∈ [0, e], then U (x, z) = T (x, z) ≤ x < y = U (y, z). If z ∈ / [0, e], then U (x, z) = z ≤ y ∨ z ≤ cl(y) ∨ cl(z) = U (y, z). (iii) x ∈ / [0, e] and y ∈ [0, e]. Since x < y, this is an impossible case. (iv) x ∈ / [0, e] and y ∈ / [0, e]. If z ∈ [0, e], then U (x, z) = x < y = U (y, z). If z ∈ / [0, e], then U (x, z) = cl(x) ∨ cl(z) ≤ cl(y) ∨ cl(z) = U (y, z). Associativity: Let x, y, z ∈ L be given. If e ∈ {x, y, z}, then it trivially holds that U (U (x, y), z) = U (x, U (y, z)). So we suppose that e ∈ / {x, y, z}. We distinguish the following four cases: (i) x < e and y < e (which imply U (x, y) = T (x, y) < e). If z < e, then U (y, z) = T (y, z) < e. Hence, U (U (x, y), z) = T (T (x, y), z) = T (x, T (y, z)) = U (x, U (y, z)). If z ∈ / [0, e], then U (U (x, y), z) = z = U (x, z) = U (x, U (y, z)). (ii) x < e and y ∈ / [0, e]. If z < e, then U (U (x, y), z) = U (y, z) = y = U (x, y) = U (x, U (y, z)). If z ∈ / [0, e], then U (y, z) ∈ / [0, e]. Hence, U (U (x, y), z) = U (y, z) = U (x, U (y, z)).

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(iii) x ∈ / [0, e] and y < e. If z < e, then U (y, z) < e. Hence, U (U (x, y), z) = U (x, z) = x = U (x, U (y, z)). If z ∈ / [0, e], then U (U (x, y), z) = U (x, z) = U (x, U (y, z)). (iv) x ∈ / [0, e] and y ∈ / [0, e] (which imply U (x, y) ∈ / [0, e]). If z < e, then U (U (x, y), z) = U (x, y) = U (x, U (y, z)). If z ∈ / [0, e], then U (y, z) ∈ / [0, e]. Hence, U (U (x, y), z) = cl(U (x, y)) ∨ cl(z) = cl(cl(x) ∨ cl(y)) ∨ cl(z) = cl(cl(x)) ∨ cl(cl(y)) ∨ cl(z) = cl(x) ∨ cl(y) ∨ cl(z). Similarly, we have U (x, U (y, z)) = cl(x) ∨ cl(y) ∨ cl(z). So U (U (x, y), z) = U (x, U (y, z)). Now we have proved that U is a commutative, monotonic and associative binary operation on L with neutral element e, i.e., U is a uninorm on L. 2 If e = 0 in Theorem 4.1, then we retrieve the corresponding result of Theorem 3.1 in [8]. Corollary 4.2. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and cl : L → L be a closure operator. Then the binary operation S : L × L → L defined by x ∨y if 0 ∈ {x, y} S(x, y) = cl(x) ∨ cl(y) otherwise is a t-conorm on L. The following theorem follows immediately from Example 3.2 and Theorem 4.1. Theorem 4.3. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice, e ∈ L be given and T : [0, e] × [0, e] → [0, e] be a t-norm. For any a ∈ L, the binary operation U : L × L → L defined by ⎧ T (x, y) if x, y ∈ [0, e] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y if x ∈ [0, e] and y ∈ / [0, e] U (x, y) = ⎪ ⎪ x if y ∈ [0, e] and x ∈ / [0, e] ⎪ ⎪ ⎪ ⎪ ⎩ x ∨ y ∨ a otherwise is a uninorm on L with neutral element e. If we put a = 1 in Theorem 4.3, then we retrieve the corresponding uninorms constructed by Karaçal and Mesiar [16].

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Table 1 A uninorm constructed via Theorem 4.3. U

0

h

k

g

e

d

f

c

a

b

1

0 h k g e d f c a b 1

0 0 0 0 0 d f c a b 1

0 h 0 h h d f c a b 1

0 0 k k k d f c a b 1

0 h k g g d f c a b 1

0 h k g e d f c a b 1

d d d d d d c c a b 1

f f f f f c c c a b 1

c c c c c c c c a b 1

a a a a a a a a a 1 1

b b b b b b b b 1 b 1

1 1 1 1 1 1 1 1 1 1 1

Corollary 4.4. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and e ∈ L \ {0, 1}. If T : [0, e]2 → [0, e] is a t-norm. Then the binary operation U : L × L → L defined by ⎧ T (x, y) ⎪ ⎪ ⎪ ⎨y U (x, y) = ⎪ x ⎪ ⎪ ⎩ 1

if x, y ∈ [0, e] if x ∈ [0, e] and y ∈ / [0, e] if y ∈ [0, e] and x ∈ / [0, e] otherwise

is a uninorm on L with neutral element e. If we put a = 0 in Theorem 4.3, then we retrieve the corresponding uninorms constructed by Çaylı, Karaçal and Mesiar [5]. Corollary 4.5. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and e ∈ L \ {0, 1}. If T : [0, e]2 → [0, e] is a t-norm. Then the binary operation U : L × L → L defined by ⎧ ⎪ ⎪ T (x, y) ⎪ ⎨y U (x, y) = ⎪x ⎪ ⎪ ⎩ x ∨y

if x, y ∈ [0, e] if x ∈ [0, e] and y ∈ / [0, e] if y ∈ [0, e] and x ∈ / [0, e] otherwise

is a uninorm on L with neutral element e. Generally, different a in Theorem 4.3 can derive different uninorms. We give an example to illustrate this observation. Example 4.6. Let L be the lattice defined in Example 3.8 and the t-norm T : [0, e] × [0, e] → [0, e] be T∧ . For a = d, the uninorm constructed via Theorem 4.3 is given by Table 1. It is not difficult to see that this uninorm is different from that constructed via Corollaries 4.4 and 4.5. The following theorem follows immediately from Theorems 3.11 and 4.1. Theorem 4.7. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice, ⇑ : L → L be defined in Definition 3.6, e ∈ L be given and T : [0, e] × [0, e] → [0, e] be a t-norm. Then the binary operation U : L × L → L defined by

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Table 2 A t-norm on [0, e] of the lattice L in Example 3.8. T

0

h

k

g

e

0 h k g e

0 0 0 0 0

0 0 0 0 h

0 0 k k k

0 0 k k g

0 h k g e

Table 3 A uninorm constructed via Theorem 4.7.

U (x, y) =

⎧ T (x, y) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y ⎪ ⎪ x ⎪ ⎪ ⎪ ⎪ ⎩ ⇑ (x)∨ ⇑ (y)

U

0

h

k

g

e

d

f

c

a

b

1

0 h k g e d f c a b 1

0 0 0 0 0 d f c a b 1

0 0 0 0 h d f c a b 1

0 0 k k k d f c a b 1

0 0 k k g d f c a b 1

0 h k g e d f c a b 1

d d d d d c c c 1 1 1

f f f f f c c c 1 1 1

c c c c c c c c 1 1 1

a a a a a 1 1 1 1 1 1

b b b b b 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1

if x, y ∈ [0, e] if x ∈ [0, e] and y ∈ / [0, e] if y ∈ [0, e] and x ∈ / [0, e] otherwise

is a uninorm on L with neutral element e. We give an example to illustrate Theorem 4.7. Example 4.8. Let L be the lattice defined in Example 3.8 and let the t-norm T : [0, e] × [0, e] → [0, e] be given by Table 2. Then the uninorm U : L × L → L constructed via Theorem 4.7 is given by Table 3. 5. Uninorms based on an interior operator We can construct uninorms via interior operators on a bounded lattice, as shown in the following theorem. Theorem 5.1. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice, int : L → L be an interior operator, e ∈ L be given and S : [e, 1] × [e, 1] → [e, 1] be a t-conorm. Then the binary operation U : L × L → L defined by ⎧ S(x, y) if x, y ∈ [e, 1] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y if x ∈ [e, 1] and y ∈ / [e, 1] U (x, y) = ⎪ ⎪ x if y ∈ [e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎪ ⎪ ⎩ int (x) ∧ int (y) otherwise is a uninorm on L with neutral element e.

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Proof. Similar to Theorem 4.1.

11

2

If e = 1 in Theorem 5.1, then we retrieve the corresponding result of Theorem 3.1 in [8]. Corollary 5.2. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and int : L → L be an interior operator. Then the binary operation T : L × L → L defined by x ∧y if 1 ∈ {x, y} T (x, y) = int (x) ∧ int (y) otherwise is a t-norm on L. The following theorem follows immediately from Example 3.13 and Theorem 5.1. Theorem 5.3. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice, e ∈ L be given and S : [e, 1] × [e, 1] → [e, 1] be a t-conorm. For any a ∈ L, the binary operation U : L × L → L defined by ⎧ S(x, y) if x, y ∈ [e, 1] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y if x ∈ [e, 1] and y ∈ / [e, 1] U (x, y) = ⎪ ⎪ x if y ∈ [e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎪ ⎪ ⎩ x ∧ y ∧ a otherwise is a uninorm on L with neutral element e. If we put a = 0 in Theorem 5.3, then we retrieve the corresponding uninorms constructed by Karaçal and Mesiar [16]. Corollary 5.4. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and e ∈ L \{0, 1}. If S : [e, 1] ×[e, 1] → [e, 1] is a t-conorm. Then the binary operation U : L × L → L defined by ⎧ S(x, y) if x, y ∈ [e, 1] ⎪ ⎪ ⎪ ⎨y if x ∈ [e, 1] and y ∈ / [e, 1] U (x, y) = ⎪ x if y ∈ [e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎩ 0 otherwise is a uninorm on L with neutral element e. If we put a = 1 in Theorem 5.3, then we retrieve the corresponding uninorms constructed by Çaylı, Karaçal and Mesiar [5]. Corollary 5.5. Let (L, ≤, ∧, ∨, 0, 1) be a bounded lattice and e ∈ L \{0, 1}. If S : [e, 1] ×[e, 1] → [e, 1] is a t-conorm. Then the binary operation U : L × L → L defined by ⎧ S(x, y) if x, y ∈ [e, 1] ⎪ ⎪ ⎪ ⎨y if x ∈ [e, 1] and y ∈ / [e, 1] U (x, y) = ⎪ x if y ∈ [e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎩ x ∧y otherwise is a uninorm on L with neutral element e. Generally, different a in Theorem 5.3 can derive different uninorms. The following theorem follows immediately from Theorems 3.17 and 5.1.

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Table 4 A t-conorm on [e, 1] of the lattice L in Example 3.8. S

e

c

a

b

1

e c a b 1

e c a b 1

c c a b 1

a a 1 1 1

b b 1 1 1

1 1 1 1 1

Table 5 A uninorm constructed via Theorem 5.6. U

0

h

k

g

d

f

e

c

a

b

1

0 h k g d f e c a b 1

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 h h h h h

0 0 0 0 0 0 k k k k k

0 0 0 g g g g g g g g

0 0 0 g g g d d d d d

0 0 0 g g g f f f f f

0 h k g d f e c a b 1

0 h k g d f c c a b 1

0 h k g d f a a 1 1 1

0 h k g d f b b 1 1 1

0 h k g d f 1 1 1 1 1

Theorem 5.6. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice, ⇓ : L → L be defined in Definition 3.15, e ∈ L be given and S : [e, 1] × [e, 1] → [e, 1] be a t-conorm. Then the binary operation U : L × L → L defined by ⎧ S(x, y) if x, y ∈ [e, 1] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y if x ∈ [e, 1] and y ∈ / [e, 1] U (x, y) = ⎪ ⎪ x if y ∈ [e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎪ ⎪ ⎩ ⇓ (x)∧ ⇓ (y) otherwise is a uninorm on L with neutral element e. We give an example to illustrate Theorem 5.6. Example 5.7. Let L be the lattice defined in Example 3.8 and let the t-conorm S : [e, 1] × [e, 1] → [e, 1] be given by Table 4. Then the uninorm U : L × L → L constructed via Theorem 5.6 is given by Table 5. The following theorem follows from Theorem 4.7 and Corollary 5.2. Theorem 5.8. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice, ⇑ : L → L be defined in Definition 3.6, ⇓ : L → L be defined in Definition 3.15 and e ∈ L be given. Then the binary operation U : L × L → L defined by ⎧ ⇓ (x)∧ ⇓ (y) if x, y ∈ [0, e) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y if x = e and y ∈ L, or if x ∈ [0, e) and y ∈ / [0, e] U (x, y) = ⎪ ⎪ x if y = e and x ∈ L, or if y ∈ [0, e) and x ∈ / [0, e] ⎪ ⎪ ⎪ ⎪ ⎩ ⇑ (x)∨ ⇑ (y) otherwise is a uninorm on L with neutral element e.

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Table 6 A uninorm constructed via Theorem 5.8. U

0

h

k

g

e

d

f

c

a

b

1

0 h k g e d f c a b 1

0 0 0 0 0 d f c a b 1

0 0 0 0 h d f c a b 1

0 0 0 0 k d f c a b 1

0 0 0 g g d f c a b 1

0 h k g e d f c a b 1

d d d d d c c c 1 1 1

f f f f f c c c 1 1 1

c c c c c c c c 1 1 1

a a a a a 1 1 1 1 1 1

b b b b b 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1

Table 7 A uninorm constructed via Theorem 5.10. U

0

h

k

g

d

f

e

c

a

b

1

0 h k g d f e c a b 1

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 h h h h h

0 0 0 0 0 0 k k k k k

0 0 0 g g g g g g g g

0 0 0 g g g d d d d d

0 0 0 g g g f f f f f

0 h k g d f e c a b 1

0 h k g d f c c 1 1 1

0 h k g d f a 1 1 1 1

0 h k g d f b 1 1 1 1

0 h k g d f 1 1 1 1 1

Example 5.9. For the lattice in Example 3.8, the uninorm constructed via Theorem 5.8 is given by Table 6. The following theorem follows from Theorem 5.6 and Corollary 4.2. Theorem 5.10. Let (L, ≤, ∧, ∨, 0, 1) be a complete lattice, ⇑ : L → L be defined in Definition 3.6, ⇓ : L → L be defined in Definition 3.15 and e ∈ L be given. Then the binary operation U : L × L → L defined by ⎧ ⇑ (x)∨ ⇑ (y) if x, y ∈ (e, 1] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨y if x = e and y ∈ L, or if x ∈ (e, 1] and y ∈ / [e, 1] U (x, y) = ⎪ ⎪ x if y = e and x ∈ L, or if y ∈ (e, 1] and x ∈ / [e, 1] ⎪ ⎪ ⎪ ⎪ ⎩ ⇓ (x)∧ ⇓ (y) otherwise is a uninorm on L with neutral element e. Example 5.11. For the lattice in Example 3.8, the uninorm constructed via Theorem 5.10 is given by Table 7. Remark 5.12. Examples 5.9 and 5.11 show that Theorems 5.8 and 5.10 provide two different constructions for uninorms on a complete lattice. 6. Concluding remarks We have proposed a rather effective method to construct uninorms on a bounded lattice L (Theorems 4.1 and 5.1). Our method is based on a closure operator (resp. an interior operator) on L and a given t-norm on [0, e] (resp. t-conorm on [e, 1]). Several known construction methods for uninorms on a bounded lattice can be derived from our method.

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When the lattice is complete, we presented a pair of interesting closure and interior operators on the lattice and constructed uninorms via this pair of operators (Theorems 4.7 and 5.6). Note that this kind of closure (interior) operators can even be defined on some special non-complete lattices and hence be applied to construct uninorms on a bounded lattice. The following example illustrates this observation. Example 6.1. Let us consider the lattice L constructed in Example 3.9. L is not complete, but we can define the mapping ⇑ on L and it is easy to see that ⇑ is a closure operator. Hence, we can apply Theorem 4.7 on L. Let T be a t-norm on [0, 12 ], then the binary operation U : L × L → L defined by ⎧ T (x, y) if x, y ∈ [0, 12 ] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ if x ∈ [0, 12 ] and y ∈ / [0, 12 ] ⎪ ⎪y ⎪ ⎨ U (x, y) = x if y ∈ [0, 12 ] and x ∈ / [0, 12 ] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x ∨y x, y ∈ ( 12 , 1) ⎪ ⎪ ⎪ ⎪ ⎩ 1+i otherwise is a uninorm on L with neutral element e = 12 . Recently, under some constraints on the lattices, Çaylı [3] used a t-norm T (with T (a, b) > 0 for any a, b > 0) and a t-conorm S (with S(a, b) < 1 for any a, b < 1) to construct uninorms on a bounded lattice. In the future, we will attempt to apply our method to the setting of [3]. Whether our method can be applied to construct other aggregation functions on a bounded lattice (nullnorms [13,15] for example) is also an interesting issue. Acknowledgement This work is supported in part by National Natural Science Foundation of China (No. 11571106 and No. 11571006). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22]

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