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Construction methods for idempotent nullnorms on bounded lattices
Applied Mathematics and Computation 366 (2020) 124746
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Construction methods for idempotent nullnorms on bounded lattices Gül Deniz Çaylı Department of Mathematics, Faculty of Science, Karadeniz Technical University, Trabzon 61080, Turkey
a r t i c l e
i n f o
Article history: Received 26 February 2018 Revised 3 September 2019 Accepted 9 September 2019
Keywords: Bounded lattice Construction method Idempotent nullnorm Nullnorm Zero element
a b s t r a c t In this paper, we discuss the presence of idempotent nullnorms on bounded lattices and investigate some properties of these operators. We introduce two new methods for constructing idempotent nullnorms on bounded lattices for a zero element under a constraint that all elements incomparable with the zero element are comparable with each other. In order to stress the role of this constraint satisfying that our constructions yield an idempotent nullnorm, some illustrative examples are added. © 2019 Elsevier Inc. All rights reserved.
1. Introduction Triangular norms (t-norms, for short) with the neutral element as 1 and triangular conorms (t-conorms, for short) with the neutral element as 0 on the real unit interval were introduced and comprehensively investigated by Schweizer and Sklar [40,41]. These operators have been extensively used in numerous practical applications such as fuzzy set theory, fuzzy logic, several branches of information sciences, multicriteria decision support and aggregation [7,23,26,27,39]. For more details concerning t-norms and t-conorms, the readers can refer to the papers [1,2,24,25,32,33,38]. Nullnorms and t-operators as a generalization of t-norms and t-conorms were first introduced in [8,35], where the zero element lies anywhere in the real unit interval rather than t-norms and t-conorms have zero elements 0 or 1, respectively. Nullnorms coincide with t-operators discussed by Mas, Mayor and Torrens [36] since both of them have the same block structures in the real unit interval, that is, once a binary operator is a t-operator, it is also a nullnorm and vice versa. Since the structure of nullnorms is a combination of t-norms and t-conorms, these operators play an important role both in theoretical investigations and in practical applications such as expert systems, neural networks, fuzzy system modelings. Nullnorms which are especially used as aggregation operators or in fuzzy logic, maintain as several logical characteristics as possible [9,20,21]. In the literature, there are some other studies about nullnorms on the real unit interval [3,16–19,28,34,37]. In [30], two methods for constructing nullnorms on a bounded lattice L with a zero element s ∈ L\{0, 1} were introduced by means of the presence of a t-norm on [s, 1]2 and a t-conorm on [0, s]2 . In [11], the existence of idempotent nullnorms on a bounded lattice L was investigated. In the same time, a new method to generate idempotent nullnorms on L with the indicated zero element s ∈ L\{0, 1} was presented under an additional assumption that there is the only one element in L incomparable with s. Following the demonstration of the existence of idempotent nullnorms on bounded lattices in [11], further researches have been promoted in this field. The fundamental target of this study is to investigate the presence E-mail address: [email protected] https://doi.org/10.1016/j.amc.2019.124746 0 096-30 03/© 2019 Elsevier Inc. All rights reserved.
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of idempotent nullnorms on a bounded lattice L with the zero element s ∈ L\{0, 1} and introduce further methods for constructing idempotent nullnorms on L. This study is organized as follows. First, some main constructions concerning nullnorms on bounded lattices and their results are shortly discussed in Section 2. In Section 3, the existence of idempotent nullnorms on an arbitrary bounded lattice L with the zero element s ∈ L\{0, 1} and some properties of them are researched. In addition, it is introduced two new methods to obtain idempotent nullnorms on L for an element s playing the role of a zero element with a constraint. Some corresponding examples are added in order to demonstrate that this constraint is necessary in our methods. Finally, some concluding remarks are given. 2. Preliminaries In this section, we recall some preliminary details concerning bounded lattices and nullnorms (t-norms and t-conorms) on them. A bounded lattice (L, ≤ ) is a lattice if it has the top and bottom elements that are written as 1 and 0, respectively; that is, there exist two elements 1, 0 ∈ L such that 0 ≤ x ≤ 1 for all x ∈ L. Definition 1 [4]. Given a bounded lattice (L, ≤ , 0, 1) and x, y ∈ L, if x and y are incomparable, in this case we use the notation xy. If x and y are comparable, in this case we use the notation xࢲy. Definition 2 [4]. Given a bounded lattice (L, ≤ , 0, 1) and x, y ∈ L, x ≤ y, a subinterval [x, y] of L is defined as
[x, y] = {a ∈ L
| x ≤ a ≤ y}.
Similarly, we define x, y = {a ∈ L | x < a ≤ y}, x, y = {a ∈ L | x ≤ a < y} and x, y = {a ∈ L | x < a < y}. Definition 3 [10,12]. Let (L, ≤ , 0, 1) be a bounded lattice. An operation T: L2 → L that is commutative, associative, increasing with respect to both variables and has the neutral element 1 such that T (x, 1 ) = x for all x ∈ L is called a t-norm on L. Definition 4 [6,10]. Let (L, ≤ , 0, 1) be a bounded lattice. An operation S: L2 → L that is commutative, associative, increasing with respect to both variables and has the neutral element 0 such that S(x, 0 ) = x for all x ∈ L is called a t-conorm on L. Definition 5 [11,30]. Let (L, ≤ , 0, 1) be a bounded lattice. If an operation V: L2 → L is commutative, associative, nondecreasing in each variable and there is an element such that s ∈ L V (x, 0 ) = x for all x ≤ s and V (x, 1 ) = x for all x ≥ s, then it is called a nullnorm on L. It is easy to see that V (x, s ) = s for all x ∈ L. So, s ∈ L is the zero element of the nullnorm V. We denote the sets Ds = [0, s] × [s, 1] ∪ [s, 1] × [0, s] and Is = {x ∈ L | x s} for an element s ∈ L\{0, 1}. The characterization of nullnorms (particularly idempotent nullnorms) on bounded lattices have attracted much attention from researchers in recent years. The main aim of this paper is to give new methods to construct idempotent nullnorms on bounded lattices. Now, we introduce some known results about nullnorms on bounded lattices, which will be used in the sequel. For more details on nullnorms on lattices, the readers can refer to [5,13–15,22,29,31,42]. Proposition 1 [17,30]. Given a bounded lattice (L, ≤ , 0, 1), s ∈ L\{0, 1} and a nullnorm V on L with the zero element s. In this case, the following statements are obtained. i) V |[0, s]2 : [0, s]2 → [0, s] is a t-conorm on [0, s]2 . ii) V |[s, 1]2 : [s, 1]2 → [s, 1] is a t-norm on [s, 1]2 . Proposition 2 [17,30]. Given a bounded lattice (L, ≤ , 0, 1), s ∈ L\{0, 1} and a nullnorm V on L with the zero element s. In this case, the following statements are obtained. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
V (x, y ) = s for all (x, y) ∈ Ds . s ≤ V(x, y) for all (x, y) ∈ [s, 1]2 ∪ [s, 1] × Is ∪ Is × [s, 1]. V(x, y) ≤ s for all (x, y) ∈ [0, s]2 ∪ [0, s] × Is ∪ Is × [0, s]. V(x, y) ≤ y for all (x, y) ∈ L × [s, 1]. V(x, y) ≤ x for all (x, y) ∈ [s, 1] × L. x ≤ V(x, y) for all (x, y) ∈ [0, s] × L. y ≤ V(x, y) for all (x, y) ∈ L × [0, s]. x ∨ y ≤ V(x, y) for all (x, y) ∈ [0, s]2 . V(x, y) ≤ x ∧ y for all (x, y) ∈ [s, 1]2 . (x ∧ s) ∨ (y ∧ s) ≤ V(x, y) for all (x, y) ∈ [0, s] × Is ∪ Is × [0, s] ∪ Is × Is . V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) for all (x, y) ∈ [s, 1] × Is ∪ Is × [s, 1] ∪ Is × Is .
Definition 6 [11,15]. Given a bounded lattice (L, ≤ , 0, 1), s ∈ L\{0, 1} and a nullnorm V on L with the zero element s. V is called an idempotent nullnorm on L if V (x, x ) = x for all x ∈ L. Proposition 3 [15]. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. In this case, the following statements are obtained.
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(i) V (x, y ) = x ∨ y for all (x, y) ∈ [0, s]2 . (ii) V (x, y ) = x ∧ y for all (x, y) ∈ [s, 1]2 . Proposition 4 [15]. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. In this case, the following statements are obtained. (i) (ii) (iii) (iv)
V (x, y ) = x ∨ (y ∧ s ) V (x, y ) = y ∨ (x ∧ s ) V (x, y ) = x ∧ (y ∨ s ) V (x, y ) = y ∧ (x ∨ s )
for for for for
all all all all
(x, y ) ∈ [0, s] × Is . (x, y ) ∈ Is × [0, s]. (x, y ) ∈ [s, 1] ∪ Is . (x, y ) ∈ Is × [s, 1].
Theorem 1. [11] Given a distributive bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. In this case, the function defined by
V∗ (x, y ) = (x ∧ y ) ∨ (x ∧ s ) ∨ (y ∧ s ) = (x ∨ y ) ∧ (x ∨ s ) ∧ (y ∨ s ) = V ∗ (x, y )
(1)
is an idempotent nullnorm on L with zero element s. Theorem 2. [11] Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. If there is only one element incomparable with s in L and it is denoted by k the element incomparable with s, in that case the function V defined by
⎧ x∨y ⎪ ⎪ ⎪ ⎪ x∧y ⎪ ⎪ ⎪ ⎪ ⎨ s x ∨ (k ∧ s ) V (x, y ) = y ∨ (k ∧ s ) ⎪ ⎪ ⎪ ⎪ x ∧ (k ∨ s ) ⎪ ⎪ ⎪ ⎪ ⎩ y ∧ (k ∨ s ) k
if if if if if if if if
(x, y ) ∈ [0, s]2 , (x, y ) ∈ [s, 1]2 , (x, y ) ∈ Ds ,
x ∈ [0, s] and y = k, x = k and y ∈ [0, s], x ∈ [s, 1] and y = k, x = k and y ∈ [s, 1], x = y = k,
(2)
is an idempotent nullnorm on L with the zero element s. 3. Idempotent nullnorms on bounded lattices In this section, considering an arbitrary bounded lattice L, we investigate the existence of idempotent nullnorms on L with the indicated zero element s ∈ L\{0, 1} and their characteristics. In addition, we enhance the methods for obtaining idempotent nullnorms on L presented in Theorems 1 and 2 and propose two new construction methods for idempotent nullnorms on L with different characteristics compared with each other. Proposition 5. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. Then (x ∧ s) ∨ (y ∧ s) ∨ (x ∧ y) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) ∧ (x ∨ y) for all x, y ∈ L. Proof. By Proposition 2 (x) and (xi), we obtain that (x ∧ s) ∨ (y ∧ s) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s). Since V is idempotent, we have that x ∧ y = V (x ∧ y, x ∧ y ) ≤ V (x, y ) ≤ V (x ∨ y, x ∨ y ) = x ∨ y. Thus, it holds (x ∧ s) ∨ (y ∧ s) ∨ (x ∧ y) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) ∧ (x ∨ y). Proposition 6. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. If x, y ∈ Is and x is comparable with y, then it holds V(x, y) ∈ Is . Proof. Let x, y ∈ Is and x be comparable with y. In this case, x ≤ y or x > y. By Proposition 5, it is (x ∧ s) ∨ (y ∧ s) ∨ (x ∧ y) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) ∧ (x ∨ y). If x ≤ y, then (y ∧ s) ∨ x ≤ V(x, y) ≤ (x ∨ s) ∧ y. Suppose that V(x, y) ≤ s. In this case, we have that x ∨ (y ∧ s) ≤ s, i.e., x ≤ s. This is a contradiction. So, it can not be V(x, y) ≤ s. Suppose that V(x, y) > s. In this case, we have that s < y ∧ (x ∨ s), i.e., s < y. This is a contradiction. So, it can not be V(x, y) > s. Hence, if x ≤ y, then V(x, y) ∈ Is . Similarly, it is showed that if x > y, then V(x, y) ∈ Is . Therefore, if x, y ∈ Is and x is comparable with y, then we obtain that V(x, y) ∈ Is . Proposition 7. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1}, V be an idempotent nullnorm on L with the zero element s and x, y ∈ Is . If there is an element z ∈ L such that x ∧ s = z, y ∧ s = z and x ∧ y = z, then V(x, y) < s or V(x, y) ∈ Is . Proof. Let x, y ∈ Is and there be z ∈ L such that x ∧ s = z, y ∧ s = z and x ∧ y = z. By Proposition 5, since (z ∧ s) ∨ (x ∧ s) ∨ (z ∧ x) ≤ V(z, x) ≤ (z ∨ s) ∧ (x ∨ s) ∧ (z ∨ x), we obtain that V (z, x ) = z. Similarly, it is obtained that V (z, y ) = z. Since V (V (z, x ), y ) = V (z, y ) = z, it holds V (z, V (x, y )) = z from the associativity of V. Suppose that V(x, y) ≥ s. Then due to the monotonicity of V and the fact that s is zero element, we have that z = V (z, V (x, y )) ≥ V (z, s ) = s, i.e., z ≥ s. This is a contradiction. So, it can not be V(x, y) ≥ s. Therefore, if there is z ∈ L such that x ∧ s = z, y ∧ s = z and x ∧ y = z, then it holds V(x, y) < s or V(x, y) ∈ Is . Proposition 8. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1}, V be an idempotent nullnorm on L with the zero element s and x, y ∈ Is . If there is an element z ∈ L such that x ∨ s = z, y ∨ s = z and x ∨ y = z, then V(x, y) > s or V(x, y) ∈ Is .
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Fig. 1. Lattice L. Table 1 Idempotent nullnorm V1 on L. V1 0 k s x y z 1
0 0 k s k k s s
k k k s k k s s
s s s s s s s s
x k k s x x z z
y k k s x y z z
z s s s z z z z
1 s s s z z z 1
z s s s z z z z
1 s s s z z z 1
Table 2 Idempotent nullnorm V2 on L. V2 0 k s x y z 1
0 0 k s k k s s
k k k s k k s s
s s s s s s s s
x k k s x y z z
y k k s y y z z
The result can be proved in a manner similar to the proof of Proposition 7. Proposition 9. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1}, V be an idempotent nullnorm on L with the zero element s and x, y ∈ Is . If there are elements k, z ∈ L such that x ∧ s = k, y ∧ s = k, x ∧ y = k, x ∨ s = z, y ∨ s = z and x ∨ y = z, then V(x, y) ∈ Is . The proof follows from Propositions 7 and 8. Example 1. Given a bounded lattice L = {0, x, y, k, z, s, 1} with the given order in Figure 1. Due to Proposition 4, Proposition 9 and the fact that the only idempotent t-norm (inf) T ∧ : [s, 1]2 → [s, 1], T∧ (x, y ) = x ∧ y and the only idempotent t-conorm (sup) S ∨ : [0, s]2 → [0, s], S∨ (x, y ) = x ∨ y, it can be defined only two idempotent nullnorms on L for the given zero element s. These idempotent nullnorms V1 and V2 on L is given by Tables 1 and 2, respectively. Consider a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. In the following Theorems 3 and 4, we propose two methods for constructing idempotent nullnorms on L for the element s playing the role of zero element under an additional assumption that all elements in L incomparable with s are comparable with each other, that is, xࢲy for all x, y ∈ Is . Note that our construction methods are more general than the methods described by Theorems 1 and 2 Lemma 1. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and x, y ∈ L. If x ≤ y, then it holds (y ∧ (x ∨ s )) ∨ s = x ∨ s. Proof. Let x ≤ y. Since ((x ∨ s ) ∧ y ) ∨ s ≤ (x ∨ s ) ∨ s = x ∨ s and ((x ∨ s ) ∧ y ) ∨ s ≥ ((x ∨ s ) ∧ x ) ∨ s = x ∨ s, it holds (y ∧ (x ∨ s )) ∨ s = x ∨ s.
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Lemma 2. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and x, y, z ∈ L. If x ≤ y ≤ z, then it holds (x ∨ (z ∧ (y ∨ s ) )) ∧ (x ∨ s ) = (z ∨ (y ∧ (x ∨ s ))) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Proof. Let x ≤ y ≤ z. 1. We show that (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Since (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) ≥ z ∧ (y ∨ s ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) and (x ∨ (z ∧ (y ∨ s))) ∧ (x ∨ s) ≤ (z ∨ (z ∧ (y ∨ s))) ∧ (x ∨ s) ≤ z ∧ (x ∨ s), then we have that (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). 2. We show that (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Since (z ∨ (y ∧ (x ∨ s )) ) ∧ (x ∨ s ) ≤ (z ∨ (z ∧ (x ∨ s )) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) and (z ∨ (y ∧ (x ∨ s))) ∧ (x ∨ s) ≥ z ∧ (x ∨ s), we obtain that (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Therefore, it holds (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) = (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) once x ≤ y ≤ z. Lemma 3. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and x, y, z ∈ L. If x ≤ z ≤ y, then it holds (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = y ∧ (x ∨ s ). Proof. Let x ≤ z ≤ y. Since (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) ≤ (y ∨ (y ∧ (x ∨ s )) ) ∧ (x ∨ s ) = y ∧ (x ∨ s ) and (z ∨ (y ∧ (x ∨ s )) ) ∧ (x ∨ s ) ≥ y ∧ (x ∨ s ) ∧ (x ∨ s ) = y ∧ (x ∨ s ), we obtain that (z ∨ (y ∧ (x ∨ s ))) ∧ (x ∨ s ) = y ∧ (x ∨ s ). Theorem 3. Let (L, ≤ , 0, 1) be a bounded lattice and s ∈ L\{0, 1}. If all elements in L incomparable with s are comparable with each other, in that case the function Vs,1 : L2 → L defined by
⎧ x∨y ⎪ ⎪ ⎪ x∧y ⎪ ⎪ ⎪ ⎪ ⎨ s x ∨ (y ∧ s ) Vs,1 (x, y ) = y ∨ (x ∧ s ) ⎪ ⎪ ⎪ x ∧ (y ∨ s ) ⎪ ⎪ ⎪ ⎪ y ∧ (x ∨ s ) ⎩ (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y )
if if if if if if if if
x, y ∈ [0, s], x, y ∈ [s, 1], x, y ∈ Ds , x ∈ [0, s] and y ∈ Is , x ∈ Is and y ∈ [0, s], x ∈ [s, 1] and y ∈ Is , x ∈ Is and y ∈ [s, 1], x, y ∈ Is ,
(3)
is an idempotent nullnorm on L with the zero element s. Proof. We have that Vs,1 (x, 0 ) = x ∨ 0 = x for all x ≤ s and Vs,1 (x, 1 ) = x ∧ 1 = x for all x ≥ s. So, s is zero element of Vs,1 . It is easy to see the commutativity of Vs,1 . Hence, we give only the proof of the monotonicity and associativity of Vs,1 . i) Monotonicity: We demonstrate that if x ≤ y, then Vs,1 (x, z) ≤ Vs,1 (y, z) for all z ∈ L. The proof is split into all possible cases. 1. Let x ≤ s. 1.1 y ≤ s, 1.1.1 z ≤ s,
Vs,1 (x, z ) = x ∨ z ≤ y ∨ z = Vs,1 (y, z ) 1.1.2. z > s,
Vs,1 (x, z ) = s = Vs,1 (y, z ) 1.1.3. zs,
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ y ∨ (z ∧ s ) = Vs,1 (y, z ) 1.2 y > s, 1.2.1. z ≤ s,
Vs,1 (x, z ) = x ∨ z ≤ s = Vs,1 (y, z ) 1.2.2. z > s,
Vs,1 (x, z ) = s ≤ y ∧ z = Vs,1 (y, z ) 1.2.3. zs,
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) 1.3. ys, 1.3.1. z ≤ s,
Vs,1 (x, z ) = x ∨ z ≤ z ∨ (y ∧ s ) = Vs,1 (y, z ) 1.3.2. z > s,
Vs,1 (x, z ) = s ≤ z ∧ (y ∨ s ) = Vs,1 (y, z )
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1.3.3. zs, If y ≤ z, then
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ (y ∨ s ) ∧ z = (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) = Vs,1 (y, z ) If z < y, then
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ (z ∨ s ) ∧ y = (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) = Vs,1 (y, z ) 2. Let x > s. Then y > s. 2.1. z ≤ s,
Vs,1 (x, z ) = s = Vs,1 (y, z ) 2.2. z > s,
Vs,1 (x, z ) = x ∧ z ≤ y ∧ z = Vs,1 (y, z ) 2.3. zs,
Vs,1 (x, z ) = x ∧ (z ∨ s ) ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) 3. Let xs. Then y > s or ys. 3.1. y > s, 3.1.1. z ≤ s,
Vs,1 (x, z ) = z ∨ (x ∧ s ) ≤ s = Vs,1 (y, z ) 3.1.2. z > s,
Vs,1 (x, z ) = z ∧ (x ∨ s ) ≤ y ∧ z = Vs,1 (y, z ) 3.1.3. zs, If x ≤ z,
Vs,1 (x, z ) = (x ∨ s ) ∧ (z ∨ s ) ∧ (x ∨ z ) = (x ∨ s ) ∧ z ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) If z < x,
Vs,1 (x, z ) = (x ∨ s ) ∧ (z ∨ s ) ∧ (x ∨ z ) = (z ∨ s ) ∧ x ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) 3.2. ys, 3.2.1. z ≤ s,
Vs,1 (x, z ) = z ∨ (x ∧ s ) ≤ z ∨ (y ∧ s ) = Vs,1 (y, z ) 3.2.2. z > s,
Vs,1 (x, z ) = z ∧ (x ∨ s ) ≤ z ∧ (y ∨ s ) = Vs,1 (y, z ) 3.2.3. zs,
Vs,1 (x, z ) = (x ∨ s ) ∧ (z ∨ s ) ∧ (x ∨ z ) ≤ (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) = Vs,1 (y, z ) ii) Associativity: We prove that Vs,1 (x, Vs,1 (y, z )) = Vs,1 (Vs,1 (x, y ), z ) for all x, y, z ∈ L. The proof is split into all possible cases. 1. Let x ≤ s, 1.1. y ≤ s, 1.1.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ z ) = x ∨ y ∨ z = Vs,1 (x ∨ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.1.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (x ∨ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.1.3 zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ (z ∧ s ) ) = x ∨ y ∨ (z ∧ s ) = Vs,1 (x ∨ y, z ) = Vs,1 (Vs,1 (x, y ), z )
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1.2. y > s, 1.2.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.2.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ z ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.2.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ (z ∨ s ) ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.3. ys, 1.3.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∨ (y ∧ s ) ) = x ∨ z ∨ (y ∧ s ) = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 1.3.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∧ (y ∨ s ) ) = s = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 1.3.3. zs, If y ≤ z, from Proposition 6, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (y ∨ s ) ∧ z ) = x ∨ ( (y ∨ s ) ∧ z ∧ s ) = x ∨ (z ∧ s ) = x ∨ (y ∧ s ) ∨ (z ∧ s ) = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) If z < y, from Proposition 6, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (z ∨ s ) ∧ y ) = x ∨ ( (z ∨ s ) ∧ y ∧ s ) = x ∨ (y ∧ s ) = x ∨ (y ∧ s ) ∨ (z ∧ s ) = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 2. Let x > s. 2.1. y ≤ s, 2.1.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ z ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.1.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.1.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ (z ∧ s ) ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z )
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2.2. y > s 2.2.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (x ∧ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.2.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ z ) = x ∧ y ∧ z = Vs,1 (x ∧ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.2.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ (z ∨ s ) ) = x ∧ y ∧ (z ∨ s ) = Vs,1 (x ∧ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.3. ys, 2.3.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∨ (y ∧ s ) ) = s = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 2.3.2. z > s,
Vs,1 (x, Vs,1 (y, z ))) = Vs,1 (x, z ∧ (y ∨ s ) ) = x ∧ z ∧ (y ∨ s ) = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 2.3.3. zs, If y ≤ z, from Proposition 6 and Lemma 1, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (y ∨ s ) ∧ z ) = x ∧ ( ( (y ∨ s ) ∧ z ) ∨ s ) = x ∧ (y ∨ s ) = x ∧ (y ∨ s ) ∧ (z ∨ s ) = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) If z < y, from Proposition 6 and Lemma 1, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (z ∨ s ) ∧ y ) = x ∧ ( ( (z ∨ s ) ∧ y ) ∨ s ) = x ∧ (z ∨ s ) = x ∧ (y ∨ s ) ∧ (z ∨ s ) = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3. Let xs. 3.1. y ≤ s, 3.1.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ z ) = y ∨ z ∨ (x ∧ s ) = Vs,1 (y ∨ (x ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.1.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (y ∨ (x ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.1.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ (z ∧ s ) ) = y ∨ (z ∧ s ) ∨ (x ∧ s ) = Vs,1 (y ∨ (x ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z )
G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746
3.2. y > s, 3.2.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (y ∧ (x ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.2.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ z ) = y ∧ z ∧ (x ∨ s ) = Vs,1 (y ∧ (x ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.2.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ (z ∨ s ) ) = y ∧ (z ∨ s ) ∧ (x ∨ s ) = Vs,1 (y ∧ (x ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3. ys, 3.3.1. If x ≤ y, 3.3.1.1. z ≤ s, from Proposition 6, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∨ (y ∧ s ) ) = z ∨ (y ∧ s ) ∨ (x ∧ s ) = z ∨ (y ∧ s ) = z ∨ ( (x ∨ s ) ∧ y ∧ s ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3.1.2. z > s, from Proposition 6 and Lemma 1, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∧ (y ∨ s ) ) = z ∧ (y ∨ s ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) = z ∧ ( (y ∧ (x ∨ s ) ) ∨ s ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3.1.3. zs, If y ≤ z, from Proposition 6 and Lemmas 1,2, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (y ∨ s ) ∧ z ) = (x ∨ s ) ∧ ( ( (y ∨ s ) ∧ z ) ∨ s ) ∧ (x ∨ ( (y ∨ s ) ∧ z ) ) = (x ∨ s ) ∧ (x ∨ ( (y ∨ s ) ∧ z ) ) = (x ∨ s ) ∧ z = (x ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = ( ( (x ∨ s ) ∧ y ) ∨ s ) ∧ (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) If z ≤ y and x ≤ z, from Proposition 6 and Lemmas 1–3, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (z ∨ s ) ∧ y ) = (x ∨ s ) ∧ ( ( (z ∨ s ) ∧ y ) ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (x ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (x ∨ s ) ∧ y = (x ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = ( ( (x ∨ s ) ∧ y ) ∨ s ) ∧ (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z )
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G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746
Fig. 2. Lattice L1 .
If z ≤ y and z ≤ x, from Proposition 6 and Lemmas 1–3, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z )) = Vs,1 (x, (z ∨ s ) ∧ y ) = (x ∨ s ) ∧ ( ( (z ∨ s ) ∧ y ) ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (z ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (z ∨ s ) ∧ y = (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = ( ( (x ∨ s ) ∧ y ) ∨ s ) ∧ (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3.2. If y ≤ x, in similar ways to 3.3.1, by using Proposition 6 and Lemmas 1–3, it can be showed that Vs,1 (x, Vs,1 (y, z )) = Vs,1 (Vs,1 (x, y ), z ). It is easy to see that Vs,1 is idempotent, that is, Vs,1 (x, x ) = x for all x ∈ L. So, Vs,1 is an idempotent nullnorm on L with the zero element s. Theorem 4. Let (L, ≤ , 0, 1) be a bounded lattice and s ∈ L\{0, 1}. If all elements in L incomparable with s are comparable with each other, in that case the function Vs,2 : L2 → L defined by
⎧x ∨ y if ⎪ ⎪ ⎪x ∧ y if ⎪ ⎪ ⎪ s if ⎪ ⎨ x ∨ (y ∧ s ) if Vs,2 (x, y ) = y ∨ (x ∧ s ) if ⎪ ⎪ ⎪ x ∧ (y ∨ s ) if ⎪ ⎪ ⎪ ⎪y ∧ (x ∨ s ) if ⎩ (x ∧ s ) ∨ (y ∧ s ) ∨ (x ∧ y ) i f
x, y ∈ [0, s], x, y ∈ [s, 1], x, y ∈ Ds , x ∈ [0, s] and y ∈ Is , x ∈ Is and y ∈ [0, s], x ∈ [s, 1] and y ∈ Is , ∈ Is and y ∈ [s, 1], x, y ∈ Is ,
(4)
is an idempotent nullnorm on L with the zero element s. The result can be proved in a manner similar to the proof of Theorem 3. Example 2. (i) The lattice L1 depicted in Hasse diagram in Fig. 2 is a positive example satisfying the constraint of Theorems 3 and 4 for the fixed zero element s, that is, xࢲy for all x, y ∈ Is . (ii) The lattice L2 depicted in Hasse diagram in Fig. 3 is a negative example that the constraint of Theorems 3 and 4 is violated for a chosen zero element s since km for the elements k, m ∈ Is .
G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746
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Fig. 3. Lattice L2 . Table 3 Idempotent nullnorm Vs,1 on L1 . Vs,1
0
n
p
s
t
m
k
q
r
1
0 n p s t m k q r 1
0 n p s n n n s s s
n n p s n n n s s s
p p p s p p p s s s
s s s s s s s s s s
n n p s t m k q r r
n n p s m m k q r r
n n p s k k k q r r
s s s s q q q q q q
s s s s r r r q r r
s s s s r r r q r 1
Table 4 Idempotent nullnorm Vs,2 on L1 . Vs,2
0
n
p
s
t
m
k
q
r
1
0 n p s t m k q r 1
0 n p s n n n s s s
n n p s n n n s s s
p p p s p p p s s s
s s s s s s s s s s
n n p s t t t q r r
n n p s t m m q r r
n n p s t m k q r r
s s s s q q q q q q
s s s s r r r q r r
s s s s r r r q r 1
Example 3. Consider a bounded lattice L1 with the given order in Fig. 2 and define the functions Vs,1 : L1 × L1 → L1 and Vs,2 : L1 × L1 → L1 by Tables 3 and 4, respectively. Then it can be seen the fact that Vs,1 and Vs,2 are idempotent nullnorms on L with the zero element s from Theorems 3 and 4. Remark 1. Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. It should be noted that in Theorems 3 and 4, the constraint that all elements in L incomparable with s are comparable with each other is not omitted, in general. In the following example, if this constraint is omitted, we demonstrate that the associativity of the functions Vs,1 and Vs,2 defined by the formulas (3) and (4), respectively, is violated. Example 4. Given a bounded lattice L2 characterized in Hasse diagram in Fig. 3 violating the constraint in Theorems 3 and 4. Considering the functions Vs,1 and Vs,2 defined by the formulas (3) and (4), respectively, we have that Vs,1 (0, Vs,1 (k, m )) = Vs,1 (0, 1 ) = s and Vs,1 (Vs,1 (0, k ), m ) = Vs,1 (0, m ) = 0 for 0, k, m ∈ L2 . So, Vs,1 is not a nullnorm on L2 since the associativity is not satisfied. Similarly, we have that Vs,2 (1, Vs,2 (k, m )) = Vs,2 (1, 0 ) = s and Vs,2 (Vs,2 (1, k ), m ) = Vs,2 (1, m ) = 1 for 1, k, m ∈ L2 . Therefore, Vs,2 also is not a nullnorm on L2 . Remark 2. It should be pointed out that in Example 4, we show that the formulas (3) and (4), respectively, defined in Theorems 3 and 4 do not generate a nullnorm on the lattice L2 that does not satisfy the constraint of Theorems 3 and 4. However, the existing construction approaches in [22,30] yield a nullnorm on L2 . But the nullnorms obtained by using these methods in [22,30] are not idempotent on L2 . In the following, we define two different nullnorms V1 : L2 × L2 → L2 and V2 : L2 × L2 → L2 on the lattice L2 by Tables 5 and 6, respectively. It can be easily seen that V1 and V2 are idempotent nullnorms
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G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746 Table 5 Idempotent nullnorm V1 on L2 . V1
0
k
m
s
1
0 k m s 1
0 0 0 s s
0 k m s 1
0 m m s 1
s s s s s
s 1 1 s 1
Table 6 Idempotent nullnorm V2 on L2 . V2
0
k
m
s
1
0 k m s 1
0 0 0 s s
0 k k s 1
0 k m s 1
s s s s s
s 1 1 s 1
Table 7 Idempotent nullnorm V on L1 . V
0
n
p
s
t
m
k
r
1
0 n p s t m k r 1
0 n p s n n n s s
n n p s n n n s s
p p p s p p p s s
s s s s s s s s s
n n p s t t t r r
n n p s t m k r r
n n p s t k k r r
s s s s r r r r r
s s s s r r r r 1
with the zero element s on L2 . Furthermore, by Propositions 3 and 4, we can see that there does not exist an idempotent nullnorm different from V1 and V2 on the lattice L2 . Consider a bounded lattice (L, ≤ , 0, 1). If we use V to denote the set of all nullnorms on L, it is a partially ordered set with the order defined as the following: For V1 , V2 ∈ V, V1 ≤ V2 ⇔V1 (x, y) ≤ V2 (x, y) for all (x, y) ∈ L2 . Let s ∈ L. In that case, each V (s ) is also a partially ordered set, where V (s ) presents the set of all nullnorms on L with the zero element s. Corollary 1. Consider a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1} such that all elements in L incomparable with s are comparable with each other. In this case, the idempotent nullnorms Vs,1 : L2 → L and Vs,2 : L2 → L defined by the formulas (3) and (4), respectively, are the greatest idempotent nullnorm and the smallest idempotent nullnorm on L with the zero element s, respectively. The proof follows from Propositions 2, 5 and 4. Remark 3. Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1} such that all elements in L incomparable with s are comparable with each other. We introduced two construction methods to obtain an idempotent nullnorm on L with the zero element s in Theorems 3 and 4. Furthermore, in Corollary 1, we see that these idempotent nullnorms are the greatest and smallest idempotent nullnorms on L with the zero element s. It should be noted that the idempotent nullnorms which are introduced in Theorems 3 and 4 slightly differ from each other. Considering Example 2, we see that the idempotent nullnorms Vs,1 and Vs,2 defined on the bounded lattice L1 in Fig. 2 are different from each other since Vs,1 (t, m ) = m = t = Vs,2 (t, m ) for the elements m, t ∈ L1 . In this case, a natural question arises: if we take a bounded lattice L such that all elements in L incomparable with s ∈ L\{0, 1} are comparable with each other, does an idempotent nullnorm on L with the zero element s have to be one of the idempotent nullnorms introduced in Theorems 3 and 4? In the following example, we give a negative example violating the above hypothesis. Example 5. Consider a bounded lattice L1 which is depicted in Hasse diagram by Fig. 2. In that case, it is possible to check the function V: L1 × L1 → L1 defined by Table 7 is an idempotent nullnorm on L1 with the indicated zero element s. In addition, the idempotent nullnorm V differs from the idempotent nullnorms Vs,1 and Vs,2 on L1 given by Tables 3 and 4,
G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746
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respectively, by using Theorems 3 and 4, respectively. Because, V (k, m ) = k = m = Vs,2 (k, m ) and V (t, m ) = t = m = Vs,1 (t, m ) for k, m, t ∈ L1 . Remark 4. Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. (i) If L is distributive such that all elements in L incomparable with s are comparable with each other, then the idempotent nullnorms obtained by the formula (3) in Theorem 3 and the formula (4) in Theorem 4 coincide with the idempotent nullnorm given by the formula (1) in Theorem 1. (ii) If there is only one element incomparable with s in L, then the idempotent nullnorms obtained by the formula (3) in Theorem 3 and the formula (4) in Theorem 4 coincide with the idempotent nullnorm given by the formulas (2) in Theorem 2. 4. Concluding remarks In previous studies, nullnorms were defined on bounded lattices and some methods for building nullnorms on bounded lattices were proposed in [22,30]. But these methods to obtain nullnorms described in [22,30] do not generate an idempotent nullnorm on general bounded lattices unless we take some special bounded lattices. For this reason, in [11], it was showed that an idempotent nullnorm on a bounded lattice L with a zero element s ∈ L\{0, 1} need not always exist. In addition, a construction method for idempotent nullnorms on a bounded lattice L with a zero element s ∈ L\{0, 1} was introduced under an additional constraint that the one element incomparable with s in L exists. In this paper, considering an arbitrary bounded lattice L, we investigated the existence of idempotent nullnorms on L with a zero element different from the top and bottom elements and some properties of these operators. Moreover, we enhanced the construction method proposed previously in [11] and introduced two additional construction methods for idempotent nullnorms on a bounded lattice L such that all elements in L incomparable with a zero element s ∈ L\{0, 1} are comparable with each other. The methods to obtain idempotent nullnorms introduced in this study are more general than the existing methods and this fact makes our methods more beneficial for practical applicati ons. Acknowledgments The author expresses her sincere thanks to the editors and reviewers for their most valuable comments and suggestions. References [1] [2] [3] [4] [5]
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Construction methods for idempotent nullnorms on bounded lattices Gül Deniz Çaylı Department of Mathematics, Faculty of Science, Karadeniz Technical University, Trabzon 61080, Turkey
a r t i c l e
i n f o
Article history: Received 26 February 2018 Revised 3 September 2019 Accepted 9 September 2019
Keywords: Bounded lattice Construction method Idempotent nullnorm Nullnorm Zero element
a b s t r a c t In this paper, we discuss the presence of idempotent nullnorms on bounded lattices and investigate some properties of these operators. We introduce two new methods for constructing idempotent nullnorms on bounded lattices for a zero element under a constraint that all elements incomparable with the zero element are comparable with each other. In order to stress the role of this constraint satisfying that our constructions yield an idempotent nullnorm, some illustrative examples are added. © 2019 Elsevier Inc. All rights reserved.
1. Introduction Triangular norms (t-norms, for short) with the neutral element as 1 and triangular conorms (t-conorms, for short) with the neutral element as 0 on the real unit interval were introduced and comprehensively investigated by Schweizer and Sklar [40,41]. These operators have been extensively used in numerous practical applications such as fuzzy set theory, fuzzy logic, several branches of information sciences, multicriteria decision support and aggregation [7,23,26,27,39]. For more details concerning t-norms and t-conorms, the readers can refer to the papers [1,2,24,25,32,33,38]. Nullnorms and t-operators as a generalization of t-norms and t-conorms were first introduced in [8,35], where the zero element lies anywhere in the real unit interval rather than t-norms and t-conorms have zero elements 0 or 1, respectively. Nullnorms coincide with t-operators discussed by Mas, Mayor and Torrens [36] since both of them have the same block structures in the real unit interval, that is, once a binary operator is a t-operator, it is also a nullnorm and vice versa. Since the structure of nullnorms is a combination of t-norms and t-conorms, these operators play an important role both in theoretical investigations and in practical applications such as expert systems, neural networks, fuzzy system modelings. Nullnorms which are especially used as aggregation operators or in fuzzy logic, maintain as several logical characteristics as possible [9,20,21]. In the literature, there are some other studies about nullnorms on the real unit interval [3,16–19,28,34,37]. In [30], two methods for constructing nullnorms on a bounded lattice L with a zero element s ∈ L\{0, 1} were introduced by means of the presence of a t-norm on [s, 1]2 and a t-conorm on [0, s]2 . In [11], the existence of idempotent nullnorms on a bounded lattice L was investigated. In the same time, a new method to generate idempotent nullnorms on L with the indicated zero element s ∈ L\{0, 1} was presented under an additional assumption that there is the only one element in L incomparable with s. Following the demonstration of the existence of idempotent nullnorms on bounded lattices in [11], further researches have been promoted in this field. The fundamental target of this study is to investigate the presence E-mail address: [email protected] https://doi.org/10.1016/j.amc.2019.124746 0 096-30 03/© 2019 Elsevier Inc. All rights reserved.
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of idempotent nullnorms on a bounded lattice L with the zero element s ∈ L\{0, 1} and introduce further methods for constructing idempotent nullnorms on L. This study is organized as follows. First, some main constructions concerning nullnorms on bounded lattices and their results are shortly discussed in Section 2. In Section 3, the existence of idempotent nullnorms on an arbitrary bounded lattice L with the zero element s ∈ L\{0, 1} and some properties of them are researched. In addition, it is introduced two new methods to obtain idempotent nullnorms on L for an element s playing the role of a zero element with a constraint. Some corresponding examples are added in order to demonstrate that this constraint is necessary in our methods. Finally, some concluding remarks are given. 2. Preliminaries In this section, we recall some preliminary details concerning bounded lattices and nullnorms (t-norms and t-conorms) on them. A bounded lattice (L, ≤ ) is a lattice if it has the top and bottom elements that are written as 1 and 0, respectively; that is, there exist two elements 1, 0 ∈ L such that 0 ≤ x ≤ 1 for all x ∈ L. Definition 1 [4]. Given a bounded lattice (L, ≤ , 0, 1) and x, y ∈ L, if x and y are incomparable, in this case we use the notation xy. If x and y are comparable, in this case we use the notation xࢲy. Definition 2 [4]. Given a bounded lattice (L, ≤ , 0, 1) and x, y ∈ L, x ≤ y, a subinterval [x, y] of L is defined as
[x, y] = {a ∈ L
| x ≤ a ≤ y}.
Similarly, we define x, y = {a ∈ L | x < a ≤ y}, x, y = {a ∈ L | x ≤ a < y} and x, y = {a ∈ L | x < a < y}. Definition 3 [10,12]. Let (L, ≤ , 0, 1) be a bounded lattice. An operation T: L2 → L that is commutative, associative, increasing with respect to both variables and has the neutral element 1 such that T (x, 1 ) = x for all x ∈ L is called a t-norm on L. Definition 4 [6,10]. Let (L, ≤ , 0, 1) be a bounded lattice. An operation S: L2 → L that is commutative, associative, increasing with respect to both variables and has the neutral element 0 such that S(x, 0 ) = x for all x ∈ L is called a t-conorm on L. Definition 5 [11,30]. Let (L, ≤ , 0, 1) be a bounded lattice. If an operation V: L2 → L is commutative, associative, nondecreasing in each variable and there is an element such that s ∈ L V (x, 0 ) = x for all x ≤ s and V (x, 1 ) = x for all x ≥ s, then it is called a nullnorm on L. It is easy to see that V (x, s ) = s for all x ∈ L. So, s ∈ L is the zero element of the nullnorm V. We denote the sets Ds = [0, s] × [s, 1] ∪ [s, 1] × [0, s] and Is = {x ∈ L | x s} for an element s ∈ L\{0, 1}. The characterization of nullnorms (particularly idempotent nullnorms) on bounded lattices have attracted much attention from researchers in recent years. The main aim of this paper is to give new methods to construct idempotent nullnorms on bounded lattices. Now, we introduce some known results about nullnorms on bounded lattices, which will be used in the sequel. For more details on nullnorms on lattices, the readers can refer to [5,13–15,22,29,31,42]. Proposition 1 [17,30]. Given a bounded lattice (L, ≤ , 0, 1), s ∈ L\{0, 1} and a nullnorm V on L with the zero element s. In this case, the following statements are obtained. i) V |[0, s]2 : [0, s]2 → [0, s] is a t-conorm on [0, s]2 . ii) V |[s, 1]2 : [s, 1]2 → [s, 1] is a t-norm on [s, 1]2 . Proposition 2 [17,30]. Given a bounded lattice (L, ≤ , 0, 1), s ∈ L\{0, 1} and a nullnorm V on L with the zero element s. In this case, the following statements are obtained. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
V (x, y ) = s for all (x, y) ∈ Ds . s ≤ V(x, y) for all (x, y) ∈ [s, 1]2 ∪ [s, 1] × Is ∪ Is × [s, 1]. V(x, y) ≤ s for all (x, y) ∈ [0, s]2 ∪ [0, s] × Is ∪ Is × [0, s]. V(x, y) ≤ y for all (x, y) ∈ L × [s, 1]. V(x, y) ≤ x for all (x, y) ∈ [s, 1] × L. x ≤ V(x, y) for all (x, y) ∈ [0, s] × L. y ≤ V(x, y) for all (x, y) ∈ L × [0, s]. x ∨ y ≤ V(x, y) for all (x, y) ∈ [0, s]2 . V(x, y) ≤ x ∧ y for all (x, y) ∈ [s, 1]2 . (x ∧ s) ∨ (y ∧ s) ≤ V(x, y) for all (x, y) ∈ [0, s] × Is ∪ Is × [0, s] ∪ Is × Is . V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) for all (x, y) ∈ [s, 1] × Is ∪ Is × [s, 1] ∪ Is × Is .
Definition 6 [11,15]. Given a bounded lattice (L, ≤ , 0, 1), s ∈ L\{0, 1} and a nullnorm V on L with the zero element s. V is called an idempotent nullnorm on L if V (x, x ) = x for all x ∈ L. Proposition 3 [15]. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. In this case, the following statements are obtained.
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(i) V (x, y ) = x ∨ y for all (x, y) ∈ [0, s]2 . (ii) V (x, y ) = x ∧ y for all (x, y) ∈ [s, 1]2 . Proposition 4 [15]. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. In this case, the following statements are obtained. (i) (ii) (iii) (iv)
V (x, y ) = x ∨ (y ∧ s ) V (x, y ) = y ∨ (x ∧ s ) V (x, y ) = x ∧ (y ∨ s ) V (x, y ) = y ∧ (x ∨ s )
for for for for
all all all all
(x, y ) ∈ [0, s] × Is . (x, y ) ∈ Is × [0, s]. (x, y ) ∈ [s, 1] ∪ Is . (x, y ) ∈ Is × [s, 1].
Theorem 1. [11] Given a distributive bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. In this case, the function defined by
V∗ (x, y ) = (x ∧ y ) ∨ (x ∧ s ) ∨ (y ∧ s ) = (x ∨ y ) ∧ (x ∨ s ) ∧ (y ∨ s ) = V ∗ (x, y )
(1)
is an idempotent nullnorm on L with zero element s. Theorem 2. [11] Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. If there is only one element incomparable with s in L and it is denoted by k the element incomparable with s, in that case the function V defined by
⎧ x∨y ⎪ ⎪ ⎪ ⎪ x∧y ⎪ ⎪ ⎪ ⎪ ⎨ s x ∨ (k ∧ s ) V (x, y ) = y ∨ (k ∧ s ) ⎪ ⎪ ⎪ ⎪ x ∧ (k ∨ s ) ⎪ ⎪ ⎪ ⎪ ⎩ y ∧ (k ∨ s ) k
if if if if if if if if
(x, y ) ∈ [0, s]2 , (x, y ) ∈ [s, 1]2 , (x, y ) ∈ Ds ,
x ∈ [0, s] and y = k, x = k and y ∈ [0, s], x ∈ [s, 1] and y = k, x = k and y ∈ [s, 1], x = y = k,
(2)
is an idempotent nullnorm on L with the zero element s. 3. Idempotent nullnorms on bounded lattices In this section, considering an arbitrary bounded lattice L, we investigate the existence of idempotent nullnorms on L with the indicated zero element s ∈ L\{0, 1} and their characteristics. In addition, we enhance the methods for obtaining idempotent nullnorms on L presented in Theorems 1 and 2 and propose two new construction methods for idempotent nullnorms on L with different characteristics compared with each other. Proposition 5. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. Then (x ∧ s) ∨ (y ∧ s) ∨ (x ∧ y) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) ∧ (x ∨ y) for all x, y ∈ L. Proof. By Proposition 2 (x) and (xi), we obtain that (x ∧ s) ∨ (y ∧ s) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s). Since V is idempotent, we have that x ∧ y = V (x ∧ y, x ∧ y ) ≤ V (x, y ) ≤ V (x ∨ y, x ∨ y ) = x ∨ y. Thus, it holds (x ∧ s) ∨ (y ∧ s) ∨ (x ∧ y) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) ∧ (x ∨ y). Proposition 6. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and V be an idempotent nullnorm on L with the zero element s. If x, y ∈ Is and x is comparable with y, then it holds V(x, y) ∈ Is . Proof. Let x, y ∈ Is and x be comparable with y. In this case, x ≤ y or x > y. By Proposition 5, it is (x ∧ s) ∨ (y ∧ s) ∨ (x ∧ y) ≤ V(x, y) ≤ (x ∨ s) ∧ (y ∨ s) ∧ (x ∨ y). If x ≤ y, then (y ∧ s) ∨ x ≤ V(x, y) ≤ (x ∨ s) ∧ y. Suppose that V(x, y) ≤ s. In this case, we have that x ∨ (y ∧ s) ≤ s, i.e., x ≤ s. This is a contradiction. So, it can not be V(x, y) ≤ s. Suppose that V(x, y) > s. In this case, we have that s < y ∧ (x ∨ s), i.e., s < y. This is a contradiction. So, it can not be V(x, y) > s. Hence, if x ≤ y, then V(x, y) ∈ Is . Similarly, it is showed that if x > y, then V(x, y) ∈ Is . Therefore, if x, y ∈ Is and x is comparable with y, then we obtain that V(x, y) ∈ Is . Proposition 7. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1}, V be an idempotent nullnorm on L with the zero element s and x, y ∈ Is . If there is an element z ∈ L such that x ∧ s = z, y ∧ s = z and x ∧ y = z, then V(x, y) < s or V(x, y) ∈ Is . Proof. Let x, y ∈ Is and there be z ∈ L such that x ∧ s = z, y ∧ s = z and x ∧ y = z. By Proposition 5, since (z ∧ s) ∨ (x ∧ s) ∨ (z ∧ x) ≤ V(z, x) ≤ (z ∨ s) ∧ (x ∨ s) ∧ (z ∨ x), we obtain that V (z, x ) = z. Similarly, it is obtained that V (z, y ) = z. Since V (V (z, x ), y ) = V (z, y ) = z, it holds V (z, V (x, y )) = z from the associativity of V. Suppose that V(x, y) ≥ s. Then due to the monotonicity of V and the fact that s is zero element, we have that z = V (z, V (x, y )) ≥ V (z, s ) = s, i.e., z ≥ s. This is a contradiction. So, it can not be V(x, y) ≥ s. Therefore, if there is z ∈ L such that x ∧ s = z, y ∧ s = z and x ∧ y = z, then it holds V(x, y) < s or V(x, y) ∈ Is . Proposition 8. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1}, V be an idempotent nullnorm on L with the zero element s and x, y ∈ Is . If there is an element z ∈ L such that x ∨ s = z, y ∨ s = z and x ∨ y = z, then V(x, y) > s or V(x, y) ∈ Is .
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Fig. 1. Lattice L. Table 1 Idempotent nullnorm V1 on L. V1 0 k s x y z 1
0 0 k s k k s s
k k k s k k s s
s s s s s s s s
x k k s x x z z
y k k s x y z z
z s s s z z z z
1 s s s z z z 1
z s s s z z z z
1 s s s z z z 1
Table 2 Idempotent nullnorm V2 on L. V2 0 k s x y z 1
0 0 k s k k s s
k k k s k k s s
s s s s s s s s
x k k s x y z z
y k k s y y z z
The result can be proved in a manner similar to the proof of Proposition 7. Proposition 9. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1}, V be an idempotent nullnorm on L with the zero element s and x, y ∈ Is . If there are elements k, z ∈ L such that x ∧ s = k, y ∧ s = k, x ∧ y = k, x ∨ s = z, y ∨ s = z and x ∨ y = z, then V(x, y) ∈ Is . The proof follows from Propositions 7 and 8. Example 1. Given a bounded lattice L = {0, x, y, k, z, s, 1} with the given order in Figure 1. Due to Proposition 4, Proposition 9 and the fact that the only idempotent t-norm (inf) T ∧ : [s, 1]2 → [s, 1], T∧ (x, y ) = x ∧ y and the only idempotent t-conorm (sup) S ∨ : [0, s]2 → [0, s], S∨ (x, y ) = x ∨ y, it can be defined only two idempotent nullnorms on L for the given zero element s. These idempotent nullnorms V1 and V2 on L is given by Tables 1 and 2, respectively. Consider a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. In the following Theorems 3 and 4, we propose two methods for constructing idempotent nullnorms on L for the element s playing the role of zero element under an additional assumption that all elements in L incomparable with s are comparable with each other, that is, xࢲy for all x, y ∈ Is . Note that our construction methods are more general than the methods described by Theorems 1 and 2 Lemma 1. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and x, y ∈ L. If x ≤ y, then it holds (y ∧ (x ∨ s )) ∨ s = x ∨ s. Proof. Let x ≤ y. Since ((x ∨ s ) ∧ y ) ∨ s ≤ (x ∨ s ) ∨ s = x ∨ s and ((x ∨ s ) ∧ y ) ∨ s ≥ ((x ∨ s ) ∧ x ) ∨ s = x ∨ s, it holds (y ∧ (x ∨ s )) ∨ s = x ∨ s.
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Lemma 2. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and x, y, z ∈ L. If x ≤ y ≤ z, then it holds (x ∨ (z ∧ (y ∨ s ) )) ∧ (x ∨ s ) = (z ∨ (y ∧ (x ∨ s ))) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Proof. Let x ≤ y ≤ z. 1. We show that (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Since (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) ≥ z ∧ (y ∨ s ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) and (x ∨ (z ∧ (y ∨ s))) ∧ (x ∨ s) ≤ (z ∨ (z ∧ (y ∨ s))) ∧ (x ∨ s) ≤ z ∧ (x ∨ s), then we have that (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). 2. We show that (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Since (z ∨ (y ∧ (x ∨ s )) ) ∧ (x ∨ s ) ≤ (z ∨ (z ∧ (x ∨ s )) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) and (z ∨ (y ∧ (x ∨ s))) ∧ (x ∨ s) ≥ z ∧ (x ∨ s), we obtain that (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ). Therefore, it holds (x ∨ (z ∧ (y ∨ s ) ) ) ∧ (x ∨ s ) = (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) once x ≤ y ≤ z. Lemma 3. Let (L, ≤ , 0, 1) be a bounded lattice, s ∈ L\{0, 1} and x, y, z ∈ L. If x ≤ z ≤ y, then it holds (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) = y ∧ (x ∨ s ). Proof. Let x ≤ z ≤ y. Since (z ∨ (y ∧ (x ∨ s ) ) ) ∧ (x ∨ s ) ≤ (y ∨ (y ∧ (x ∨ s )) ) ∧ (x ∨ s ) = y ∧ (x ∨ s ) and (z ∨ (y ∧ (x ∨ s )) ) ∧ (x ∨ s ) ≥ y ∧ (x ∨ s ) ∧ (x ∨ s ) = y ∧ (x ∨ s ), we obtain that (z ∨ (y ∧ (x ∨ s ))) ∧ (x ∨ s ) = y ∧ (x ∨ s ). Theorem 3. Let (L, ≤ , 0, 1) be a bounded lattice and s ∈ L\{0, 1}. If all elements in L incomparable with s are comparable with each other, in that case the function Vs,1 : L2 → L defined by
⎧ x∨y ⎪ ⎪ ⎪ x∧y ⎪ ⎪ ⎪ ⎪ ⎨ s x ∨ (y ∧ s ) Vs,1 (x, y ) = y ∨ (x ∧ s ) ⎪ ⎪ ⎪ x ∧ (y ∨ s ) ⎪ ⎪ ⎪ ⎪ y ∧ (x ∨ s ) ⎩ (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y )
if if if if if if if if
x, y ∈ [0, s], x, y ∈ [s, 1], x, y ∈ Ds , x ∈ [0, s] and y ∈ Is , x ∈ Is and y ∈ [0, s], x ∈ [s, 1] and y ∈ Is , x ∈ Is and y ∈ [s, 1], x, y ∈ Is ,
(3)
is an idempotent nullnorm on L with the zero element s. Proof. We have that Vs,1 (x, 0 ) = x ∨ 0 = x for all x ≤ s and Vs,1 (x, 1 ) = x ∧ 1 = x for all x ≥ s. So, s is zero element of Vs,1 . It is easy to see the commutativity of Vs,1 . Hence, we give only the proof of the monotonicity and associativity of Vs,1 . i) Monotonicity: We demonstrate that if x ≤ y, then Vs,1 (x, z) ≤ Vs,1 (y, z) for all z ∈ L. The proof is split into all possible cases. 1. Let x ≤ s. 1.1 y ≤ s, 1.1.1 z ≤ s,
Vs,1 (x, z ) = x ∨ z ≤ y ∨ z = Vs,1 (y, z ) 1.1.2. z > s,
Vs,1 (x, z ) = s = Vs,1 (y, z ) 1.1.3. zs,
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ y ∨ (z ∧ s ) = Vs,1 (y, z ) 1.2 y > s, 1.2.1. z ≤ s,
Vs,1 (x, z ) = x ∨ z ≤ s = Vs,1 (y, z ) 1.2.2. z > s,
Vs,1 (x, z ) = s ≤ y ∧ z = Vs,1 (y, z ) 1.2.3. zs,
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) 1.3. ys, 1.3.1. z ≤ s,
Vs,1 (x, z ) = x ∨ z ≤ z ∨ (y ∧ s ) = Vs,1 (y, z ) 1.3.2. z > s,
Vs,1 (x, z ) = s ≤ z ∧ (y ∨ s ) = Vs,1 (y, z )
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1.3.3. zs, If y ≤ z, then
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ (y ∨ s ) ∧ z = (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) = Vs,1 (y, z ) If z < y, then
Vs,1 (x, z ) = x ∨ (z ∧ s ) ≤ (z ∨ s ) ∧ y = (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) = Vs,1 (y, z ) 2. Let x > s. Then y > s. 2.1. z ≤ s,
Vs,1 (x, z ) = s = Vs,1 (y, z ) 2.2. z > s,
Vs,1 (x, z ) = x ∧ z ≤ y ∧ z = Vs,1 (y, z ) 2.3. zs,
Vs,1 (x, z ) = x ∧ (z ∨ s ) ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) 3. Let xs. Then y > s or ys. 3.1. y > s, 3.1.1. z ≤ s,
Vs,1 (x, z ) = z ∨ (x ∧ s ) ≤ s = Vs,1 (y, z ) 3.1.2. z > s,
Vs,1 (x, z ) = z ∧ (x ∨ s ) ≤ y ∧ z = Vs,1 (y, z ) 3.1.3. zs, If x ≤ z,
Vs,1 (x, z ) = (x ∨ s ) ∧ (z ∨ s ) ∧ (x ∨ z ) = (x ∨ s ) ∧ z ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) If z < x,
Vs,1 (x, z ) = (x ∨ s ) ∧ (z ∨ s ) ∧ (x ∨ z ) = (z ∨ s ) ∧ x ≤ y ∧ (z ∨ s ) = Vs,1 (y, z ) 3.2. ys, 3.2.1. z ≤ s,
Vs,1 (x, z ) = z ∨ (x ∧ s ) ≤ z ∨ (y ∧ s ) = Vs,1 (y, z ) 3.2.2. z > s,
Vs,1 (x, z ) = z ∧ (x ∨ s ) ≤ z ∧ (y ∨ s ) = Vs,1 (y, z ) 3.2.3. zs,
Vs,1 (x, z ) = (x ∨ s ) ∧ (z ∨ s ) ∧ (x ∨ z ) ≤ (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) = Vs,1 (y, z ) ii) Associativity: We prove that Vs,1 (x, Vs,1 (y, z )) = Vs,1 (Vs,1 (x, y ), z ) for all x, y, z ∈ L. The proof is split into all possible cases. 1. Let x ≤ s, 1.1. y ≤ s, 1.1.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ z ) = x ∨ y ∨ z = Vs,1 (x ∨ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.1.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (x ∨ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.1.3 zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ (z ∧ s ) ) = x ∨ y ∨ (z ∧ s ) = Vs,1 (x ∨ y, z ) = Vs,1 (Vs,1 (x, y ), z )
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1.2. y > s, 1.2.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.2.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ z ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.2.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ (z ∨ s ) ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 1.3. ys, 1.3.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∨ (y ∧ s ) ) = x ∨ z ∨ (y ∧ s ) = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 1.3.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∧ (y ∨ s ) ) = s = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 1.3.3. zs, If y ≤ z, from Proposition 6, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (y ∨ s ) ∧ z ) = x ∨ ( (y ∨ s ) ∧ z ∧ s ) = x ∨ (z ∧ s ) = x ∨ (y ∧ s ) ∨ (z ∧ s ) = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) If z < y, from Proposition 6, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (z ∨ s ) ∧ y ) = x ∨ ( (z ∨ s ) ∧ y ∧ s ) = x ∨ (y ∧ s ) = x ∨ (y ∧ s ) ∨ (z ∧ s ) = Vs,1 (x ∨ (y ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 2. Let x > s. 2.1. y ≤ s, 2.1.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ z ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.1.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.1.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ (z ∧ s ) ) = s = Vs,1 (s, z ) = Vs,1 (Vs,1 (x, y ), z )
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2.2. y > s 2.2.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (x ∧ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.2.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ z ) = x ∧ y ∧ z = Vs,1 (x ∧ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.2.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ (z ∨ s ) ) = x ∧ y ∧ (z ∨ s ) = Vs,1 (x ∧ y, z ) = Vs,1 (Vs,1 (x, y ), z ) 2.3. ys, 2.3.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∨ (y ∧ s ) ) = s = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 2.3.2. z > s,
Vs,1 (x, Vs,1 (y, z ))) = Vs,1 (x, z ∧ (y ∨ s ) ) = x ∧ z ∧ (y ∨ s ) = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 2.3.3. zs, If y ≤ z, from Proposition 6 and Lemma 1, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (y ∨ s ) ∧ z ) = x ∧ ( ( (y ∨ s ) ∧ z ) ∨ s ) = x ∧ (y ∨ s ) = x ∧ (y ∨ s ) ∧ (z ∨ s ) = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) If z < y, from Proposition 6 and Lemma 1, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (z ∨ s ) ∧ y ) = x ∧ ( ( (z ∨ s ) ∧ y ) ∨ s ) = x ∧ (z ∨ s ) = x ∧ (y ∨ s ) ∧ (z ∨ s ) = Vs,1 (x ∧ (y ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3. Let xs. 3.1. y ≤ s, 3.1.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ z ) = y ∨ z ∨ (x ∧ s ) = Vs,1 (y ∨ (x ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.1.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (y ∨ (x ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.1.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∨ (z ∧ s ) ) = y ∨ (z ∧ s ) ∨ (x ∧ s ) = Vs,1 (y ∨ (x ∧ s ), z ) = Vs,1 (Vs,1 (x, y ), z )
G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746
3.2. y > s, 3.2.1. z ≤ s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, s ) = s = Vs,1 (y ∧ (x ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.2.2. z > s,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ z ) = y ∧ z ∧ (x ∨ s ) = Vs,1 (y ∧ (x ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.2.3. zs,
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, y ∧ (z ∨ s ) ) = y ∧ (z ∨ s ) ∧ (x ∨ s ) = Vs,1 (y ∧ (x ∨ s ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3. ys, 3.3.1. If x ≤ y, 3.3.1.1. z ≤ s, from Proposition 6, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∨ (y ∧ s ) ) = z ∨ (y ∧ s ) ∨ (x ∧ s ) = z ∨ (y ∧ s ) = z ∨ ( (x ∨ s ) ∧ y ∧ s ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3.1.2. z > s, from Proposition 6 and Lemma 1, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, z ∧ (y ∨ s ) ) = z ∧ (y ∨ s ) ∧ (x ∨ s ) = z ∧ (x ∨ s ) = z ∧ ( (y ∧ (x ∨ s ) ) ∨ s ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3.1.3. zs, If y ≤ z, from Proposition 6 and Lemmas 1,2, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (y ∨ s ) ∧ z ) = (x ∨ s ) ∧ ( ( (y ∨ s ) ∧ z ) ∨ s ) ∧ (x ∨ ( (y ∨ s ) ∧ z ) ) = (x ∨ s ) ∧ (x ∨ ( (y ∨ s ) ∧ z ) ) = (x ∨ s ) ∧ z = (x ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = ( ( (x ∨ s ) ∧ y ) ∨ s ) ∧ (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) If z ≤ y and x ≤ z, from Proposition 6 and Lemmas 1–3, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z ) ) = Vs,1 (x, (z ∨ s ) ∧ y ) = (x ∨ s ) ∧ ( ( (z ∨ s ) ∧ y ) ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (x ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (x ∨ s ) ∧ y = (x ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = ( ( (x ∨ s ) ∧ y ) ∨ s ) ∧ (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z )
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Fig. 2. Lattice L1 .
If z ≤ y and z ≤ x, from Proposition 6 and Lemmas 1–3, it holds
Vs,1 (x, Vs,1 (y, z )) = Vs,1 (x, (y ∨ s ) ∧ (z ∨ s ) ∧ (y ∨ z )) = Vs,1 (x, (z ∨ s ) ∧ y ) = (x ∨ s ) ∧ ( ( (z ∨ s ) ∧ y ) ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (z ∨ s ) ∧ (x ∨ ( (z ∨ s ) ∧ y ) ) = (z ∨ s ) ∧ y = (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = ( ( (x ∨ s ) ∧ y ) ∨ s ) ∧ (z ∨ s ) ∧ ( ( (x ∨ s ) ∧ y ) ∨ z ) = Vs,1 ( (x ∨ s ) ∧ y, z ) = Vs,1 ( (x ∨ s ) ∧ (y ∨ s ) ∧ (x ∨ y ), z ) = Vs,1 (Vs,1 (x, y ), z ) 3.3.2. If y ≤ x, in similar ways to 3.3.1, by using Proposition 6 and Lemmas 1–3, it can be showed that Vs,1 (x, Vs,1 (y, z )) = Vs,1 (Vs,1 (x, y ), z ). It is easy to see that Vs,1 is idempotent, that is, Vs,1 (x, x ) = x for all x ∈ L. So, Vs,1 is an idempotent nullnorm on L with the zero element s. Theorem 4. Let (L, ≤ , 0, 1) be a bounded lattice and s ∈ L\{0, 1}. If all elements in L incomparable with s are comparable with each other, in that case the function Vs,2 : L2 → L defined by
⎧x ∨ y if ⎪ ⎪ ⎪x ∧ y if ⎪ ⎪ ⎪ s if ⎪ ⎨ x ∨ (y ∧ s ) if Vs,2 (x, y ) = y ∨ (x ∧ s ) if ⎪ ⎪ ⎪ x ∧ (y ∨ s ) if ⎪ ⎪ ⎪ ⎪y ∧ (x ∨ s ) if ⎩ (x ∧ s ) ∨ (y ∧ s ) ∨ (x ∧ y ) i f
x, y ∈ [0, s], x, y ∈ [s, 1], x, y ∈ Ds , x ∈ [0, s] and y ∈ Is , x ∈ Is and y ∈ [0, s], x ∈ [s, 1] and y ∈ Is , ∈ Is and y ∈ [s, 1], x, y ∈ Is ,
(4)
is an idempotent nullnorm on L with the zero element s. The result can be proved in a manner similar to the proof of Theorem 3. Example 2. (i) The lattice L1 depicted in Hasse diagram in Fig. 2 is a positive example satisfying the constraint of Theorems 3 and 4 for the fixed zero element s, that is, xࢲy for all x, y ∈ Is . (ii) The lattice L2 depicted in Hasse diagram in Fig. 3 is a negative example that the constraint of Theorems 3 and 4 is violated for a chosen zero element s since km for the elements k, m ∈ Is .
G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746
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Fig. 3. Lattice L2 . Table 3 Idempotent nullnorm Vs,1 on L1 . Vs,1
0
n
p
s
t
m
k
q
r
1
0 n p s t m k q r 1
0 n p s n n n s s s
n n p s n n n s s s
p p p s p p p s s s
s s s s s s s s s s
n n p s t m k q r r
n n p s m m k q r r
n n p s k k k q r r
s s s s q q q q q q
s s s s r r r q r r
s s s s r r r q r 1
Table 4 Idempotent nullnorm Vs,2 on L1 . Vs,2
0
n
p
s
t
m
k
q
r
1
0 n p s t m k q r 1
0 n p s n n n s s s
n n p s n n n s s s
p p p s p p p s s s
s s s s s s s s s s
n n p s t t t q r r
n n p s t m m q r r
n n p s t m k q r r
s s s s q q q q q q
s s s s r r r q r r
s s s s r r r q r 1
Example 3. Consider a bounded lattice L1 with the given order in Fig. 2 and define the functions Vs,1 : L1 × L1 → L1 and Vs,2 : L1 × L1 → L1 by Tables 3 and 4, respectively. Then it can be seen the fact that Vs,1 and Vs,2 are idempotent nullnorms on L with the zero element s from Theorems 3 and 4. Remark 1. Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. It should be noted that in Theorems 3 and 4, the constraint that all elements in L incomparable with s are comparable with each other is not omitted, in general. In the following example, if this constraint is omitted, we demonstrate that the associativity of the functions Vs,1 and Vs,2 defined by the formulas (3) and (4), respectively, is violated. Example 4. Given a bounded lattice L2 characterized in Hasse diagram in Fig. 3 violating the constraint in Theorems 3 and 4. Considering the functions Vs,1 and Vs,2 defined by the formulas (3) and (4), respectively, we have that Vs,1 (0, Vs,1 (k, m )) = Vs,1 (0, 1 ) = s and Vs,1 (Vs,1 (0, k ), m ) = Vs,1 (0, m ) = 0 for 0, k, m ∈ L2 . So, Vs,1 is not a nullnorm on L2 since the associativity is not satisfied. Similarly, we have that Vs,2 (1, Vs,2 (k, m )) = Vs,2 (1, 0 ) = s and Vs,2 (Vs,2 (1, k ), m ) = Vs,2 (1, m ) = 1 for 1, k, m ∈ L2 . Therefore, Vs,2 also is not a nullnorm on L2 . Remark 2. It should be pointed out that in Example 4, we show that the formulas (3) and (4), respectively, defined in Theorems 3 and 4 do not generate a nullnorm on the lattice L2 that does not satisfy the constraint of Theorems 3 and 4. However, the existing construction approaches in [22,30] yield a nullnorm on L2 . But the nullnorms obtained by using these methods in [22,30] are not idempotent on L2 . In the following, we define two different nullnorms V1 : L2 × L2 → L2 and V2 : L2 × L2 → L2 on the lattice L2 by Tables 5 and 6, respectively. It can be easily seen that V1 and V2 are idempotent nullnorms
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G.D. Çaylı / Applied Mathematics and Computation 366 (2020) 124746 Table 5 Idempotent nullnorm V1 on L2 . V1
0
k
m
s
1
0 k m s 1
0 0 0 s s
0 k m s 1
0 m m s 1
s s s s s
s 1 1 s 1
Table 6 Idempotent nullnorm V2 on L2 . V2
0
k
m
s
1
0 k m s 1
0 0 0 s s
0 k k s 1
0 k m s 1
s s s s s
s 1 1 s 1
Table 7 Idempotent nullnorm V on L1 . V
0
n
p
s
t
m
k
r
1
0 n p s t m k r 1
0 n p s n n n s s
n n p s n n n s s
p p p s p p p s s
s s s s s s s s s
n n p s t t t r r
n n p s t m k r r
n n p s t k k r r
s s s s r r r r r
s s s s r r r r 1
with the zero element s on L2 . Furthermore, by Propositions 3 and 4, we can see that there does not exist an idempotent nullnorm different from V1 and V2 on the lattice L2 . Consider a bounded lattice (L, ≤ , 0, 1). If we use V to denote the set of all nullnorms on L, it is a partially ordered set with the order defined as the following: For V1 , V2 ∈ V, V1 ≤ V2 ⇔V1 (x, y) ≤ V2 (x, y) for all (x, y) ∈ L2 . Let s ∈ L. In that case, each V (s ) is also a partially ordered set, where V (s ) presents the set of all nullnorms on L with the zero element s. Corollary 1. Consider a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1} such that all elements in L incomparable with s are comparable with each other. In this case, the idempotent nullnorms Vs,1 : L2 → L and Vs,2 : L2 → L defined by the formulas (3) and (4), respectively, are the greatest idempotent nullnorm and the smallest idempotent nullnorm on L with the zero element s, respectively. The proof follows from Propositions 2, 5 and 4. Remark 3. Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1} such that all elements in L incomparable with s are comparable with each other. We introduced two construction methods to obtain an idempotent nullnorm on L with the zero element s in Theorems 3 and 4. Furthermore, in Corollary 1, we see that these idempotent nullnorms are the greatest and smallest idempotent nullnorms on L with the zero element s. It should be noted that the idempotent nullnorms which are introduced in Theorems 3 and 4 slightly differ from each other. Considering Example 2, we see that the idempotent nullnorms Vs,1 and Vs,2 defined on the bounded lattice L1 in Fig. 2 are different from each other since Vs,1 (t, m ) = m = t = Vs,2 (t, m ) for the elements m, t ∈ L1 . In this case, a natural question arises: if we take a bounded lattice L such that all elements in L incomparable with s ∈ L\{0, 1} are comparable with each other, does an idempotent nullnorm on L with the zero element s have to be one of the idempotent nullnorms introduced in Theorems 3 and 4? In the following example, we give a negative example violating the above hypothesis. Example 5. Consider a bounded lattice L1 which is depicted in Hasse diagram by Fig. 2. In that case, it is possible to check the function V: L1 × L1 → L1 defined by Table 7 is an idempotent nullnorm on L1 with the indicated zero element s. In addition, the idempotent nullnorm V differs from the idempotent nullnorms Vs,1 and Vs,2 on L1 given by Tables 3 and 4,
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respectively, by using Theorems 3 and 4, respectively. Because, V (k, m ) = k = m = Vs,2 (k, m ) and V (t, m ) = t = m = Vs,1 (t, m ) for k, m, t ∈ L1 . Remark 4. Given a bounded lattice (L, ≤ , 0, 1) and s ∈ L\{0, 1}. (i) If L is distributive such that all elements in L incomparable with s are comparable with each other, then the idempotent nullnorms obtained by the formula (3) in Theorem 3 and the formula (4) in Theorem 4 coincide with the idempotent nullnorm given by the formula (1) in Theorem 1. (ii) If there is only one element incomparable with s in L, then the idempotent nullnorms obtained by the formula (3) in Theorem 3 and the formula (4) in Theorem 4 coincide with the idempotent nullnorm given by the formulas (2) in Theorem 2. 4. Concluding remarks In previous studies, nullnorms were defined on bounded lattices and some methods for building nullnorms on bounded lattices were proposed in [22,30]. But these methods to obtain nullnorms described in [22,30] do not generate an idempotent nullnorm on general bounded lattices unless we take some special bounded lattices. For this reason, in [11], it was showed that an idempotent nullnorm on a bounded lattice L with a zero element s ∈ L\{0, 1} need not always exist. In addition, a construction method for idempotent nullnorms on a bounded lattice L with a zero element s ∈ L\{0, 1} was introduced under an additional constraint that the one element incomparable with s in L exists. In this paper, considering an arbitrary bounded lattice L, we investigated the existence of idempotent nullnorms on L with a zero element different from the top and bottom elements and some properties of these operators. Moreover, we enhanced the construction method proposed previously in [11] and introduced two additional construction methods for idempotent nullnorms on a bounded lattice L such that all elements in L incomparable with a zero element s ∈ L\{0, 1} are comparable with each other. The methods to obtain idempotent nullnorms introduced in this study are more general than the existing methods and this fact makes our methods more beneficial for practical applicati ons. Acknowledgments The author expresses her sincere thanks to the editors and reviewers for their most valuable comments and suggestions. References [1] [2] [3] [4] [5]
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