New structures for uninorms on bounded lattices 1

Abstract

In this article, we present new methods for constructing uninorms on bounded lattices under the additional constraints and prove that some of these constraints are sufficient and necessary for the uninorms. Moreover, some illustrative examples for the construction of uninorms are provided. At last, we show that the additional constraints on t-norms (t-conorms) and t-subnorms (t-subconorms) of some uninorms in the literature are exactly sufficient and necessary.

Keywords

Bounded lattices uninorms

1 Introduction

In [5], as a function from [0, 1] × [0, 1] to [0, 1], the aggregative operator is for the theory of evaluation, which “sheds new fight on such well-known concepts as membership, conjunction and disjunction and seems to be a very promising tool to handle representation problems as they grow from the fields of theory of fuzzy set and its many applications of human decision making and of multi-criteria analysis” [5].

Triangular norms (t-norms for short) and triangular conorms (t-conorms for short) on the unit interval [0, 1], introduced by Menger [26], Schweizer and Sklar [30, 32], play an important role in many fields, such as fuzzy set theory, fuzzy logic and so on (see, e.g., [4, 18, 25, 31, 34, 40]). Uninorms on [0, 1], as an important aggregative operator generalizing the notions of t-norms and t-conorms, are introduced by Yager and Rybalov [37] and then applied to various fields [14–16, 28, 29], such as fuzzy logic, fuzzy set theory, expert systems, neural networks and so on.

Recently, on one hand, because of “For general systems, where we cannot always expect real (or comparable) data, this extension of the underlying career together with its structure is significant...” [9], the researchers widely study uninorms on the bounded lattices instead of [0, 1]. On the other hand, In [17], fuzzy sets are extended to L-fuzzy sets, where L is a lattice. Then many fuzzy theories are generalized to L-fuzzy theories, such as L-fuzzy topology. From the point of view of generalization, it is necessary to study uninorms on the bounded lattices. This generalization may makes uninorms on lattices to be applied in L-fuzzy theory, lattice-valued logic and so on.

About the methods for the construction of uninorms, they mainly focus on t-norms (t-conorms) [1, 3, 6–8, 10, 11, 13, 23, 35], t-subnorms (t-subconorms) [19, 22, 38], closure operators (interior operators) [9, 20, 27, 39], additive generators [21] and uninorms [36].

For the extension, as shown in many existing results, yielding uninorms on the bounded lattice always needs some additional constraints on the lattice, t-norms (resp. t-conorms), t-subnorms (resp. t-subconorms) or closure operators (resp. interior operators) to guarantee that the methods work well. However, we may ask the following question: are these additional constraints necessary?

In this paper, we present new methods for constructing uninorms on the bounded lattices under some additional constraints and show that some constraints are necessary, that is, these constraints are sufficient and necessary for the uninorms. Moreover, some sufficient conditions of some uninorms are modified to be sufficient and necessary.

2 Preliminaries

In this section, we recall some preliminary details concerning bounded lattice and results related to aggregation functions on bounded lattices.

Definition 2.1. ([2]) A lattice (L, ≤) is bounded if it has top and bottom elements, which are written as 1 and 0, respectively, that is, there exit two elements 1, 0 ∈ L such that 0 ≤ x ≤ 1 for all x ∈ L.

Throughout this article, unless stated otherwise, we denote L as a bounded lattice with the top and bottom elements 1 and 0, respectively.

Definition 2.2. ([2]) Let L be a bounded lattice, a, b ∈ L with a ≤ b. A subinterval [a, b] of L is defined as $[a, b] = {x \in L : a \leq x \leq b} .$ Similarly, we can define [a, b [= {x ∈ L : a ≤ x < b} , ] a, b] = {x ∈ L : a < x ≤ b} and ] a, b [= {x ∈ L : a < x < b}. If a and b are incomparable, then we use the notation a ∥ b. In the following, I_a denotes the set of all incomparable elements with a, that is, I_a = {x ∈ L ∣ x ∥ a}.

Definition 2.3. ([24]) Let (L, ≤ , 0, 1) be a bounded lattice.

(i) An operation T : L² → L is called a t-norm on L if it is commutative, associative, and increasing with respect to both variables, and it has the neutral element 1 ∈ L, that is, T (1, x) = x for all x ∈ L. The greatest t-norm T_∧ on L is given as T_∧ (x, y) = x ∧ y;

(ii) An operation S : L² → L is called a t-conorm on L if it is commutative, associative, and increasing with respect to both variables, and it has the neutral element 0 ∈ L, that is, S (0, x) = x for all x ∈ L. The smallest t-conorm S_∨ on L is given as S_∨ (x, y) = x ∨ y.

Definition 2.4. ([23]) Let (L, ≤ , 0, 1) be a bounded lattice. An operation U : L² → L is called a uninorm on L (a uninorm if L is fixed) if it is commutative, associative, and increasing with respect to both variables, and it has the neutral element e ∈ L, that is, U (e, x) = x for all x ∈ L.

Proposition 2.5. ([23]) Let (L, ≤ , 0, 1) be a bounded lattice and U be a uninorm on L with a neutral element e ∈ L \ {0, 1}. Then we have the following:

(i) T_e = U ∣ [0, e] ² : [0, e] ² → [0, e] is a t-norm on [0, e];

(ii) S_e = U ∣ [e, 1] ² : [e, 1] ² → [e, 1] is a t-conorm on [e, 1].

T_e and S_e given in proposition 2.5 are called the underlying t-norm and t-conorm of a uninorm U on a bounded lattice L with the neutral element e, respectively. Throughout this study, we denote T_e as the underlying t-norm and S_e as the underlying t-conorm of a given uninorm U on L.

Definition 2.6. ([24]) Let (L, ≤ , 0, 1) be a bounded lattice.

(i) An operation F : L² → L is called a t-subnorm on L if it is commutative, associative, increasing with respect to both variables, and F (x, y) ≤ x ∧ y for all x, y ∈ L.

(ii) An operation R : L² → L is called a t-subconorm on L if it is commutative, associative, increasing with respect to both variables, and R (x, y) ≥ x ∨ y for all x, y ∈ L.

Remark 2.7. By lemma 1 and lemma 2 of [7], we can get the following results.

(i)If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then R (x, y) ∈ I_e or R (x, y) =1 for all x, y ∈ I_e.

(ii)If x ∥ y for all x ∈ I_e and y ∈]0, e], then F (x, y) ∈ I_e or F (x, y) =0 for all x, y ∈ I_e.

3 New construction of uninorms on bounded lattices

In this section, we present new construction of uninorms on some appropriate bounded lattices L, which are constructed by means of t-conorms (t-norms) and t-subconorms (t-subnorms). Moreover, the sufficient and necessary conditions are provided for these uninorms.

Theorem 3.1. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}, T_e be a t-norm on [0, e] and S_e be a t-conorm on [e, 1]. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then the function U₁ : L² → L defined by

$U_{1} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ S_{e} (x, y) & if (x, y) \in [e, 1]^{2}, \\ x & if (x, y) \in I_{e} \times [e, 1 [ \\ \cup I_{e} \times [0, e [, \\ y & if (x, y) \in [e, 1 [\times I_{e} \\ \cup [0, e [\times I_{e}, \\ x \lor y & if (x, y) \in {I_{e}}^{2} \cup I_{e} \times {1} \\ \cup {1} \times I_{e} \cup [0, e [\times {1} \\ \cup {1} \times [0, e [, \\ x \land y & otherwise, \end{matrix}$

is a uninorm on L with e ∈ L \ {0, 1} iff S_e (x, y) <1 for all x, y < 1.

Proof. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then we give the proof of the fact that U₁ is a uninorm iff S_e (x, y) <1 for all x, y < 1.

Necessity. Let U₁ be a uninorm on L with the identity e. Then we prove that S_e (x, y) <1 for all x, y < 1. Assume that there are some elements x ∈] e, 1 [ and y ∈] e, 1 [ such that S_e (x, y) =1. Then U₁ (x, U₁ (y, 0)) = U₁ (x, 0) =0 and U₁ (U₁ (x, y) , 0) = U₁ (S_e (x, y) , 0) = U₁ (1, 0) =1. Since 0 ≠ 1, the associativity property of U₁ is violated. Then U₁ is not a uninorm on L, which is a contradiction. Hence, S_e (x, y) <1 for all x, y < 1.

Sufficiency. To prove the sufficiency, we compare U₁ with $U_{1}^{e}$ in Theorem 5 of [7] and find that besides the condition that T_e (x, y) >0 for all x, y > 0, the values of U₁ and $U_{1}^{e}$ are different on I_e × {0} ∪ {1} × {0}, {0} × I_e ∪ {0} × {1} and ] 0, e [×]0, e [. So, combined with Theorem 5 of [7], the proof of the sufficiency can be reduced to the above areas. The commutativity of U₁ and the fact that e is a neutral element are evident. Hence, we prove only the monotonicity and the associativity of U₁.

I. Monotonicity: We prove that if x ≤ y, then for all z ∈ L, U₁ (x, z) ≤ U₁ (y, z).

If one of x, y, z is equal to 1, then it is clear that U₁ (x, z) ≤ U₁ (y, z). There is no 1 in any of the following cases.

1. Let x ∈ [0, e [.

1.1. y ∈ [0, e [,

1.1.1. z ∈ I_e,

U₁ (x, z) = z = U₁ (y, z)

1.2. y ∈ [e, 1 [,

1.2.1. z ∈ I_e,

U₁ (x, z) = z = U₁ (y, z)

1.3. y ∈ I_e,

1.3.1. z ∈ [0, e [,

U₁ (x, z) = T_e (x, z) ≤ y = U₁ (y, z)

1.3.2. z ∈ [e, 1 [,

U₁ (x, z) = x ≤ y = U₁ (y, z)

1.3.3. z ∈ I_e,

U₁ (x, z) = z ≤ y ∨ z = U₁ (y, z)

2. Let x ∈ I_e.

2.1. y ∈ I_e,

2.1.1. z ∈ [0, e [,

U₁ (x, z) = x ≤ y = U₁ (y, z)

II. Associativity: We prove that U₁ (x, U₁ (y, z)) = U₁ (U₁ (x, y) , z) for all x, y, z ∈ L.

If one of x, y, z is equal to 1, then it is clear that U₁ (x, U₁ (y, z)) =1 = U₁ (U₁ (x, y) , z). So there is no 1 in any of the following cases.

1. Let x ∈ [0, e [.

1.1. y ∈ [0, e [,

1.1.1. z ∈ I_e,

U₁ (x, U₁ (y, z)) = U₁ (x, z) = z = U₁ (T_e (x, y) , z) = U₁ (U₁ (x, y) , z)

1.2. y ∈ [e, 1 [,

1.2.1. z ∈ I_e,

U₁ (x, U₁ (y, z)) = U₁ (x, z) = z = U₁ (x, z) U₁ (U₁ (x, y) , z)

1.3. y ∈ I_e,

1.3.1. z ∈ [0, e [,

U₁ (x, U₁ (y, z)) = U₁ (x, y) = y = U₁ (y, z) = U₁ (U₁ (x, y) , z)

1.3.2. z ∈ [e, 1 [,

U₁ (x, U₁ (y, z)) = U₁ (x, y) = y = U₁ (y, z) = U₁ (U₁ (x, y) , z)

1.3.3. z ∈ I_e,

1.3.3.1. y ∨ z ∈ I_e,

U₁ (x, U₁ (y, z)) = U₁ (x, y ∨ z) = y ∨ z = U₁ (y, z) = U₁ (U₁ (x, y) , z)

1.3.3.2. y ∨ z = 1,

U₁ (x, U₁ (y, z)) = U₁ (x, y ∨ z) =1 = y ∨ z = U₁ (y, z) = U₁ (U₁ (x, y) , z)

2. Let x ∈ [e, 1 [.

2.1. y ∈ [0, e [,

2.1.1. z ∈ I_e,

U₁ (x, U₁ (y, z)) = U₁ (x, z) = z = U₁ (y, z) = U₁ (U₁ (x, y) , z)

2.2. y ∈ I_e,

2.2.1. z ∈ [0, e [,

U₁ (x, U₁ (y, z)) = U₁ (x, y) = y = U₁ (y, z) = U₁ (U₁ (x, y) , z)

3. Let x ∈ I_e.

3.1. y ∈ [0, e [,

3.1.1. z ∈ [0, e [,

U₁ (x, U₁ (y, z)) = U₁ (x, T_e (y, z)) = x = U₁ (x, z) = U₁ (U₁ (x, y) , z)

3.1.2. z ∈ [e, 1 [,

U₁ (x, U₁ (y, z)) = U₁ (x, y) = x = U₁ (x, z) = U₁ (U₁ (x, y) , z)

3.1.3. z ∈ I_e,

U₁ (x, U₁ (y, z)) = U₁ (x, z) = x ∨ z = U₁ (x, z) = U₁ (U₁ (x, y) , z)

3.2. y ∈ [e, 1 [,

3.2.1. z ∈ [0, e [,

U₁ (x, U₁ (y, z)) = U₁ (x, z) = x = U₁ (x, z) = U₁ (U₁ (x, y) , z)

3.3. y ∈ I_e,

3.3.1. z ∈ [0, e [,

3.3.1.1. x ∨ y ∈ I_e,

U₁ (x, U₁ (y, z)) = U₁ (x, y) = x ∨ y = U₁ (x ∨ y, z) = U₁ (U₁ (x, y) , z)

3.3.1.2. x ∨ y = 1,

U₁ (x, U₁ (y, z)) = U₁ (x, y) = x ∨ y = 1 = U₁ (x ∨ y, z) = U₁ (U₁ (x, y) , z)

Hence, U₁ is a uninorm on L with a neutral element e.

Theorem 3.2. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}, T_e be a t-norm on [0, e] and S_e be a t-conorm on [e, 1]. If x ∥ y for all x ∈ I_e and y ∈]0, e], then the function U₂ : L² → L defined by

$U_{2} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ S_{e} (x, y) & if (x, y) \in [e, 1]^{2}, \\ x & if (x, y) \in I_{e} \times] e, 1] \\ \cup I_{e} \times] 0, e], \\ y & if (x, y) \in] e, 1] \times I_{e} \\ \cup] 0, e] \times I_{e}, \\ x \land y & if (x, y) \in {I_{e}}^{2} \cup I_{e} \times {0} \\ \cup {0} \times I_{e} \cup] e, 1] \times {0} \\ \cup {0} \times] e, 1], \\ x \lor y & otherwise, \end{matrix}$ is a uninorm on L with e ∈ L \ {0, 1} iff T_e (x, y) >0 for all x, y > 0.

Proof. It can be proved in an analogous way to the proof of Theorem 3.1. □

It is worth pointing out that if the condition that x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ in Theorem 3.1 is not satisfied, then U₁ may not be a uninorm on L₁. Similarly, if the condition that x ∥ y for all x ∈ I_e and y ∈]0, e] in Theorem 3.1 is not satisfied, then U₂ may not be a uninorm on L₁. The following example will show the above facts.

Example 3.3. Consider the bounded lattice L₁ = {0, a, e, c, f, g, 1} depicted by the Hasse diagram in Fig. 1, where a ∈ [0, e [, g ∈] e, 1], c ∈ I_e, f ∈ I_e, g > c, g > f,a < c and a < f. Take T_e (x, y) = T_∧ (x, y) on [0, e] and S_e (x, y) = S_∨ (x, y) on [e, 1]. The function U₁ on L₁, shown in Table 1, is not a uninorm on L₁ with the neutral element e. In fact, the associativity is not satisfied, since U₁ (a, U₁ (c, f)) = U₁ (a, g) = a and U₁ (U₁ (a, c) , f) = U₁ (c, f) = g. Similarly, the function U₂ on L₁, shown in Table 2, is not a uninorm on L₁ with the neutral element e. In fact, the associativity is not satisfied, since U₂ (g, U₂ (c, f)) = U₂ (g, a) = g and U₂ (U₂ (g, c) , f) = U₂ (c, f) = a.

Fig. 1

The lattice L

Table 1

The function U₁ on L₁ given in Fig. 1

U ₁	0	a	e	c	f	g	1
0	0	0	0	c	f	0	1
a	0	a	a	c	f	a	1
e	0	a	e	c	f	g	1
c	c	c	c	c	g	c	1
f	f	f	f	g	f	f	1
g	0	a	g	c	f	g	1
1	1	1	1	1	1	1	1

Table 2

The function U₂ on L₁ given in Fig. 1

U ₂	a	e	c	f	g	1
0	0	0	0	0	0	0
a	a	a	c	f	g	1
e	a	e	c	f	g	1
c	c	c	c	a	c	c
f	f	f	a	f	f	f
g	g	g	c	f	g	1
1	1	1	c	f	1	1

Theorem 3.4. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let T_e be a t-norm on [0, e] and R be a t-subconorm on L. Let U₃ : L² → L be a function defined as follows:

$U_{3} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ x & if (x, y) \in I_{e} \times] 0, e] \cup \\ I_{e} \times] e, 1 [, \\ y & if (x, y) \in] 0, e] \times I_{e} \cup \\ ] e, 1 [\times I_{e}, \\ R (x, y) & if (x, y) \in I_{e}^{2} \cup] e, 1]^{2}, \\ x \lor y & if (x, y) \in I_{e} \times {1} \cup {1} \\ \times I_{e} \cup] 0, e [\times {1} \cup {1} \\ \times] 0, e [\cup] e, 1] \times {e} \cup \\ {e} \times] e, 1], \\ x \land y & otherwise . \end{matrix}$

(1) If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and R (x, y) <1 for all x, y < 1, then U₃ is a uninorm on L with e ∈ L \ {0, 1} iff T_e (x, y) >0 for all x, y > 0.

(2) If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and I_e∪]0, e [≠ ∅, then U₃ is a uninorm on L with e ∈ L \ {0, 1} iff T_e (x, y) >0 for all x, y > 0 and R (x, y) <1 for all x, y < 1.

Proof. (1) Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and R (x, y) <1 for all x, y < 1, then we give the proof of the fact that U₃ is a uninorm iff T_e (x, y) >0 for all x, y > 0.

Necessity. Let the function U₃ be a uninorm on L with the identity e. Then we will prove that T_e (x, y) >0 for all x, y > 0. Assume that there are some elements x ∈]0, e [ and y ∈]0, e [ such that T_e (x, y) =0. Then U₃ (x, U₃ (y, 1)) = U₃ (x, 1) =1 and U₃ (U₃ (x, y) , 1) = U₃ (T_e (x, y) , 1) = U₃ (0, 1) =0. Since 0 ≠ 1, the associativity property is violated. Hence, U₃ is not a uninorm on L which is a contradiction. Therefore, the condition that T_e (x, y) >0 for all x, y > 0 is necessary.

Sufficiency. To prove the sufficiency, we compare U₃ with $U_{R}^{e}$ in Theorem 3.1 of [7] and find that besides the condition that T_e (x, y) >0 for all x, y > 0, the values of U₃ and $U_{R}^{e}$ are different on I_e × {0} ∪ {1} × {0}, {0} × I_e ∪ {0} × {1} and ] 0, e [×]0, e [. So, combined with Theorem 3.1 of [7], the proof of the sufficiency can be reduced to the above areas.

The commutativity of U₃ and the fact that e is a neutral element are evident. Hence, we prove only the monotonicity and the associativity of U₃.

I. Monotonicity: We prove that if x ≤ y, then for all z ∈ L, U₃ (x, z) ≤ U₃ (y, z).

1.x = 0 or y = 0 or z = 0

U₃ (x, z) =0 ≤ U₃ (y, z)

II. Associativity: We prove that U₃ (x, U₃ (y, z)) = U₃ (U₃ (x, y) , z) for all x, y, z ∈ L.

If one of x, y, z is equal to 0, then it is clear that U₃ (x, U₃ (y, z)) = U₃ (U₃ (x, y) , z).

Hence, U₃ is a uninorm on L with a neutral element e.

(2) Next we just prove that if x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and I_e∪]0, e [≠ ∅, then the condition R (x, y) <1 for all x, y < 1 is necessary.

Let the function U₃ be a uninorm on L with the identity e. Then we will prove that R (x, y) <1 for all x, y < 1. Assume that there are some elements x ∈] e, 1 [ and y ∈] e, 1 [ such that R (x, y) =1. If z ∈ I_e ∪]0, e [, then we obtain U₃ (x, U₃ (y, z)) = U₃ (x, z) = z and U₃ (U₃ (x, y) , z) = U₃ (R (x, y) , z) = U₃ (1, z) =1. Since z ≠ 1, the associativity property is violated. Hence, U₃ is not a uninorm on L which is a contradiction. Therefore, the condition that R (x, y) <1 for all x, y < 1 is necessary.

Theorem 3.5. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let F be a t-subnorm on L and S_e be a t-conorm on [e, 1]. Let U₄ : L² → L be a function defined as follows:

$U_{4} (x, y) = {\begin{matrix} S_{e} (x, y) & if (x, y) \in [e, 1]^{2}, \\ x & if (x, y) \in I_{e} \times [e, 1 [\cup \\ I_{e} \times] 0, e [, \\ y & if (x, y) \in [e, 1 [\times I_{e} \cup \\ ] 0, e [\times I_{e}, \\ F (x, y) & if (x, y) \in I_{e}^{2} \cup [0, e [^{2}, \\ x \land y & if (x, y) \in I_{e} \times {0} \cup {0} \\ \times I_{e} \cup] e, 1 [\times {0} \cup {0} \\ \times] e, 1 [\cup [0, e [\times {e} \cup \\ {e} \times [0, e [, \\ x \lor y & otherwise . \end{matrix}$

(1) If x ∥ y for all x ∈ I_e and y ∈]0, e] and F (x, y) >0 for all x, y > 0, then U₄ is a uninorm on L with e ∈ L \ {0, 1} iff S_e (x, y) <1 for all x, y < 1.

(2) If x ∥ y for all x ∈ I_e and y ∈]0, e] and I_e∪] e, 1 [≠ ∅, then U₄ is a uninorm on L with e ∈ L \ {0, 1} iff F (x, y) >0 for all x, y > 0 and S_e (x, y) <1 for all x, y < 1.

Proof. It can be proved in an analogous way to the proof of Theorem 3.4. □

It is worth pointing out that if the condition that x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ in Theorem 3.4 is not satisfied, then U₃ may not be a uninorm on L₁. Similarly, if the condition that x ∥ y for all x ∈ I_e and y ∈]0, e] in Theorem 3.5 is not satisfied, then U₄ may not be a uninorm on L₁.

Example 3.6. Consider the bounded lattice L₁ = {0, a, e, c, f, g, 1} depicted by the Hasse diagram in Fig. 1. (1) If we take T_e (x, y) = T_∧ (x, y) on [0, e] and R (x, y) = x ∨ y, then the function U₃ on L₁, shown in Table 3, is not a uninorm on L₁ with the neutral element e. In fact, the associativity is not satisfied, since U₃ (a, U₃ (c, f)) = U₃ (a, g) = a and U₃ (U₃ (a, c) , f) = U₃ (c, f) = g.

Table 3

The function U₃ on L₁ given in Fig. 1

U ₃	a	e	c	f	g	1
0	0	0	0	0	0	0
a	a	a	c	f	a	1
e	a	e	c	f	g	1
c	c	c	c	g	c	1
f	f	f	g	f	f	1
g	a	g	c	f	g	1
1	1	1	1	1	1	1

(2) If we take S_e (x, y) = S_∨ (x, y) on [e, 1] and F (x, y) = x ∧ y, then the function U₄ on L₁, shown in Table 4, is not a uninorm on L₁ with the neutral element e. In fact, the associativity is not satisfied, since U₄ (g, U₄ (c, f)) = U₄ (g, a) = g and U₄ (U₄ (g, c) , f) = U₄ (c, f) = a.

Table 4

The function U₄ on L₁ given in Fig. 1

U ₄	0	a	e	c	f	g	1
0	0	0	0	0	0	0	1
a	0	a	a	c	f	g	1
e	0	a	e	c	f	g	1
c	0	c	c	c	a	c	1
f	0	f	f	a	f	f	1
g	0	g	g	c	f	g	1
1	1	1	1	1	1	1	1

Theorem 3.7. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let T_e be a t-norm on [0, e], and R be a t-subconorm on L. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then the function U₅ : L² → L defined by

$U_{5} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ x \land y & if (x, y) \in [0, e [\times I_{e} \cup \\ I_{e} \times [0, e [\cup [0, e [\times] e, 1] \cup \\ ] e, 1] \times [0, e [, \\ y & if (x, y) \in {e} \times I_{e} \cup \\ {e} \times] e, 1], \\ x & if (x, y) \in I_{e} \times {e} \cup \\ ] e, 1] \times {e}, \\ R (x, y) & if (x, y) \in {I_{e}}^{2}, \\ 1 & if (x, y) \in I_{e} \times] e, 1] \cup \\ ] e, 1] \times I_{e} \cup] e, 1]^{2}, \end{matrix}$

is a uninorm on L with the neutral element e iff x > y for all x ∈ I_e and y ∈ [0, e [.

Proof. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then we give the proof of the fact that U₅ is a uninorm iff x > y for all x ∈ I_e and y ∈ [0, e [.

Necessity. Let the function U₅ be a uninorm on L with the identity e. Then we will prove that x > y for all x ∈ I_e and y ∈ [0, e [. Assume that there are some elements x ∈ I_e and y ∈ [0, e [ such that x ∥ y. Then U₅ (y, U₅ (x, 1)) = U₅ (y, 1) = y and U₅ (U₅ (y, x) , 1) = U₅ (x ∧ y, 1) = x ∧ y. Since y ≠ x ∧ y, the associativity property is violated. Then U₅ is not a uninorm on L which is a contradiction. Hence, x > y for all x ∈ I_e and y ∈ [0, e [.

Sufficiency. Let x > y for all x ∈ I_e and y ∈ [0, e [, with x ∥ y for all x ∈ I_e and y ∈ [e, 1 [. The commutativity of U₅ and the fact that e is a neutral element are evident. Hence, we prove only the monotonicity and the associativity of U₅.

I. Monotonicity: We prove that if x ≤ y, then for all z ∈ L, U₅ (x, z) ≤ U₅ (y, z). The proof is split into all possible cases. If one of x, y, z is equal to e, then it is clear that U₅ (x, z) = z = U₅ (y, z)

1. Let z = e.

U₅ (x, z) = x ≤ y = U₅ (y, z)

2. Let x = e.

2.1. y = e,

U₅ (x, z) = z = U₅ (y, z)

2.2. y ∈] e, 1],

2.2.1. z ∈ [0, e [,

U₅ (x, z) = z = U₅ (y, z)

2.2.2. z ∈] e, 1],

U₅ (x, z) = z ≤ 1 = U₅ (y, z)

2.2.3. z ∈ I_e,

U₅ (x, z) = z ≤ 1 = U₅ (y, z)

3. Let y = e.

3.1. x ∈ [0, e [,

3.1.1. z ∈ [0, e [,

U₅ (x, z) = T_e (x, z) ≤ z = U₅ (y, z)

3.1.2. z ∈] e, 1],

U₅ (x, z) = x ≤ z = U₅ (y, z)

3.1.3. z ∈ I_e,

U₅ (x, z) = x ∧ z = x ≤ z = U₅ (y, z)

There is no e in any of the following cases.

1. Let x ∈ [0, e [.

1.1. y ∈ [0, e [,

1.1.1. z ∈ [0, e [,

U₅ (x, z) = T_e (x, z) ≤ T_e (y, z) = U₅ (y, z)

1.1.2. z ∈] e, 1],

U₅ (x, z) = x ∧ z = x ≤ y = y ∧ z = U₅ (y, z)

1.1.3. z ∈ I_e,

U₅ (x, z) = x ∧ z = x ≤ y = y ∧ z = U₅ (y, z)

1.2. y ∈] e, 1],

1.2.1. z ∈ [0, e [,

U₅ (x, z) = T_e (x, z) ≤ z = y ∧ z = U₅ (y, z)

1.2.2. z ∈] e, 1],

U₅ (x, z) = x ∧ z = x ≤ 1 = U₅ (y, z)

1.2.3. z ∈ I_e,

U₅ (x, z) = x ∧ z = x ≤ 1 = U₅ (y, z)

1.3. y ∈ I_e,

1.3.1. z ∈ [0, e [,

U₅ (x, z) = T_e (x, z) ≤ z = y ∧ z = U₅ (y, z)

1.3.2. z ∈] e, 1],

U₅ (x, z) = x ∧ z = x ≤ 1 = U₅ (y, z)

1.3.3. z ∈ I_e,

U₅ (x, z) = x ∧ z = x ≤ R (y, z) = U₅ (y, z)

2. Let x ∈] e, 1].

2.1. y ∈] e, 1],

2.1.1. z ∈ [0, e [,

U₅ (x, z) = x ∧ z = z = y ∧ z = U₅ (y, z)

2.1.2. z ∈] e, 1],

U₅ (x, z) =1 = U₅ (y, z)

2.1.3. z ∈ I_e,

U₅ (x, z) =1 = U₅ (y, z)

3. Let x ∈ I_e.

3.1. y ∈ I_e,

3.1.1. z ∈ [0, e [,

U₅ (x, z) = x ∧ z = z = y ∧ z = U₅ (y, z)

3.1.2. z ∈] e, 1],

U₅ (x, z) =1 = U₅ (y, z)

3.1.3. z ∈ I_e,

U₅ (x, z) = R (x, z) ≤ R (y, z) = U₅ (y, z)

II. Associativity: We need to prove that U₅ (x, U₅ (y, z))

= U₅ (U₅ (x, y) , z) for all x, y, z ∈ L. The proof is split into all possible cases. If one of x, y, z is equal to e, then it is clear that U₅ (x, U₅ (y, z)) = U₅ (U₅ (x, y) , z). So there is no e in any of the following cases. By Theorem 3.12 of [19], we only need to consider the following cases:

1. Let x ∈ [0, e [.

1.1. y ∈ [0, e [,

1.1.1. z ∈ [0, e [,

U₅ (x, U₅ (y, z)) = T_e (x, T_e (y, z)) = T_e (T_e (x, y) , z) = U₅ (U₅ (x, y) , z)

1.1.2. z ∈] e, 1],

U₅ (x, U₅ (y, z)) = U₅ (x, y ∧ z) = T_e (x, y) = U₅ (T_e (x, y) , z) =

U₅ (U₅ (x, y) , z)

1.1.3. z ∈ I_e,

U₅ (x, U₅ (y, z)) = U₅ (x, y ∧ z) = T_e (x, y) = U₅ (T_e (x, y) , z) =

U₅ (U₅ (x, y) , z)

1.2. y ∈] e, 1],

1.2.1. z ∈] e, 1],

U₅ (x, U₅ (y, z)) = U₅ (x, 1) = x = U₅ (x, z) = U₅ (U₅ (x, y) , z)

1.2.2. z ∈ I_e,

U₅ (x, U₅ (y, z)) = U₅ (x, 1) = x, U₅ (U₅ (x, y) , z) = U₅ (x, z) = x

and U₅ (y, U₅ (x, z)) = U₅ (y, x) = x

1.3. y ∈ I_e,

1.3.1. z ∈ I_e,

U₅ (x, U₅ (y, z)) = U₅ (x, R (y, z)) = x = U₅ (x, z) = U₅ (U₅ (x, y) , z)

2. Let x ∈] e, 1].

2.1. y ∈] e, 1],

2.1.1. z ∈] e, 1],

U₅ (x, U₅ (y, z)) = U₅ (x, 1) =1 = U₅ (1, z) = U₅ (U₅ (x, y) , z)

2.1.2. z ∈ I_e,

U₅ (x, U₅ (y, z)) = U₅ (x, 1) =1 = U₅ (1, z) = U₅ (U₅ (x, y) , z)

2.2. y ∈ I_e,

2.2.1. z ∈ I_e,

U₅ (x, U₅ (y, z)) = U₅ (x, R (y, z)) =1 = U₅ (1, z) = U₅ (U₅ (x, y) , z)

3. Let x ∈ I_e.

3.1. y ∈ I_e,

3.1.1. z ∈ I_e,

3.1.1.1. R (x, y) =1, R (y, z) =1,

U₅ (x, U₅ (y, z)) = U₅ (x, R (y, z)) =1 = U₅ (R (x, y) , z) = U₅ (U₅ (x, y) , z)

3.1.1.2. R (x, y) =1, R (y, z) ∈ I_e, $\begin{matrix} U_{5} (x, U_{5} (y, z)) & = & U_{5} (x, R (y, z)) = R (x, R (y, z)) \\ = & R (R (x, y), z) = R (1, z) = 1 \\ = & U_{5} (1, z) = U_{5} (R (x, y), z) \\ = & U_{5} (U_{5} (x, y), z) (by Remark 2.7 (3)) \end{matrix}$

3.1.1.3. R (x, y) ∈ I_e, R (y, z) =1, $\begin{matrix} U_{5} (x, U_{5} (y, z)) & = & U_{5} (x, R (y, z)) = U_{5} (x, 1) \\ = & 1 = R (x, 1) = R (x, R (y, z)) \\ = & R (R (x, y), z) = U_{5} (R (x, y), z) \\ = & U_{5} (U_{5} (x, y), z) (by Remark 2.7 (3)) \end{matrix}$

3.1.1.4. R (x, y) ∈ I_e, R (y, z) ∈ I_e,

U₅ (x, U₅ (y, z)) = R (x, R (y, z)) = R (R (x, y) , z) = U₅ (U₅ (x, y) , z)

Hence, by Theorem 3.12 of [19], U₅ (x, U₅ (y, z)) = U₅ (U₅ (x, y) , z) for all x, y, z ∈ L. Therefore, U₅ is a uninorm on L with the neutral element e.

Theorem 3.8. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let S_e be a t-conorm on [e, 1], and F be a t-subnorm on L. If x ∥ y for all x ∈ I_e and y ∈]0, e], then the function U₆ : L₂ → L defined by .

Proof. It can be proved in an analogous way to the proof of Theorem 3.7. □

Corollary 3.9. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}, x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, x > y for all x ∈ I_e and y ∈ [0, e [. Suppose that the uninorm U₅ in Theorem 3.7, the uninorm U_R in Theorem 5.9 of [22] and the uninorm U₁ in Corollary 3.2 of [9], then U₅ ≥ U_R and U₅ ≥ U₁.

Corollary 3.10. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}, x ∥ y for all x ∈ I_e and y ∈]0, e], x < y for all x ∈ I_e and y ∈] e, 1]. Suppose that the uninorm U₆ in Theorem 3.8, the uninorm U_F in Theorem 5.7 of [22] and the uninorm U₄ in Corollary 3.11 of [9], then U₆ ≤ U_F and U₆ ≤ U₄.

It is worth pointing out that if the condition that x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ in Theorem 3.7 is not satisfied, then U₅ may not be a uninorm on L₂ with the neutral element e. Similarly, if the condition that x ∥ y for all x ∈ I_e and y ∈]0, e] in Theorem 3.8 is not satisfied, then U₆ may not be a uninorm on L₂ with the neutral element e. The following example will show the above facts.

Example 3.11. Consider the bounded lattice L₂ = {0, a, e, c, d, f, g, 1} depicted by the Hasse diagram in Fig. 2, where a ∈ [0, e [, g ∈] e, 1], c ∈ I_e, d ∈ I_e, f ∈ I_e, g > c, g > d, g > f, a < c, a < d and a < f. If we take T_e (x, y) = T_∧ (x, y) on [0, e] and R (x, y) = x ∨ y, then the function U₅ on L₂, shown in Table 5, is not a uninorm on L₂ with the neutral element e. In fact, the associativity is not satisfied, since U₅ (f, U₅ (c, d)) = U₅ (f, d) = g and U₅ (U₅ (f, c) , d) = U₅ (g, d) =1. Similarly, if we take S_e (x, y) = S_∨ (x, y) on [e, 1] and F (x, y) = x ∧ y, then the function U₆ on L₂, shown in Table 6, is not a uninorm on L₂ with the neutral element e. In fact, the associativity is not satisfied, since U₆ (f, U₆ (c, d)) = U₆ (f, c) = a and U₆ (U₆ (f, c) , d) = U₆ (a, d) =0.

Fig. 2

The lattice L₂

Table 5

The function U₅ on L₂ given in Fig. 2

U ₅	a	e	c	d	f	g	1
0	0	0	0	0	0	0	0
a	a	a	a	a	a	a	a
e	a	e	c	d	f	g	1
c	a	c	c	d	g	1	1
d	a	d	d	d	g	1	1
f	a	f	g	g	f	1	1
g	a	g	1	1	1	1	1
1	a	1	1	1	1	1	1

Table 6

The function U₆ on L₂ given in Fig. 2

U ₆	0	a	e	c	d	f	g	1
0	0	0	0	0	0	0	g	1
a	0	0	a	0	0	0	g	1
e	0	a	e	c	d	f	g	1
c	0	0	c	c	c	a	g	1
d	0	0	d	c	d	a	g	1
f	0	0	f	a	a	f	g	1
g	g	g	g	g	g	g	g	1
1	1	1	1	1	1	1	1	1

4 Modifications of some uninorms

In this section, we mainly modify some uninorms, in which some constraints are actually necessary and sufficient.

Theorem 4.1. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let T_e be a t-norm on [0, e] and S_e be a t-conorm on [e, 1]. Let $U_{1}^{e} : L^{2} \to L$ be a function defined by $U_{1}^{e} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ S_{e} (x, y) & if (x, y) \in [e, 1]^{2}, \\ x & if (x, y) \in I_{e} \times [e, 1 [\cup \\ I_{e} \times] 0, e [, \\ y & if (x, y) \in [e, 1 [\times I_{e} \cup \\ ] 0, e [\times I_{e}, \\ x \lor y & if (x, y) \in I_{e}^{2} \cup I_{e} \times {1} \cup \\ {1} \times I_{e} \cup] 0, e [\times {1} \cup \\ {1} \times] 0, e [, \\ x \land y & otherwise . \end{matrix}$

(1) If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and S_e (x, y) <1 for all x, y < 1, then $U_{1}^{e}$ is a uninorm on L with e ∈ L \ {0, 1} iff T_e (x, y) >0 for all x, y > 0.

(2) If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and I_e∪]0, e [≠ ∅, then $U_{1}^{e}$ is a uninorm on L with e ∈ L \ {0, 1} iff T_e (x, y) >0 for all x, y > 0 and S_e (x, y) <1 for all x, y < 1.

Proof. (1) If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and S_e (x, y) <1 for all x, y < 1, then we give the proof of the fact that $U_{1}^{e}$ is a uninorm iff T_e (x, y) >0 for all x, y > 0.

Necessity. Let $U_{1}^{e}$ be a uninorm on L with the identity e. Next we will prove that T_e (x, y) >0 for all x, y > 0. Assume that there exist x ∈]0, e [ and y ∈]0, e [ such that T_e (x, y) =0. Then $U_{1}^{e} (x, U_{1}^{e} (y, 1)) = U_{1}^{e} (x, 1) = 1$ and, since T_e (x, y) =0, $U_{1}^{e} (U_{1}^{e} (x, y), z) = U_{1}^{e} (T_{e} (x, y), z) = U_{1}^{e} (0, 1) = 0$ . Hence, the associativity property is violated. So, $U_{1}^{e}$ is not a uninorm on L which is a contradiction. Therefore, the condition that T_e (x, y) >0 for all x, y > 0 is necessary.

Sufficiency. This proof is the same as Theorem 5 of [7].

(2) Next we just prove that if x ∥ y for all x ∈ I_e and y ∈ [e, 1 [ and I_e∪]0, e [≠ ∅, then the condition S_e (x, y) <1 for all x, y < 1 is necessary.

Assume that there exist x ∈] e, 1 [ and y ∈] e, 1 [ such that S_e (x, y) =1. Take z ∈ I_e ∪]0, e [. Then $U_{1}^{e} (x, U_{1}^{e} (y, z)) = U_{1}^{e} (x, z) = z$ and, since S_e (x, y) =1, $U_{1}^{e} (U_{1}^{e} (x, y), z) = U_{1}^{e} (S_{e} (x, y), z) = U_{1}^{e} (1, z) = 1$ . Hence, the associativity property is violated. So, $U_{1}^{e}$ is not a uninorm on L which is a contradiction. Therefore, the condition S_e (x, y) <1 for all x, y < 1 is necessary. □

Theorem 4.2. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let T_e be a t-norm on [0, e] and S_e be a t-conorm on [e, 1]. Let $U_{2}^{e} : L^{2} \to L$ be a function defined by

$U_{2}^{e} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ S_{e} (x, y) & if (x, y) \in [e, 1]^{2}, \\ x & if (x, y) \in I_{e} \times] e, 1 [\cup \\ I_{e} \times] 0, e], \\ y & if (x, y) \in] e, 1 [\times I_{e} \cup \\ ] 0, e] \times I_{e}, \\ x \land y & if (x, y) \in I_{e}^{2} \cup I_{e} \times {0} \cup \\ {0} \times I_{e} \cup] e, 1 [\times {0} \cup \\ {0} \times] e, 1 [, \\ x \lor y & otherwise . \end{matrix}$

(1) If x ∥ y for all x ∈ I_e and y ∈]0, e] and T_e (x, y) >0 for all x, y > 0, then $U_{2}^{e}$ is a uninorm on L with e ∈ L \ {0, 1} iff S_e (x, y) <1 for all x, y < 1.

(2) If x ∥ y for all x ∈ I_e and y ∈]0, e] and I_e∪] e, 1 [≠ ∅, then $U_{2}^{e}$ is a uninorm on L with e ∈ L \ {0, 1} iff T_e (x, y) >0 for all x, y > 0 and S_e (x, y) <1 for all x, y < 1.

Proof. It can be proved in an analogous way to the proof of Theorem 4.1. □

Remark 4.3. In Theorem 5 and Theorem 6 of [7], the conditions that T_e is a t-norm on [0, e] such that T_e (x, y) >0 for all x, y > 0 and S_e is a t-conorm on [e, 1] such that S_e (x, y) <1 for all x, y < 1 are sufficient for $U_{1}^{e}$ and $U_{2}^{e}$ , respectively. In fact, these conditions are also necessary for $U_{1}^{e}$ in Theorem 4.1 and $U_{2}^{e}$ in Theorem 4.2, respectively. And we find that S_e is a t-conorm on [e, 1] such that S_e (x, y) <1 for all x, y < 1 and T_e is a t-norm on [0, e] such that T_e (x, y) >0 for all x, y > 0 are necessary for $U_{1}^{e}$ and $U_{2}^{e}$ under some additional constraints, respectively.

Remark 4.4. Let U₁ is the uninorm on L defined as in Theorem 3.1 and U₂ is the uninorm on L defined as in Theorem 3.1, and let $U_{1}^{e}$ is the uninorm on L defined as in Theorem 4.1 and $U_{2}^{e}$ is the uninorm on L defined as in Theorem 4.2.

(1) It is easy to see that the values of U₁ and $U_{1}^{e}$ are different on I_e × {0} ∪ {1} × {0} and {0} × I_e ∪ {0} × {1}. In fact, U₁ = x and $U_{1}^{e} = 0$ when (x, y) ∈ I_e × {0} ∪ {1} × {0}; U₁ = y, $U_{1}^{e} = 0$ when (x, y) ∈ {0} × I_e ∪ {0} × {1}. The values of U₁ could be 0 on ] 0, e [×]0, e [; however, the values of $U_{1}^{e}$ are not equal to 0 on ] 0, e [×]0, e [.

(2) The additional constraints of U₁ and $U_{1}^{e}$ are different, that is, the condition that T_e (x, y) >0 for all x, y > 0 is not needed for U₁.

According to (1) and (2), by changing the values of $U_{1}^{e}$ , we can obtain a new method to construct the nuinorm U₁ and then additional constraints are different for $U_{1}^{e}$ and U₁.

Similarly, the relationship between U₂ and $U_{2}^{e}$ is same as that of U₁ and $U_{1}^{e}$ .

Theorem 4.5. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let T_e be a t-norm on [0, e] and R be a t-subconorm on L. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then the function $U_{R}^{e} : L^{2} \to L$ defined by∥ $U_{R}^{e} (x, y) = {\begin{matrix} T_{e} (x, y) & if (x, y) \in [0, e]^{2}, \\ x & if (x, y) \in I_{e} \times [0, e] \cup \\ I_{e} \times [e, 1 [, \\ y & if (x, y) \in [0, e] \times I_{e} \cup \\ e, 1]^{2}, \\ x \lor y & if (x, y) \in I_{e} \times {1} \cup {1} \\ \times I_{e} \cup [0, e] \times {1} \cup {1} \\ \times [0, e] \cup [e, 1] \times {e} \cup \\ {e} \times [e, 1], \\ x \land y & otherwise, \end{matrix}$

is a uninorm on L with e ∈ L \ {0, 1} iff R (x, y) <1 for all x, y < 1.

Proof. If x ∥ y for all x ∈ I_e and y ∈ [e, 1 [, then we give the proof of the fact that $U_{R}^{e}$ is a uninorm iff R (x, y) <1 for all x, y < 1.

Necessity. Let $U_{R}^{e}$ be a uninorm on L with the identity e. Then we will prove that R (x, y) <1 for all x, y < 1. Assume that there exist x ∈] e, 1 [ and y ∈] e, 1 [ such that R (x, y) =1. Then $U_{R}^{e} (x, U_{R}^{e} (y, 0)) = U_{R}^{e} (x, 0) = 0$ . Since R (x, y) =1, $U_{R}^{e} (U_{R}^{e} (x, y), 0) = U_{R}^{e} (R (x, y), 0) = U_{R}^{e} (1, 0) = 1$ . Hence, the associativity property of $U_{R}^{e}$ is violated. So, $U_{R}^{e}$ is not a uninorm on L which is a contradiction. Therefore, the condition that R (x, y) <1 for all x, y < 1 is necessary.

Sufficiency. This proof is the same as Theorem 3.1 of [7]. □

Theorem 4.6. Let (L, ≤ , 0, 1) be a bounded lattice with e ∈ L \ {0, 1}. Let S_e be a t-conorm on [e, 1] and F be a t-subnorm on L. If x ∥ y for all x ∈ I_e and y ∈]0, e], then the function $U_{F}^{e} : L^{2} \to L$ defined by∥ $U_{F}^{e} (x, y) = {\begin{matrix} S_{e} (x, y) & if (x, y) \in [e, 1]^{2}, \\ x & if (x, y) \in I_{e} \times] 0, e] \cup \\ I_{e} \times [e, 1], \\ y & if (x, y) \in] 0, e] \times I_{e} \cup \\ [e, 1] \times I_{e}, \\ F (x, y) & if (x, y) \in I_{e}^{2} \cup [0, e [^{2}, \\ x \land y & if (x, y) \in I_{e} \times {0} \cup {0} \\ \times I_{e} \cup [e, 1] \times {0} \cup {0} \\ \times [e, 1] \cup [0, e] \times {e} \cup \\ {e} \times [0, e], \\ x \lor y & otherwise, \end{matrix}$ is a uninorm on L with e ∈ L \ {0, 1} iff F (x, y) >0 for all x, y > 0.

Proof. It can be proved in an analogous way to the proof of Theorem 4.5. □

Remark 4.7. In Theorem 3.1 of [7], the condition that R is a t-subconorm on [e, 1] such that R (x, y) <1 for all x, y < 1 is sufficient for $U_{R}^{e}$ . In Theorem 3.9 of [7], the condition that F is a t-subnorm on [0, e] such that F (x, y) >0 for all x, y > 0 is sufficient for $U_{F}^{e}$ , respectively. In fact, the conditions are also necessary for $U_{F}^{e}$ in Theorem 4.5 and $U_{R}^{e}$ in Theorem 4.6, respectively.

Remark 4.8. Let U₃ is the uninorm on L defined as in Theorem 3.4 and U₄ is the uninorm on L defined as in Theorem 3.5, and let $U_{R}^{e}$ is the uninorm on L defined as in Theorem 4.5 and $U_{F}^{e}$ is the uninorm on L defined as in Theorem 4.6.

(1) It is easy to see that the values of U₃ and $U_{R}^{e}$ are different on I_e × {0} ∪ {1} × {0} and {0} × I_e ∪ {0} × {1}. In fact, U₃ = 0 and $U_{R}^{e} = x$ when (x, y) ∈ I_e × {0} ∪ {1} × {0}; U₃ = 0 and $U_{R}^{e} = y$ when (x, y) ∈ {0} × I_e ∪ {0} × {1}. The values of U₃ are not equal to 0 on ] 0, e [×]0, e [; however, the values of $U_{R}^{e}$ could be 0 on ] 0, e [×]0, e [.

(2) The additional constraints of U₃ and $U_{R}^{e}$ are different, that is, the condition that T_e (x, y) >0 for all x, y > 0 is needed for U₃.

According to (1) and (2), by changing the values of $U_{R}^{e}$ , we can obtain a new method to construct the nuinorm U₃ and then additional constraints are different for $U_{R}^{e}$ and U₃.

Similarly, the relationship between U₄ and $U_{F}^{e}$ is same as that of U₃ and $U_{R}^{e}$ .

5 Conclusion

In this article, we investigate the constructions of uninorms on some appropriate bounded lattices with e ∈ L \ {0, 1}, based on some additional constraints. In this progress, we mainly focus on the rationality of these constraints and then show that some of these constraints are sufficient and necessary for the construction of uninorms. Meanwhile, we study the constraints of some existing uninorms and find that some constraints are also necessary. In the future research, when considering the new construction of uninorms, if the constraints are needed, we expect they are sufficient and necessary. Moreover, as we can see, the constructions of uninorms on bounded lattices had been investigated widely. Then it may be an interesting work that how to apply uninorms on bounded lattices to L-fuzzy theory, lattice-valued logic and so on.

References

Aşıcı

and Mesiar

, On the construction of uninorms on bounded lattices, Fuzzy Sets and Systems 408 (2021), 65–85.

Birkhoff

, Lattice theory, 3rd Edition, Amer. Math. Soc., Rhode Island, 1967.

Bodjanova

and Kalina

, Construction of uninorms on bounded lattices, in: IEEE 12th International Symposium on Intelligent Systems and Informatics, SISY 2014, September 11–13, Subotica, Serbia, 61–66

Baczyński

and Jayaram

, Fuzzy Implications, Springer, Berlin, 2008.

Dombi

, Basic concepts for a theory of evaluation: The aggregative operator, European Journal of Operational Research 10 (1982), 282–293.

Çaylı

G.D.

, New methods to construct uninorms on bounded lattices, International Journal of Approximate Reasoning 115 (2019), 254–264.

Çaylı

G.D.

, On the structure of uninorms on bounded lattices, Fuzzy Sets and Systems 357 (2019), 2–26.

Çaylı

G.D.

, Uninorms on bounded lattices with the underlying t-norms and t-conorms, Fuzzy Sets and Systems 395 (2020), 107–129.

Çaylı

G.D.

, New construction approaches of uninorms on bounded lattices, International Journal of General Systems 50 (2021), 139–158.

10.

Çaylı

G.D.

and Karaçal

, Construction of uninorms on bounded lattices, Kybernetika 53 (2017), 394–417.

11.

Çaylı

G.D.

, Karaçal

and Mesiar

, On a new class of uninorms on bounded lattices, Information Sciences 367 (2016), 221–231.

12.

Dan

Y.X.

and Hu

B.Q.

, A new structure for uninorms on bounded lattices, Fuzzy Sets and Systems 386 (2020), 77–94.

13.

Dan

Y.X.

, Hu

B.Q.

and Qiao

J.S.

, New constructions of uninorms on bounded lattices, International Journal of Approximate Reasoning 110 (2019), 185–209.

14.

De Baets

and Van Fodor

, Melle’s combining function in MYCIN is a representable uninorm: an alternative proof, Fuzzy Sets and Systems 104 (1999), 133–136.

15.

Grabisch

, Marichal

J.L.

, Mesiar

and Pap

, Aggregation functions, Cambridge University Press, 2009.

16.

Grabisch

, Marichal

J.L.

, Mesiar

and Pap

, Aggregation functions: construction methods, conjunctive, disjunctive and mixed classes, Information Sciences 181 (2011), 23–43.

17.

Goguen

, L-fuzzy sets, Journal of Mathematical Analysis and Applications 18 (1967), 145–147.

18.

Höhle

, Commutative, residuated l-monoids, in: U. Höhle, E.Klement (Eds.), Non-Classical Logics and Their Applications to Fuzzy Subsets: A Handbook on the Mathematical Foundations of Fuzzy Set Theory, Kluwer, Dordrecht, 1995.

19.

Hua

X.-J.

and Ji

, Uninorms on bounded lattices constructed by t-norms and t-subconorms, Fuzzy Sets and Systems 427 (2022), 109–131.

20.

Hua

X.-J.

, Zhang

H.-P.

and Ouyang

, Note on “Construction of uninorms on bounded lattices”, Kybenetika 57(2) (2021), 372–382.

21.

and Wang

X.-P.

, Constructing uninorms on bounded lattices by using additive generators, International Journal of Approximate Reasoning 136 (2021), 1–13.

22.

, Constructions of uninorms on bounded lattices by means of t-subnorms and t-subconorms, Fuzzy Sets and Systems 403 (2021), 38–55.

23.

Karaçal

and Mesiar

, Uninorms on bounded lattices, Fuzzy Sets and Systems 261 (2015), 33–43.

24.

Klement

E.P.

, Mesiar

and Pap

, Triangular norms. Position paper I: basic analytical and algebraic properties, Fuzzy Sets and Systems 143 (2004), 5–26.

25.

Liang

and Pedrycz

, Logic-based fuzzy networks: a study in system modeling with triangular norms and uninorms, Fuzzy Sets and Systems 160 (2009), 3475–3502.

26.

Menger

, Statistical metrics, Proc Natl Acad Sci USA 8 (1942), 535–537.

27.

Ouyang

and Zhang

H.P.

, Constructing uninorms via closure operators on a bounded lattice, Fuzzy Sets and Systems 395 (2020), 93–106.

28.

Pedrycz

and Hirota

, Uninorm-based logic neurons as adaptive and interpretable processing constructs, Soft Computing 11 (2007), 41–52.

29.

Quesada

F.J.

, Palomares

and Martinez

, Managing experts behavior in large-scale consensus reaching processes with uninorm aggregation operators, Applied Soft Computing 35 (2015), 873–887.

30.

Schweizer

and Sklar

, Statistical metric spaces, Pacific Journal of Mathematics 10 (1960), 313–334.

31.

Schweizer

and Sklar

, Associative functions and statistical triangular inequalities, Publicationes mathematicae 8 (1961), 169–186.

32.

Schweizer

and Sklar

, Probabilistic Metric Spaces, North-Holland, New York, 1983.

33.

Saminger

, On ordinal sums of triangular norms on bounded lattices, Fuzzy Sets and Systems 157 (2006), 1403–1416.

34.

Wang

, TL-filters of integral residuated l-monoids, Information Sciences 177 (2007), 887–896.

35.

Xie

A.F.

and Li

S.J.

, On constructing the largest and smallest uninorms on bounded lattices, Fuzzy Sets and Systems 386 (2020), 95–104.

36.

Xiu

Z.Y.

and Zheng

, New construction methods of uninorms on bounded lattices via uninorms, Fuzzy Sets and Systems 2023. doi.org/10.1016/j.fss.2023.108535.

37.

Yager

R.R.

and Rybalov

, Uninorm aggregation operators, Fuzzy sets and systems 80 (1996), 111–120.

38.

Zhang

H.P.

, Wu

, Wang

, Ouyang

and De Baets

, A characterization of the classes Umin and Umax of uninorms on a bounded lattice, Fuzzy Sets and Systems 423 (2021), 107–121.

39.

Zhao

and Wu

, Some further results about uninorms on bounded lattices, International Journal of Approximate Reasoning 130 (2021), 22–49.

40.

Zimmermann

H.J.

, Fuzzy Set Theory and Its Applications, fourth ed., Kluwer, Aachen, 2001.