On a class of left-continuous uninorms constructed from the representable uninorm

Abstract

Uninorms are an important generalization of triangular norms and triangular conorms, having a neutral element lying anywhere in the unit interval. Many different classes of uninorms have been discussed in literature. In this paper, a construction method for a class of left-continuous uninorms is proposed based on the representable uninorm. The idea stems from the rotation method. Some illustrative examples of such uninorms are provided.

Keywords

Uninorm representable uninorm additive generator left continuity

1. Introduction

Uninorms were introduced in 1996 by Yager and Rybalov [28] and generalize the notions of both triangular norms and triangular conorms [2, 19] by allowing the neutral element to lie anywhere in the unit interval rather than at one or zero as in the case of triangular norms (t-norms for short) and triangular conorms (t-conorms for short). The first deep study by Fodor et al. [13] revealed the structure of uninorms. Later on it was shown that uninorms are useful in many fields like expert systems [6, 29], fuzzy decision making [30], fuzzy integrals [20] and fuzzy morphology [3]. On the other hand, the theoretical study of uninorms has been even more extensive. Many different classes of uninorms [24] have been discussed in the literature, including $U_{min}, U_{max}$ [13], representable uninorms [13], idempotent uninorms [4], uninorms locally internal in the area $A (e)$ [11], uninorms continuous in the open unit square [15], uninorms with continuous underlying operators, left-continuous uninorms [12 , 26] and uninorms locally internal on the boundary [9, 23]. Moreover, some construction methods for uninorms were also proposed in [16, 21]. In this paper, we focus on the construction of uninorms. We will achieve our goal by generalizing the rotation construction.

2. Preliminaries

It is assumed that the reader is familiar with the theory of t-norms, t-conorms and fuzzy negations (see for instance [2, 19]). We recall here only some basic facts and necessary notations on uninorms which will be used in this paper.

A negation N on [0, 1] is a decreasing unary function on [0, 1] satisfying N (0) =1, N (1) =0. A negation is called strong if it is involutive, i.e. if N (N (x)) = x for any x ∈ [0, 1].

A uninorm is a commutative, associative, increasing function U : [0, 1] ² → [0, 1] such that U (e, x) = x for all x ∈ [0, 1], where e ∈ [0, 1] is called the neutral element. Note that uninorm can be defined on any interval I ⊆ [0,1] by an isomorphic transformation from I to [0, 1].

We summarize some fundamental results from [13]. It is clear that uninorm U becomes a t-norm when e = 1 and a t-conorm when e = 0. For any uninorm we have U (0, 1) ∈ {0, 1}. A uninorm U such that U (0, 1) =0 is called conjunctive and if U (0, 1) =1 then it is called disjunctive.

Throughout this paper, we exclusively consider uninorms with a neutral element e ∈]0, 1 [.

With any uninorm U with neutral element e ∈]0, 1 [, we can associate two binary operations T_U, S_U : [0, 1] ² → [0, 1] defined by $\begin{matrix} T_{U} (x, y) = \frac{1}{e} \cdot U (ex, ey) \end{matrix}$ and $\begin{matrix} S_{U} (x, y) = \frac{1}{1 - e} (U (e + (1 - e) x, e + (1 - e) y) - e) . \end{matrix}$

It is easy to see that T_U is a t-norm and that S_U is a t-conorm. In other words, on [0, e] ² any uninorm U is determined by a t-norm T_U, and on [e, 1] ² any uninorm U is determined by a t-conorm S_U; T_U is called the underlying t-norm, and S_U is called the underlying t-conorm. Let us denote the remaining part of the unit square by A (e), i.e., $A (e) = [0, 1]^{2} ∖ ([0, e]^{2} \cup [e, 1]^{2}) .$

On the set $A (e)$ , any uninorm U is bounded by the minimum and the maximum of its arguments, i.e., for any $(x, y) \in A (e)$ it holds that $min (x, y) \leq U (x, y) \leq max (x, y) .$ (1)

The residual implication I_U [5] corresponding to conjunctive, left-continuous U is the binary function on [0, 1] defined by $\begin{matrix} I_{U} (x, y) = sup {t \in [0, 1] : U (x, t) \leq y} . \end{matrix}$

Note that the residual implication corresponding to left-continuous t-norm can be defined similarly. For any x, y, z ∈ [0, 1] the following equivalence (called residuation property) holds: $\begin{matrix} U (x, z) \leq y \Leftrightarrow I_{U} (x, y) \geq z . \end{matrix}$

Some class of left-continuous uninorms and the corresponding residual implications have been characterized in [1]. Their applications in fuzzy logic, fuzzy control were discussed in [27].

It is well known that there exist many construction methods of left-continuous t-norms which are summarized in [17], for example, rotation construction, rotation-annihilation construction and so on. In [16], Jenei and De Baets generalized these methods to construct the left-continuous uninorms. First, we recall the main results in [16] about the rotation of uninorms.

Theorem 1. Consider a strong negation N with unique fixpoint t and a left-continuous uninorm U with neutral element e. Let U₁ (acting on [t, 1] ²) be the linear transformation of U (acting on [0, 1] ²) into [t, 1] and denote the image of e under this linear transformation by e^* = t + e (1 - t). Then the binary operation M on [0, 1] ² defined by $M (x, y) = {\begin{matrix} U_{1} (x, y) & (x, y) \in] t, 1]^{2}, \\ N (I_{U_{1}} (x, N (y))) & (x, y) \in] t, 1] \times [0, t], \\ N (I_{U_{1}} (y, N (x))) & (x, y) \in [0, t] \times] t, 1], \\ 0 & (x, y) \in [0, t]^{2}, \end{matrix}$ (2) is a left-continuous uninorm with neutral element e^* if the left-continuous underlying t-norm T_U of U satisfies one of the following conditions:

(C1) T_U has no zero divisors;

(C2) T_U has zero divisors, and there exists c ∈]0, 1] such that for any zero divisor x (i.e. there exists y ∈]0, 1 [ such that T_U (x, y) =0) we have I_{T
_U} (x, 0) = c.

Note that the uninorm M constructed in Theorem 1 satisfies the rotation invariance property with respect to the strong negation N, i.e., for any x, y, z ∈ [0, 1], $M (x, y) \leq z \Leftrightarrow M (N (z), x) \leq N (y) .$

Some illustrated examples about the uninorm M were presented in [16] (see Example 3 and Example 4 in [16]).

3. Main results

In this section, we will focus on the condition (C1) in Theorem 1.

If the left-continuous underlying t-norm T_U of uninorm U satisfies condition (C1) in Theorem 1, then T_U (x, y) >0 for all (x, y) ∈]0, 1] ². It is obvious that the representable uninorm U satisfies condition (C1). The first result of this type is due to Klement et al. [18] (compare also the paper of Dombi [10]). Moreover, in papers [5 , 14], many properties of the representable uninorm were also discussed. In the following proposition, some related results used in this section are presented.

Proposition 1. Let U be a conjunctive representable uninorm with neutral element e ∈]0, 1 [. Then the following statements hold:

(i) There exists a strictly increasing continuous function h : [0, 1] → [- ∞, + ∞] with h (0) = -∞, h (e) =0, h (1) = + ∞ such that, for all x, y ∈ [0, 1] ² \ {(0, 1), (1, 0)}, $\begin{matrix} U (x, y) = h^{- 1} (h (x) + h (y)), \end{matrix}$ and the residual implication is $I_{U} (x, y) = h^{- 1} (h (y) - h (x)),$ for (x, y) ∈ [0, 1] ² \ {(0, 1), (1, 0)}, and $\begin{matrix} I_{U} (0, 1) = I_{U} (1, 0) = 1 . \end{matrix}$

(ii) U (x, 1) = U (1, x) =1 for all x ∈]0, 1].

(iii) U (x, y) = U (y, x) >0 for all (x, y) ∈]0, 1] ².

(iv) U is strictly increasing and continuous on ] 0, 1 [²;

(v) The underlying operators T_U, S_U of U are strict, i.e. continuous and strictly increasing in the region ] 0, 1] ²;

(vi) There exists an strong negation with fixpoint e,i.e. N (e) = e, such that $\begin{matrix} U (x, N (x)) = e, \forall x, y \in] 0, 1 [^{2} . \end{matrix}$

Note that the statement (v) in Proposition 1 is from Theorem 5 in [14] (compare also Theorem 2 in [18]), the other statements are from Theorem 3 in [13] and Proposition 6 in [5], moreover, the statement about the residual implication is from Theorem 7 in [5]. By Proposition 1, it is known that for fixed x₀ ∈ [0, t [, N (I_{U
₁} (y, N (x₀))) in Theorem 1 is a strictly increasing function about y ∈] t, 1] and for fixed y₀ ∈] t, 1], N (I_{U
₁} (y₀, N (x))) in Theorem 1 is a strictly increasing function about x ∈ [0, t]. A problem emerges: Can these strictly increasing functions be replaced with arbitrary functions? So, we want to offer a construction method of left-continuous uninorm based on the representable uninorm, i.e. our aim is to construct a conjunctive left-continuous uninorm such that $U (x, y) = {\begin{matrix} U_{r} (x, y) & (x, y) \in] t, 1]^{2}, \\ 0 & (x, y) \in [0, t]^{2} or \\ min (x, y) = 0, \end{matrix}$ (3) where U_r is a representable uninorm defined on [t, 1] ², t ∈]0, 1 [ with neutral element e ∈] t, 1 [.

Standing assumption: throughout this section, notation U will denote the conjunctive left-continuous uninorm with form (3) when no explicit statement is mentioned.

Proposition 2. Let U be a uninorm of form (3) for some fixed t ∈]0, 1 [. Then U (t, 1) = U (1, t) = t.

Proof: Since t ∈]0, e [, U (t, 1) = U (1, t) ∈ [t, 1] by Equation (1). On the contrary, suppose that U (t, 1) ∈] t, 1]. Then U (U (t, 1), U (t, 1)) = U_r (U (t, 1), U (t, 1)) > t and U (U (t, t), U (1, 1)) = U (0, 1) =0, which is a contradiction with the associativity and commutativity of U. □

For arbitrary x ∈]0, t [, we denote by U_x : [t, 1] → [0, 1] the partial function U_x (y) = U (x, y). For arbitrary y ∈] t, 1], we denote by U^y : [0, t] → [0, t] the partial function U^y (x) = U (x, y).

First, we discuss some properties about the partial function U^y : [0, t] → [0, t]. Let us denote the range of U^y by Ran (U^y).

Proposition 3. Let y₁, y₂ ∈] t, 1 [. Then Ran (U^{y
₁}) = Ran (U^{y
₂}).

Proof: If z₁ ∈ Ran (U^{y
₁}) then $\begin{matrix} U^{y_{1}} (x_{1}) = U (x_{1}, y_{1}) = z_{1} \end{matrix}$ for some x₁ ∈ [0, t]. Since U_r is representable, by Proposition 1 there exists y₀ ∈] t, 1 [ such that $\begin{matrix} U (y_{0}, y_{2}) = U_{r} (y_{0}, y_{2}) = y_{1} . \end{matrix}$

So, $\begin{matrix} z_{1} = U (x_{1}, y_{1}) = U ((x_{1}, U (y_{0}, y_{2})) \\ = U (U (x_{1}, y_{0}), y_{2}) = U^{y_{2}} (U (x_{1}, y_{0})) \end{matrix}$ by the associativity of uninorm U.

Now, we prove that U (x₁, y₀) ∈ [0, t]. On the contrary, suppose that U (x₁, y₀) > t. Then by Proposition 1 and the associativity of uninorm U, $\begin{matrix} U (U (x_{1}, y_{0}), U (x_{1}, y_{0})) \\ = U_{r} (U (x_{1}, y_{0}), U (x_{1}, y_{0})) > 0 \end{matrix}$ and $\begin{matrix} U (U (x_{1}, y_{0}), U (x_{1}, y_{0})) \\ = U (U (x_{1}, x_{1}), U (y_{0}, y_{0})) \\ = U (0, U (y_{0}, y_{0})) = 0, \end{matrix}$ which is a contradiction. Thus, U (x₁, y₀) ∈ [0, t], z₁ = U^{y
₂} (U (x₁, y₀)) ∈ Ran (U^{y
₂}) and Ran (U^{y
₁}) ⊆ Ran (U^{y
₂}). By the similar proof, we can proved that Ran (U^{y
₂}) ⊆ Ran (U^{y
₁}). Hence, the result holds.□

Proposition 4. For arbitrary y₀ ∈] t, 1 [, there exists no interval of constantness of the partial function U^{y
₀}.

Proof: On the contrary, suppose that there exist x₁, x₂ ∈]0, t [ such that U^{y
₀} (x₁) = U^{y
₀} (x₂). Since U_r is representable, by Proposition 1 there exists y₂ ∈] t, 1 [ such that $\begin{matrix} U (y_{0}, y_{2}) = U_{r} (y_{0}, y_{2}) = e . \end{matrix}$

So, by the associativity of U, we have $\begin{array}{l} x_{1} = U (x_{1}, e) = U (x_{1}, U (y_{0}, y_{2})) \\ = U (U (x_{1}, y_{0}), y_{2}) = U (U (x_{2}, y_{0}), y_{2}) \\ = U (x_{2}, U (y_{0}, y_{2})) = U (x_{2}, e) = x_{2} . □ \end{array}$

By Proposition 4, we know that the partial function U^y : [0, t] → [0, t] is strict increasing.

Proposition 5. For arbitrary y₀ ∈] t, 1], U (t, y₀) = U (y₀, t) = t.

Proof: By Proposition 2 and the monotonicity of U, U (t, y₀) ≤ U (t, 1) = t. The proof requires the consideration of three different cases.

(i) y₀ = e. Then it is obvious that the result holds.

(ii) y₀ ∈] e, 1]. By Equation (1), $\begin{matrix} (t, y_{0}) = U (y_{0}, t) = t . \end{matrix}$

(iii) y₀ ∈] t, e [. Then there exists y₁ ∈] e, 1 [ such that U (y₁, y₀) = e. Hence, by the associativity of U and case (ii), we have $\begin{array}{l} t = U (t, e) = U (t, U (y_{1}, y_{0})) \\ = U (U (t, y_{1}), y_{0}) = U (t, y_{0}) . □ \end{array}$

Proposition 6. If U (x₀, y₀) = x₀ for some (x₀, y₀) ∈]0, t [×] t, e [∪] e, 1 [, then there exists an interval I ⊆] t, 1] such that e is the interior point of I and U (x₀, y) = x₀ for all y ∈ I. Furthermore, if U (x₀, 1) = x₀ then U (x₀, y) = x₀ for all y ∈] t, 1].

Proof: Since U_r is representable, by Proposition 1 there exists y₁ ∈] t, e [∪] e, 1 [ such that min(y₀, y₁) < e < max(y₀, y₁) and U (y₀, y₁) = e. So, $\begin{matrix} U (x_{0}, U (y_{0}, y_{1})) = U (x_{0}, e) = x_{0} \end{matrix}$ and $\begin{matrix} U (U (x_{0}, y_{0}), y_{1}) = U (x_{0}, y_{1}) . \end{matrix}$

By the associativity of U, we have U (x₀, y₁) = x₀. By the monotonicity of U, U (x₀, y) = x₀ for the interval I = [min(y₀, y₁), max(y₀, y₁)].

If U (x₀, 1) = x₀, then for arbitrary positive number ɛ ∈]0, 1 - t], $\begin{matrix} U (x_{0}, U (1, t + ɛ)) = U (x_{0}, 1) = x_{0} \end{matrix}$ and $\begin{matrix} U (U (x_{0}, 1), t + ɛ) = U (x_{0}, t + ɛ) . \end{matrix}$

By the associativity of U, we have U (x₀, t + ɛ) = x₀. So, by the monotonicity of U, U (x₀, y) = x₀ for all y ∈] t, 1].□

From Proposition 6, we know that the uninorm U (x, y) = min(x, y), (x, y) ∈]0, t [×] t, 1 [ whenever U (x, 1) = min(x, 1) for all x ∈]0, t [.

Now, one result of Clifford [7] is introduced in the following theorem.

Theorem 2. Let A≠ ∅ be a totally ordered set and (G_α) _α∈A with G_α = (X_α, ∗ _α) be a family of semigroups. Assume that for all α, β ∈ A with α < β the sets X_α and X_β are either disjoint or that X_α ∩ X_β = {x_αβ}, where x_αβ is both the neutral element of G_α and the annihilator of G_β and where for each γ ∈ A with α < γ < β we have X_γ = {x_αβ}. Put X = ⋃ _α∈AX_α and define the binary operation ∗ on X by $x * y = {\begin{array}{l} x *_{α} y & i f (x, y) \in X_{α} \times X_{α}, \\ x & i f (x, y) \in X_{α} \times X_{β} a n d α < β, \\ y & i f (x, y) \in X_{α} \times X_{β} a n d β < α . \end{array}$

Then G = (X, ∗) is a semigroup. The semigroup G is commutative if and only if for each α ∈ A the semigroup G_α is commutative.

Theorem 3. The binary function $\begin{matrix} U (x, y) = {\begin{matrix} U_{r} (x, y) & (x, y) \in] t, 1]^{2}, \\ 0 & (x, y) \in [0, t]^{2}, \\ min (x, y) & otherwise . \end{matrix} \end{matrix}$ where U_r is a representable uninorm defined on [t, 1] ², t ∈]0, 1 [ with neutral element e ∈] t, 1 [, is a conjunctive left-continuous uninorm with neutral element e.

Proof: The result obviously holds by Theorem 2.□

Remark 1. By taking the underlying t-norm T_U of uninorm U in Theorem 3, this procedure results in a class of left-continuous t-norms which coincide with those of Theorem 5 in [8].

Example 1. Consider the “Three Pi” operator [13] $\begin{matrix} M (x, y) = {\begin{matrix} 0 & (x, y) \in {(0, 1), (1, 0)}, \\ \frac{xy}{xy + (1 - x) (1 - y)} & otherwise . \end{matrix} \end{matrix}$

It is a representable uninorm with neutral element $\frac{1}{2}$ . By the linear transformation, we can obtain the representable uninorm $\begin{matrix} U_{r} (x, y) = \frac{1}{2} + \frac{1}{2} \cdot M (\frac{x - \frac{1}{2}}{1 - \frac{1}{2}}, \frac{y - \frac{1}{2}}{1 - \frac{1}{2}}) \\ = \frac{1}{2} + \frac{1}{2} \cdot \frac{(2 x - 1) (2 y - 1)}{(2 x - 1) (2 y - 1) + 4 xy}, \end{matrix}$ which is defined on $[\frac{1}{2}, 1]^{2}$ having neutral element $\frac{3}{4}$ . Let us construct the binary function $\begin{matrix} U (x, y) = {\begin{matrix} U_{r} (x, y) & (x, y) \in] \frac{1}{2}, 1]^{2}, \\ 0 & (x, y) \in [0, \frac{1}{2}]^{2}, \\ min (x, y) & otherwise . \end{matrix} \end{matrix}$

It is obvious that U is a left-continuous uninorm with neutral element $\frac{3}{4}$ .

For arbitrary x ∈]0, t [, we discuss the partial function U_x : [t, 1] → [0, 1].

Proposition 7. Let x₁, x₂ ∈]0, t [. If x₂ ∈ Ran (U_{x
₁}) then the partial function U_{x
₂} is uniquely deriving from U_{x
₁}.

Proof: If x₂ ∈ Ran (U_{x
₁}) then there exists y₁ ∈ [t, 1] such that U_{x
₁} (y₁) = U (x₁, y₁) = x₂. So, for arbitrary y ∈] t, 1], we have $\begin{matrix} U_{x_{2}} (y) = U (x_{2}, y) \\ = U (U (x_{1}, y_{1}), y) = U (x_{1}, U (y_{1}, y)) . \end{matrix}$

Since U (y₁, y) ≥ t, U_{x
₂} (y) = U_{x
₁} (U (y₁, y)).□

Corollary 1. Let x₁, x₂ ∈]0, t [. If U_{x
₁} (1) = t, x₂ ∈ Ran (U_{x
₁}) then U (x₂, 1) = t.

Proposition 8. Let x₁ ∈]0, t [. Assume that I is a maximal interval of constantness of U_{x
₁}. Then one of the following statements holds.

(i) e > y for arbitrary y ∈ I. Then the interval I is also an constantness interval of the partial function U_{x
₂}, x₂ ∈ Ran (U_{x
₁}), x₂ ≤ x₁.

(ii) e < y for arbitrary y ∈ I. Then the interval I is also an constantness interval of the partial function U_{x
₂}, x₂ ∈ Ran (U_{x
₁}), x₂ ≥ x₁.

(iii) e ∈ I. Then the interval I is also an constantness interval of the partial function U_{x
₂}, x₂ ∈ Ran (U_{x
₁}).

Proof: Since the partial function U_{x
₁} is increasing, the maximal interval of constantness of U_{x
₁} is a connected set.

(i) If e > y for arbitrary y ∈ I, then I ⊆]0, e [. Assume that U_{x
₁} (y) = c < x₁ for all y ∈ I. Since x₂ ∈ Ran (U_{x
₁}), x₂ ≤ x₁, there exists y₁ ∈] t, e] such that U (x₁, y₁) = x₂. Hence, for arbitrary y ∈ I, $\begin{matrix} U_{x_{2}} (y) = U (x_{2}, y) = U (U (x_{1}, y_{1}), y) \\ = U (U (x_{1}, y), y_{1}) = U (c, y_{1}) . \end{matrix}$

So, the result holds.

(ii) and (iii) The proof is similar to (i).□

Corollary 2. Let x₁ ∈]0, t [. If the partial function U_{x
₁} is continuous, strictly increasing and U_{x
₁} (1) = t, then for all x ∈]0, t [ the partial function U_x is continuous, strictly increasing and U_x (1) = t.

Proof: By Proposition 7, Corollary 1 and Proposition 8, the result holds.□

Theorem 4. Let t ∈]0, 1 [, x₀ ∈]0, t [, e ∈] t, 1 [ and f : [t, 1] → [0, t] be continuous, strictly increasing unary function fulfilling f (t) =0, f (e) = x₀ and f (1) = t. Then the binary function: U (x, y) $= {\begin{matrix} U_{r} (x, y) & (x, y) \in] t, 1]^{2}, \\ 0 & (x, y) \in [0, t]^{2} or \\ min (x, y) = 0, \\ f (h^{- 1} (h (y_{0}) + h (y))) & (x, y) \in] 0, t] \times] t, 1] \\ and x = f (y_{0}), \\ f (h^{- 1} (h (y_{0}) + h (x))) & (x, y) \in] t, 1] \times] 0, t] \\ and y = f (y_{0}), \end{matrix}$ (4) where U_r is a conjunctive representable uninorm defined on [t, 1] ² with neutral element e and the additive generator h : [t, 1] → [- ∞, + ∞], is a conjunctive left-continuous uninorm.

Proof: Because h is the additive generator of U_r, h (t) = - ∞, h (e) =0, h (1) = + ∞, U_r (x, y) = h^-1 (h (x) + h (y)), (x, y) ∈] t, 1] ². So, if x = f (y₀), y ∈] t, 1], then $\begin{matrix} U (x, y) = f (h^{- 1} (h (y_{0}) + h (y))) \leq t \end{matrix}$ and $\begin{matrix} U (x, e) = f (h^{- 1} (h (y_{0}) + h (e))) \\ = f (h^{- 1} (h (y_{0}))) \\ = f (y_{0}) = x \end{matrix}$ for all x ∈]0, t]. Since e is the neutral element of U_r and U (0, e) =0, U has the neutral element e. It is obvious that U is commutative, monotone. We only need to prove the associativity of U. Let x, y, z ∈ [0, 1]. We prove that the following equation holds $\begin{matrix} U (U (x, y), z) = U (x, U (y, z)) . \end{matrix}$

Without loss of generality, suppose that x ≤ y ≤ z. Only several cases need to be considered.

(i) x ≤ y ≤ t, z > t. Then $\begin{matrix} U (U (x, y), z) = U (0, z) = 0 \end{matrix}$

and $\begin{matrix} U (y, z) \leq t, U (x, U (y, z)) = 0 . \end{matrix}$

(ii) x ≤ t < y ≤ z. If x = f (y₀) then $\begin{matrix} U (x, y) = f (h^{- 1} (h (y_{0}) + h (y))), \\ U (U (x, y), z) \\ = f (h^{- 1} (h \cdot h^{- 1} (h (y_{0}) + h (y)) + h (z))) \\ = f (h^{- 1} (h (y_{0}) + h (y) + h (z))), \end{matrix}$ and $\begin{matrix} U (y, z) = h^{- 1} (h (y) + h (z)) > t, \\ U (x, U (y, z)) \\ = f (h^{- 1} (h (y_{0}) + h \cdot h^{- 1} (h (y) + h (z)))) \\ = f (h^{- 1} (h (y_{0}) + h (y) + h (z))) . \end{matrix}$ So, U (U (x, y), z) = U (x, U (y, z)).

(iii) The other cases. Then $\begin{matrix} U (U (x, y), z) = U (x, U (y, z)) = 0 \end{matrix}$ or the associativity of U follows by the associativity of U_r.

The left-continuity of U can be obtained by the continuity of f, h, h^-1 and Proposition 1.□

Remark 2. (i) Note that U (x, 1) = t for all x ∈]0, t] for uninorm U in Theorem 4, whereas M (x, 1) =1, x ∈]0, t] for uninorm M in Theorem 1. So, uninorm U in Theorem 4 does not belong to the class of uninorms locally internal on the boundary [23].

(ii) By taking the underlying t-norm T_U of uninorm U in Theorem 4, this procedure results in a class of left-continuous t-norms which are different from those of Theorem 5 in [8].

Example 2. Consider the “Three Pi” operator in Example 1 $\begin{matrix} M (x, y) = {\begin{matrix} 0 & (x, y) \in {(0, 1), (1, 0)}, \\ \frac{xy}{xy + (1 - x) (1 - y)} & otherwise . \end{matrix} \end{matrix}$

It is a representable uninorm with neutral element $\frac{1}{2}$ and the additive generator $h_{M} (x) = log \frac{x}{1 - x}$ . Let $t = \frac{1}{2}, x_{0} = \frac{1}{4}$ and the linear transformation of M defined by $\begin{matrix} U_{r} (x, y) = \frac{1}{2} + \frac{1}{2} \cdot M (\frac{x - \frac{1}{2}}{1 - \frac{1}{2}}, \frac{y - \frac{1}{2}}{1 - \frac{1}{2}}) \\ = \frac{1}{2} + \frac{1}{2} \cdot \frac{(2 x - 1) (2 y - 1)}{(2 x - 1) (2 y - 1) + 4 xy}, \end{matrix}$

It is obvious that U_r is a representable uninorm defined on $[\frac{1}{2}, 1]^{2}$ with neutral element $\frac{3}{4}$ whose additive generator $h (x) = log \frac{2 x - 1}{2 - 2 x}$ . Let f be the (piecewise) linear function satisfying $f (\frac{1}{2}) = 0, f (\frac{3}{4}) = \frac{1}{4}, f (1) = \frac{1}{2}$ , i.e., $f (x) = x - \frac{1}{2}$ .

Then, we can construct the left-continuous uninorm $\begin{matrix} U (x, y) = {\begin{matrix} U_{r} (x, y) & (x, y) \in] \frac{1}{2}, 1]^{2}, \\ 0 & (x, y) \in [0, \frac{1}{2}]^{2} or \\ min (x, y) = 0, \\ \frac{2 xy - x}{8 xy - 6 x - 2 y + 2} & (x, y) \in] 0, \frac{1}{2}] \times] \frac{1}{2}, 1], \\ \frac{2 xy - y}{8 xy - 6 y - 2 x + 2} & (x, y) \in] \frac{1}{2}, 1] \times] 0, \frac{1}{2}] . \end{matrix} \end{matrix}$

The result of Theorem 4 can be generalized.

Theorem 5. Let t ∈]0, 1 [, e ∈] t, 1 [ and 0 = x₀ < x₁ < x₂ <... < x_2k+1 < x_2k = t, k ∈ N be the given finite fixed points. Let f_i : [t, 1] → [x_2i, x_2i+2], 0 ≤ i ≤ (k - 1) be continuous, strictly increasing unary functions fulfilling f_i (t) = x_2i, f_i (e) = x_2i+1 and f_i (1) = x_2i+2. Then the binary function: U (x, y) $\begin{matrix} = {\begin{matrix} U_{r} (x, y) & (x, y) \in] t, 1]^{2}, \\ 0 & (x, y) \in [0, t]^{2} or \\ min (x, y) = 0, \\ f_{i} (h^{- 1} (h (y_{0}) + h (y))) & (x, y) \in] x_{2 i}, x_{2 i + 2}] \times] t, 1] \\ and x = f_{i} (y_{0}), \\ f_{i} (h^{- 1} (h (y_{0}) + h (x))) & (x, y) \in] t, 1] \times] x_{2 i}, x_{2 i + 2}] \\ and y = f_{i} (y_{0}), \end{matrix} \end{matrix}$ (5) where U_r is a conjunctive representable uninorm defined on [t, 1] ² with neutral element e ∈] t, 1 [ and the additive generator h : [t, 1] → [- ∞, + ∞], is a conjunctive left-continuous uninorm.

Proof: The proof is similar to that of Theorem 4. Hence, the proof is omitted here.

Remark 3. (i) Notice that U (x, 1) = x_2i+2 for all x ∈] x_2i, x_2i+2], 0 ≤ i ≤ (k - 1) for uninorm U in Theorem 5. The uninorm U does not belong to the class of uninorms locally internal on the boundary [23].

(ii) By taking the underlying t-norm T_U of uninorm U in Theorem 5, this procedure results in another class of left-continuous t-norms which are different from those of Theorem 5 in [8].

Example 3. Consider the “Three Pi” operator M and the linear transformation U_r in Example 2. Let $t = \frac{1}{2}, 0 = x_{0} < x_{1} = \frac{1}{8} < x_{2} = \frac{1}{4} < x_{3} = \frac{3}{8} < x_{4} = \frac{1}{2}$ . Assume that f₀ is the (piecewise) linear function satisfying $f_{0} (\frac{1}{2}) = 0, f (\frac{3}{4}) = \frac{1}{8}, f (1) = \frac{1}{4}$ , i.e., $\begin{matrix} f_{0} (x) = \frac{1}{2} \cdot x - \frac{1}{4} . \end{matrix}$ and f₁ is the (piecewise) linear function satisfying $f_{1} (\frac{1}{2}) = \frac{1}{4}, f (\frac{3}{4}) = \frac{3}{8}, f (1) = \frac{1}{2}$ , i.e., $\begin{matrix} f_{1} (x) = \frac{1}{2} \cdot x . \end{matrix}$

Then, we can construct the left-continuous uninorm

$\begin{matrix} U (x, y) = {\begin{matrix} U_{r} (x, y) & (x, y) \in] \frac{1}{2}, 1]^{2}, \\ 0 & (x, y) \in [0, \frac{1}{2}]^{2} or \\ min (x, y) = 0, \\ \frac{2 xy - x}{16 xy - 12 x - 2 y + 2} & (x, y) \in] 0, \frac{1}{4}] \times] \frac{1}{2}, 1], \\ \frac{12 xy - 8 x - 4 y + 3}{32 xy - 24 x - 12 y + 10} & (x, y) \in] \frac{1}{4}, \frac{1}{2}] \times] \frac{1}{2}, 1], \\ \frac{2 xy - y}{16 xy - 12 y - 2 x + 2} & (x, y) \in] \frac{1}{2}, 1] \times] 0, \frac{1}{4}], \\ \frac{12 xy - 8 y - 4 x + 3}{32 xy - 24 y - 12 x + 10} & (x, y) \in] \frac{1}{2}, 1] \times] \frac{1}{4}, \frac{1}{2}] . \end{matrix} \end{matrix}$

At last of this section, we focus on the (generalized) Modus Ponens which comes from approximate reasoning. In fuzzy rules based systems, any fuzzy conditional: p → q is often represented by a residual implication. If the residual implications are used in inference processes, where a concrete conjunction (usually a left-continuous t-norm) is selected, then it is essential that such residual implications satisfy the Modus Ponens with respect to the selected t-norm.

Let I_U be an residual implication from the conjunctive left-continuous uninorm U and T a t-norm. It is said that I_U satisfies the Modus Ponens property with respect to T or that I_U is a T-conditional if $\begin{matrix} T (x, I_{U} (x, y)) \leq y, x, y \in [0, 1] . \end{matrix}$

A well known general result on T-conditionality was proved in [25].

Proposition 9. Let U be a conjunctive left-continuous uninorm with neutral element e ∈]0, 1 [ and T a t-norm. Consider T_U its underlying t-norm, and let I_U be the corresponding residual implication. Then the following items are equivalent:

(i) I_U is a T-conditional.

(ii) The inequality $\begin{matrix} T (x, {eI}_{T_{U}} (\frac{x}{e}, \frac{y}{e})) \leq y \end{matrix}$ holds for all x, y such that y < x < e, where I_{T
_U} denotes the residual implication of the t-norm T_U.

Consider the conjunctive representable uninorm U_r defined on [t, 1] ² with neutral element e ∈] t, 1 [ and the additive generator h : [t, 1] → [- ∞, + ∞]. Let denote the corresponding uninorms in Theorem 3 by U₃.

Proposition 10. Let T be a t-norm. Consider the residual implication I_{U
₃} corresponding to uninorm U₃. Then I_{U
₃} is a T-conditional if and only if T (x, y) =0 for all x, y ∈ [0, t] and T (x, h^-1 (h (y) - h (x))) ≤ y for all t < y < x < e.

Proof: By the simple computation, the residual implication I_{T
_{U
₃}} derived from the underlying t-norm T_{U
₃} of uninorm U₃ is partially defined by $\begin{matrix} I_{T_{U_{3}}} (x, y) = {\begin{matrix} \frac{t}{e} & 0 \leq y < x \leq \frac{t}{e}, \\ y & y \leq \frac{t}{e} < x < 1, \\ \frac{h^{- 1} (h (ey) - h (ex))}{e} & \frac{t}{e} < y < x < 1, \end{matrix} \end{matrix}$

So, $T (x, {eI}_{T_{U_{3}}} (\frac{x}{e}, \frac{y}{e}))$ $\begin{matrix} = {\begin{matrix} T (x, t) & 0 \leq y < x \leq t, \\ T (x, y) & y \leq t < x < e, \\ T (x, h^{- 1} (h (y) - h (x))) & t < y < x < e . \end{matrix} \end{matrix}$ $T (x, {eI}_{T_{U_{3}}} (\frac{x}{e}, \frac{y}{e})) \leq y$ for all y < x < e, if and only if $\begin{matrix} T (x, t) \leq y for all 0 \leq y < x \leq t, \\ T (x, y) \leq y for all y \leq t < x < e, \\ T (x, h^{- 1} (h (y) - h (x))) \leq y for all t < y < x < e . \end{matrix}$

Hence, by the monotonicity of T and T (x, y) ≤ min(x, y), t-norm T must satisfy the following conditions: $\begin{matrix} T (x, y) = 0 for all (x, y) \in [0, t]^{2}, \\ T (x, h^{- 1} (h (y) - h (x))) \leq y for all t < y < x < e . \end{matrix}$

By Proposition 10, the result holds.□

Remark 4. (i) Notice that the underlying t-norm T_{U
₃} of uninorm U₃ is just left-continuous. However, in [25], Mas et al. focus on the uninorms with continuous underlying t-norm. So, the result of Proposition 10 is different to those in [25].

(ii) Consider the t-norm T defined by $\begin{matrix} T (x, y) = {\begin{matrix} 0 & (x, y) \in [0, t], \\ h^{- 1} (h (x) + h (y)) & (x, y) \in] t, e]^{2}, \\ min (x, y) & otherwise . \end{matrix} \end{matrix}$

By Theorem 5 in [8], T is left-continuous t-norm. T satisfies the condition in Proposition 10. Hence, I_{U
₃} is a T-conditional.

4. Conclusion

In this paper, one construction method of left-continuous uninorm has been proposed based on the representable uninorm. The full characterization of residual implications corresponding to them is our future work.

Footnotes

Acknowledgements

The authors were supported by the National Natural Science Foundation of China (Grant No. 61403220, 61573211).

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