In this paper, we introduce the notion of T-distributivity for any t-norm on a bounded lattice. We determine a relation between the t-norms T and , where is a T-distributive t-norm. Also, for an arbitrary t-norm T, we give a necessary and sufficient condition for to be T-distributive and for T to be -distributive. Moreover, we investigate the relation between the T-distributivity and the concepts of the T-partial order, the divisibility of t-norms. We also determine that the T-distributivity is preserved under the isomorphism. Finally, we construct a family of t-norms which are not distributive over each other with the help of incomparable elements in a bounded lattice.
MSC:03B52, 03E72.
1 Introduction
Triangular norms based on a notion used by Menger [1] were introduced by Schweizer and Sklar [2] in the framework of probabilistic metric spaces, and they play a fundamental role in several branches of mathematics like in fuzzy logics and their applications [3, 4], the games theory [5], the non-additive measures and integral theory [6–8].
A triangular norm (t-norm for short) is a commutative, associative, non-decreasing operation on with a neutral element 1. The four basic t-norms on are the minimum , the product , the Łukasiewicz t-norm and the drastic product given by, respectively, , , and
Recall that for any t-norms and , is called weaker than if for every , .
T-norms are defined on a bounded lattice in a similar way, and then extremal t-norms as well as on L are defined similarly and on . For more details on t-norms on bounded lattices, we refer to [9–17]. Also, the order between t-norms on a bounded lattice is defined similarly.
In the present paper, we introduce the notion of T-distributivity for any t-norms on a bounded lattice . The aim of this study is to discuss the properties of T-distributivity. The paper is organized as follows. Firstly, we recall some basic notions in Section 2. In Section 3, we define the T-distributivity for any t-norm on a bounded lattice. For any two t-norms and , where is -distributive, we show that is weaker than and give an example illustrating the converse of this need not be true. Also, we prove that the only t-norm T, where every t-norm is T-distributive, is the infimum t-norm when the lattice L is especially a chain. If L is not a chain, we give an example illustrating any t-norm need not be . Also, we show that for any t-norm T on a bounded lattice, is T-distributive. Moreover, we show that the T-distributivity is preserved under the isomorphism. For any two t-norms and such that is -distributive, we prove that the divisibility of t-norm requires the divisibility of t-norm . Also, we obtain that for any two t-norms and , where is -distributive, the -partial order implies -partial order. Finally, we construct a family of t-norms which are not distributive over each other with the help of incomparable elements in a bounded lattice.
2 Notations, definitions and a review of previous results
Let be a bounded lattice. A triangular norm T (t-norm for short) is a binary operation on L which is commutative, associative, monotone and has a neutral element 1.
Let
Then is a t-norm on L. Since it holds that for any t-norm T on L, is the smallest t-norm on L.
The largest t-norm on a bounded lattice is given by .
A t-norm T on L is divisible if the following condition holds:
A basic example of a non-divisible t-norm on any bounded lattice (i.e., ) is the weakest t-norm . Trivially, the infimum is divisible: is equivalent to .
are obviously t-norms on L such that is -distributive.
Proposition 1Letbe a bounded lattice andandbe twot-norms onL. Ifis -distributive, thenis weaker than .
Proof Since all t-norms coincide on the boundary of , it is sufficient to show that for all . By the -distributivity of , it is obtained that
Thus, , i.e., is weaker than . □
Remark 1 The converse of Proposition 1 need not be true. Namely, for any two t-norms and , even if is weaker than , may not be -distributive. Now, let us investigate the following example.
Example 2 Consider the product and the Łukasiewicz t-norm . It is clear that . Since
and
is not -distributive.
Corollary 1LetLbe a bounded lattice andandbe any twot-norms onL. If bothis -distributive andis -distributive, then .
Proposition 2LetLbe a bounded chain andbe at-norm onL. For everyt-normT, Tis -distributive if and only if .
Proof :⇒ Let T be an arbitrary t-norm on L such that -distributive. By Proposition 1, it is obvious that for any t-norm T. Thus, .
⇐: Since L is a chain, for any , either or . Suppose that . By using the monotonicity of any t-norm T, it is obtained that for any , . Then
holds. Thus, for any ,
is satisfied, which shows that any t-norm T is -distributive. □
Remark 2 In Proposition 2, if L is not a chain, then the left-hand side of Proposition 2 may not be satisfied. Namely, if L is not a chain, then any t-norm T need not be -distributive. Moreover, even if L is a distributive lattice, any t-norm on L may not be -distributive. Now, let us investigate the following example.
Example 3 Consider the lattice as displayed in Figure 2.
Remark 3 The fact that any t-norm T is -distributive means that T is ∧-distributive.
Theorem 1Letbe a bounded lattice. For anyt-normTonL, isT-distributive.
Proof Let T be an arbitrary t-norm on L. We must show that the equality
holds for every element x, y, z of L with or . Suppose that . If , the desired equality holds since and . Let . Then or . If , since and , the equality holds again. Now, let . Since and , . Then and , whence the equality holds. Thus, is T-distributive for any t-norm T on L. □
IfTis at-norm andis a strictly increasing bijection, then the operationgiven by
is at-norm which is isomorphic toT. Thist-norm is calledφ-transform ofT.
Let and be any two t-norms on and let φ be a strictly increasing bijection from to . Denote the φ-transforms of the t-norms and by and , respectively.
Theorem 2Letandbe anyt-norms onand letφbe a strictly increasing bijection fromto . is -distributive if and only ifis -distributive.
Proof Let be -distributive. We must show that for every with or ,
Since is a strictly increasing bijection, for every element with or , it must be or . By using -distributivity of , we obtain that the equality
holds. Thus, is -distributive.
Conversely, let be -distributive. We will show that for every element with or . Since is the φ-transform of the t-norm , for every , . Since φ is a bijection, it is clear that
(1)
holds. Also, by using (1), it is obtained that
(2)
From (2), it follows
(3)
Also, the similar equalities for t-norm can be written. Since or for every with or , by using -distributivity of , it is obtained that the following equalities:
hold. Thus, is -distributive. □
Proposition 4Letbe a bounded lattice andandbe twot-norms onLsuch thatis -distributive. Ifis divisible, thenis also divisible.
Proof Consider two elements x, y of L with . If , then would be always a divisible t-norm since . Let . Since is divisible, there exists an element of L such that . Then, by using -distributivity of , it is obtained that
Thus, for any elements x, y of L with and , since there exists an element such that , is a divisible t-norm. □
Corollary 2Letbe a bounded lattice andandbe twot-norms onL. Ifis -distributive, then the -partial order implies the -partial order.
Proof Let for any . If , then it would be since for the element . Now, suppose that but . Then there exists an element such that . Since , it must be . Then . Since is -distributive, it is obtained that
for elements with , whence . So, we obtain that . □
Remark 4 For any t-norms and , if is -distributive, then we show that is weaker than in Proposition 1 and the -partial order implies the -partial order in Proposition 2. Although is weaker than , that does not require the -partial order to imply the -partial order. Let us investigate the following example illustrating this case.
Example 4 Consider the drastic product and the function defined as follows:
It is clear that the function is a t-norm such that , but . Indeed.
First, let us show that . Suppose that . Then, for some ,
For , either or . Let . Since , it is obtained that , which contradicts . Then it must be . Since , which is a contradiction. Thus, it is obtained that . On the other hand, since means that there exists an element ℓ of L such that and , we have that . So, it is obtained that .
Now, let us construct a family of t-norms which are not distributive over each other with the help of incomparable elements in a bounded lattice.
Theorem 3LetLbe a complete lattice andbe a nonempty family of nonempty sets consisting of the elements inLwhich are all incomparable to each other with respect to the order onL. If for any element , is comparable to every element inL, then the familydefined by
is a family oft-norms which are not distributive over each other. Namely, for any , neitheris -distributive noris -distributive.
Proof Firstly, let us show that for every , each function is a t-norm.
(i)
Since , for every element , . Then it follows from , that is, the boundary condition is satisfied.
(ii)
It can be easily shown that the commutativity holds.
(iii)
Considering the monotonicity, suppose that for . Let be arbitrary. Then there are the following possible conditions for the couples , .
Let . Then we get clearly the equality
Let and . Then . Clearly, and . Since and , we obtain . By , we get , whence .
Let and . Then it is clear that . In this case,
By and , it is clear that . Since is comparable to every element in L, either or . If , it would be from , a contradiction. Thus, it must be . Since , . Thus, the inequality
holds.
Let . By , we have that
So, the monotonicity holds.
(iv)
Now let us show that for every , the equality holds.
Let . Then
and
whence the equality holds.
If and , then it must be . Here, there are two choices for z: either or .
Let . Then . By the inequality , it is clear that . Since , the following inequalities:
hold, that is, . Thus, we have that
and
So, the equality holds again.
Let . Then there exists at least an element v in such that v is comparable to the element z; i.e., either or . Let . Since , it is clear that . Also, from the inequalities and , it follows , i.e., it is obtained that . Thus,
and
Thus, the equality is satisfied.
Now, suppose that . If , it would be , which is a contradiction. Thus, either or z and u are not comparable. If , then it must be since is comparable to every element in L and . Thus, we have that
and
whence the equality holds.
Let z and u be not comparable. Since is comparable to every element in L, either or . If , it would be , a contradiction. Then it must be . By , it is obtained that . Then the equalities
and
In this case, the equality is satisfied.
Similarly, one can show that the equality holds when and .
Now, let us investigate the last condition. If , then it is obvious that
and
whence the equality holds.
Consequently, we prove that is a family of t-norms on L. Now, we will show that for every , and are not distributive t-norms over each other.
Suppose that is -distributive. By Proposition 1, it must be , that is, for every , . Since m and n are not comparable, it is clear that and . Then n must not be in . Thus,
On the other hand, since ,
Then we have that . Otherwise, we obtain that , which is a contradiction. So, we have that contradicts . Thus, is not -distributive. Similarly, it can be shown that is not -distributive. So, the family given above is a family of t-norms which are not distributive over each other. □
To explain how the family in Theorem 3 can be determined, let us investigate the following example.
Example 5 Let be a bounded lattice as shown in Figure 3.
For the family of , there are two choices: one of them must be and the other must be . Then, by Theorem 3, for every and , the following functions:
and
are two families of t-norms.
Remark 5 In Theorem 3, if the condition that is comparable to every element in L is canceled, then for any element , is not a t-norm. The following is an example showing that is not a t-norm when the condition that for any element , is comparable to every element in L is canceled.
Example 6 Let be a bounded lattice as displayed in Figure 4.
From Figure 4, it is clear that is not comparable to b. However, for the set , the function defined by
does not satisfy the associativity since and . So, is not a t-norm.
4 Conclusions
In this paper, we introduced the notion of T-distributivity for any t-norm on a bounded lattice and discussed some properties of T-distributivity. We determined a necessary and sufficient condition for to be T-distributive and for T to be -distributive. We obtained that T-distributivity is preserved under the isomorphism. We proved that the divisibility of t-norm requires the divisibility of t-norm for any two t-norms and where is -distributive. Also, we constructed a family of t-norms which are not distributive over each other with the help of incomparable elements in a bounded lattice.
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