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On the property of T-distributivity

Fixed Point Theory and Applications20132013:32

https://doi.org/10.1186/1687-1812-2013-32

Received: 4 December 2012

Accepted: 31 January 2013

Published: 15 February 2013

Abstract

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 T , where T is a T-distributive t-norm. Also, for an arbitrary t-norm T, we give a necessary and sufficient condition for T D to be T-distributive and for T to be T -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.

Keywords

triangular normbounded latticeT-partial orderdivisibilitydistributivity

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 [68].

A triangular norm (t-norm for short) T : [ 0 , 1 ] 2 [ 0 , 1 ] is a commutative, associative, non-decreasing operation on [ 0 , 1 ] with a neutral element 1. The four basic t-norms on [ 0 , 1 ] are the minimum T M , the product T P , the Łukasiewicz t-norm T L and the drastic product T D given by, respectively, T M ( x , y ) = min ( x , y ) , T P ( x , y ) = x y , T L ( x , y ) = max ( 0 , x + y 1 ) and
T D ( x , y ) = { x , if  y = 1 , y , if  x = 1 , 0 , otherwise .

Recall that for any t-norms T 1 and T 2 , T 1 is called weaker than T 2 if for every ( x , y ) [ 0 , 1 ] 2 , T 1 ( x , y ) T 2 ( x , y ) .

T-norms are defined on a bounded lattice ( L , , 0 , 1 ) in a similar way, and then extremal t-norms T D as well as T on L are defined similarly T D and T M on [ 0 , 1 ] . For more details on t-norms on bounded lattices, we refer to [917]. 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 ( L , , 0 , 1 ) . 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 T 1 and T 2 , where T 1 is T 2 -distributive, we show that T 1 is weaker than T 2 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 T 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 T . Also, we show that for any t-norm T on a bounded lattice, T D is T-distributive. Moreover, we show that the T-distributivity is preserved under the isomorphism. For any two t-norms T 1 and T 2 such that T 1 is T 2 -distributive, we prove that the divisibility of t-norm T 1 requires the divisibility of t-norm T 2 . Also, we obtain that for any two t-norms T 1 and T 2 , where T 1 is T 2 -distributive, the T 1 -partial order implies T 2 -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

Definition 1 [14]

Let ( L , , 0 , 1 ) 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
T D ( x , y ) = { x , if  y = 1 , y , if  x = 1 , 0 , otherwise .

Then T D is a t-norm on L. Since it holds that T D T for any t-norm T on L, T D is the smallest t-norm on L.

The largest t-norm on a bounded lattice ( L , , 0 , 1 ) is given by T ( x , y ) = x y .

Definition 2 [18]

A t-norm T on L is divisible if the following condition holds:
x , y L  with  x y ,  there is a  z L  such that x = T ( y , z ) .

A basic example of a non-divisible t-norm on any bounded lattice (i.e., card L > 2 ) is the weakest t-norm T D . Trivially, the infimum T is divisible: x y is equivalent to x y = x .

Definition 3 [12]

Let L be a bounded lattice, T be a t-norm on L. The order defined as follows is called a T-partial order (triangular order) for a t-norm T.
x T y : T ( , y ) = x for some  L .

Definition 4 [19]

  1. (i)
    A t-norm T on a lattice L is called -distributive if
    T ( a , b 1 b 2 ) = T ( a , b 1 ) T ( a , b 2 )

    for every a , b 1 , b 2 L .

     
  2. (ii)
    A t-norm T on a complete lattice ( L , , 0 , 1 ) is called infinitely -distributive if
    T ( a , I b τ ) = I T ( a , b τ )

    for every subset { a , b τ L , τ I } of L.

     

3 T-distributivity

Definition 5 Let ( L , , 0 , 1 ) be a bounded lattice and T 1 and T 2 be two t-norms on L. For every x , y , z L such that at least one of the elements y, z is not 1, if the condition
T 1 ( x , T 2 ( y , z ) ) = T 2 ( T 1 ( x , y ) , T 1 ( x , z ) )

is satisfied, then T 1 is called T 2 -distributive or we say that T 1 is distributive over T 2 .

Example 1 Let ( L = { 0 , a , b , c , 1 } , , 0 , 1 ) be a bounded lattice whose lattice diagram is displayed in Figure 1.
Figure 1
Figure 1

( L = { 0 , a , b , c , 1 } , , 0 , 1 ) .

The functions T 1 and T 2 on the lattice L defined by
T 1 ( x , y ) = { 0 , if  x = a , y = a , b , if  x = c , y = c , x y , otherwise
and
T 2 ( x , y ) = { b , if  x = c , y = c , x y , otherwise

are obviously t-norms on L such that T 1 is T 2 -distributive.

Proposition 1 Let ( L , , 0 , 1 ) be a bounded lattice and T 1 and T 2 be two t-norms on L. If T 1 is T 2 -distributive, then T 1 is weaker than T 2 .

Proof Since all t-norms coincide on the boundary of L 2 , it is sufficient to show that T 1 T 2 for all x , y , z L { 0 , 1 } . By the T 2 -distributivity of T 1 , it is obtained that
T 1 ( x , y ) = T 1 ( T 2 ( x , 1 ) , y ) = T 2 ( T 1 ( x , y ) , T 1 ( 1 , y ) ) = T 2 ( T 1 ( x , y ) , y ) T 2 ( x , y ) .

Thus, T 1 T 2 , i.e., T 1 is weaker than T 2 . □

Remark 1 The converse of Proposition 1 need not be true. Namely, for any two t-norms T 1 and T 2 , even if T 1 is weaker than T 2 , T 1 may not be T 2 -distributive. Now, let us investigate the following example.

Example 2 Consider the product T P and the Łukasiewicz t-norm T L . It is clear that T L < T P . Since
T L ( 3 4 , T P ( 5 8 , 1 2 ) ) = T L ( 3 4 , 5 16 ) = 1 16
and
T P ( T L ( 3 4 , 5 8 ) , T L ( 3 4 , 1 2 ) ) = T P ( 3 8 , 1 4 ) = 3 32

T L is not T P -distributive.

Corollary 1 Let L be a bounded lattice and T 1 and T 2 be any two t-norms on L. If both T 1 is T 2 -distributive and T 2 is T 1 -distributive, then T 1 = T 2 .

Proposition 2 Let L be a bounded chain and T be a t-norm on L. For every t-norm T, T is T -distributive if and only if T = T .

Proof : Let T be an arbitrary t-norm on L such that T -distributive. By Proposition 1, it is obvious that T T for any t-norm T. Thus, T = T .

: Since L is a chain, for any y , z L , either y z or z y . Suppose that y z . By using the monotonicity of any t-norm T, it is obtained that for any x L , T ( x , y ) T ( x , z ) . Then
T ( x , y ) = T ( x , y ) T ( x , z )
holds. Thus, for any x , y , z L ,
T ( x , T ( y , z ) ) = T ( x , y ) = T ( x , y ) T ( x , z ) = T ( T ( x , y ) , T ( x , z ) )

is satisfied, which shows that any t-norm T is T -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 T -distributive. Moreover, even if L is a distributive lattice, any t-norm on L may not be T -distributive. Now, let us investigate the following example.

Example 3 Consider the lattice ( L = { 0 , x , y , z , a , 1 } , ) as displayed in Figure 2.
Figure 2
Figure 2

( L = { 0 , x , y , z , a , 1 } , ) .

Obviously, L is a distributive lattice. Define the function T on L as shown in Table 1.
Table 1

T -norm on the lattice ( L = { 0 , x , y , z , a , 1 } , )

T

0

x

y

z

a

1

0

0

0

0

0

0

0

x

0

0

0

0

0

x

y

0

0

y

0

y

y

z

0

0

0

z

z

z

a

0

0

y

z

a

a

1

0

x

y

z

a

1

One can easily check that T is a t-norm. Since
T ( a , T ( y , z ) ) = T ( a , x ) = 0
and
T ( T ( a , y ) , T ( a , z ) ) = T ( y , z ) = x ,

T is not T -distributive.

Remark 3 The fact that any t-norm T is T -distributive means that T is -distributive.

Theorem 1 Let ( L , , 0 , 1 ) be a bounded lattice. For any t-norm T on L, T D is T-distributive.

Proof Let T be an arbitrary t-norm on L. We must show that the equality
T D ( x , T ( y , z ) ) = T ( T D ( x , y ) , T D ( x , z ) )

holds for every element x, y, z of L with y 1 or z 1 . Suppose that z 1 . If x = 1 , the desired equality holds since T D ( x , T ( y , z ) ) = T ( y , z ) and T ( T D ( x , y ) , T D ( x , z ) ) = T ( y , z ) . Let x 1 . Then y = 1 or y 1 . If y = 1 , since T D ( x , T ( y , z ) ) = T D ( x , z ) = 0 and T ( T D ( x , y ) , T D ( x , z ) ) = T ( x , 0 ) = 0 , the equality holds again. Now, let y 1 . Since T ( y , z ) y 1 and y 1 , T ( y , z ) 1 . Then T D ( x , T ( y , z ) ) = 0 and T ( T D ( x , y ) , T D ( x , z ) ) = T ( 0 , 0 ) = 0 , whence the equality holds. Thus, T D is T-distributive for any t-norm T on L. □

Proposition 3 [20]

If T is a t-norm and φ : [ 0 , 1 ] [ 0 , 1 ] is a strictly increasing bijection, then the operation T φ : [ 0 , 1 ] 2 [ 0 , 1 ] given by
T φ ( x , y ) = φ 1 ( T ( φ ( x ) , φ ( y ) ) )

is a t-norm which is isomorphic to T. This t-norm is called φ-transform of T.

Let T 1 and T 2 be any two t-norms on [ 0 , 1 ] and let φ be a strictly increasing bijection from [ 0 , 1 ] to [ 0 , 1 ] . Denote the φ-transforms of the t-norms T 1 and T 2 by T φ 1 and T φ 2 , respectively.

Theorem 2 Let T 1 and T 2 be any t-norms on [ 0 , 1 ] and let φ be a strictly increasing bijection from [ 0 , 1 ] to [ 0 , 1 ] . T 1 is T 2 -distributive if and only if T φ 1 is T φ 2 -distributive.

Proof Let T 1 be T 2 -distributive. We must show that for every x , y , z [ 0 , 1 ] with y 1 or z 1 ,
T φ 1 ( x , T φ 2 ( y , z ) ) = T φ 2 ( T φ 1 ( x , y ) , T φ 1 ( x , z ) ) .
Since φ : [ 0 , 1 ] [ 0 , 1 ] is a strictly increasing bijection, for every element y , z [ 0 , 1 ] with y 1 or z 1 , it must be φ ( y ) 1 or φ ( z ) 1 . By using T 2 -distributivity of T 1 , we obtain that the equality
T φ 1 ( x , T φ 2 ( y , z ) ) = φ 1 ( T 1 ( φ ( x ) , φ ( T φ 2 ( y , z ) ) ) ) = φ 1 ( T 1 ( φ ( x ) , φ ( φ 1 ( T 2 ( φ ( y ) , φ ( z ) ) ) ) ) ) = φ 1 ( T 1 ( φ ( x ) , T 2 ( φ ( y ) , φ ( z ) ) ) ) = φ 1 ( T 2 ( T 1 ( φ ( x ) , φ ( y ) ) , T 1 ( φ ( x ) , φ ( z ) ) ) ) = φ 1 ( T 2 ( ( φ φ 1 ) T 1 ( φ ( x ) , φ ( y ) ) , ( φ φ 1 ) T 1 ( φ ( x ) , φ ( z ) ) ) ) = φ 1 ( T 2 ( φ ( φ 1 ( T 1 ( φ ( x ) , φ ( y ) ) ) ) , φ ( φ 1 ( T 1 ( φ ( x ) , φ ( z ) ) ) ) ) ) = φ 1 ( T 2 ( φ ( T φ 1 ( x , y ) ) , φ ( T φ 1 ( x , z ) ) ) ) = T φ 2 ( T φ 1 ( x , y ) , T φ 1 ( x , z ) )

holds. Thus, T φ 1 is T φ 2 -distributive.

Conversely, let T φ 1 be T φ 2 -distributive. We will show that T 1 ( x , T 2 ( y , z ) ) = T 2 ( T 1 ( x , y ) , T 1 ( x , z ) ) for every element x , y , z [ 0 , 1 ] with y 1 or z 1 . Since T φ 1 is the φ-transform of the t-norm T 1 , for every x , y [ 0 , 1 ] , T φ 1 ( x , y ) = φ 1 ( T 1 ( φ ( x ) , φ ( y ) ) ) . Since φ is a bijection, it is clear that
T 1 ( φ ( x ) , φ ( y ) ) = φ ( T φ 1 ( x , y ) )
(1)
holds. Also, by using (1), it is obtained that
T 1 ( x , y ) = T 1 ( φ ( φ 1 ( x ) ) , φ ( φ 1 ( y ) ) ) = φ ( T φ 1 ( φ 1 ( x ) , φ 1 ( y ) ) )
(2)
From (2), it follows
T φ 1 ( φ 1 ( x ) , φ 1 ( y ) ) = φ 1 ( T 1 ( x , y ) ) .
(3)
Also, the similar equalities for t-norm T 2 can be written. Since φ 1 ( y ) 1 or φ 1 ( z ) 1 for every y , z [ 0 , 1 ] with y 1 or z 1 , by using T φ 2 -distributivity of T φ 1 , it is obtained that the following equalities:
T 1 ( x , T 2 ( y , z ) ) = ( 2 ) T 1 ( x , φ ( T φ 2 ( φ 1 ( y ) , φ 1 ( z ) ) ) ) = ( 2 ) φ ( T φ 1 ( φ 1 ( x ) , φ 1 ( φ ( T φ 2 ( φ 1 ( y ) , φ 1 ( z ) ) ) ) ) ) = φ ( T φ 1 ( φ 1 ( x ) , T φ 2 ( φ 1 ( y ) , φ 1 ( z ) ) ) ) = φ ( T φ 2 ( T φ 1 ( φ 1 ( x ) , φ 1 ( y ) ) , T φ 1 ( φ 1 ( x ) , φ 1 ( z ) ) ) ) = ( 3 ) φ ( T φ 2 ( φ 1 ( T 1 ( x , y ) ) , φ 1 ( T 1 ( x , z ) ) ) ) = ( 2 ) φ ( φ 1 ( T 2 ( T 1 ( x , y ) , T 1 ( x , z ) ) ) ) = T 2 ( T 1 ( x , y ) , T 1 ( x , z ) )

hold. Thus, T 1 is T 2 -distributive. □

Proposition 4 Let ( L , , 0 , 1 ) be a bounded lattice and T 1 and T 2 be two t-norms on L such that T 1 is T 2 -distributive. If T 1 is divisible, then T 2 is also divisible.

Proof Consider two elements x, y of L with x y . If x = y , then T 2 would be always a divisible t-norm since T 2 ( y , 1 ) = y = x . Let x y . Since T 1 is divisible, there exists an element 1 z of L such that T 1 ( y , z ) = x . Then, by using T 2 -distributivity of T 1 , it is obtained that
x = T 1 ( y , z ) = T 1 ( y , T 2 ( z , 1 ) ) = T 2 ( T 1 ( y , z ) , T 1 ( y , 1 ) ) = T 2 ( T 1 ( y , z ) , y ) .

Thus, for any elements x, y of L with x y and x y , since there exists an element T 1 ( y , z ) L such that x = T 2 ( T 1 ( y , z ) , y ) , T 2 is a divisible t-norm. □

Corollary 2 Let ( L , , 0 , 1 ) be a bounded lattice and T 1 and T 2 be two t-norms on L. If T 1 is T 2 -distributive, then the T 1 -partial order implies the T 2 -partial order.

Proof Let a T 1 b for any a , b L . If a = b , then it would be a T 2 b since T 2 ( b , 1 ) = b = a for the element 1 L . Now, suppose that a T 1 b but a b . Then there exists an element L such that T 1 ( b , ) = a . Since a b , it must be 1 . Then T 1 ( b , T 2 ( , 1 ) ) = T 1 ( b , ) = a . Since T 1 is T 2 -distributive, it is obtained that
a = T 1 ( b , T 2 ( , 1 ) ) = T 2 ( T 1 ( b , ) , T 1 ( b , 1 ) ) = T 2 ( a , b ) .

for elements b , , 1 L with 1 , whence a T 2 b . So, we obtain that T 1 T 2 . □

Remark 4 For any t-norms T 1 and T 2 , if T 1 is T 2 -distributive, then we show that T 1 is weaker than T 2 in Proposition 1 and the T 1 -partial order implies the T 2 -partial order in Proposition 2. Although T 1 is weaker than T 2 , that does not require the T 1 -partial order to imply the T 2 -partial order. Let us investigate the following example illustrating this case.

Example 4 Consider the drastic product T P and the function defined as follows:
T ( x , y ) = { x y , if  ( x , y ) [ 0 , 1 2 ] 2 , min ( x , y ) , otherwise .

It is clear that the function T is a t-norm such that T P T , but T P T . Indeed.

First, let us show that 3 8 T 1 2 . Suppose that 3 8 T 1 2 . Then, for some [ 0 , 1 ] ,
T ( , 1 2 ) = 3 8 .

For [ 0 , 1 ] , either 1 2 or > 1 2 . Let 1 2 . Since 3 8 = T ( , 1 2 ) = 1 2 , it is obtained that = 3 4 , which contradicts 1 2 . Then it must be > 1 2 . Since 3 8 = T ( , 1 2 ) = min ( , 1 2 ) = 1 2 , which is a contradiction. Thus, it is obtained that 3 8 T 1 2 . On the other hand, since x T P y means that there exists an element of L such that T p ( , y ) = y = x and T P ( 1 2 , 3 4 ) = 3 8 , we have that 3 8 T P 1 2 . So, it is obtained that T P T .

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 3 Let L be a complete lattice and { S α | α I } be a nonempty family of nonempty sets consisting of the elements in L which are all incomparable to each other with respect to the order on L. If for any element u S α , inf { u μ i | μ i S α } is comparable to every element in L, then the family ( T u ) u S α defined by
T u ( x , y ) = { inf { u μ i | μ i S α } , if ( x , y ) [ inf { u μ i | μ i S α } , u ] 2 , x y , otherwise

is a family of t-norms which are not distributive over each other. Namely, for any , q S α , neither T is T q -distributive nor T q is T -distributive.

Proof Firstly, let us show that for every u S α , each function T u is a t-norm.
  1. (i)

    Since x 1 , for every element x L , 1 S α . Then it follows T u ( x , 1 ) = x 1 = x from ( x , 1 ) [ inf { u μ i | μ i S α } , u ] 2 , that is, the boundary condition is satisfied.

     
  2. (ii)

    It can be easily shown that the commutativity holds.

     
  3. (iii)

    Considering the monotonicity, suppose that x y for x , y L . Let z L be arbitrary. Then there are the following possible conditions for the couples ( x , z ) , ( y , z ) .

     
  • Let ( x , z ) , ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 . Then we get clearly the equality
    T u ( x , z ) = inf { u μ i | μ i S α } = T u ( y , z ) .
  • Let ( x , z ) [ inf { u μ i | μ i S α } , u ] 2 and ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 . Then y [ inf { u μ i | μ i S α } , u ] . Clearly, T u ( x , z ) = inf { u μ i | μ i S α } and T u ( y , z ) = y z . Since x [ inf { u μ i | μ i S α } , u ] and x y , we obtain inf { u μ i | μ i S α } y . By inf { u μ i | μ i S α } z , we get inf { u μ i | μ i S α } y z , whence T u ( x , z ) T u ( y , z ) .

  • Let ( x , z ) [ inf { u μ i | μ i S α } , u ] 2 and ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 . Then it is clear that x [ inf { u μ i | μ i S α } , u ] . In this case,
    T u ( x , z ) = x z and T u ( y , z ) = inf { u μ i | μ i S α } .
By x y and y u , it is clear that x u . Since inf { u μ i | μ i S α } is comparable to every element in L, either x inf { u μ i | μ i S α } or inf { u μ i | μ i S α } x . If inf { u μ i | μ i S α } x , it would be x [ inf { u μ i | μ i S α } , u ] from x u , a contradiction. Thus, it must be x inf { u μ i | μ i S α } . Since z [ inf { u μ i | μ i S α } , u ] , x z = x . Thus, the inequality
T u ( x , z ) = x z = x inf { u μ i | μ i S α } = T u ( y , z )
holds.
  • Let ( x , z ) , ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 . By x y , we have that
    T u ( x , z ) = x z y z = T u ( y , z ) .
So, the monotonicity holds.
  1. (iv)

    Now let us show that for every x , y , z L , the equality T u ( x , T u ( y , z ) ) = T u ( T u ( x , y ) , z ) holds.

     
  • Let ( x , y ) , ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 . Then
    T u ( x , T u ( y , z ) ) = inf { u μ i | μ i S α }
and
T u ( T u ( x , y ) , z ) = inf { u μ i | μ i S α } ,
whence the equality holds.
  • If ( x , y ) [ inf { u μ i | μ i S α } , u ] 2 and ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 , then it must be z [ inf { u μ i | μ i S α } , u ] . Here, there are two choices for z: either z S α or z S α .

Let z S α . Then inf { u μ i | μ i S α } z . By the inequality inf { u μ i | μ i S α } u , it is clear that inf { u μ i | μ i S α } u z . Since inf { u μ i | μ i S α } y u , the following inequalities:
inf { u μ i | μ i S α } = inf { u μ i | μ i S α } z y z y u
hold, that is, y z [ inf { u μ i | μ i S α } , u ] . Thus, we have that
T u ( x , T u ( y , z ) ) = T u ( x , y z ) = inf { u μ i | μ i S α }
and
T u ( T u ( x , y ) , z ) = T u ( inf { u μ i | μ i S α } , z ) = inf { u μ i | μ i S α } z = inf { u μ i | μ i S α } .

So, the equality holds again.

Let z S α . Then there exists at least an element v in S α such that v is comparable to the element z; i.e., either z v or v z . Let v z . Since u , v S α , it is clear that inf { u μ i | μ i S α } u v u z u . Also, from the inequalities inf { u μ i | μ i S α } y and inf { u μ i | μ i S α } v z , it follows inf { u μ i | μ i S α } y z y u , i.e., it is obtained that y z [ inf { u μ i | μ i S α } , u ] . Thus,
T u ( x , T u ( y , z ) ) = T u ( x , y z ) = inf { u μ i | μ i S α }
and
T u ( T u ( x , y ) , z ) = T u ( inf { u μ i | μ i S α } , z ) = inf { u μ i | μ i S α } z = inf { u μ i | μ i S α } .

Thus, the equality is satisfied.

Now, suppose that z v . If u z , it would be u v , which is a contradiction. Thus, either z < u or z and u are not comparable. If z < u , then it must be z < inf { u μ i | μ i S α } since inf { u μ i | μ i S α } is comparable to every element in L and z [ inf { u μ i | μ i S α } , u ] . Thus, we have that
T u ( T u ( x , y ) , z ) = T u ( inf { u μ i | μ i S α } , z ) = inf { u μ i | μ i S α } z = z
and
T u ( x , T u ( y , z ) ) = T u ( x , y z ) = T u ( x , z ) = x z = z ,

whence the equality holds.

Let z and u be not comparable. Since inf { u μ i | μ i S α } is comparable to every element in L, either inf { u μ i | μ i S α } < z or inf { u μ i | μ i S α } > z . If inf { u μ i | μ i S α } > z , it would be z < u , a contradiction. Then it must be inf { u μ i | μ i S α } < z . By inf { u μ i | μ i S α } = inf { u μ i | μ i S α } y < y z < y < u , it is obtained that y z [ inf { u μ i | μ i S α } , u ] . Then the equalities
T u ( x , T u ( y , z ) ) = T u ( x , y z ) = inf { u μ i | μ i S α }
and
T u ( T u ( x , y ) , z ) = T u ( inf { u μ i | μ i S α } , z ) = inf { u μ i | μ i S α } z = inf { u μ i | μ i S α } .

In this case, the equality is satisfied.

Similarly, one can show that the equality T u ( x , T u ( y , z ) ) = T u ( T u ( x , y ) , z ) holds when ( x , y ) [ inf { u μ i | μ i S α } , u ] 2 and ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 .
  • Now, let us investigate the last condition. If ( x , y ) , ( y , z ) [ inf { u μ i | μ i S α } , u ] 2 , then it is obvious that
    T u ( x , T u ( y , z ) ) = T u ( x , y z ) = x ( y z )
and
T u ( T u ( x , y ) , z ) = T u ( x y , z ) = ( x y ) z ,

whence the equality holds.

Consequently, we prove that ( T u ) u S α is a family of t-norms on L. Now, we will show that for every m , n S α , T m and T n are not distributive t-norms over each other.

Suppose that T m is T n -distributive. By Proposition 1, it must be T m T n , that is, for every x , y L , T m ( x , y ) T n ( x , y ) . Since m and n are not comparable, it is clear that n m and m n . Then n must not be in [ inf { m μ i | μ i S α } , m ] . Thus,
T m ( n , n ) = n n = n .
On the other hand, since n [ inf { n μ i | μ i S α } , n ] ,
T n ( n , n ) = inf { n μ i | μ i S α } .

Then we have that T n ( n , n ) T m ( n , n ) . Otherwise, we obtain that n m , which is a contradiction. So, we have that T n ( n , n ) < T m ( n , n ) contradicts T m T n . Thus, T m is not T n -distributive. Similarly, it can be shown that T n is not T m -distributive. So, the family given above is a family of t-norms which are not distributive over each other. □

To explain how the family ( S α ) α I in Theorem 3 can be determined, let us investigate the following example.

Example 5 Let ( L = { 0 , a , b , c , d , e , 1 } , , 0 , 1 ) be a bounded lattice as shown in Figure 3.
Figure 3
Figure 3

( L = { 0 , a , b , c , d , e , 1 } , , 0 , 1 ) .

For the family of ( S α ) α I , there are two choices: one of them must be S α 1 = { c , d , e } and the other must be S α 2 = { b , e } . Then, by Theorem 3, for every u S α 1 and v S α 2 , the following functions:
T u ( x , y ) = { a , if  ( x , y ) [ a , u ] 2 , x y , otherwise
and
T v ( x , y ) = { a , if  ( x , y ) [ a , v ] 2 , x y , otherwise

are two families of t-norms.

Remark 5 In Theorem 3, if the condition that inf { u μ i | μ i S α } is comparable to every element in L is canceled, then for any element u S α , T u is not a t-norm. The following is an example showing that T u is not a t-norm when the condition that for any element u S α , inf { u μ i | μ i S α } is comparable to every element in L is canceled.

Example 6 Let ( L = { 0 , a , b , c , d , e , f , g , h , j , 1 } , , 0 , 1 ) be a bounded lattice as displayed in Figure 4.
Figure 4
Figure 4

( L = { 0 , a , b , c , d , e , f , g , h , j , 1 } , , 0 , 1 ) .

From Figure 4, it is clear that inf { j , e , f } = a is not comparable to b. However, for the set S = { j , e , f } , the function defined by
T e ( x , y ) = { a , if  ( x , y ) [ a , e ] 2 , x y , otherwise

does not satisfy the associativity since T e ( T e ( c , d ) , b ) = 0 and T e ( c , T e ( d , b ) ) = b . So, T e 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 T D to be T-distributive and for T to be T -distributive. We obtained that T-distributivity is preserved under the isomorphism. We proved that the divisibility of t-norm T 1 requires the divisibility of t-norm T 2 for any two t-norms T 1 and T 2 where T 1 is T 2 -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.

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Authors’ Affiliations

(1)
Department of Mathematics, Recep Tayyip Erdoğan University, Rize, Turkey

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© Kesicioğlu; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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