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On the property of T-distributivity

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.

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)=xy, T L (x,y)=max(0,x+y1) 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)=xy.

Definition 2 [18]

A t-norm T on L is divisible if the following condition holds:

x,yL with xy, there is a zL such thatx=T(y,z).

A basic example of a non-divisible t-norm on any bounded lattice (i.e., cardL>2) is the weakest t-norm T D . Trivially, the infimum T is divisible: xy is equivalent to xy=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)=xfor 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,τI} 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,zL 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,zL{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,zL, either yz or zy. Suppose that yz. By using the monotonicity of any t-norm T, it is obtained that for any xL, T(x,y)T(x,z). Then

T(x,y)=T(x,y)T(x,z)

holds. Thus, for any x,y,zL,

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},)

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 y1 or z1. Suppose that z1. 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 x1. Then y=1 or y1. 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 y1. Since T(y,z)y1 and y1, 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 y1 or z1,

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 y1 or z1, 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 y1 or z1. 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 y1 or z1, 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 xy. If x=y, then T 2 would be always a divisible t-norm since T 2 (y,1)=y=x. Let xy. Since T 1 is divisible, there exists an element 1z 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 xy and xy, 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,bL. If a=b, then it would be a T 2 b since T 2 (b,1)=b=a for the element 1L. Now, suppose that a T 1 b but ab. Then there exists an element L such that T 1 (b,)=a. Since ab, 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,,1L 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 |αI} 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 x1, for every element xL, 1 S α . Then it follows T u (x,1)=x1=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 xy for x,yL. Let zL 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)=yz. Since x[inf{u μ i | μ i S α },u] and xy, 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 }yz, 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)=xzand T u (y,z)=inf{u μ i | μ i S α }.

By xy and yu, it is clear that xu. Since inf{u μ i | μ i S α } is comparable to every element in L, either xinf{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 xu, a contradiction. Thus, it must be xinf{u μ i | μ i S α }. Since z[inf{u μ i | μ i S α },u], xz=x. Thus, the inequality

T u (x,z)=xz=xinf{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 xy, we have that

    T u (x,z)=xzyz= T u (y,z).

So, the monotonicity holds.

  1. (iv)

    Now let us show that for every x,y,zL, 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 }uz. Since inf{u μ i | μ i S α }yu, the following inequalities:

inf{u μ i | μ i S α }=inf{u μ i | μ i S α }zyzyu

hold, that is, yz[inf{u μ i | μ i S α },u]. Thus, we have that

T u ( x , T u ( y , z ) ) = T u (x,yz)=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 zv or vz. Let vz. Since u,v S α , it is clear that inf{u μ i | μ i S α }uvuzu. Also, from the inequalities inf{u μ i | μ i S α }y and inf{u μ i | μ i S α }vz, it follows inf{u μ i | μ i S α }yzyu, i.e., it is obtained that yz[inf{u μ i | μ i S α },u]. Thus,

T u ( x , T u ( y , z ) ) = T u (x,yz)=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 zv. If uz, it would be uv, 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<yz<y<u, it is obtained that yz[inf{u μ i | μ i S α },u]. Then the equalities

T u ( x , T u ( y , z ) ) = T u (x,yz)=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,yz)=x(yz)

and

T u ( T u ( x , y ) , z ) = T u (xy,z)=(xy)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,yL, T m (x,y) T n (x,y). Since m and n are not comparable, it is clear that nm and mn. Then n must not be in [inf{m μ i | μ i S α },m]. Thus,

T m (n,n)=nn=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 nm, 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.

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

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Kesicioğlu, M.N. On the property of T-distributivity. Fixed Point Theory Appl 2013, 32 (2013). https://doi.org/10.1186/1687-1812-2013-32

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