# 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 ${T}^{\prime }$, where ${T}^{\prime }$ 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 ${T}_{\wedge }$-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  were introduced by Schweizer and Sklar  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 , the non-additive measures and integral theory .

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

Recall that for any t-norms ${T}_{1}$ and ${T}_{2}$, ${T}_{1}$ is called weaker than ${T}_{2}$ if for every $\left(x,y\right)\in {\left[0,1\right]}^{2}$, ${T}_{1}\left(x,y\right)\le {T}_{2}\left(x,y\right)$.

T-norms are defined on a bounded lattice $\left(L,\le ,0,1\right)$ in a similar way, and then extremal t-norms ${T}_{D}$ as well as ${T}_{\wedge }$ on L are defined similarly ${T}_{D}$ and ${T}_{M}$ on $\left[0,1\right]$. For more details on t-norms on bounded lattices, we refer to . 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 $\left(L,\le ,0,1\right)$. 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 ${T}_{\wedge }$ 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 ${T}_{\wedge }$. 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 

Let $\left(L,\le ,0,1\right)$ 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 ${T}_{D}$ is a t-norm on L. Since it holds that ${T}_{D}\le T$ for any t-norm T on L, ${T}_{D}$ is the smallest t-norm on L.

The largest t-norm on a bounded lattice $\left(L,\le ,0,1\right)$ is given by ${T}_{\wedge }\left(x,y\right)=x\wedge y$.

Definition 2 

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., $cardL>2$) is the weakest t-norm ${T}_{D}$. Trivially, the infimum ${T}_{\wedge }$ is divisible: $x\le y$ is equivalent to $x\wedge y=x$.

Definition 3 

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.

Definition 4 

1. (i)

A t-norm T on a lattice L is called -distributive if

$T\left(a,{b}_{1}\wedge {b}_{2}\right)=T\left(a,{b}_{1}\right)\wedge T\left(a,{b}_{2}\right)$

for every $a,{b}_{1},{b}_{2}\in L$.

2. (ii)

A t-norm T on a complete lattice $\left(L,\le ,0,1\right)$ is called infinitely -distributive if

$T\left(a,{\wedge }_{I}{b}_{\tau }\right)={\wedge }_{I}T\left(a,{b}_{\tau }\right)$

for every subset $\left\{a,{b}_{\tau }\in L,\tau \in I\right\}$ of L.

## 3 T-distributivity

Definition 5 Let $\left(L,\le ,0,1\right)$ be a bounded lattice and ${T}_{1}$ and ${T}_{2}$ be two t-norms on L. For every $x,y,z\in L$ such that at least one of the elements y, z is not 1, if the condition

${T}_{1}\left(x,{T}_{2}\left(y,z\right)\right)={T}_{2}\left({T}_{1}\left(x,y\right),{T}_{1}\left(x,z\right)\right)$

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

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

The functions ${T}_{1}$ and ${T}_{2}$ on the lattice L defined by

and

are obviously t-norms on L such that ${T}_{1}$ is ${T}_{2}$-distributive.

Proposition 1 Let $\left(L,\le ,0,1\right)$ 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 ${T}_{1}\le {T}_{2}$ for all $x,y,z\in L\mathrm{\setminus }\left\{0,1\right\}$. By the ${T}_{2}$-distributivity of ${T}_{1}$, it is obtained that

${T}_{1}\left(x,y\right)={T}_{1}\left({T}_{2}\left(x,1\right),y\right)={T}_{2}\left({T}_{1}\left(x,y\right),{T}_{1}\left(1,y\right)\right)={T}_{2}\left({T}_{1}\left(x,y\right),y\right)\le {T}_{2}\left(x,y\right).$

Thus, ${T}_{1}\le {T}_{2}$, 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

${T}_{L}\left(\frac{3}{4},{T}_{P}\left(\frac{5}{8},\frac{1}{2}\right)\right)={T}_{L}\left(\frac{3}{4},\frac{5}{16}\right)=\frac{1}{16}$

and

${T}_{P}\left({T}_{L}\left(\frac{3}{4},\frac{5}{8}\right),{T}_{L}\left(\frac{3}{4},\frac{1}{2}\right)\right)={T}_{P}\left(\frac{3}{8},\frac{1}{4}\right)=\frac{3}{32}$

${T}_{L}$ 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 ${T}^{\prime }$ be a t-norm on L. For every t-norm T, T is ${T}^{\prime }$-distributive if and only if ${T}^{\prime }={T}_{\wedge }$.

Proof : Let T be an arbitrary t-norm on L such that ${T}^{\prime }$-distributive. By Proposition 1, it is obvious that $T\le {T}^{\prime }$ for any t-norm T. Thus, ${T}^{\prime }={T}_{\wedge }$.

: Since L is a chain, for any $y,z\in L$, either $y\le z$ or $z\le y$. Suppose that $y\le z$. By using the monotonicity of any t-norm T, it is obtained that for any $x\in L$, $T\left(x,y\right)\le T\left(x,z\right)$. Then

$T\left(x,y\right)=T\left(x,y\right)\wedge T\left(x,z\right)$

holds. Thus, for any $x,y,z\in L$,

$\begin{array}{rcl}T\left(x,{T}_{\wedge }\left(y,z\right)\right)& =& T\left(x,y\right)\\ =& T\left(x,y\right)\wedge T\left(x,z\right)\\ =& {T}_{\wedge }\left(T\left(x,y\right),T\left(x,z\right)\right)\end{array}$

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

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

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

One can easily check that T is a t-norm. Since

$T\left(a,{T}_{\wedge }\left(y,z\right)\right)=T\left(a,x\right)=0$

and

${T}_{\wedge }\left(T\left(a,y\right),T\left(a,z\right)\right)={T}_{\wedge }\left(y,z\right)=x,$

T is not ${T}_{\wedge }$-distributive.

Remark 3 The fact that any t-norm T is ${T}_{\wedge }$-distributive means that T is -distributive.

Theorem 1 Let $\left(L,\le ,0,1\right)$ 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

${T}_{D}\left(x,T\left(y,z\right)\right)=T\left({T}_{D}\left(x,y\right),{T}_{D}\left(x,z\right)\right)$

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

Proposition 3 

If T is a t-norm and $\phi :\left[0,1\right]\to \left[0,1\right]$ is a strictly increasing bijection, then the operation ${T}_{\phi }:{\left[0,1\right]}^{2}\to \left[0,1\right]$ given by

${T}_{\phi }\left(x,y\right)={\phi }^{-1}\left(T\left(\phi \left(x\right),\phi \left(y\right)\right)\right)$

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 $\left[0,1\right]$ and let φ be a strictly increasing bijection from $\left[0,1\right]$ to $\left[0,1\right]$. Denote the φ-transforms of the t-norms ${T}_{1}$ and ${T}_{2}$ by ${T}_{\phi }^{1}$ and ${T}_{\phi }^{2}$, respectively.

Theorem 2 Let ${T}_{1}$ and ${T}_{2}$ be any t-norms on $\left[0,1\right]$ and let φ be a strictly increasing bijection from $\left[0,1\right]$ to $\left[0,1\right]$. ${T}_{1}$ is ${T}_{2}$-distributive if and only if ${T}_{\phi }^{1}$ is ${T}_{\phi }^{2}$-distributive.

Proof Let ${T}_{1}$ be ${T}_{2}$-distributive. We must show that for every $x,y,z\in \left[0,1\right]$ with $y\ne 1$ or $z\ne 1$,

${T}_{\phi }^{1}\left(x,{T}_{\phi }^{2}\left(y,z\right)\right)={T}_{\phi }^{2}\left({T}_{\phi }^{1}\left(x,y\right),{T}_{\phi }^{1}\left(x,z\right)\right).$

Since $\phi :\left[0,1\right]\to \left[0,1\right]$ is a strictly increasing bijection, for every element $y,z\in \left[0,1\right]$ with $y\ne 1$ or $z\ne 1$, it must be $\phi \left(y\right)\ne 1$ or $\phi \left(z\right)\ne 1$. By using ${T}_{2}$-distributivity of ${T}_{1}$, we obtain that the equality

$\begin{array}{rcl}{T}_{\phi }^{1}\left(x,{T}_{\phi }^{2}\left(y,z\right)\right)& =& {\phi }^{-1}\left({T}_{1}\left(\phi \left(x\right),\phi \left({T}_{\phi }^{2}\left(y,z\right)\right)\right)\right)\\ =& {\phi }^{-1}\left({T}_{1}\left(\phi \left(x\right),\phi \left({\phi }^{-1}\left({T}_{2}\left(\phi \left(y\right),\phi \left(z\right)\right)\right)\right)\right)\right)\\ =& {\phi }^{-1}\left({T}_{1}\left(\phi \left(x\right),{T}_{2}\left(\phi \left(y\right),\phi \left(z\right)\right)\right)\right)\\ =& {\phi }^{-1}\left({T}_{2}\left({T}_{1}\left(\phi \left(x\right),\phi \left(y\right)\right),{T}_{1}\left(\phi \left(x\right),\phi \left(z\right)\right)\right)\right)\\ =& {\phi }^{-1}\left({T}_{2}\left(\left(\phi \circ {\phi }^{-1}\right){T}_{1}\left(\phi \left(x\right),\phi \left(y\right)\right),\left(\phi \circ {\phi }^{-1}\right){T}_{1}\left(\phi \left(x\right),\phi \left(z\right)\right)\right)\right)\\ =& {\phi }^{-1}\left({T}_{2}\left(\phi \left({\phi }^{-1}\left({T}_{1}\left(\phi \left(x\right),\phi \left(y\right)\right)\right)\right),\phi \left({\phi }^{-1}\left({T}_{1}\left(\phi \left(x\right),\phi \left(z\right)\right)\right)\right)\right)\right)\\ =& {\phi }^{-1}\left({T}_{2}\left(\phi \left({T}_{\phi }^{1}\left(x,y\right)\right),\phi \left({T}_{\phi }^{1}\left(x,z\right)\right)\right)\right)\\ =& {T}_{\phi }^{2}\left({T}_{\phi }^{1}\left(x,y\right),{T}_{\phi }^{1}\left(x,z\right)\right)\end{array}$

holds. Thus, ${T}_{\phi }^{1}$ is ${T}_{\phi }^{2}$-distributive.

Conversely, let ${T}_{\phi }^{1}$ be ${T}_{\phi }^{2}$-distributive. We will show that ${T}_{1}\left(x,{T}_{2}\left(y,z\right)\right)={T}_{2}\left({T}_{1}\left(x,y\right),{T}_{1}\left(x,z\right)\right)$ for every element $x,y,z\in \left[0,1\right]$ with $y\ne 1$ or $z\ne 1$. Since ${T}_{\phi }^{1}$ is the φ-transform of the t-norm ${T}_{1}$, for every $x,y\in \left[0,1\right]$, ${T}_{\phi }^{1}\left(x,y\right)={\phi }^{-1}\left({T}_{1}\left(\phi \left(x\right),\phi \left(y\right)\right)\right)$. Since φ is a bijection, it is clear that

${T}_{1}\left(\phi \left(x\right),\phi \left(y\right)\right)=\phi \left({T}_{\phi }^{1}\left(x,y\right)\right)$
(1)

holds. Also, by using (1), it is obtained that

${T}_{1}\left(x,y\right)={T}_{1}\left(\phi \left({\phi }^{-1}\left(x\right)\right),\phi \left({\phi }^{-1}\left(y\right)\right)\right)=\phi \left({T}_{\phi }^{1}\left({\phi }^{-1}\left(x\right),{\phi }^{-1}\left(y\right)\right)\right)$
(2)

From (2), it follows

${T}_{\phi }^{1}\left({\phi }^{-1}\left(x\right),{\phi }^{-1}\left(y\right)\right)={\phi }^{-1}\left({T}_{1}\left(x,y\right)\right).$
(3)

Also, the similar equalities for t-norm ${T}_{2}$ can be written. Since ${\phi }^{-1}\left(y\right)\ne 1$ or ${\phi }^{-1}\left(z\right)\ne 1$ for every $y,z\in \left[0,1\right]$ with $y\ne 1$ or $z\ne 1$, by using ${T}_{\phi }^{2}$-distributivity of ${T}_{\phi }^{1}$, it is obtained that the following equalities:

$\begin{array}{rcl}{T}_{1}\left(x,{T}_{2}\left(y,z\right)\right)& \stackrel{\left(\text{2}\right)}{=}& {T}_{1}\left(x,\phi \left({T}_{\phi }^{2}\left({\phi }^{-1}\left(y\right),{\phi }^{-1}\left(z\right)\right)\right)\right)\\ \stackrel{\left(\text{2}\right)}{=}& \phi \left({T}_{\phi }^{1}\left({\phi }^{-1}\left(x\right),{\phi }^{-1}\left(\phi \left({T}_{\phi }^{2}\left({\phi }^{-1}\left(y\right),{\phi }^{-1}\left(z\right)\right)\right)\right)\right)\right)\\ =& \phi \left({T}_{\phi }^{1}\left({\phi }^{-1}\left(x\right),{T}_{\phi }^{2}\left({\phi }^{-1}\left(y\right),{\phi }^{-1}\left(z\right)\right)\right)\right)\\ =& \phi \left({T}_{\phi }^{2}\left({T}_{\phi }^{1}\left({\phi }^{-1}\left(x\right),{\phi }^{-1}\left(y\right)\right),{T}_{\phi }^{1}\left({\phi }^{-1}\left(x\right),{\phi }^{-1}\left(z\right)\right)\right)\right)\\ \stackrel{\left(\text{3}\right)}{=}& \phi \left({T}_{\phi }^{2}\left({\phi }^{-1}\left({T}_{1}\left(x,y\right)\right),{\phi }^{-1}\left({T}_{1}\left(x,z\right)\right)\right)\right)\\ \stackrel{\left(\text{2}\right)}{=}& \phi \left({\phi }^{-1}\left({T}_{2}\left({T}_{1}\left(x,y\right),{T}_{1}\left(x,z\right)\right)\right)\right)\\ =& {T}_{2}\left({T}_{1}\left(x,y\right),{T}_{1}\left(x,z\right)\right)\end{array}$

hold. Thus, ${T}_{1}$ is ${T}_{2}$-distributive. □

Proposition 4 Let $\left(L,\le ,0,1\right)$ 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 $x\le y$. If $x=y$, then ${T}_{2}$ would be always a divisible t-norm since ${T}_{2}\left(y,1\right)=y=x$. Let $x\ne y$. Since ${T}_{1}$ is divisible, there exists an element $1\ne z$ of L such that ${T}_{1}\left(y,z\right)=x$. Then, by using ${T}_{2}$-distributivity of ${T}_{1}$, it is obtained that

$\begin{array}{rcl}x& =& {T}_{1}\left(y,z\right)={T}_{1}\left(y,{T}_{2}\left(z,1\right)\right)\\ =& {T}_{2}\left({T}_{1}\left(y,z\right),{T}_{1}\left(y,1\right)\right)\\ =& {T}_{2}\left({T}_{1}\left(y,z\right),y\right).\end{array}$

Thus, for any elements x, y of L with $x\le y$ and $x\ne y$, since there exists an element ${T}_{1}\left(y,z\right)\in L$ such that $x={T}_{2}\left({T}_{1}\left(y,z\right),y\right)$, ${T}_{2}$ is a divisible t-norm. □

Corollary 2 Let $\left(L,\le ,0,1\right)$ 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 $a,b\in L$. If $a=b$, then it would be $a{⪯}_{{T}_{2}}b$ since ${T}_{2}\left(b,1\right)=b=a$ for the element $1\in L$. Now, suppose that $a{⪯}_{{T}_{1}}b$ but $a\ne b$. Then there exists an element $\ell \in L$ such that ${T}_{1}\left(b,\ell \right)=a$. Since $a\ne b$, it must be $\ell \ne 1$. Then ${T}_{1}\left(b,{T}_{2}\left(\ell ,1\right)\right)={T}_{1}\left(b,\ell \right)=a$. Since ${T}_{1}$ is ${T}_{2}$-distributive, it is obtained that

$\begin{array}{rcl}a& =& {T}_{1}\left(b,{T}_{2}\left(\ell ,1\right)\right)={T}_{2}\left({T}_{1}\left(b,\ell \right),{T}_{1}\left(b,1\right)\right)\\ =& {T}_{2}\left(a,b\right).\end{array}$

for elements $b,\ell ,1\in L$ with $\ell \ne 1$, whence $a{⪯}_{{T}_{2}}b$. So, we obtain that ${⪯}_{{T}_{1}}\subseteq {⪯}_{{T}_{2}}$. □

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:

It is clear that the function ${T}^{\ast }$ is a t-norm such that ${T}_{P}\le {T}^{\ast }$, but ${⪯}_{{T}_{P}}\phantom{\rule{0.2em}{0ex}}⊈\phantom{\rule{0.2em}{0ex}}{⪯}_{{T}^{\ast }}$. Indeed.

First, let us show that $\frac{3}{8}{⋠}_{{T}^{\ast }}\frac{1}{2}$. Suppose that $\frac{3}{8}{⪯}_{{T}^{\ast }}\frac{1}{2}$. Then, for some $\ell \in \left[0,1\right]$,

${T}^{\ast }\left(\ell ,\frac{1}{2}\right)=\frac{3}{8}.$

For $\ell \in \left[0,1\right]$, either $\ell \le \frac{1}{2}$ or $\ell >\frac{1}{2}$. Let $\ell \le \frac{1}{2}$. Since $\frac{3}{8}={T}^{\ast }\left(\ell ,\frac{1}{2}\right)=\frac{1}{2}\ell$, it is obtained that $\ell =\frac{3}{4}$, which contradicts $\ell \le \frac{1}{2}$. Then it must be $\ell >\frac{1}{2}$. Since $\frac{3}{8}={T}^{\ast }\left(\ell ,\frac{1}{2}\right)=min\left(\ell ,\frac{1}{2}\right)=\frac{1}{2}$, which is a contradiction. Thus, it is obtained that $\frac{3}{8}{⋠}_{{T}^{\ast }}\frac{1}{2}$. On the other hand, since $x{⪯}_{{T}_{P}}y$ means that there exists an element of L such that ${T}_{p}\left(\ell ,y\right)=\ell y=x$ and ${T}_{P}\left(\frac{1}{2},\frac{3}{4}\right)=\frac{3}{8}$, we have that $\frac{3}{8}{⪯}_{{T}_{P}}\frac{1}{2}$. So, it is obtained that ${⪯}_{{T}_{P}}\phantom{\rule{0.2em}{0ex}}⊈\phantom{\rule{0.2em}{0ex}}{⪯}_{{T}^{\ast }}$.

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 $\left\{{S}_{\alpha }|\alpha \in I\right\}$ 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 $u\in {S}_{\alpha }$, $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ is comparable to every element in L, then the family ${\left({T}_{u}\right)}_{u\in {S}_{\alpha }}$ defined by

${T}_{u}\left(x,y\right)=\left\{\begin{array}{cc}inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},\hfill & \mathit{\text{if}}\phantom{\rule{0.5em}{0ex}}\left(x,y\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2},\hfill \\ x\wedge y,\hfill & \mathit{\text{otherwise}}\hfill \end{array}$

is a family of t-norms which are not distributive over each other. Namely, for any $\ell ,q\in {S}_{\alpha }$, neither ${T}_{\ell }$ is ${T}_{q}$-distributive nor ${T}_{q}$ is ${T}_{\ell }$-distributive.

Proof Firstly, let us show that for every $u\in {S}_{\alpha }$, each function ${T}_{u}$ is a t-norm.

1. (i)

Since $x\le 1$, for every element $x\in L$, $1\notin {S}_{\alpha }$. Then it follows ${T}_{u}\left(x,1\right)=x\wedge 1=x$ from $\left(x,1\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$, 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 $x\le y$ for $x,y\in L$. Let $z\in L$ be arbitrary. Then there are the following possible conditions for the couples $\left(x,z\right)$, $\left(y,z\right)$.

• Let $\left(x,z\right),\left(y,z\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$. Then we get clearly the equality

${T}_{u}\left(x,z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}={T}_{u}\left(y,z\right).$
• Let $\left(x,z\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$ and $\left(y,z\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$. Then $y\notin \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. Clearly, ${T}_{u}\left(x,z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ and ${T}_{u}\left(y,z\right)=y\wedge z$. Since $x\in \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$ and $x\le y$, we obtain $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le y$. By $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le z$, we get $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le y\wedge z$, whence ${T}_{u}\left(x,z\right)\le {T}_{u}\left(y,z\right)$.

• Let $\left(x,z\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$ and $\left(y,z\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$. Then it is clear that $x\notin \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. In this case,

${T}_{u}\left(x,z\right)=x\wedge z\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{T}_{u}\left(y,z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}.$

By $x\le y$ and $y\le u$, it is clear that $x\le u$. Since $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ is comparable to every element in L, either $x\le inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ or $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le x$. If $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le x$, it would be $x\in \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$ from $x\le u$, a contradiction. Thus, it must be $x\le inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$. Since $z\in \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$, $x\wedge z=x$. Thus, the inequality

${T}_{u}\left(x,z\right)=x\wedge z=x\le inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}={T}_{u}\left(y,z\right)$

holds.

• Let $\left(x,z\right),\left(y,z\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$. By $x\le y$, we have that

${T}_{u}\left(x,z\right)=x\wedge z\le y\wedge z={T}_{u}\left(y,z\right).$

So, the monotonicity holds.

1. (iv)

Now let us show that for every $x,y,z\in L$, the equality ${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)={T}_{u}\left({T}_{u}\left(x,y\right),z\right)$ holds.

• Let $\left(x,y\right),\left(y,z\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$. Then

${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$

and

${T}_{u}\left({T}_{u}\left(x,y\right),z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},$

whence the equality holds.

• If $\left(x,y\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$ and $\left(y,z\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$, then it must be $z\notin \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. Here, there are two choices for z: either $z\in {S}_{\alpha }$ or $z\notin {S}_{\alpha }$.

Let $z\in {S}_{\alpha }$. Then $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le z$. By the inequality $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le u$, it is clear that $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le u\wedge z$. Since $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le y\le u$, the following inequalities:

$inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\wedge z\le y\wedge z\le y\le u$

hold, that is, $y\wedge z\in \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. Thus, we have that

${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)={T}_{u}\left(x,y\wedge z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$

and

$\begin{array}{rcl}{T}_{u}\left({T}_{u}\left(x,y\right),z\right)& =& {T}_{u}\left(inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},z\right)\\ =& inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\wedge z=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}.\end{array}$

So, the equality holds again.

Let $z\notin {S}_{\alpha }$. Then there exists at least an element v in ${S}_{\alpha }$ such that v is comparable to the element z; i.e., either $z\le v$ or $v\le z$. Let $v\le z$. Since $u,v\in {S}_{\alpha }$, it is clear that $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le u\wedge v\le u\wedge z\le u$. Also, from the inequalities $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le y$ and $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le v\le z$, it follows $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\le y\wedge z\le y\le u$, i.e., it is obtained that $y\wedge z\in \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. Thus,

${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)={T}_{u}\left(x,y\wedge z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$

and

$\begin{array}{rcl}{T}_{u}\left({T}_{u}\left(x,y\right),z\right)& =& {T}_{u}\left(inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},z\right)\\ =& inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\wedge z\\ =& inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}.\end{array}$

Thus, the equality is satisfied.

Now, suppose that $z\le v$. If $u\le z$, it would be $u\le v$, which is a contradiction. Thus, either $z or z and u are not comparable. If $z, then it must be $z since $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ is comparable to every element in L and $z\notin \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. Thus, we have that

$\begin{array}{rcl}{T}_{u}\left({T}_{u}\left(x,y\right),z\right)& =& {T}_{u}\left(inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},z\right)\\ =& inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\wedge z\\ =& z\end{array}$

and

$\begin{array}{rcl}{T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)& =& {T}_{u}\left(x,y\wedge z\right)\\ =& {T}_{u}\left(x,z\right)\\ =& x\wedge z=z,\end{array}$

whence the equality holds.

Let z and u be not comparable. Since $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ is comparable to every element in L, either $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\} or $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}>z$. If $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}>z$, it would be $z, a contradiction. Then it must be $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}. By $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\wedge y, it is obtained that $y\wedge z\in \left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]$. Then the equalities

${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)={T}_{u}\left(x,y\wedge z\right)=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$

and

$\begin{array}{rcl}{T}_{u}\left({T}_{u}\left(x,y\right),z\right)& =& {T}_{u}\left(inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},z\right)\\ =& inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}\wedge z=inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}.\end{array}$

In this case, the equality is satisfied.

Similarly, one can show that the equality ${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)={T}_{u}\left({T}_{u}\left(x,y\right),z\right)$ holds when $\left(x,y\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$ and $\left(y,z\right)\in {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$.

• Now, let us investigate the last condition. If $\left(x,y\right),\left(y,z\right)\notin {\left[inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},u\right]}^{2}$, then it is obvious that

${T}_{u}\left(x,{T}_{u}\left(y,z\right)\right)={T}_{u}\left(x,y\wedge z\right)=x\wedge \left(y\wedge z\right)$

and

${T}_{u}\left({T}_{u}\left(x,y\right),z\right)={T}_{u}\left(x\wedge y,z\right)=\left(x\wedge y\right)\wedge z,$

whence the equality holds.

Consequently, we prove that ${\left({T}_{u}\right)}_{u\in {S}_{\alpha }}$ is a family of t-norms on L. Now, we will show that for every $m,n\in {S}_{\alpha }$, ${T}_{m}$ 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 ${T}_{m}\le {T}_{n}$, that is, for every $x,y\in L$, ${T}_{m}\left(x,y\right)\le {T}_{n}\left(x,y\right)$. Since m and n are not comparable, it is clear that $n\nleqq m$ and $m\nleqq n$. Then n must not be in $\left[inf\left\{m\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},m\right]$. Thus,

${T}_{m}\left(n,n\right)=n\wedge n=n.$

On the other hand, since $n\in \left[inf\left\{n\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\},n\right]$,

${T}_{n}\left(n,n\right)=inf\left\{n\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}.$

Then we have that ${T}_{n}\left(n,n\right)\ne {T}_{m}\left(n,n\right)$. Otherwise, we obtain that $n\le m$, which is a contradiction. So, we have that ${T}_{n}\left(n,n\right)<{T}_{m}\left(n,n\right)$ contradicts ${T}_{m}\le {T}_{n}$. 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 ${\left({S}_{\alpha }\right)}_{\alpha \in I}$ in Theorem 3 can be determined, let us investigate the following example.

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

For the family of ${\left({S}_{\alpha }\right)}_{\alpha \in I}$, there are two choices: one of them must be ${S}_{{\alpha }_{1}}=\left\{c,d,e\right\}$ and the other must be ${S}_{{\alpha }_{2}}=\left\{b,e\right\}$. Then, by Theorem 3, for every $u\in {S}_{{\alpha }_{1}}$ and $v\in {S}_{{\alpha }_{2}}$, the following functions:

and

are two families of t-norms.

Remark 5 In Theorem 3, if the condition that $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ is comparable to every element in L is canceled, then for any element $u\in {S}_{\alpha }$, ${T}_{u}$ 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 $u\in {S}_{\alpha }$, $inf\left\{u\wedge {\mu }_{i}|{\mu }_{i}\in {S}_{\alpha }\right\}$ is comparable to every element in L is canceled.

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

From Figure 4, it is clear that $inf\left\{j,e,f\right\}=a$ is not comparable to b. However, for the set $S=\left\{j,e,f\right\}$, the function defined by

does not satisfy the associativity since ${T}_{e}\left({T}_{e}\left(c,d\right),b\right)=0$ and ${T}_{e}\left(c,{T}_{e}\left(d,b\right)\right)=b$. 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 ${T}_{\wedge }$-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.

## References

1. Menger K: Statistical metrics. Proc. Natl. Acad. Sci. USA 1942, 28: 535–537. 10.1073/pnas.28.12.535

2. Schweizer B, Sklar A: Probabilistic Metric Spaces. Elsevier, Amsterdam; 1983.

3. Höhle U: Commutative: residuated -monoids. In Non-Classical Logics and Their Applications to Fuzzy Subsets: a Handbook on the Math., Foundations of Fuzzy Set Theory. Edited by: Höhle U, Klement EP. Kluwer Academic, Dordrecht; 1995.

4. Liang X, Pedrycz W: Logic-based fuzzy networks: a study in system modeling with triangular norms and uninorms. Fuzzy Sets Syst. 2009, 160: 3475–3502. 10.1016/j.fss.2009.04.014

5. Butnariu D, Klement EP: Triangular Norm-Based Measures and Games with Fuzzy Coalitions. Kluwer Academic, Dordrecht; 1993.

6. Klement EP, Mesiar R, Pap E: Integration with respect to decomposable measures, based on a conditionally distributive semiring on the unit interval. Int. J. Uncertain. Fuzziness Knowl.-Based Syst. 2000, 8: 701–717.

7. Klement EP, Weber S: An integral representation for decomposable measures of measurable functions. Aequ. Math. 1994, 47: 255–262. 10.1007/BF01832963

8. Kolesárová A: On the integral representation of possibility measures of fuzzy events. J. Fuzzy Math. 1997, 5: 759–766.

9. Birkhoff G: Lattice Theory. 3rd edition. Am. Math. Soc., Providence; 1967.

10. De Baets B, Mesiar R: Triangular norms on product lattices. Fuzzy Sets Syst. 1999, 104: 61–75. 10.1016/S0165-0114(98)00259-0

11. Jenei S, De Baets B: On the direct decomposability of t -norms on product lattices. Fuzzy Sets Syst. 2003, 139: 699–707. 10.1016/S0165-0114(03)00125-8

12. Karaçal F, Kesicioğlu MN: A T -partial order obtained from t -norms. Kybernetika 2011, 47: 300–314.

13. Karaçal F, Sağıroğlu Y: Infinitely -distributive t -norm on complete lattices and pseudo-complements. Fuzzy Sets Syst. 2009, 160: 32–43. 10.1016/j.fss.2008.03.022

14. Karaçal F, Khadjiev D: -distributive and infinitely -distributive t -norms on complete lattice. Fuzzy Sets Syst. 2005, 151: 341–352. 10.1016/j.fss.2004.06.013

15. Karaçal F: On the direct decomposability of strong negations and S -implication operators on product lattices. Inf. Sci. 2006, 176: 3011–3025. 10.1016/j.ins.2005.12.010

16. Saminger S: On ordinal sums of triangular norms on bounded lattices. Fuzzy Sets Syst. 2006, 157: 1403–1416. 10.1016/j.fss.2005.12.021

17. Saminger-Platz S, Klement EP, Mesiar R: On extensions of triangular norms on bounded lattices. Indag. Math. 2009, 19: 135–150.

18. Casasnovas J, Mayor G: Discrete t -norms and operations on extended multisets. Fuzzy Sets Syst. 2008, 159: 1165–1177. 10.1016/j.fss.2007.12.005

19. Wang Z, Yu Y: Pseudo t -norms and implication operators: direct product and direct product decompositions. Fuzzy Sets Syst. 2003, 139: 673–683. 10.1016/S0165-0114(02)00503-1

20. Klement EP, Mesiar R, Pap E: Triangular Norms. Kluwer Academic, Dordrecht; 2000.

## Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Author information

Authors

### Corresponding author

Correspondence to Mücahide Nesibe Kesicioğlu.

### Competing interests

The author declares that they have no competing interests.

## Authors’ original submitted files for images

Below are the links to the authors’ original submitted files for images.

## Rights and permissions

Reprints and Permissions

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

• Accepted:

• Published:

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

### Keywords

• triangular norm
• bounded lattice
• T-partial order
• divisibility
• distributivity 