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 [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) $T:{[0,1]}^2\to [0,1]$ 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, $T_M(x,y)=min(x,y)$, $T_P(x,y)=xy$, $T_L(x,y)=max(0,x+y-1)$ and

$$T_D(x,y)=\{\begin{array}{cc}x,\hfill & \text{if }y=1,\hfill \\ y,\hfill & \text{if }x=1,\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

Recall that for any t-norms $T_1$ and $T_2$, $T_1$ is called weaker than $T_2$ if for every $(x,y)\in {[0,1]}^2$, $T_1(x,y)\le T_2(x,y)$.

T-norms are defined on a bounded lattice $(L,\le ,0,1)$ 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 $[0,1]$. 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 $(L,\le ,0,1)$. 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 $T_{\wedge }$ 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 [14]

Let $(L,\le ,0,1)$ 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

$$T_D(x,y)=\{\begin{array}{cc}x,\hfill & \text{if }y=1,\hfill \\ y,\hfill & \text{if }x=1,\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

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 $(L,\le ,0,1)$ is given by $T_{\wedge }(x,y)=x\wedge y$.

Definition 2 [18]

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

$$\mathrm{\forall }x,y\in L\text{ with }x\le y,\text{ there is a }z\in L\text{ such that}x=T(y,z).$$

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 $T_{\wedge }$ is divisible: $x\le y$ is equivalent to $x\wedge y=x$.

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.

$$x{⪯}_{T}y:\iff T(\ell ,y)=x\text{for some }\ell \in L.$$

Definition 4 [19]

• (i) A t-norm T on a lattice L is called ∧-distributive if

  $$T(a,b_1\wedge b_2)=T(a,b_1)\wedge T(a,b_2)$$

  for every $a,b_1,b_2\in L$.

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

  $$T(a,{\wedge }_I{b}_{\tau })={\wedge }_{I}T(a,b_{\tau })$$

  for every subset $\{a,b_{\tau }\in L,\tau \in I\}$ of L.

3 T-distributivity

Definition 5 Let $(L,\le ,0,1)$ 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

$$T_1(x,T_2(y,z))=T_2(T_1(x,y),T_1(x,z))$$

is satisfied, then $T_1$ is called $T_2$-distributive or we say that $T_1$ is distributive over $T_2$.

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

Figure 1

$\mathbf{(}\mathit{L}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathbf{1}\mathbf{\}}\mathbf{,}\mathbf{\le }\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$ .

The functions $T_1$ and $T_2$ on the lattice L defined by

$$T_1(x,y)=\{\begin{array}{cc}0,\hfill & \text{if }x=a,y=a,\hfill \\ b,\hfill & \text{if }x=c,y=c,\hfill \\ x\wedge y,\hfill & \text{otherwise}\hfill \end{array}$$

and

$$T_2(x,y)=\{\begin{array}{cc}b,\hfill & \text{if }x=c,y=c,\hfill \\ x\wedge y,\hfill & \text{otherwise}\hfill \end{array}$$

are obviously t-norms on L such that $T_1$ is $T_2$-distributive.

Proposition 1 Let $(L,\le ,0,1)$ 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 $T_1\le T_2$ for all $x,y,z\in L\mathrm{\setminus }\{0,1\}$. By the $T_2$-distributivity of $T_1$, it is obtained that

$$T_1(x,y)=T_1(T_2(x,1),y)=T_2(T_1(x,y),T_1(1,y))=T_2(T_1(x,y),y)\le T_2(x,y).$$

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

$$T_L(\frac{3}{4},T_P(\frac{5}{8},\frac{1}{2}))=T_L(\frac{3}{4},\frac{5}{16})=\frac{1}{16}$$

and

$$T_P(T_L(\frac{3}{4},\frac{5}{8}),T_L(\frac{3}{4},\frac{1}{2}))=T_P(\frac{3}{8},\frac{1}{4})=\frac{3}{32}$$

$T_L$ 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 $x\in L$, $T(x,y)\le T(x,z)$. Then

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

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

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

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 $(L=\{0,x,y,z,a,1\},\le )$ as displayed in Figure 2.

Figure 2

$\mathbf{(}\mathit{L}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{,}\mathit{a}\mathbf{,}\mathbf{1}\mathbf{\}}\mathbf{,}\mathbf{\le }\mathbf{)}$ .

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

Table 1 T -norm on the lattice $\mathbf{(}\mathit{L}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{,}\mathit{a}\mathbf{,}\mathbf{1}\mathbf{\}}\mathbf{,}\mathbf{\le }\mathbf{)}$

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

$$T(a,T_{\wedge }(y,z))=T(a,x)=0$$

and

$$T_{\wedge }(T(a,y),T(a,z))=T_{\wedge }(y,z)=x,$$

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 $(L,\le ,0,1)$ 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 $y\ne 1$ or $z\ne 1$. Suppose that $z\ne 1$. 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 $x\ne 1$. Then $y=1$ or $y\ne 1$. 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 $y\ne 1$. Since $T(y,z)\le y\le 1$ and $y\ne 1$, $T(y,z)\ne 1$. 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 $\phi :[0,1]\to [0,1]$ is a strictly increasing bijection, then the operation $T_{\phi }:{[0,1]}^2\to [0,1]$ given by

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

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 $[0,1]$ 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 $T_{\phi }^1$ and $T_{\phi }^2$, 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 $T_{\phi }^1$ is $T_{\phi }^2$-distributive.

Proof Let $T_1$ be $T_2$-distributive. We must show that for every $x,y,z\in [0,1]$ with $y\ne 1$ or $z\ne 1$,

$$T_{\phi }^1(x,T_{\phi }^2(y,z))=T_{\phi }^2(T_{\phi }^1(x,y),T_{\phi }^1(x,z)).$$

Since $\phi :[0,1]\to [0,1]$ is a strictly increasing bijection, for every element $y,z\in [0,1]$ with $y\ne 1$ or $z\ne 1$, it must be $\phi (y)\ne 1$ or $\phi (z)\ne 1$. By using $T_2$-distributivity of $T_1$, we obtain that the equality

$$\begin{array}{rcl}{T}_{\phi }^1(x,T_{\phi }^2(y,z))& =& {\phi }^{-1}\left(T_1(\phi (x),\phi (T_{\phi }^2(y,z)))\right)\\ =& {\phi }^{-1}\left(T_1(\phi (x),\phi \left({\phi }^{-1}\left(T_2(\phi (y),\phi (z))\right)\right))\right)\\ =& {\phi }^{-1}\left(T_1(\phi (x),T_2(\phi (y),\phi (z)))\right)\\ =& {\phi }^{-1}\left(T_2(T_1(\phi (x),\phi (y)),T_1(\phi (x),\phi (z)))\right)\\ =& {\phi }^{-1}\left(T_2((\phi \circ {\phi }^{-1})T_1(\phi (x),\phi (y)),(\phi \circ {\phi }^{-1})T_1(\phi (x),\phi (z)))\right)\\ =& {\phi }^{-1}\left(T_2(\phi \left({\phi }^{-1}\left(T_1(\phi (x),\phi (y))\right)\right),\phi \left({\phi }^{-1}\left(T_1(\phi (x),\phi (z))\right)\right))\right)\\ =& {\phi }^{-1}\left(T_2(\phi (T_{\phi }^1(x,y)),\phi (T_{\phi }^1(x,z)))\right)\\ =& T_{\phi }^2(T_{\phi }^1(x,y),T_{\phi }^1(x,z))\end{array}$$

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

Conversely, let $T_{\phi }^1$ be $T_{\phi }^2$-distributive. We will show that $T_1(x,T_2(y,z))=T_2(T_1(x,y),T_1(x,z))$ for every element $x,y,z\in [0,1]$ with $y\ne 1$ or $z\ne 1$. Since $T_{\phi }^1$ is the φ-transform of the t-norm $T_1$, for every $x,y\in [0,1]$, $T_{\phi }^1(x,y)={\phi }^{-1}(T_1(\phi (x),\phi (y)))$. Since φ is a bijection, it is clear that

$$T_1(\phi (x),\phi (y))=\phi (T_{\phi }^1(x,y))$$

(1)

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

$$T_1(x,y)=T_1(\phi ({\phi }^{-1}(x)),\phi ({\phi }^{-1}(y)))=\phi \left(T_{\phi }^1({\phi }^{-1}(x),{\phi }^{-1}(y))\right)$$

(2)

From (2), it follows

$$T_{\phi }^1({\phi }^{-1}(x),{\phi }^{-1}(y))={\phi }^{-1}(T_1(x,y)).$$

(3)

Also, the similar equalities for t-norm $T_2$ can be written. Since ${\phi }^{-1}(y)\ne 1$ or ${\phi }^{-1}(z)\ne 1$ for every $y,z\in [0,1]$ 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:

$$\begin{array}{rcl}{T}_1(x,T_2(y,z))& \stackrel{(\text{2})}{=}& T_1(x,\phi \left(T_{\phi }^2({\phi }^{-1}(y),{\phi }^{-1}(z))\right))\\ \stackrel{(\text{2})}{=}& \phi \left(T_{\phi }^1({\phi }^{-1}(x),{\phi }^{-1}\left(\phi \left(T_{\phi }^2({\phi }^{-1}(y),{\phi }^{-1}(z))\right)\right))\right)\\ =& \phi \left(T_{\phi }^1({\phi }^{-1}(x),T_{\phi }^2({\phi }^{-1}(y),{\phi }^{-1}(z)))\right)\\ =& \phi \left(T_{\phi }^2(T_{\phi }^1({\phi }^{-1}(x),{\phi }^{-1}(y)),T_{\phi }^1({\phi }^{-1}(x),{\phi }^{-1}(z)))\right)\\ \stackrel{(\text{3})}{=}& \phi \left(T_{\phi }^2({\phi }^{-1}(T_1(x,y)),{\phi }^{-1}(T_1(x,z)))\right)\\ \stackrel{(\text{2})}{=}& \phi \left({\phi }^{-1}\left(T_2(T_1(x,y),T_1(x,z))\right)\right)\\ =& T_2(T_1(x,y),T_1(x,z))\end{array}$$

hold. Thus, $T_1$ is $T_2$-distributive. □

Proposition 4 Let $(L,\le ,0,1)$ 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 $x\le y$. If $x=y$, then $T_2$ would be always a divisible t-norm since $T_2(y,1)=y=x$. Let $x\ne y$. Since $T_1$ is divisible, there exists an element $1\ne z$ of L such that $T_1(y,z)=x$. Then, by using $T_2$-distributivity of $T_1$, it is obtained that

$$\begin{array}{rcl}x& =& T_1(y,z)=T_1(y,T_2(z,1))\\ =& T_2(T_1(y,z),T_1(y,1))\\ =& T_2(T_1(y,z),y).\end{array}$$

Thus, for any elements x, y of L with $x\le y$ and $x\ne y$, since there exists an element $T_1(y,z)\in L$ such that $x=T_2(T_1(y,z),y)$, $T_2$ is a divisible t-norm. □

Corollary 2 Let $(L,\le ,0,1)$ 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 $a,b\in L$. If $a=b$, then it would be $a{⪯}_{T_2}b$ since $T_2(b,1)=b=a$ 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 $T_1(b,\ell )=a$. Since $a\ne b$, it must be $\ell \ne 1$. Then $T_1(b,T_2(\ell ,1))=T_1(b,\ell )=a$. Since $T_1$ is $T_2$-distributive, it is obtained that

$$\begin{array}{rcl}a& =& T_1(b,T_2(\ell ,1))=T_2(T_1(b,\ell ),T_1(b,1))\\ =& T_2(a,b).\end{array}$$

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:

$$T^{\ast }(x,y)=\{\begin{array}{cc}xy,\hfill & \text{if }(x,y)\in {[0,\frac{1}{2}]}^2,\hfill \\ min(x,y),\hfill & \text{otherwise}.\hfill \end{array}$$

It is clear that the function $T^{\ast }$ is a t-norm such that $T_P\le T^{\ast }$, but ${⪯}_{T_P}⊈{⪯}_{T^{\ast }}$. 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 $\ell \in [0,1]$,

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

For $\ell \in [0,1]$, either $\ell \le \frac{1}{2}$ or $\ell >\frac{1}{2}$. Let $\ell \le \frac{1}{2}$. Since $\frac{3}{8}=T^{\ast }(\ell ,\frac{1}{2})=\frac{1}{2}\ell$, it is obtained that $\ell =\frac{3}{4}$, which contradicts $\ell \le \frac{1}{2}$. Then it must be $\ell >\frac{1}{2}$. Since $\frac{3}{8}=T^{\ast }(\ell ,\frac{1}{2})=min(\ell ,\frac{1}{2})=\frac{1}{2}$, 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 ℓ of L such that $T_p(\ell ,y)=\ell y=x$ and $T_P(\frac{1}{2},\frac{3}{4})=\frac{3}{8}$, we have that $\frac{3}{8}{⪯}_{T_P}\frac{1}{2}$. So, it is obtained that ${⪯}_{T_P}⊈{⪯}_{T^{\ast }}$.

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 $\{S_{\alpha }|\alpha \in I\}$ 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 $u\in S_{\alpha }$, $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$ is comparable to every element in L, then the family ${(T_u)}_{u\in S_{\alpha }}$ defined by

$$T_u(x,y)=\{\begin{array}{cc}inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},\hfill & \mathit{\text{if}}(x,y)\in {[inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]}^2,\hfill \\ x\wedge y,\hfill & \mathit{\text{otherwise}}\hfill \end{array}$$

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.

• (i) Since $x\le 1$, for every element $x\in L$, $1\notin S_{\alpha }$. Then it follows $T_u(x,1)=x\wedge 1=x$ from $(x,1)\notin {[inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]}^2$, that is, the boundary condition is satisfied.

• (ii) It can be easily shown that the commutativity holds.

• (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 $(x,z)$, $(y,z)$.

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

  $$T_u(x,z)=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}=T_u(y,z).$$

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

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

  $$T_u(x,z)=x\wedge z\text{and}{T}_u(y,z)=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}.$$

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

$$T_u(x,z)=x\wedge z=x\le inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}=T_u(y,z)$$

holds.

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

  $$T_u(x,z)=x\wedge z\le y\wedge z=T_u(y,z).$$

So, the monotonicity holds.

• (iv) Now let us show that for every $x,y,z\in L$, the equality $T_u(x,T_u(y,z))=T_u(T_u(x,y),z)$ holds.

• Let $(x,y),(y,z)\in {[inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]}^2$. Then

  $$T_u(x,T_u(y,z))=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$$

and

$$T_u(T_u(x,y),z)=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},$$

whence the equality holds.

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

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

$$inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\wedge z\le y\wedge z\le y\le u$$

hold, that is, $y\wedge z\in [inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]$. Thus, we have that

$$T_u(x,T_u(y,z))=T_u(x,y\wedge z)=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$$

and

$$\begin{array}{rcl}{T}_u(T_u(x,y),z)& =& T_u(inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},z)\\ =& inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\wedge z=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}.\end{array}$$

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 $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\le u\wedge v\le u\wedge z\le u$. Also, from the inequalities $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\le y$ and $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\le v\le z$, it follows $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\le y\wedge z\le y\le u$, i.e., it is obtained that $y\wedge z\in [inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]$. Thus,

$$T_u(x,T_u(y,z))=T_u(x,y\wedge z)=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$$

and

$$\begin{array}{rcl}{T}_u(T_u(x,y),z)& =& T_u(inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},z)\\ =& inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\wedge z\\ =& inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}.\end{array}$$

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 $z<u$ or z and u are not comparable. If $z<u$, then it must be $z<inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$ since $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$ is comparable to every element in L and $z\notin [inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]$. Thus, we have that

$$\begin{array}{rcl}{T}_u(T_u(x,y),z)& =& T_u(inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},z)\\ =& inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\wedge z\\ =& z\end{array}$$

and

$$\begin{array}{rcl}{T}_u(x,T_u(y,z))& =& T_u(x,y\wedge z)\\ =& T_u(x,z)\\ =& x\wedge z=z,\end{array}$$

whence the equality holds.

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

$$T_u(x,T_u(y,z))=T_u(x,y\wedge z)=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$$

and

$$\begin{array}{rcl}{T}_u(T_u(x,y),z)& =& T_u(inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},z)\\ =& inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}\wedge z=inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}.\end{array}$$

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 $(x,y)\notin {[inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]}^2$ and $(y,z)\in {[inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},u]}^2$.

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

  $$T_u(x,T_u(y,z))=T_u(x,y\wedge z)=x\wedge (y\wedge z)$$

and

$$T_u(T_u(x,y),z)=T_u(x\wedge y,z)=(x\wedge y)\wedge z,$$

whence the equality holds.

Consequently, we prove that ${(T_u)}_{u\in S_{\alpha }}$ 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 $x,y\in L$, $T_m(x,y)\le T_n(x,y)$. Since m and n are not comparable, it is clear that $n\nleqq m$ and $m\nleqq n$. Then n must not be in $[inf\{m\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},m]$. Thus,

$$T_m(n,n)=n\wedge n=n.$$

On the other hand, since $n\in [inf\{n\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\},n]$,

$$T_n(n,n)=inf\{n\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}.$$

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

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

Figure 3

$\mathbf{(}\mathit{L}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{,}\mathit{e}\mathbf{,}\mathbf{1}\mathbf{\}}\mathbf{,}\mathbf{\le }\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$ .

For the family of ${(S_{\alpha })}_{\alpha \in I}$, there are two choices: one of them must be $S_{{\alpha }_1}=\{c,d,e\}$ and the other must be $S_{{\alpha }_2}=\{b,e\}$. Then, by Theorem 3, for every $u\in S_{{\alpha }_1}$ and $v\in S_{{\alpha }_2}$, the following functions:

$$T_u(x,y)=\{\begin{array}{cc}a,\hfill & \text{if }(x,y)\in {[a,u]}^2,\hfill \\ x\wedge y,\hfill & \text{otherwise}\hfill \end{array}$$

and

$$T_v(x,y)=\{\begin{array}{cc}a,\hfill & \text{if }(x,y)\in {[a,v]}^2,\hfill \\ x\wedge y,\hfill & \text{otherwise}\hfill \end{array}$$

are two families of t-norms.

Remark 5 In Theorem 3, if the condition that $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$ 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 $u\in S_{\alpha }$, $inf\{u\wedge {\mu }_i|{\mu }_i\in S_{\alpha }\}$ is comparable to every element in L is canceled.

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

Figure 4

$\mathbf{(}\mathit{L}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{,}\mathit{e}\mathbf{,}\mathit{f}\mathbf{,}\mathit{g}\mathbf{,}\mathit{h}\mathbf{,}\mathit{j}\mathbf{,}\mathbf{1}\mathbf{\}}\mathbf{,}\mathbf{\le }\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$ .

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

$$T_e(x,y)=\{\begin{array}{cc}a,\hfill & \text{if }(x,y)\in {[a,e]}^2,\hfill \\ x\wedge y,\hfill & \text{otherwise}\hfill \end{array}$$

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 $T_{\wedge }$-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.