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

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

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.

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}\phantom{\rule{1em}{0ex}}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.

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

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

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

Corollary 1LetLbe a bounded lattice and{T}_{1}and{T}_{2}be any twot-norms onL. If both{T}_{1}is{T}_{2}-distributive and{T}_{2}is{T}_{1}-distributive, then{T}_{1}={T}_{2}.

Proposition 2LetLbe a bounded chain and{T}^{\prime}be at-norm onL. For everyt-normT, Tis{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

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.

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(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 1Let(L,\le ,0,1)be a bounded lattice. For anyt-normTonL, {T}_{D}isT-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. □

is at-norm which is isomorphic toT. Thist-norm is calledφ-transform ofT.

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 2Let{T}_{1}and{T}_{2}be anyt-norms on[0,1]and letφbe 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,

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

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

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:

Proposition 4Let(L,\le ,0,1)be a bounded lattice and{T}_{1}and{T}_{2}be twot-norms onLsuch 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

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 2Let(L,\le ,0,1)be a bounded lattice and{T}_{1}and{T}_{2}be twot-norms onL. If{T}_{1}is{T}_{2}-distributive, then the{T}_{1}-partial order implies the{T}_{2}-partial order.

Proof Let a{\u2aaf}_{{T}_{1}}b for any a,b\in L. If a=b, then it would be a{\u2aaf}_{{T}_{2}}b since {T}_{2}(b,1)=b=a for the element 1\in L. Now, suppose that a{\u2aaf}_{{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

for elements b,\ell ,1\in L with \ell \ne 1, whence a{\u2aaf}_{{T}_{2}}b. So, we obtain that {\u2aaf}_{{T}_{1}}\subseteq {\u2aaf}_{{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 {\u2aaf}_{{T}_{P}}\phantom{\rule{0.2em}{0ex}}\u2288\phantom{\rule{0.2em}{0ex}}{\u2aaf}_{{T}^{\ast}}. Indeed.

First, let us show that \frac{3}{8}{\u22e0}_{{T}^{\ast}}\frac{1}{2}. Suppose that \frac{3}{8}{\u2aaf}_{{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}{\u22e0}_{{T}^{\ast}}\frac{1}{2}. On the other hand, since x{\u2aaf}_{{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}{\u2aaf}_{{T}_{P}}\frac{1}{2}. So, it is obtained that {\u2aaf}_{{T}_{P}}\phantom{\rule{0.2em}{0ex}}\u2288\phantom{\rule{0.2em}{0ex}}{\u2aaf}_{{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 3LetLbe a complete lattice and\{{S}_{\alpha}|\alpha \in I\}be a nonempty family of nonempty sets consisting of the elements inLwhich are all incomparable to each other with respect to the order onL. If for any elementu\in {S}_{\alpha}, inf\{u\wedge {\mu}_{i}|{\mu}_{i}\in {S}_{\alpha}\}is comparable to every element inL, then the family{({T}_{u})}_{u\in {S}_{\alpha}}defined by

is a family oft-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

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,

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

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

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,

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

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

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

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],

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.

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:

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.

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

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.

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