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On the property of Tdistributivity
Fixed Point Theory and Applications volume 2013, Article number: 32 (2013)
Abstract
In this paper, we introduce the notion of Tdistributivity for any tnorm on a bounded lattice. We determine a relation between the tnorms T and {T}^{\prime}, where {T}^{\prime} is a Tdistributive tnorm. Also, for an arbitrary tnorm T, we give a necessary and sufficient condition for {T}_{D} to be Tdistributive and for T to be {T}_{\wedge}distributive. Moreover, we investigate the relation between the Tdistributivity and the concepts of the Tpartial order, the divisibility of tnorms. We also determine that the Tdistributivity is preserved under the isomorphism. Finally, we construct a family of tnorms 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 nonadditive measures and integral theory [6–8].
A triangular norm (tnorm for short) T:{[0,1]}^{2}\to [0,1] is a commutative, associative, nondecreasing operation on [0,1] with a neutral element 1. The four basic tnorms on [0,1] are the minimum {T}_{M}, the product {T}_{P}, the Łukasiewicz tnorm {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
Recall that for any tnorms {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).
Tnorms are defined on a bounded lattice (L,\le ,0,1) in a similar way, and then extremal tnorms {T}_{D} as well as {T}_{\wedge} on L are defined similarly {T}_{D} and {T}_{M} on [0,1]. For more details on tnorms on bounded lattices, we refer to [9–17]. Also, the order between tnorms on a bounded lattice is defined similarly.
In the present paper, we introduce the notion of Tdistributivity for any tnorms on a bounded lattice (L,\le ,0,1). The aim of this study is to discuss the properties of Tdistributivity. The paper is organized as follows. Firstly, we recall some basic notions in Section 2. In Section 3, we define the Tdistributivity for any tnorm on a bounded lattice. For any two tnorms {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 tnorm T, where every tnorm is Tdistributive, is the infimum tnorm {T}_{\wedge} when the lattice L is especially a chain. If L is not a chain, we give an example illustrating any tnorm need not be {T}_{\wedge}. Also, we show that for any tnorm T on a bounded lattice, {T}_{D} is Tdistributive. Moreover, we show that the Tdistributivity is preserved under the isomorphism. For any two tnorms {T}_{1} and {T}_{2} such that {T}_{1} is {T}_{2}distributive, we prove that the divisibility of tnorm {T}_{1} requires the divisibility of tnorm {T}_{2}. Also, we obtain that for any two tnorms {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 tnorms 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 (tnorm 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 tnorm on L. Since it holds that {T}_{D}\le T for any tnorm T on L, {T}_{D} is the smallest tnorm on L.
The largest tnorm on a bounded lattice (L,\le ,0,1) is given by {T}_{\wedge}(x,y)=x\wedge y.
Definition 2 [18]
A tnorm T on L is divisible if the following condition holds:
A basic example of a nondivisible tnorm on any bounded lattice (i.e., cardL>2) is the weakest tnorm {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 tnorm on L. The order defined as follows is called a Tpartial order (triangular order) for a tnorm T.
Definition 4 [19]

(i)
A tnorm 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 tnorm 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 Tdistributivity
Definition 5 Let (L,\le ,0,1) be a bounded lattice and {T}_{1} and {T}_{2} be two tnorms 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
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.
The functions {T}_{1} and {T}_{2} on the lattice L defined by
and
are obviously tnorms 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 tnorms on L. If {T}_{1} is {T}_{2}distributive, then {T}_{1} is weaker than {T}_{2}.
Proof Since all tnorms 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 tnorms {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 tnorm {T}_{L}. It is clear that {T}_{L}<{T}_{P}. Since
and
{T}_{L} is not {T}_{P}distributive.
Corollary 1 Let L be a bounded lattice and {T}_{1} and {T}_{2} be any two tnorms 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 tnorm on L. For every tnorm T, T is {T}^{\prime}distributive if and only if {T}^{\prime}={T}_{\wedge}.
Proof :⇒ Let T be an arbitrary tnorm on L such that {T}^{\prime}distributive. By Proposition 1, it is obvious that T\le {T}^{\prime} for any tnorm 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 tnorm T, it is obtained that for any x\in L, T(x,y)\le T(x,z). Then
holds. Thus, for any x,y,z\in L,
is satisfied, which shows that any tnorm T is {T}_{\wedge}distributive. □
Remark 2 In Proposition 2, if L is not a chain, then the lefthand side of Proposition 2 may not be satisfied. Namely, if L is not a chain, then any tnorm T need not be {T}_{\wedge}distributive. Moreover, even if L is a distributive lattice, any tnorm 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 tnorm. Since
and
T is not {T}_{\wedge}distributive.
Remark 3 The fact that any tnorm T is {T}_{\wedge}distributive means that T is ∧distributive.
Theorem 1 Let (L,\le ,0,1) be a bounded lattice. For any tnorm T on L, {T}_{D} is Tdistributive.
Proof Let T be an arbitrary tnorm on L. We must show that the equality
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 Tdistributive for any tnorm T on L. □
Proposition 3 [20]
If T is a tnorm 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
is a tnorm which is isomorphic to T. This tnorm is called φtransform of T.
Let {T}_{1} and {T}_{2} be any two tnorms on [0,1] and let φ be a strictly increasing bijection from [0,1] to [0,1]. Denote the φtransforms of the tnorms {T}_{1} and {T}_{2} by {T}_{\phi}^{1} and {T}_{\phi}^{2}, respectively.
Theorem 2 Let {T}_{1} and {T}_{2} be any tnorms 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,
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 φtransform of the tnorm {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
holds. Also, by using (1), it is obtained that
From (2), it follows
Also, the similar equalities for tnorm {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:
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 tnorms 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 tnorm 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 tnorm. □
Corollary 2 Let (L,\le ,0,1) be a bounded lattice and {T}_{1} and {T}_{2} be two tnorms on L. 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 tnorms {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 tnorm 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],
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 tnorms 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
is a family of tnorms 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 tnorm.

(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\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{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
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
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:
hold, that is, y\wedge z\in [inf\{u\wedge {\mu}_{i}{\mu}_{i}\in {S}_{\alpha}\},u]. Thus, we have that
and
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,
and
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
and
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
and
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
whence the equality holds.
Consequently, we prove that {({T}_{u})}_{u\in {S}_{\alpha}} is a family of tnorms on L. Now, we will show that for every m,n\in {S}_{\alpha}, {T}_{m} and {T}_{n} are not distributive tnorms 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,
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 tnorms 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:
and
are two families of tnorms.
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 tnorm. The following is an example showing that {T}_{u} is not a tnorm 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 tnorm.
4 Conclusions
In this paper, we introduced the notion of Tdistributivity for any tnorm on a bounded lattice and discussed some properties of Tdistributivity. We determined a necessary and sufficient condition for {T}_{D} to be Tdistributive and for T to be {T}_{\wedge}distributive. We obtained that Tdistributivity is preserved under the isomorphism. We proved that the divisibility of tnorm {T}_{1} requires the divisibility of tnorm {T}_{2} for any two tnorms {T}_{1} and {T}_{2} where {T}_{1} is {T}_{2}distributive. Also, we constructed a family of tnorms which are not distributive over each other with the help of incomparable elements in a bounded lattice.
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Dedicated to Professor Hari M Srivastava.
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Kesicioğlu, M.N. On the property of Tdistributivity. Fixed Point Theory Appl 2013, 32 (2013). https://doi.org/10.1186/16871812201332
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DOI: https://doi.org/10.1186/16871812201332