On the property of Tdistributivity
 Mücahide Nesibe Kesicioğlu^{1}Email author
https://doi.org/10.1186/16871812201332
© Kesicioğlu; licensee Springer. 2013
Received: 4 December 2012
Accepted: 31 January 2013
Published: 15 February 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.
Keywords
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].
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.
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 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]
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
is satisfied, then ${T}_{1}$ is called ${T}_{2}$distributive or we say that ${T}_{1}$ is distributive over ${T}_{2}$.
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 ${T}_{2}$.
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.
${T}_{L}$ 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, ${T}^{\prime}={T}_{\wedge}$.
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.
T norm on the lattice $\mathbf{(}\mathit{L}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{,}\mathit{a}\mathbf{,}\mathbf{1}\mathbf{\}}\mathbf{,}\mathbf{\le}\mathbf{)}$
T  0  x  y  z  a  1 

0  0  0  0  0  0  0 
x  0  0  0  0  0  x 
y  0  0  y  0  y  y 
z  0  0  0  z  z  z 
a  0  0  y  z  a  a 
1  0  x  y  z  a  1 
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.
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]
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.
holds. Thus, ${T}_{\phi}^{1}$ is ${T}_{\phi}^{2}$distributive.
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.
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.
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.
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.
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.
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.
 (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,${T}_{u}(x,z)=x\wedge z\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{T}_{u}(y,z)=inf\{u\wedge {\mu}_{i}{\mu}_{i}\in {S}_{\alpha}\}.$

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

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 $z\notin {S}_{\alpha}$.
So, the equality holds again.
Thus, the equality is satisfied.
whence the equality holds.
In this case, the equality is satisfied.

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${T}_{u}(x,{T}_{u}(y,z))={T}_{u}(x,y\wedge z)=x\wedge (y\wedge z)$
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.
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.
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.
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.
Declarations
Acknowledgements
Dedicated to Professor Hari M Srivastava.
Authors’ Affiliations
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