# Some notes on *T*-partial order

- Funda Karaçal
^{1}and - Emel Aşıcı
^{1}Email author

**2013**:219

https://doi.org/10.1186/1029-242X-2013-219

© Karaçal and Aşıcı; licensee Springer 2013

**Received: **4 December 2012

**Accepted: **18 April 2013

**Published: **30 April 2013

## Abstract

In this study, by means of the *T*-partial order defined in Karaçal and Kesicioğlu (Kybernetika 47:300-314, 2011), an equivalence relation on the class of *t*-norms on $([0,1],\le ,0,1)$ is defined. Then, it is showed that the equivalence class of the weakest *t*-norm ${T}_{D}$ on $[0,1]$ contains a *t*-norm which is different from ${T}_{D}$. Finally, defining the sets of some incomparable elements with any $x\in (0,1)$ according to ${\u2aaf}_{T}$, these sets are studied.

**MSC:**03E72, 03B52.

### Keywords

triangular norm*T*-partial order

## 1 Introduction

Triangular norms were originally studied in the framework of probabilistic metric spaces [1–4] aiming at an extension of the triangle inequality and following some ideas of Menger [5]. Later on, they turned out to be interpretations of the conjunction in many-valued logics [6–8], in particular in fuzzy logics, where the unit interval serves as a set of truth values.

*t*-norms on a bounded lattice was introduced. For any elements

*x*,

*y*of a bounded lattice

where *T* is a *t*-norm. This order ${\u2aaf}_{T}$ is called a *t*-partial order of *T*. Moreover, some connections between the natural order and the *t*-partial order ${\u2aaf}_{T}$ were studied.

In [10], it was investigated that ${\u2aaf}_{T}$ implies the natural order but its converse needs not be true. It was showed that a partially ordered set is not a lattice with respect to ${\u2aaf}_{T}$. Some sets which are lattices with respect to ${\u2aaf}_{T}$ under some special conditions were determined. For more details on *t*-norms on bounded lattices, we refer to [11–17].

In the present paper, we introduce an equivalence on the class of *t*-norms on $([0,1],\le ,0,1)$ based on the equality of the sets of all incomparable elements with respect to ${\u2aaf}_{T}$. The paper is organized as follows. We shortly recall some basic notions in Section 2. In Section 3, we define an equivalence on the class of *t*-norms on $([0,1],\le ,0,1)$, and we determine the equivalence class of the weakest *t*-norm ${T}_{D}$ on $[0,1]$. Thus, we obtain that the equivalence class of the weakest *t*-norm ${T}_{D}$ contains a *t*-norm which is different from a *t*-norm ${T}_{D}$. We obtain that for arbitrary $m\in {K}_{T}$, there exists an element ${y}_{m}\in {K}_{T}$ such that $T(m,\cdot ):[0,1]\to [0,m]$ or $T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}]$ is not continuous.

## 2 Basic definitions and properties

**Definition 2.1** [18]

A triangular norm (*t*-norm for short) is a binary operation *T* on the unit interval $[0,1]$, *i.e.*, a function $T:[0,1]\times [0,1]\to [0,1]$, such that for all $x,y,z\in [0,1]$ the following four axioms are satisfied:

(T1) $T(x,y)=T(y,x)$ (commutativity);

(T2) $T(x,T(y,z))=T(T(x,y),z)$ (associativity);

(T3) $T(x,y)\le T(x,z)$ whenever $y\le z$ (monotonicity);

(T4) $T(x,1)=x$ (boundary condition).

**Example 2.1** [18]

*t*-norms ${T}_{M}$, ${T}_{P}$, ${T}_{L}$, ${T}_{D}$ given by, respectively,

Also, *t*-norms on a bounded lattice $(L,\le ,0,1)$ are defined in a similar way, and then extremal *t*-norms ${T}_{W}$ as well as ${T}_{\wedge}$ on *L* are defined as ${T}_{D}$ and ${T}_{M}$ on $[0,1]$.

**Remark 2.1** [18]

- (i)Directly from Definition 2.1, we can deduce that, for all $x\in [0,1]$, each
*t*-norm*T*satisfies the following additional boundary conditions:$\begin{array}{c}T(0,x)=0,\hfill \\ T(1,x)=x.\hfill \end{array}$Therefore, all

*t*-norms coincide on the boundary of the unit square $[0,1]\times [0,1]$. - (ii)The monotonicity of a
*t*-norm*T*in its second component described by (T3) is, together with the commutativity (T1), equivalent to the monotonicity in both components,*i.e.*, to$T({x}_{1},{y}_{1})\le T({x}_{2},{y}_{2})\phantom{\rule{1em}{0ex}}\text{whenever}\phantom{\rule{1em}{0ex}}{x}_{1}\le {x}_{2}\text{and}{y}_{1}\le {y}_{2}.$(2.1)

**Definition 2.2** [18]

**Proposition 2.1** [18]

*A function* $F:[0,1]\times [0,1]\to [0,1]$ *which is non*-*decreasing*, *i*.*e*., *which satisfies* (2.1), *is continuous if and only if it is continuous in each component*, *i*.*e*., *if for all* ${x}_{0},{y}_{0}\in [0,1]$, *both the vertical section* $F({x}_{0},\cdot ):[0,1]\to [0,1]$ *and the horizontal section* $F(\cdot ,{y}_{0}):[0,1]\to [0,1]$ *are continuous functions in one variable*.

**Proposition 2.2** [18]

*A non*-

*decreasing function*$F:[0,1]\times [0,1]\to [0,1]$

*is lower semicontinuous if and only if it is left*-

*continuous in each component*,

*i*.

*e*.,

*if for all*${x}_{0},{y}_{0}\in [0,1]$

*and for all sequences*${({x}_{n})}_{n\in \mathbb{N}},{({y}_{n})}_{n\in \mathbb{N}}\in {[0,1]}^{\mathbb{N}}$,

*we have*

By the same token, the upper semicontinuity of a non-decreasing function $F:[0,1]\times [0,1]\to [0,1]$ is equivalent to its right-continuity in each component.

**Proposition 2.3** [7]

*For a non*-

*decreasing function*$F:{\mathbb{I}}^{n}\to \mathbb{R}$ ($\mathbb{I}$

*an interval*),

*the following conditions are equivalent*:

- (i)
*F**is continuous*; - (ii)
*F**is continuous in each variable*,*i*.*e*.,*for any*$x\in {\mathbb{I}}^{n}$*and any*$i\in \{1,2,\dots ,n\}$,*the unary function*$u\to F({x}_{1},\dots ,{x}_{i-1},u,{x}_{i+1},\dots ,{x}_{n})$*is continuous*; - (iii)
*F**has the intermediate value property*:*For any*$x,y\in {\mathbb{I}}^{n}$,*with*$x\le y$,*and any*$c\in [F(x),F(y)]$,*there exists*$z\in {\mathbb{I}}^{n}$,*with*$x\le z\le y$,*such that*$F(z)=c$.

**Definition 2.3** [10]

*L*be a bounded lattice, let

*T*be a

*t*-norm on

*L*. The order defined as follows is called a

*t*-order (triangular order) for a

*t*-norm

*T*.

**Example 2.2** [10]

*t*-norm ${T}_{W}$. Then $a\le b$, but $a{\u22e0}_{{T}_{W}}b$. We suppose that $a{\u2aaf}_{{T}_{W}}b$. Then there exists an element $\ell \in L$ such that ${T}_{W}(\ell ,b)=a$. If $\ell =0$, then it is obtained that $a=0$, a contradiction. If $\ell =a,b$ or

*c*, then we have that ${T}_{W}(\ell ,b)=0=a$, a contradiction. If $\ell =1$, then it is obtained that ${T}_{W}(1,b)=b=a$, which is not possible. So, there does not exist any element $\ell \in L$ satisfying ${T}_{W}(\ell ,b)=a$. Thus, $a{\u22e0}_{{T}_{W}}b$. Then the order ${\u2aaf}_{{T}_{W}}$ on

*L*is as in Figure 2.

**Proposition 2.4** [10]

*Let* *L* *be a bounded lattice*, *let* *T* *be a* *t*-*norm on* *L*. *Then the binary relation* ${\preccurlyeq}_{T}$ *is a partial order on* *L*.

**Definition 2.4** [10]

This partial order ${\preccurlyeq}_{T}$ is called a *T*-partial order on *L*.

**Definition 2.5**Let

*T*be a

*t*-norm on $[0,1]$ and let ${K}_{T}$ be defined by

We will use the notation ${K}_{T}$ to denote the set of all incomparable elements with respect to ${\u2aaf}_{T}$.

## 3 The equivalence of any two *t*-norms

Let *L* be a lattice and let *T* be any *t*-norm on *L*. In [10], a partial order for a *t*-norm *T* on *L* was defined. In this section, we define an equivalence relation with the help of the sets of all incomparable elements with respect to ${\u2aaf}_{T}$. The above introduced *T*-partial order allows us to introduce the next equivalence relation on the class of all *t*-norms on $([0,1],\le ,0,1)$.

**Definition 3.1**Let $([0,1],\le ,0,1)$ be the unit interval. Define a relation ∼ on the class of all

*t*-norms on $([0,1],\le ,0,1)$ by ${T}_{1}\sim {T}_{2}$ if and only if the set of all incomparable elements with respect to the ${T}_{1}$-partial order is equal to the set of all incomparable elements with respect to the ${T}_{2}$-partial order, that is,

**Proposition 3.1** *The relation* ∼ *given in Definition * 3.1 *is an equivalence relation*.

*Proof* Let ${T}_{1}$, ${T}_{2}$ and ${T}_{3}$ be *t*-norms on $([0,1],\le ,0,1)$. Since ${K}_{{T}_{1}}={K}_{{T}_{1}}$, it is obtained that ${T}_{1}\sim {T}_{1}$. Thus, the reflexivity is satisfied.

Let ${T}_{1}\sim {T}_{2}$. Then we have that ${K}_{{T}_{1}}={K}_{{T}_{2}}$, and since ${K}_{{T}_{2}}={K}_{{T}_{1}}$, it is obtained that ${T}_{2}\sim {T}_{1}$. Thus, the symmetry is satisfied. Let ${T}_{1}\sim {T}_{2}$ and ${T}_{2}\sim {T}_{3}$. Then we have ${K}_{{T}_{1}}={K}_{{T}_{2}}$ and ${K}_{{T}_{2}}={K}_{{T}_{3}}$. Since ${K}_{{T}_{1}}={K}_{{T}_{3}}$, it is obtained that ${T}_{1}\sim {T}_{3}$. This means that the relation ∼ satisfies the transitivity. So, we have that ∼ is an equivalence relation. □

**Definition 3.2**For a given

*t*-norm

*T*on $([0,1],\le ,0,1)$, we denote by $\overline{T}$ the ∼ equivalence class linked to

*T*,

*i.e.*,

Proposition 3.2 shows that the equivalence class of the *t*-norm ${T}_{D}$ contains a *t*-norm which is different from ${T}_{D}$.

**Proposition 3.2** *Let the* *t*-*norm* ${T}_{D}$ *be on* $[0,1]$. *Then* $\overline{{T}_{D}}\ne \{{T}_{D}\}$.

We give a contrary example as follows for the proof of Proposition 3.2.

**Example 3.1**Consider the

*t*-norm $T:[0,1]\times [0,1]\to [0,1]$ defined by

Then ${K}_{T}={K}_{{T}_{D}}$. Firstly, let us show that ${K}_{T}=(0,1)$. Let $x\in (0,1)$ and $y=\frac{2x}{3}$. Then $y<x$, but $y{\u22e0}_{T}x$. Suppose that $y{\preccurlyeq}_{T}x$. Then, for some *ℓ*, $T(\ell ,x)=\frac{2x}{3}$. Since $x\ne y$, it is not possible $\ell =1$. Then $\frac{2x}{3}=T(\ell ,x)=\frac{\ell x}{2}$, whence it is obtained that $\ell =\frac{4}{3}$, a contradiction. Since for any $x\in (0,1)$ there exists an element $y=\frac{2x}{3}$ such that $\frac{2x}{3}<x$ but $\frac{2x}{3}{\u22e0}_{T}x$, $x\in {K}_{T}$. Conversely, for any *t*-norm *T*, it is clear that ${K}_{T}\subseteq (0,1)$. So, it is obtained that ${K}_{T}=(0,1)$.

Now, we will show that ${K}_{{T}_{D}}=(0,1)$. Let $x\in (0,1)$. For any $y\in (0,1)$ with $x<y$, it is obvious that $x{\u22e0}_{{T}_{D}}y$. Otherwise, it would be ${T}_{D}(\ell ,y)=x$ for some *ℓ*. Since $x\ne y$, $\ell \ne 1$. Thus, it must be $x=0$ for $\ell ,y\in (0,1)$, a contradiction. Since there is an element y with $x<y$ such that $x{\u22e0}_{{T}_{D}}y$, $x\in {K}_{{T}_{D}}$. This shows that ${K}_{{T}_{D}}=(0,1)$. So, it is obtained that ${K}_{T}={K}_{{T}_{D}}$.

**Definition 3.3**Let

*T*be a

*t*-norm on $[0,1]$ and let ${K}_{x\downarrow}$, ${K}_{x\uparrow}$ be defined by

**Lemma 3.1** *Let* *T* *be a* *t*-*norm on* $[0,1]$ *and* $x\in {K}_{T}$ *be arbitrarily chosen*. *If* *T* *is continuous at* $(x,y)$ *for all* $y\in [0,1]$, *then* ${K}_{x\downarrow}=\mathrm{\varnothing}$.

*Proof* Let *T* be a *t*-norm on $[0,1]$ and let $x\in {K}_{T}$ be arbitrarily chosen. Suppose that ${K}_{x\downarrow}\ne \mathrm{\varnothing}$. Then there exists an element ${y}_{0}\in [0,1]$ such that ${y}_{0}\le x$, but ${y}_{0}{\u22e0}_{T}x$. Since the *t*-norm $T(x,\cdot ):[0,1]\to [0,x]$ is continuous, there exists an element $z\in [0,1]$ such that $T(x,z)={y}_{0}$ for ${y}_{0}\in [0,x]$ by Proposition 2.3. So, it is obtained that ${y}_{0}{\u2aaf}_{T}x$, a contradiction. Therefore we have that ${K}_{x\downarrow}=\mathrm{\varnothing}$. □

**Lemma 3.2** *Let* *T* *be a* *t*-*norm on* $[0,1]$ *and the function* $T({x}_{0},\cdot ):[0,1]\to [0,{x}_{0}]$ *be continuous*. *Then*, *for all* $y\in [0,1]$ *with* $y\le {x}_{0}$, *we have that* $y{\u2aaf}_{T}{x}_{0}$.

*Proof* Let *T* be a *t*-norm on $[0,1]$ and the function $T({x}_{0},\cdot ):[0,1]\to [0,{x}_{0}]$ be continuous. Suppose that there exists an element ${y}_{0}\in [0,1]$ such that ${y}_{0}\le {x}_{0}$ and ${y}_{0}{\u22e0}_{T}{x}_{0}$. Since $T({x}_{0},\cdot ):[0,1]\to [0,{x}_{0}]$ is continuous, there exists an element $t\in [0,1]$ such that $T({x}_{0},t)={y}_{0}$ for ${y}_{0}\in [0,{x}_{0}]$ by Proposition 2.3. Thus, it is obtained that ${y}_{0}{\u2aaf}_{T}{x}_{0}$, a contradiction. Therefore, for all $y\in [0,1]$ with $y\le {x}_{0}$, we have that $y{\u2aaf}_{T}{x}_{0}$. □

**Theorem 3.1** *Let* *T* *be a* *t*-*norm on* $[0,1]$ *and* ${K}_{T}\ne \mathrm{\varnothing}$. *Then*, *for arbitrary* $m\in {K}_{T}$, *there exists an element* ${y}_{m}\in {K}_{T}$ *such that* $T(m,\cdot ):[0,1]\to [0,m]$ *or* $T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}]$ *is not continuous*.

*Proof* Let *T* be a *t*-norm on $[0,1]$ and ${K}_{T}\ne \mathrm{\varnothing}$. Suppose that $T(x,\cdot ):[0,1]\to [0,x]$ is continuous for $x\in {K}_{T}$. Choose $m\in {K}_{T}$ arbitrarily. Then there exists an element ${y}_{m}\in {K}_{T}$ such that $m<{y}_{m}$ but $m{\u22e0}_{T}{y}_{m}$, or ${y}_{m}<m$ but ${y}_{m}{\u22e0}_{T}m$. Let $m<{y}_{m}$ but $m{\u22e0}_{T}{y}_{m}$. Since $T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}]$ is continuous, then it is obtained that $m{\u2aaf}_{T}{y}_{m}$ by Lemma 3.2, a contradiction. Let ${y}_{m}<m$ but ${y}_{m}{\u22e0}_{T}m$. Since $T(m,\cdot ):[0,1]\to [0,m]$ is continuous, then it is obtained that ${y}_{m}{\u2aaf}_{T}m$ by Lemma 3.2, a contradiction. Therefore, for arbitrary $m\in {K}_{T}$, there exists an element ${y}_{m}\in {K}_{T}$ such that $T(m,\cdot ):[0,1]\to [0,m]$ or $T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}]$ is not continuous. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Authors’ Affiliations

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