# Pseudo triangular norms on bounded trellises

Abdelkrim Mehenni<EMAIL_ADDRESS>LA3C Laboratory, Mathematics Faculty,
Department of Algebra and Number Theory, University of Science and Technology
Houari Boumediene, Bp 32, Bab Ezzouar, Algiers, Algeria  Lemnaouar Zedam
<EMAIL_ADDRESS>Laboratory of Pure and Applied Mathematics,
Department of Mathematics, University of M’sila, Algeria

###### Abstract

In this paper, we introduce the notion of pseudo-t-norm on bounded trellises
(also known as weakly associative lattices) as an extension of meet and join
operations (resp. t-norm) on bounded trellises, and provide some basic
examples. We provide a first generic construction method that allows extending
a pseudo-t-norm on bounded trellises. Also, we introduce the notion of
T-distributivity for any pseudo-t-norm $T$ on bounded trellises. Moreover, We
determine the relationship between pseudo-t-norms and isomorphisms on bounded
trellises.

###### keywords:

Binary operation, trellis, t-norm, pseudo-t-norm, T-distributive.

##  1 Introduction

Binary operations play an important role in many of the technological tasks
scientists are faced with nowadays. They are specifically important in many
problems related to the fusion of information. More generally, binary
operations are widely used in pure mathematics (e.g., group theory, monoids
theory) (see, e.g., [5, 14, 15]). Binary operations have become essential
tools in the unit interval and lattices and its applications, several notions
and properties (see, e.g., [17, 21]).

Triangular norms (t-norms) (as specific binary operations) were introduced by
Karl Menger [8] with the goal of constructing metric spaces using
probabilistic distributions (and therefore values in the interval $[0,1]$),
instead of using real numbers, to describe the distance between two elements.
Besides, the original proposal is very weak even including triangular conorms
(t-conorms). However, only with the work of Berthold Schweizer and Abe Sklar
in [18] gave an axiomatic for t-norms as they are used today. Also, they play
an important role in theories of fuzzy sets and fuzzy logic [1] as they
generalize the basic connectives between fuzzy sets. Thus, the main
characteristic of the binary operations is that they are used in a large
number of areas and disciplines.

In [3, 4] they were generalized the notion of t-norm on bounded partially
ordered sets, which is a more general structure than bounded lattice. In [16]
it was considered an extension of t-norm for bounded lattice which coincides
with the one given by [4] and [3].

In 1970, E. Fried [7] introduced the notions of pseudo-ordered sets and
trellises (also called weakly associative lattices or WA-lattices). Trellises
are generalization of lattices by considering sets with a reflexive and
antisymmetric order, but not necessarily transitive. In 1971, H. X. Skala [19]
investigated some properties of this notion and provide some examples in
particular classes of trellis and the notion of the trellis itself can be
interpreted in terms of binary operations on it (see, e.g., [8, 21]).

The aim of the present paper is to introduce the notion of a pseudo-triangular
norm (pseudo-t-norm, for short) and a pseudo-triangular conorm (pseudo-t-
conorm, for short) on a bounded trellis as a generalization of triangular norm
and triangular conorm on a bounded lattice and bounded trellis (see, e.g., [3,
4, 6, 13, 2, 10]), and we investigate their fundamental properties and some
constructions of pseudo-t-norms (resp. pseudo-t-conorms). More specifically,
we show necessary and sufficient conditions under which a given binary
operation on a trellis coincides with its meet- and its join-operation.
Mourover, we characterize pseudo-t-norms (resp. pseudo-t-conorms) on bounded
trellises with respect to the $F$-distributivity. Furthermore, we study the
relationship among pseudo-t-norms and isomorphisms on a bounded trellises.

This paper is organized as follows. We briefly recall some basic concepts in
Section 2. In section 3, we introduce the notion of pseudo-t-norm and pseudo-
t-conorm on bounded trellises and investigate their properties. In section 4,
we construct some elements, some constructions and showing necessary and
sufficient conditions under which a given binary operation (resp. pseudo-t-
norm and pseudo-t-conorm) on a trellis (resp. bounded trellis) coincide with
its meet- and its join-operation. In section 5, we introduce the notion of
T-distributivity for any pseudo-t-norm (resp. pseudo-t-conorm) on bounded
trellises and characterize some properties. In section 6, we show that any
isomorphism act on pseudo-t-norms generating also a new pseudo-t-norms on
bounded trellises. Finally, we present some conclusions and discuss future
research in Section 7.

##  2 Basic concepts

This section serves an introductory purpose. First, we recall some definitions
and properties related to pseudo-ordered sets and trellises. Second, we
present some specific elements of a trellis that will be needed throughout
this paper.

###  2.1 Pseudo-ordered sets and trellises

In this subsection, we recall the notions of pseudo-ordered sets and
trellises; more information can be found in [7, 19, 20]. A pseudo-order
(relation) $\unlhd$ on a set $X$ is a binary relation on $X$ that is reflexive
(i.e., $x\unlhd x$, for any $x\in X$) and antisymmetric (i.e., $x\unlhd y$ and
$y\unlhd x$ implies $x=y$, for any $x,y\in X$). A set $X$ equipped with a
pseudo-order relation $\unlhd$ is called a pseudo-ordered set (psoset, for
short) and denoted by $(X,\unlhd)$. For any two elements $a,b\in X$, if
$a\unlhd b$ and $a\neq b$, then we write $a\lhd b$; if $a\unlhd b$ does not
hold, then we also write $a\ntrianglelefteq b$. Similarly as for partially
ordered sets, a finite pseudo-ordered set can be represented by a Hasse-type
diagram with the following difference: if $x$ and $y$ are not related, while
in a partially ordered set this would be implied by transitivity, then $x$ and
$y$ are joined by a dashed curve.

######  Example 2.1.

Consider the pseudo-order relation $\unlhd$ on $X=\\{a,b,c,d,e,f\\}$
represented by the Hasse-type diagram in Fig. 1. Here, $b\unlhd d$, $c\unlhd
d$, $d\unlhd e$, while $b\ntrianglelefteq e$ and $c\unlhd e$.

$a$$b$$c$$d$$e$$f$ Figure 1: Hasse-type diagram of $(X,\unlhd)$.

The notions of minimal/maximal element, smallest/greatest element, lower/upper
bound, greatest lower bound (or infimum), least upper bound (or supremum) for
psosets are defined in the same way as the corresponding notions for partially
ordered sets. For a subset $A$ of a psoset $(X,\unlhd)$, the antisymmetry of
the pseudo-order implies that if $A$ has an infimum (resp. supremum), then it
is unique, and is denoted by $\bigwedge A$ (resp. $\bigvee A$). If
$A=\\{a,b\\}$, then we write $a\wedge b$ (called meet) instead of
$\bigwedge\\{a,b\\}$ and $a\vee b$ (called join) instead of
$\bigvee\\{a,b\\}$.

######  Definition 2.1.

[20] Let $(X,\unlhd)$ be a psoset. For $x,y\in X$, we write $x\lesssim y$ if
there exists a finite sequence $(x_{1},\ldots,x_{n})$ such that $x\unlhd
x_{1}\unlhd\ldots\unlhd x_{n}\unlhd y$.

Note that the relation $\lesssim$ is a pre-order relation, i.e., it is
reflexive and transitive, but not necessarily antisymmetric. If for any
$x,y\in X$, it holds that $x\lesssim y$ or $y\lesssim x$, then $(X,\unlhd)$ is
called a pseudo-chain.

######  Definition 2.2.

[9] A $\wedge$-semi-trellis (resp. $\vee$-semi-trellis) is a psoset
$(X,\unlhd)$ such that $x\wedge y$ (resp. $x\vee y$) exists for all $x,y\in
X$.

######  Definition 2.3.

[19] A trellis is a psoset that is both a $\wedge$-semi-trellis and a
$\vee$-semi-trellis. In other words, a trellis is an algebra
$(X,\wedge,\vee)$, where $X$ is a nonempty set with the binary operations
$\wedge$ and $\vee$ satisfying the following properties, for any $a,b,c\in X$:

  1. (i)

$a\vee b=b\vee a$ and $a\wedge b=b\wedge a$ (commutativity) ;

  2. (ii)

$a\vee(b\wedge a)=a=a\wedge(b\vee a)$ (absorption) ;

  3. (iii)

$a\vee((a\wedge b)\vee(a\wedge c))=a=a\wedge((a\vee b)\wedge(a\vee c))$ (part-
preservation) .

######  Theorem 2.1.

[19] A set $X$ with two commutative, absorptive, and part-preserving
operations $\wedge$ and $\vee$ is a trellis if $a\unlhd b$ is defined as
$a\wedge b=a$ and/or $a\vee b=b$. The operations are also idempotent (i.e.,
$x\wedge x=x\vee x=x$, for any $x\in X$).

######  Remark 2.1.

One can observe that the difference between the notions of a lattice and a
trellis is that the operations $\wedge$ and $\vee$ are not required to be
associative in the case of a trellis.

A bounded trellis is a trellis $(X,\unlhd,\wedge,\vee)$ that additionally has
a smallest element denoted by $0$ and a greatest element denoted by $1$
satisfying $0\unlhd x\unlhd 1$, for any $x\in X$. For a bounded trellis, the
notation $(X,\unlhd,\wedge,\vee,0,1)$ is used. Also, a trellis
$(X,\unlhd,\wedge,\vee)$ is called complete if every subset of $X$ has an
infimum and a supremum.

Let $(X,\unlhd,\wedge,\vee)$ and $(Y,\sqsubseteq,\sqcap,\sqcup)$ be two
trellises. A mapping $\varphi:X\to Y$ is called a homomorphism, if it
satisfies $\varphi(x\wedge y)=\varphi(x)\sqcap\varphi(y)$ and $\varphi(x\vee
y)=\varphi(x)\sqcup\varphi(y)$, for any $x,y\in X$. An isomorphism is a
bijective homomorphism.

######  Definition 2.4.

[9] Let $(X,\unlhd,\wedge,\vee)$ be a trellis and $A\subseteq X$. Then

  1. (i)

$A$ is called a sub-trellis of $X$ if $x\wedge y\in A$ and $x\vee y\in A$, for
any $x,y\in A$;

  2. (ii)

$A$ is called a sub-lattice of $X$ if is a sub-trellis and $\unlhd$ is
transitive on $A$.

######  Theorem 2.2.

[20] Let $(X,\unlhd,\wedge,\vee)$ be a trellis. The following statements are
equivalent:

  1. (i)

$\unlhd$ is transitive;

  2. (ii)

the meet $\wedge$ and the join $\vee$ operations are associative;

  3. (iii)

one of the operations meet $\wedge$ or join $\vee$ is associative.

######  Definition 2.5.

[20] A trellis $(X,\unlhd,\wedge,\vee)$ is said to be modular, if $x\unlhd z$
implies that $x\vee(y\wedge z)=(x\vee y)\wedge z$, for any $y\in X$.

We will also use the following results.

######  Proposition 2.1.

[19] Let $(X,\unlhd,\wedge,\vee)$ be a modular trellis and $x,y,z\in X$. If
$x\unlhd y\unlhd z$, then $x\wedge z\unlhd y\unlhd x\vee z$.

######  Proposition 2.2.

[19] Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded modular trellis. If
$x\unlhd z$ and $x\vee y=1$, then $x\wedge y\unlhd z$, for any $x,y,z\in X$.

###  2.2 Specific elements in a trellis

In this subsection, we present some specific elements in a trellis that will
play an important role in this paper.

######  Definition 2.6.

[20] Let $(X,\unlhd,\wedge,\vee)$ be a trellis. An element $\alpha\in X$ is
called:

  1. (i)

right-transitive, if $\alpha\unlhd x\unlhd y$ implies $\alpha\unlhd y$, for
any $x,y\in X$;

  2. (ii)

left-transitive, if $x\unlhd y\unlhd\alpha$ implies $x\unlhd\alpha$, for any
$x,y\in X$;

  3. (iii)

middle-transitive, if $x\unlhd\alpha\unlhd y$ implies $x\unlhd y$, for any
$x,y\in X$;

  4. (iv)

transitive, if it is right-, left- and middle-transitive.

######  Definition 2.7.

[20] Let $(X,\unlhd,\wedge,\vee)$ be a trellis.

  1. (i)

A 3-tuple $(x,y,z)\in X^{3}$ is called $\wedge$-associative (resp.
$\vee$-associative), if $(x\wedge y)\wedge z=x\wedge(y\wedge z)$ (resp.
$(x\vee y)\vee z=x\vee(y\vee z)$);

  2. (ii)

An element $\alpha\in X$ is called $\wedge$-associative (resp.
$\vee$-associative), if any 3-tuple in $X$ including $\alpha$ is
$\wedge$-associative (resp. $\vee$-associative);

  3. (iii)

An element $\alpha\in X$ is called associative if it is both $\wedge$\- and
$\vee$-associative.

Note that for the different notions of associative element $\alpha\in X$, due
to the commutativity of the meet and the join operations it is sufficient to
consider only 3-tuples if the type $(\alpha,x,y)$.

The following results show the links between the above notions.

######  Proposition 2.3.

[20] Let $(X,\unlhd,\wedge,\vee)$ be a trellis. Any $\wedge$-associative or
$\vee$-associative element is transitive, but the converse does not hold.

######  Theorem 2.3.

[20] Let $(X,\unlhd,\wedge,\vee)$ be a pseudo-chain or a modular trellis. Then
it holds that an element is associative if and only if it is transitive.

######  Theorem 2.4.

[9] A trellis of finite length is complete if and only if every cycle has the
meet and the join.

##  3 Pseudo-triangular norms on bounded trellises

This section is devoted to introduce the notions of pseudo-triangular norm on
a bounded trellis and to investigate their various properties and to present
some interesting examples in bounded trellises. These notions are inspired
from triangular norms and triangular conorms on bounded lattices and bounded
trellises (see, e.g., [3, 4, 12, 13]). Also, we provide a construction to
obtain new pseudo-t-norms on bounded trellises. In particular, we give
necessary and sufficient conditions under which a pseudo-t-norm on a bounded
trellis coincides with its meet ($\wedge$) operation.

###  3.1 Binary operations on trellises

In this subsection, we present some basic definitions and properties of binary
operations on a psoset or trellis. Some of them are adopted from the
corresponding notions on a poset or lattice (see, e.g., [8, 11, 21]). A binary
operation $F$ on a psoset $(X,\unlhd)$ is called:

  1. (i)

commutative, if $F(x,y)=F(y,x)$, for any $x,y\in X$;

  2. (ii)

associative, if $F(x,F(y,z))=F(F(x,y),z)$, for any $x,y,z\in X$;

  3. (iii)

idempotent, if $F(x,x)=x$, for any $x\in X$;

  4. (iv)

increasing, if $x\unlhd y$ implies $F(x,z)\unlhd F(y,z)$, for any $z\in X$.

An element $e\in X$ is called a neutral element of $F$, if $F(e,x)=F(x,e)=x$,
for any $x\in X$.

A binary operation $F$ on a trellis $(X,\unlhd,\wedge,\vee)$ is called:

  1. (i)

conjunctive, if $F(x,y)\unlhd x\wedge y$, for any $x,y\in X$;

  2. (ii)

disjunctive, if $x\vee y\unlhd F(x,y)$, for any $x,y\in X$.

######  Remark 3.1.

Consider a trellis $(X,\unlhd,\wedge,\vee)$.Then the meet $\wedge$ (resp. join
$\vee$) is conjunctive (resp. disjunctive).

######  Notation 3.1.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis. We denote by:

  1. (i)

$X^{tr}$ : the set of all transitive elements of $X$;

  2. (ii)

$X^{\wedge-ass}$: the set of all $\wedge$-associative elements of $X$;

  3. (iii)

$X^{\vee-ass}$: the set of all $\vee$-associative elements of $X$;

  4. (iv)

$X^{ass}$: the set of all associative elements of $X$;

  5. (v)

$X^{dis}$: the set of all distributive elements of $X$.

######  Notation 3.2.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis, $A\in X$ and $x_{1},\cdots,x_{n}\in
X$, for any $n\geq 1$. If $\\{x_{1},\cdots,x_{n}\\}\cap A\neq\emptyset$, then
we said that $[x_{1},\cdots,x_{n}]\in A$.

The following proposition is immediate.

######  Proposition 3.1.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis. Then it holds that

  1. (i)

$x\unlhd y\text{ implies }x\wedge z\unlhd y\wedge z\text{ and }z\wedge x\unlhd
z\wedge y,\text{ for any }([x,y]\in X^{\mathrm{tr}}\text{ and }z\in X);$

  2. (ii)

$x\unlhd y\text{ implies }x\vee z\unlhd y\vee z\text{ and }z\vee x\unlhd z\vee
y,\text{ for any }([x,y]\in X^{\mathrm{tr}}\text{ and }z\in X);$

  3. (iii)

$(x\wedge y)\wedge z=x\wedge(y\wedge z)\text{, for any }([x,y,z]\in X^{\wedge-
ass}\text{ or }[x,y,z]\in X^{\wedge-ass})\ ;$

  4. (iv)

$(x\vee y)\vee z=x\vee(y\vee z)\text{, for any }([x,y,z]\in X^{\wedge-
ass}\text{ or }[x,y,z]\in X^{\vee-ass})\ .$

Next, we extends the increasingness and associativity properties of the meet
and the join operations on bounded trellises and leads the following
definition.

######  Definition 3.1.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis and $F$ a binary operation on $X$.

  1. (i)

$F$ is called weakly-increasing if it satisfies:

$x\unlhd y\Rightarrow F(x,z)\unlhd F(y,z),\text{ for any }([x,y]\in
X^{\mathrm{tr}}\text{ and }z\in X);$

  2. (ii)

$F$ is weakly-associative if it satisfies:

$F(x,F(y,z))=F(F(x,y),z)\text{, for any }([x,y,z]\in X^{\wedge-ass}\text{ or
}[x,y,z]\in X^{\vee-ass})\ .$

Next, we illustrate the previous definition weakly-increasing and weakly-
associative operations on a bounded trellis.

######  Example 3.1.

Let $(X=\\{0,a,b,c,1\\},\unlhd,\wedge,\vee)$ be a trellis given by the Hasse
diagram in Figure 2 and $F,G$ two binary operations defined by the following
tables:

$F(x,y)$ | $0$ | $a$ | $b$ | $c$ | $1$
---|---|---|---|---|---
$0$ | $a$ | $a$ | $b$ | $c$ | $1$
$a$ | $b$ | $b$ | $c$ | $c$ | $1$
$b$ | $b$ | $b$ | $c$ | $c$ | $1$
$c$ | $c$ | $1$ | $1$ | $1$ | $1$
$1$ | $1$ | $1$ | $1$ | $1$ | $1$

$G(x,y)$ $0$ $a$ $b$ $c$ $1$ $0$ $0$ $0$ $0$ $0$ $0$ $a$ $0$ $a$ $b$ $c$ $1$
$b$ $0$ $a$ $c$ $c$ $1$ $c$ $0$ $0$ $1$ $b$ $c$ $1$ $0$ $a$ $c$ $c$ $1$

One easily verifies that $F$ is weakly-increasing and $G$ is weakly-
associative.

$0$$a$$b$$c$$1$ Figure 2: Hasse diagram of the trellis
$(X=\\{0,a,b,c,1\\},\unlhd)$.

###  3.2 Triangular norms on bounded trellises

In this subsection, we introduce the notion of triangular norm and on a
bounded trellis and we present some illustrative examples.

######  Definition 3.2.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. A binary operation
$T:X^{2}\to X$ is called a triangular norm (t-norm, for short), if it is
commutative, increasing, associative and has $1$ as neutral element, i.e.,
$T(1,x)=x$, for any $x\in X$.

Analogously, we define a triangular conorm on a bounded trellis.

######  Definition 3.3.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. A binary operation
$S:X^{2}\to X$ is called a pseudo-triangular conorm (pseudo-t-conorm, for
short), if it is commutative, increasing, associative and has $0$ as neutral
element, i.e., $S(0,x)=x$, for any $x\in X$.

######  Example 3.2.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $T_{D}$ a binary
operation defined on $X$ as follow:
$T_{D}(x,y)=\left\\{\begin{array}[]{ll}{x\wedge y}&{\text{ if
}x=1\leavevmode\nobreak\ \text{or}\leavevmode\nobreak\ y=1},\\\ {0}&{\text{
otherwise, }}\end{array}\right.$ and
$S_{D}(x,y)=\left\\{\begin{array}[]{ll}{x\vee y}&{\text{ if
}x=0\leavevmode\nobreak\ \text{or}\leavevmode\nobreak\ y=0},\\\ {1}&{\text{
otherwise. }}\end{array}\right.$

One easily verifies that $T_{D}$ is the smallest t-norm and $S_{D}$ is the
greatest t-conorm on $X$.

######  Example 3.3.

Let $(X=\\{0,a,b,c,1\\},\unlhd,\wedge,\vee)$ be a bounded trellis given by the
Hasse-type diagram in Figure 2 and $T$ a binary operation defined by the
following table:

$T(x,y)$ | $0$ | $a$ | $b$ | $c$ | $1$
---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $0$ | $0$ | $0$ | $a$
$b$ | $0$ | $0$ | $b$ | $b$ | $b$
$c$ | $0$ | $0$ | $b$ | $c$ | $c$
$1$ | $0$ | $a$ | $b$ | $c$ | $1$

One easily verifies that $T$ is a t-norm on $X$ such that
$T_{D}\vartriangleleft T$.

######  Example 3.4.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis such that for any $i\in$
$Coatom(X)$ and $j\in$ $Atom(X)$, the binary operations $T_{i}$ and $S_{j}$
defined as:

$T_{i}(x,y)=\left\\{\begin{array}[]{ll}{i}&{\text{ if }(x,y)=(i,i)},\\\
{T_{D}(x,y)}&{\text{ otherwise, }}\end{array}\right.$

is a t-norm on $X$. Moreover, $T_{D}\vartriangleleft T_{i}$.

######  Notation 3.3.

Let $(X,\unlhd,\wedge,\vee)$ be a bounded trellis. We denote by:

  1. (i)

$\mathcal{AO}_{1}(X)$: the class (or the set) of all t-norms on $X$;

  2. (ii)

$\mathcal{AO}_{0}(X)$: the class (or the set) of all t-conorms on $X$.

######  Remark 3.2.

In a bounded trellis $(X,\unlhd,\wedge,\vee,0,1)$ wish has not a structure of
a bounded lattice, there is not exist any idempotent t-norm on $X$.

It is natural that the trellis structure has cycles and dashed curves at the
same time. In the following results, we present relationship among cycles and
t-norms. First, we need the following definition of atom and coatom on a
trellis. This definition is a natural generalization of the same notions on a
lattice (see, e.g. [11]).

######  Definition 3.4.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. An element $\alpha\in
X$ is called:

  1. (i)

atom, if it is a minimal element of the set $X\backslash\\{0\\}$;

  2. (ii)

coatom, if it is a maximal element of the set $X\backslash\\{1\\}$.

We denote by $Atom(X)$ (resp. $Coatom(X)$), the set of all atoms (resp.
coatoms) of $X$.

The following propositions are immediate.

######  Proposition 3.2.

For a given bounded trellis $(X,\unlhd,\wedge,\vee,0,1)$ has a non-trivial
cycle $C$. Then $\\{0,1\\}\notin C$.

######  Proposition 3.3.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. If $X$ has a non-
trivial cycle $C$, then $x\notin Atom(X)\cup Coatom(X)$, for any $x\in C$.

######  Proposition 3.4.

Let $(X,\unlhd,\wedge,\vee)$ be a modular trellis. Then $X$ does not have any
non-trivial cycle.

###### Proof.

Let $C=\\{x,y,z\\}$ be a non-trivial cycle on $X$ such that $x\unlhd y\unlhd
z\unlhd x$. Since $z\unlhd x$, it holds from the modularity of $X$ that
$(z\vee y)\wedge x=z\wedge x=z$ and $z\vee(y\wedge x)=z\vee x=x$. Then $x=z$.
Consequently, $x=y=z$, contradiction. Hence, $X$ does not have any non-trivial
cycle. ∎

######  Proposition 3.5.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis has a non-trivial cycle $C$. Then it
holds that

$C\cap\\{X^{r-tr}\cup X^{\ell-tr}\\}=\emptyset$.

###### Proof.

Let $C=\\{x_{1},\cdots,x_{n}\\}$ be a non-trivial cycle such that $x_{1}\unlhd
x_{2}\unlhd\cdots\unlhd x_{n}\unlhd x_{1}$ and $n\geq 3$. Suppose that
$x_{i}\in X^{r-tr}$, for some $i\in\\{1,\cdots,n\\}$. Then $x_{i}\unlhd
x_{i-1}$. Since $x_{i-1}\unlhd x_{i}$, it holds that $x_{i}=x_{i-1}$.
Consequently, $x_{i}=x_{i-1}=x_{i-2}=\cdots=x_{1}=x_{n}=\cdots=x_{i+1}$. Thus,
$|C|=1$, a contradiction. Hence, $C\cap X^{r-tr}=\emptyset$. In similar way,
$C\cap X^{\ell-tr}=\emptyset$. Therefore, $C\cap\\{X^{r-tr}\cup X^{\ell-
tr}\\}=\emptyset$. ∎

######  Proposition 3.6.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis and $\\{C_{i}\\}_{i\in I}$ is the
set of all non-trivial cycles of three elements of $X$. If $T$ is a t-norm on
$X$, then $T(x,y)\notin C_{i}$, for any $x,y\in C_{i}$ and $i\in I$ and .
Moreover, $T(x,y)=T(z,t)$, for any $x,y,z,t\in C_{i}$ and $i\in I$ (i.e.,
$\bigcup_{a,b\in C_{i}}T(a,b)$ is a trivial cycle).

###### Proof.

Let $C=\\{x,y,z\\}\subseteq\bigcup_{i\in I}C_{i}$ be a non-trivial cycle such
that $x\unlhd y\unlhd z\unlhd x$. Suppose that $a,b\in C$ such that
$T(a,b)=x$. Since $T$ is a t-norm, it holds that $T(a,b)=x\unlhd a$ and
$T(a,b)=x\unlhd b$. Thus, $a\in\\{x,y\\}$ and $b\in\\{x,y\\}$. Then the
commutativity of $T$ implies that two cases:

  1. (i)

If $a=x$ and $b=y$, then $T(a,b)=T(x,y)=x$. Since $T(x,y)\unlhd T(y,z)$, it
follows two possible cases:

    1. (i)

If $T(y,z)=x$, then $T(y,z)=x\unlhd T(1,z)=z$. Thus, $x\unlhd z$. Hence,
$x=y=z$.

    2. (ii)

If $T(y,z)=y$, then $T(y,z)\unlhd T(y,x)=T(x,y)=x$. Thus, $y\unlhd x$. Hence,
$x=y=z$.

  2. (ii)

If $a=b=x$, then $T(a,b)=T(x,x)=x$. Since $T(x,x)\unlhd T(y,y)$, it holds that
two cases:

    1. (i)

If $T(y,y)=x$, then $T(x,x)=T(y,y)=x$. Since $T(x,x)\unlhd T(x,y)\unlhd
T(y,y)$, it holds that $T(x,y)=x$, a contradiction.

    2. (ii)

If $T(y,y)=y$, Since $T(x,x)\unlhd T(x,y)\unlhd T(y,y)$, it holds that
$T(x,y)=x$ or $T(x,y)=y$. Thus, $T(x,y)=x$ is a contradiction and
$T(x,y)=y\unlhd T(x,1)=x$. Hence, $y\unlhd x$. Hence, $x=y=z$.

  3. (iii)

If $a=b=y$, then $T(a,b)=T(y,y)=x$. Since $T(x,x)\unlhd T(y,y)$, it holds that
$T(x,x)=x$ or $T(x,x)=z$. Thus, $T(x,x)=x$ is a contradiction and
$T(x,x)=z\unlhd T(x,y)\unlhd T(y,y)=x$ implies that $T(x,y)=x$ or $T(x,y)=z$,
a contradiction.

In similar way, if $T(a,b)=y$ or $T(a,b)=z$, it follows that $x=y=z$.
Therefore, $T(a,b)\notin C$, for any $a,b\in C$. Next, let
$C^{\prime}=\\{\alpha,\beta,\gamma\\}$ an other non-trivial cycle such that
$|\bigcup_{a,b\in C}T(a,b)\cap C^{{}^{\prime}}|\geq 2$. Suppose that
$T(x,x)=\alpha$, $T(y,y)=\beta$. Then $T(z,z)=\gamma$. The fact that $T$ is a
t-norm implies $T(x,x)\unlhd T(x,y)\unlhd T(y,y)$. Then it holds that two
cases:

  1. (i)

If $T(x,y)=\alpha$, then from $T$ is a t-norm, it holds that $T(x,y)\unlhd
T(y,z)\unlhd T(z,z)$. Thus, $T(y,z)=\beta$ . The fact that
$T(y,z)=T(z,y)\unlhd T(x,y)$ implies $T(x,y)=T(y,z)$. Then $\alpha=\beta$.
Hence, $\alpha=\beta=\gamma$.

  2. (ii)

If $T(x,y)=\beta$, then from $T$ is a t-norm, it holds that
$T(x,y)=\beta\unlhd T(x,z)\unlhd T(x,x)=\alpha$. Thus, $T(x,z)=\gamma$. Since
$T(x,z)=T(z,x)\unlhd T(x,y)$, it holds that $\gamma\unlhd\beta$. Hence,
$\alpha=\beta=\gamma$.

In similar way, if $T(a,b)=\alpha$ and $T(c,d)=\beta$, for any $a,b,c,d\in C$,
it follows that $\alpha=\beta=\gamma$. Hence, $\bigcup_{a,b\in C}T(a,b)$ is a
trivial cycle (i.e., $T(x,y)=T(z,t)$, for any $x,y,z,t\in C$. ∎

######  Proposition 3.7.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis and $C_{n}$ is a non-trivial cycle
of $n$ elements of $X$. If $T$ is a t-norm on $X$, then $T(x,y)\notin C_{n}$,
for any $x,y\in C_{n}$. Moreover, $T(x,y)=T(z,t)$, for any $x,y,z,t\in C_{n}$
(i.e., $\bigcup_{a,b\in C_{n}}T(a,b)$ is a trivial cycle).

###### Proof.

Let $T$ is a t-norm on $X$ and $C_{3}$ is anon-trivial cycles of three
elements, then Proposition 3.6 guarantees that $T(x,y)\notin C_{3}$, for any
$x,y\in C_{3}$ and $T(x,y)=T(z,t)$, for any $x,y,z,t\in C_{3}$. Suppose that
$C_{n}=\\{x_{1},\cdots,x_{n}\\}$ be a non-trivial cycle such that
$T(x,y)\notin C_{n}$, for any $x,y\in C_{n}$ and $T(x,y)=T(z,t)$, for any
$x,y,z,t\in C_{n}$. Next, we prove that for
$C_{n+1}=\\{x_{1},\cdots,x_{n},x_{n+1}\\}$. First, let $T(x,y)\in C_{n+1}$,
for any $x,y\in C_{n}$. Since $T(x,y)\notin C_{n}$, for any $x,y\in C_{n}$, it
holds that $T(x,y)=x_{n+1}$. Suppose that $x=y=x_{n}$, then
$T(x,y)=T(x_{n},x_{n})=x_{n+1}$. Since $T$ is a t-norm, it holds that
$T(x_{n},x_{n})\unlhd x_{n}$. Thus, $x_{n}\unlhd x_{n+1}=T(x_{n},x_{n})\unlhd
x_{n}$. Hence, $T(x_{n},x_{n})=x_{n}\in C_{n}$, a contradiction. Thus,
$T(x,y)\notin C_{n+1}$, for any $x,y\in C_{n}$. Second, we prove that
$T(x,y)=T(x_{n+1},z)=T(x_{n+1},x_{n+1})$, for any $x,y,z\in C_{n}$. On the one
hand, since $T$ is a t-norm, it follows that $T(x_{n},x_{n})\unlhd
T(x_{n+1},x_{n+1})\unlhd T(x_{1},x_{1})=T(x_{n},x_{n})$. Then
$T(x_{n+1},x_{n+1})=T(x_{n},x_{n})$. Thus, $T(x_{n+1},x_{n+1})=T(x,y)$, for
any $x,y\in C_{n}$ (using our hypothesis). On the other hand,
$T(x_{n},z)\unlhd T(x_{n+1},z)\unlhd T(x_{1},z)=T(x,y)$, for any $x,y,z\in
C_{n}$. Thus, $T(x_{n+1},z)=T(x,y)$, for any $x,y,z\in C_{n}$. Hence,
$T(x,y)=T(x_{n+1},z)=T(x_{n+1},x_{n+1})$, for any $x,y,z\in C_{n}$ (i.e.,
$T(x,y)=T(z,t)$, for any $x,y,z,t\in C_{n+1}$). Since $T(x,y)\notin C_{n+1}$,
for any $x,y\in C_{n}$, it follows that $T(x_{n+1},z)=T(x_{n+1},x_{n+1})\notin
C_{n+1}$, for any $z\in C_{n+1}$. Consequently, $T(x,y)\notin C_{n+1}$, for
any $x,y\in C_{n+1}$. ∎

Theorem 2.4 and Proposition 3.7 leads to the following corollary.

######  Corollary 3.1.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a finite bounded trellis and complete. If
$T$ is a t-norm on $X$ such that $\\{C_{i}\\}_{i\in I}$ is the set of all non-
trivial cycles on $X$ which contains $x$ and $y$, then it holds that
$T(x,y)\unlhd T(\bigwedge(\bigcup_{i\in I}C_{i}),\bigwedge(\bigcup_{i\in
I}C_{i}))$.

In the following illustrative example, we give all t-norms on a given bounded
trellis has one cycle.

######  Example 3.5.

Let $(X=\\{0,a,b,c,1\\},\unlhd,\wedge,\vee)$ be a bounded trellis given by the
Hasse-type diagram in Figure 3. Then the only t-norm on $X$ is $T_{W}$.

$0$$a$$b$$c$$1$ Figure 3: The Hasse-type diagram of the trellis
$(X=\\{0,a,b,c,1\\},\unlhd)$.

######  Proposition 3.8.

The smallest pseudo-chain have at most two dashed curve, are isomorphic to one
of the following pseudo-chains defined as follows:

$0$$a$$b$$c$$1$$0$$a$$b$$c$$d$$1$$0$$a$$b$$c$$d$$1$$0$$a$$b$$c$$d$$e$$1$$0$$a$$b$$c$$d$$1$$0$$a$$b$$c$$d$$e$$1$

######  Remark 3.3.

The bounded pseudo-chains have at most two dashed curve defined in Proposition
3.8 has the greatest t-norm (using Matlab Program). In general, the greatest
t-norm (resp. the smallest t-conorm) on a arbitrary bounded pseudo-chain does
not necessarily exist. Indeed, let
$(X=\\{0,a,b,c,d,e,f,1\\},\unlhd,\wedge,\vee)$ be a bounded pseudo-chain have
three dashed curve given by the Hasse-type diagram in Figure 4 and $T_{1}$ a
binary operation on $X$ defined by the following table:

$T_{1}$ | $0$ | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ | $1$
---|---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $a$
$b$ | $0$ | $0$ | $b$ | $b$ | $b$ | $b$ | $b$ | $b$
$c$ | $0$ | $0$ | $b$ | $c$ | $c$ | $c$ | $c$ | $c$
$d$ | $0$ | $0$ | $b$ | $c$ | $c$ | $c$ | $c$ | $d$
$e$ | $0$ | $0$ | $b$ | $c$ | $c$ | $e$ | $e$ | $e$
$f$ | $0$ | $0$ | $b$ | $c$ | $c$ | $e$ | $f$ | $f$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ | $1$

One easily verifies that $T_{1}$ is a maximal t-norm on $X$ (using Matlab
Program). On other hand, for guarantees that $T_{1}$ is not the greatest
t-norm on $X$, it is enough to find one t-norm on $X$ such that
$T\ntrianglelefteq T_{1}$. Let $T_{2}$ be a binary operation on $X$ defined by
the following table:

$T_{2}$ | $0$ | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ | $1$
---|---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $a$
$b$ | $0$ | $0$ | $0$ | $0$ | $0$ | $b$ | $b$ | $b$
$c$ | $0$ | $0$ | $0$ | $0$ | $0$ | $b$ | $b$ | $c$
$d$ | $0$ | $0$ | $0$ | $0$ | $a$ | $b$ | $c$ | $d$
$e$ | $0$ | $0$ | $b$ | $b$ | $b$ | $e$ | $e$ | $e$
$f$ | $0$ | $0$ | $b$ | $b$ | $c$ | $e$ | $e$ | $f$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $e$ | $f$ | $1$

One easily verifies that $T_{2}$ is a t-norm on $X$ (Matlab program). Let
$x,y\in X$ such that $x=y=d$. Then $T_{1}(d,d)=c$ and $T_{2}(d,d)=a$. Since
$a$ and $c$ are incomparable, it holds that $T_{1}$ and $T_{2}$ are
incomparable t-norms. Hence, $T_{1}$ is not the greatest t-norm on $X$.

$0$$a$$b$$c$$d$$e$$f$$1$ Figure 4: The Hasse-type diagram of the pseudo-chain
$(X=\\{0,a,b,c,d,e,f,1\\},\unlhd)$.

Moreover, by a computer program (Matlab program), the number of all t-norms is
159.

###  3.3 Pseudo-triangular norms on bounded trellises

In this subsection, we introduce the notions of pseudo-triangular norms on
bounded trellises and present some illustrative examples.

######  Problem 3.1.

Remark 3.2 leads to research a new pseudo triangular norms on bounded
trellises as t-norms on bounded lattices including the meet operation of
bounded trellises. In particular, an idempotent pseudo triangular norms on
bounded trellises. Moreover, the pseudo triangular norms on bounded trellises
extends triangular norms on bounded lattices.

######  Definition 3.5.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. A binary operation
$T:X^{2}\to X$ is called a pseudo-triangular norm (pseudo-t-norm, for short),
if it is commutative, weakly-increasing, weakly-associative and has $1$ as
neutral element, i.e., $T(1,x)=x$, for any $x\in X$.

Analogously, we define a pseudo-triangular conorm on a bounded trellis.

######  Definition 3.6.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. A binary operation
$S:X^{2}\to X$ is called a pseudo-triangular conorm (pseudo-t-conorm, for
short), if it is commutative, weakly-increasing, weakly-associative and has
$0$ as neutral element, i.e., $S(0,x)=x$, for any $x\in X$.

Next, we give some examples of pseudo-t-norms and pseudo-t-conorms bounded
trellises.

######  Example 3.6.

Let $(X,\unlhd,\wedge,\vee)$ be a bounded trellis. It holds that

  1. (i)

$\wedge$ is a pseudo-t-norm on $X$;

  2. (ii)

$\vee$ is a pseudo-t-conorm on $X$;

  3. (iii)

The binary operations $T_{D}$ (resp. $S_{D}$) defined in Example 3.2 is a
pseudo-t-norm (resp. pseudo-t-conorm).

######  Notation 3.4.

Let $(X,\unlhd,\wedge,\vee)$ be a bounded trellis. We denote by:

  1. (i)

$\mathcal{WAO}_{1}(X)$: the class (or the set) of all pseudo-t-norms on $X$;

  2. (ii)

$\mathcal{WAO}_{0}(X)$: the class (or the set) of all pseudo-t-conorms on $X$.

######  Remark 3.4.

  1. (i)

$\mathcal{WAO}_{1}(X)$ (resp. $\mathcal{WAO}_{0}(X)$) extends the class of all
t-norms (resp. the class of all t-conorms) on the bounded trellis $X$.

  2. (ii)

In general, one can easily observe that
$\mathcal{AO}_{e}(X)\subseteq\mathcal{WAO}_{e}(X)$, for any $e\in\\{0,1\\}$.

###  3.4 Properties of pseudo-triangular norms on bounded trellises

In this subsection, we investigate some properties of $\mathcal{WAO}_{1}(X)$
and $\mathcal{WAO}_{0}(X)$.

The following Proposition shows the duality between the two classes
$\mathcal{WAO}_{1}(X)$ and $\mathcal{WAO}_{0}(X)$. We recall that for a given
bounded trellis $(X,\unlhd,\wedge,\vee,0,1)$, its dual bounded trellis is
defined as $(X^{*},\unlhd^{*},\wedge^{*},\vee^{*},0^{*},1^{*})$, where
$X^{*}=X$, $x\unlhd^{*}y$ if and only if $y\unlhd x$, $0^{*}=1$ and $1^{*}=0$.

######  Proposition 3.9.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $F$ a binary
operation on $X$. Then the following implications hold:

  1. (i)

If $F\in\mathcal{WAO}_{1}(X)$, then $F\in\mathcal{WAO}_{0}(X^{*})$;

  2. (ii)

If $F\in\mathcal{WAO}_{0}(X)$, then $F\in\mathcal{WAO}_{1}(X^{*})$.

###### Proof.

The proof is straightforward. ∎

######  Proposition 3.10.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. The following
implications hold:

  1. (i)

Any element of $\mathcal{WAO}_{1}(X)$ is conjunctive;

  2. (ii)

Any element of $\mathcal{WAO}_{0}(X)$ is disjunctive.

###### Proof.

  1. (i)

Let $T\in\mathcal{WAO}_{1}(X)$ and $x,y\in X$. Since $1\in X^{tr}$ and $T$ is
weakly-increasing and commutative, it follows that $T(x,y)\unlhd T(1,y)$ and
$T(x,y)\unlhd T(x,1)$. The fact that $1$ is the neutral element of $T$ implies
that $T(x,y)\unlhd y$ and $T(x,y)\unlhd x$. Thus, $T(x,y)\unlhd x\wedge y$.
Therefore, $T$ is conjunctive.

  2. (ii)

The proof is dual to that of (i).

∎

######  Proposition 3.11.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $F$ a binary
operation on $X$. Then the following implications hold:

  1. (i)

If $F\in\mathcal{WAO}_{1}(X)$, then $F(x,0)=0$, for any $x\in X$;

  2. (ii)

If $F\in\mathcal{WAO}_{0}(X)$, then $F(x,1)=1$, for any $x\in X$.

###### Proof.

  1. (i)

Suppose that $F\in\mathcal{WAO}_{1}(X)$ and $x\in X$. Since $1\in X^{tr}$,
$x\unlhd 1$ and $F$ is weakly-increasing, it follows that $F(x,0)\unlhd
F(1,0)=0$. Thus, $F(x,0)=0$, for any $x\in X$.

  2. (ii)

The proof is dual to that of (i).

∎

######  Remark 3.5.

If the cardinal of $X$ is greater than $1$ (i.e., $|X|>1$), then

$\mathcal{WAO}_{0}(X)\cap\mathcal{WAO}_{1}(X)=\emptyset.$

###  3.5 Psoset structures of $\mathcal{WAO}_{0}(X)$ and
$\mathcal{WAO}_{1}(X)$

In this subsection, we discuss the bounded psoset structures of
$\mathcal{WAO}_{1}(X)$ and $\mathcal{WAO}_{0}(X)$.

For any $F_{1},F_{2}\in\mathcal{WAO}_{e}(X)$, we define:

$F_{1}\unlhd_{\mathcal{WAO}}F_{2}\text{ if and only if }\ F_{1}(x,y)\unlhd
F_{2}(x,y),\text{ for any }x,y\in X.$

The following result is a natural generalization to that of triangular norms
on the trellis.

######  Proposition 3.12.

Let $(X,\unlhd_{X},\wedge_{X},\vee_{X},0,1)$ be a bounded trellis. Then it
holds that:

  1. (i)

$T_{D}\unlhd_{\mathcal{WAO}}F\unlhd_{\mathcal{WAO}}\wedge$, for any
$F\in\mathcal{WAO}_{1}(X)$ ;

  2. (ii)

$\vee\unlhd_{\mathcal{WAO}}F\unlhd_{\mathcal{WAO}}S_{D}$, for any
$F\in\mathcal{WAO}_{0}(X)$ .

###### Proof.

  1. (i)

On the one hand, Proposition 3.10 guarantees that
$F\unlhd_{\mathcal{WAO}}\wedge$, for any $F\in\mathcal{WAO}_{1}(X)$. On the
other hand, $T_{D}(x,y)=0\unlhd T(x,y)$, for any
$(x,y)\in(X\setminus\\{1\\})^{2}$. If $x=1$ (resp. $y=1$), it holds that
$T_{D}(1,y)=y=F(1,y)$ (resp. $T_{D}(x,1)=x=F(x,1)$ ). Hence, $T_{D}(x,y)\unlhd
F(x,y)$, for any $x,y\in X$. Thus,
$T_{D}\unlhd_{\mathcal{WAO}}F\unlhd_{\mathcal{WAO}}\wedge$, for any
$F\in\mathcal{WAO}_{1}(X)$ .

  2. (ii)

The proof is dual to that of (i).

∎

In a bounded Trellis $(X,\unlhd_{X},\wedge_{X},\vee_{X},0,1)$, the structures
$(\mathcal{WAO}_{1}(X),\unlhd_{\mathcal{WAO}},T_{D},\wedge)$ and
$(\mathcal{WAO}_{0}(X),\unlhd_{\mathcal{WAO}},\vee,S_{D})$ are bounded
psosets.

######  Remark 3.6.

The bounded psosets
$(\mathcal{WAO}_{1}(X),\unlhd_{\mathcal{WAO}},T_{D},\wedge)$ and
$(\mathcal{WAO}_{0}(X),\unlhd_{\mathcal{WAO}},\vee,S_{D})$ are not necessary
bounded trellises, since the meet (resp. the join) of any two elements is not
necessary an element of $\mathcal{WAO}_{1}(X)$ or $\mathcal{WAO}_{0}(X)$.

The following proposition shows a case when an element of
$\mathcal{WAO}_{1}(X)$ (resp. an element of $\mathcal{WAO}_{0}(X)$) coincides
with the meet (resp. the join) operation. It is particular case of the weaker
types of increasing binary operations on a bounded trellis that coincide with
the meet (resp. the join) operation.

######  Proposition 3.13.

Let $(X,\unlhd,\wedge,\vee)$ be a bounded trellis and $F$ a binary operation
on $X$. The following statements hold:

  1. (i)

If $F\in\mathcal{WAO}_{1}(X)$, idempotent and satisfying $F(x\wedge y,x\wedge
y)\unlhd F(x,y)$, for any $x,y\in X$, then $F$ is the meet $(\wedge)$
operation of $X$;

  2. (ii)

If $F\in\mathcal{WAO}_{0}(X)$, idempotent and satisfying $F(x,y)\unlhd F(x\vee
y,x\vee y)$, for any $x,y\in X$, then $F$ is the join $(\vee)$ operation of
$X$.

###### Proof.

  1. (i)

On the one hand, since $F\in\mathcal{WAO}_{1}(X)$ which means that $F$ is
conjunctive, it holds that $F(x,y)\unlhd x\wedge y$, for any $x,y\in X$. On
the other hand, the fact that $F$ is idempotent and satisfying $F(x\wedge
y,x\wedge y)\unlhd F(x,y)$, for any $x,y\in X$ implies that $x\wedge
y=F(x\wedge y,x\wedge y)\unlhd F(x,y)$. Thus, $F$ is the meet operation
$(\wedge)$ of $X$.

  2. (ii)

The proof is dual to that of (i).

∎

######  Remark 3.7.

The converse of the above Proposition 3.13 is immediate.

##  4 Constructions of some elements of $\mathcal{WAO}_{0}(X)$ and
$\mathcal{WAO}_{1}(X)$

In this section, we construct some elements of $\mathcal{WAO}_{1}(X)$ and
$\mathcal{WAO}_{0}(X)$ on bounded trellises.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $e\in X$. Let
$T_{e}$ and $S_{e}$ two binary operations on $X$ defined as follows:

$T_{e}(x,y)=\left\\{\begin{array}[]{ll}{x\wedge y}&\text{if }x=1\text{ or
}y=1;\\\ {(x\wedge y)\wedge e}&\text{otherwise;}\end{array}\right.$

and

$S_{e}(x,y)=\left\\{\begin{array}[]{ll}{x\vee y}&\text{if }x=0\text{ or
}y=0;\\\ {(x\vee y)\vee e}&\text{otherwise.}\end{array}\right.$

######  Remark 4.1.

In general, $T_{e}$ (resp. $S_{e}$) is not necessarily an element of
$\mathcal{WAO}_{1}(X)$ (resp. $\mathcal{WAO}_{0}(X)$). Indeed, let
$(X=\\{0,a,b,c,d,e,f,1\\},\unlhd,\wedge,\vee,0,1)$ be a bounded trellis given
by the Hasse diagram in Figure 5.

$0$$a$$b$$c$$d$$e$$f$$1$ Figure 5: Hasse diagram of the trellis
$(X=\\{0,a,b,c,d,e,f,1\\},\unlhd)$.

Sitting $x=f$ and $y=d$, then $x\unlhd y$ and $(x,y)\in(X^{tr})^{2}$. Since
$T_{e}(f,c)=(f\wedge c)\wedge e=a\ntrianglelefteq T_{e}(d,c)=(d\wedge c)\wedge
e=0$, it follows that $T_{e}$ is not weakly-increasing. Therefore,
$T_{e}\notin\mathcal{WAO}_{1}(X)$.

In view of remark 4.1, we give sufficient conditions under which the binary
operation $T_{e}$ is an element of $\mathcal{WAO}_{1}(X)$.

######  Proposition 4.1.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis. The following
implications hold:

  1. (i)

If $e\in X^{\wedge-ass}$, then $T_{e}\in\mathcal{WAO}_{1}(X)$;

  2. (ii)

If $e\in X^{\vee-ass}$, then $S_{e}\in\mathcal{WAO}_{0}(X)$.

${1}$${a}$${b}$${0}$

###### Proof.

We only give the proof of (i), as the proof of (ii) is similar. One easily
verifies that $T_{e}$ is commutative and satisfies the boundary condition.
Now, let $(x,y)\in X\times X^{tr}$ such that $x\unlhd y$ and $z\in X$. Then we
discuss the following two possible cases:

  1. (i)

If $z=1$, then $T_{e}(x,z)=x\unlhd y=T_{e}(y,z)$.

  2. (ii)

If $z\neq 1$, then we have three possible cases:

    1. (i)

If $x=1$, then $y=1$ and $T_{e}(x,z)=z\unlhd z=T_{e}(y,z)$.

    2. (ii)

If $x\neq 1$ and $y=1$, then the fact that $e\in X^{\wedge-ass}$ implies that
$T_{e}(x,z)=(x\wedge z)\wedge e=(x\wedge e)\wedge z\unlhd z=T_{e}(y,z)$. Thus,
$T_{e}(x,z)\unlhd T_{e}(y,z)$.

    3. (iii)

If $y\neq 1$, then $T_{e}(x,z)=(x\wedge z)\wedge e$ and $T_{e}(y,z)=(y\wedge
z)\wedge e$. Since $x\unlhd y$ and $y\in X^{tr}$, it follows that $x\wedge
z\unlhd y\wedge z$. The fact that $e\in X^{\wedge-ass}$ implies $(x\wedge
z)\wedge e\unlhd(y\wedge z)\wedge e$. Thus, $T_{e}(x,z)\unlhd T_{e}(y,z)$.

Therefore, $T_{e}$ is weakly-increasing.

Now, we prove that $T_{e}$ is weakly-associative. Let $x,y,z\in X$ such that
$[x,y,z]\in X^{\wedge-ass}$. Since $e\in X^{\wedge-ass}$, it holds that

$\displaystyle T_{e}(x,T_{e}(y,z))$ $\displaystyle=(x\wedge((y\wedge z)\wedge
e))\wedge e$ $\displaystyle=((x\wedge(y\wedge z))\wedge e)\wedge e$
$\displaystyle=(((x\wedge y)\wedge z)\wedge e)\wedge e$
$\displaystyle=(((x\wedge y)\wedge e)\wedge z)\wedge e$
$\displaystyle=(T_{e}(x,y)\wedge z)\wedge e$
$\displaystyle=T_{e}(T_{e}(x,y),z)\,.$

Hence, $T_{e}$ is weakly-associative. Therefore,
$T_{e}\in\mathcal{WAO}_{1}(X)$. ∎

######  Remark 4.2.

Particular cases: since $0,1\in X^{\wedge-ass}$, we recognize that

  1. (i)

$T_{0}=T_{D}$ and $T_{1}=\wedge$;

  2. (ii)

$S_{0}=\vee$ and $S_{1}=S_{D}$.

######  Proposition 4.2.

Let $\mathbb{X}=(X,\unlhd,\wedge,\vee,0,1)$ be a bounded modular trellis and
the binary operations $Z$ and $Z^{*}$ defined as follows:

$Z(x,y)=\left\\{\begin{array}[]{ll}{x\wedge y}&{\text{ if }x\vee y=1};\\\
{0}&{\text{ otherwise; }}\end{array}\right.$ and
$Z^{*}(x,y)=\left\\{\begin{array}[]{ll}{x\vee y}&{\text{ if }x\wedge y=0};\\\
{1}&{\text{ otherwise;}}\end{array}\right.$

Then, $Z\in\mathcal{WAO}_{1}(X)$ and $Z^{*}\in\mathcal{WAO}_{0}(X)$.

###### Proof.

The proof is similar to that of $Z^{*}$. One easily verifies that
$T_{\mathrm{Z}}$ is commutative and satisfies the boundary conditions. Now,
let $(x,y)\in X\times X^{tr}$ such that $x\unlhd y$. Then we discuss the
following two possible cases:

  1. (i)

If $T_{\mathrm{Z}}(x,z)=0$, then $T_{\mathrm{Z}}(x,z)=0\unlhd
T_{\mathrm{Z}}(y,z)$, for any $z\in X$.

  2. (ii)

If $T_{\mathrm{Z}}(x,z)=x\wedge z$, then $x\vee z=1$. Proposition 3.1
guarantees that $x\vee z\unlhd y\vee z$, for any $z\in X$. Thus, $y\vee z=1$
and $T_{\mathrm{Z}}(y,z)=y\wedge z$. Since $y\in X^{tr}$, it holds that
$T_{\mathrm{Z}}(x,z)=x\wedge z\unlhd y\wedge z=T_{\mathrm{Z}}(y,z)$, for any
$z\in X$.

Hence, $T_{\mathrm{Z}}$ is weakly-increasing. Now, we prove that
$T_{\mathrm{Z}}$ is weakly-associative. Let $x,y,z\in X$ such that $[x,y,z]\in
X^{\wedge}$. On the one hand, we have that
$T_{\mathrm{Z}}\left(x,T_{\mathrm{Z}}(y,z)\right)$
$\quad=\left\\{\begin{array}[]{ll}{x\wedge y\wedge z}&{\text{ if }y\vee
z=1\text{ and }x\vee(y\wedge z)=1},\\\ {0}&{\text{ otherwise.
}}\end{array}\right.$
On the other hand, it holds that
$T_{\mathrm{Z}}\left(T_{\mathrm{Z}}(x,y),z\right)$
$\quad=\left\\{\begin{array}[]{ll}{x\wedge y\wedge z}&{\text{ if }x\vee
y=1\text{ and }z\vee(x\wedge y)=1},\\\ {0}&{\text{ otherwise.
}}\end{array}\right.$
We will show that $y\vee z=1$ and $x\vee(y\wedge z)=1$ implies $x\vee y=1$ and
$z\vee(x\wedge y)=1$. The proof of the converse implication is similar.
Firstly, since $y\unlhd x\vee y$ and $y\vee z=1$. Proposition 2.1 guarantees
that $y\wedge z\unlhd x\vee y$. Thus, $1=x\vee(y\wedge z)\unlhd x\vee y$.
Hence, $x\vee y=1$. Secondly, since $x\vee(y\wedge z)=1$, it holds that
$y=y\wedge(x\vee(y\wedge z))$. The fact that $\mathbb{X}$ is modular implies
that $y=y\wedge(x\vee(y\wedge z))=(x\wedge y)\vee(y\wedge z)=(x\wedge y)\vee
z)\wedge y$. Thus, $y\unlhd(x\wedge y)\vee z$. On the other hand, since
$z\unlhd(x\wedge y)\vee z$, it follows that $y\vee z\unlhd(x\wedge y)\vee z$.
Hence, $(x\wedge y)\vee z=1$. Since $[x,y,z]\in X^{\wedge}$, it holds that
$x\wedge(y\wedge z)=(x\wedge y)\wedge z$. Hence, $T_{\mathrm{Z}}$ is weakly-
associative. Therefore, $T_{\mathrm{Z}}$ is a pseudo-t-norm on $X$ (i.e.,
$Z\in\mathcal{WAO}_{1}(X)$). ∎

In the following result, we propose a new ordinal sum construction of
$\mathcal{WAO}_{1}(X)$ and $\mathcal{WAO}_{0}(X)$ on bounded trellises
according to [6]. We start by the following immediate proposition.

######  Proposition 4.3.

Let $(X,\unlhd,\wedge,\vee)$ be a trellis and $a,b\in X^{ass}$ such that
$a\unlhd b$. The following subintervals of $X$ defined as:

$[a,b]=\\{x\in X|a\unlhd x\unlhd b\\},$ $(a,b]=\\{x\in X|a\lhd x\unlhd b\\},$
$[a,b)=\\{x\in X|a\unlhd x\lhd b\\},$ $(a,b)=\\{x\in X|a\lhd x\lhd b\\},$

are subtrellises of $X$.

######  Theorem 4.1.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $a\in
X^{ass}\backslash\\{0,1\\}$. If $V:[a,1]^{2}\rightarrow[a,1]$ an element of
$\mathcal{WAO}_{1}([a,1])$ and $W:[0,a]^{2}\rightarrow[0,a]$ an element of
$\mathcal{WAO}_{0}([0,a])$, then the binary operations $T:X^{2}\rightarrow X$
and $S:X^{2}\rightarrow X$ defined as follows:

$T(x,y)=\left\\{\begin{array}[]{ll}{x\wedge y}&{\text{if }x=1\text{ or
}y=1};\\\ {V(x,y)}&{\text{if }x,y\in[a,1)};\\\ {x\wedge y\wedge
a}&{\text{otherwise;}}\end{array}\right.$

and

$S(x,y)=\left\\{\begin{array}[]{ll}{x\vee y}&{\text{if }x=0\text{ or }y=0};\\\
{W(x,y)}&{\text{if }x,y\in(0,a]};\\\ {x\vee y\vee
a}&{\text{otherwise;}}\end{array}\right.$

are elements of $\mathcal{WAO}_{1}(X)$ and $\mathcal{WAO}_{0}(X)$,
respectively.

###### Proof.

The proof is similar to that of $S$. One easily verifies that $T$ is
commutative and satisfies the boundary conditions. Now, let $x,y\in X\times
X^{tr}$ such that $x\unlhd y$. Then we discuss the following possible cases:

  1. (i)

If $x=1$ or $z=1$, then $T(x,z)=T(y,z)$.

  2. (ii)

If $x,z\in[a,1)$, then, also $a\unlhd y$ and $T(x,z)=V(x,z)\unlhd
V(y,z)=T(y,z)$.

  3. (iii)

If $x\notin[a,1)$ and $z\in[a,1)$, it holds that $T(x,z)=x\wedge a$ and we
have three possible cases:

    1. (i)

If $y=1$, then $T(y,z)=z$. Since $a\in X^{ass}$, then $T(x,z)=x\wedge a\unlhd
z=T(y,z)$.

    2. (ii)

If $y\in[a,1)$, then $T(y,z)=V(y,z)\in[a,1)$. Since, $a\in X^{ass}$, it
follows that $T(x,z)=x\wedge a\unlhd V(y,z)=T(y,z)$.

    3. (iii)

If $y\notin[a,1]$, then $T(y,z)=y\wedge a$. Since $a\in X^{ass}$, then it
follows that $T(x,z)=x\wedge a\unlhd y\wedge a=T(y,z)$.

  4. (iv)

If $x\notin[a,1]$ and $z\notin[a,1]$, then $T(x,z)=x\wedge z\wedge a$ and
$T(y,z)=y\wedge z\wedge a$. Thus, Proposition 4.1 guarantees that
$T(x,z)=x\wedge z\wedge a\unlhd y\wedge z\wedge a=T(y,z)$.

Hence, $T$ is weakly-increasing. Next, we prove that $T$ is weakly-
associative. Let $x,y,z\in X$ such that $[x,y,z]\in X^{\wedge}$. The proof is
split into all possible cases.

  1. (i)

If $x,y\in[a,1)$, then we have two cases:

    1. (a)

If $z\in[a,1)$, then:

$\begin{aligned} T(x,T(y,z))&=T(x,V(y,z))\\\ &=V(x,V(y,z))\\\ &=V(V(x,y),z)\\\
&=T(V(x,y),z)\\\ &=T(T(x,y),z).\end{aligned}$

    2. (b)

If $z\in X\setminus[a,1)$, then:

$\begin{aligned} T(x,T(y,z))&=T(x,y\wedge z\wedge a)\\\ &=x\wedge(y\wedge
z\wedge a)\wedge a\\\ &=z\wedge a\text{\quad(car, }a\in X^{ass})\\\
&=V(x,y)\wedge z\wedge a\\\ &=T(V(x,y),z)\\\ &=T(T(x,y),z).\end{aligned}$

  2. (ii)

If $x\in[a,1)$ and $y\in X\setminus[a,1)$, then we have two cases:

    1. (a)

If $z\in[a,1)$, then this case have been studied in $(i.b)$.

    2. (b)

If $z\in X\setminus[a,1)$, then:

$\begin{aligned} T(x,T(y,z))&=T(x,y\wedge z\wedge a)\\\ &=x\wedge(y\wedge
z\wedge a)\wedge a\\\ &=(x\wedge y\wedge a)\wedge z\wedge a\text{\quad(car,
}a\in X^{ass})\\\ &=T(x\wedge y\wedge a,z)\\\ &=T(T(x,y),z).\end{aligned}$

  3. (iii)

If $x,y\in X\setminus[a,1)$, then we have two cases:

    1. (a)

If $z\in[a,1)$, then this case have been studied in $(ii.b)$.

    2. (b)

If $z\in X\setminus[a,1)$, then:

$\begin{aligned} T(x,T(y,z))&=T(x,y\wedge z\wedge a)\\\ &=x\wedge y\wedge
z\wedge a\\\ &=T(x\wedge y\wedge a,z)\\\ &=T(T(x,y),z).\end{aligned}$

Hence, $T$ is weakly-associative on $X$. Therefore,
$T\in\mathcal{WAO}_{1}(X)$. ∎

One easily Observes that $T$ and $S$ on a bounded trellis considered in
Theorem 4.1 can be described as follows:

$T(x,y)=\left\\{\begin{array}[]{ll}{V(x,y)}&{\text{ if }(x,y)\in[a,1)^{2}};\\\
{y\wedge a}&{\text{ if }x\in[a,1),y\|a};\\\ {x\wedge a}&{\text{ if
}y\in[a,1),x\|a};\\\ {x\wedge y\wedge a}&{\text{ if }x\|a,y\|a};\\\ {x\wedge
y}&{\text{ otherwise; }}\end{array}\right.$

and

$S(x,y)=\left\\{\begin{array}[]{ll}{W(x,y)}&{\text{ if }(x,y)\in(0,a]^{2}};\\\
{y\vee a}&{\text{ if }x\in(0,a],y\|a};\\\ {x\vee a}&{\text{ if
}y\in(0,a],x\|a};\\\ {x\vee y\vee a}&{\text{ if }x\|a,y\|a};\\\ {x\vee
y}&{\text{ otherwise.}}\end{array}\right.$

Thus, we get $T$ and $S$ by the next figures:

##  5 T-distributivity on bounded trellises

In this section, we introduce the notion of $F$-distributivity for an
arbitrary binary operation $F$ on a bounded trellis and we determine a
relationship between t-norms and pseudo-t-norms. Moreover, we characterize
pseudo-t-norms (resp. pseudo-t-conorms) on a bounded trellis with respect to
the $F$-distributivity.

######  Definition 5.1.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $F_{1}$ and $F_{2}$
two binary operations on $X$. If
$F_{1}\left(x,F_{2}(y,z)\right)=F_{2}\left(F_{1}(x,y),F_{1}(x,z)\right)$, for
any $x,y,z\in X$ where at least one of the elements $y,z$ is not $1$ or not
$0$ (i.e., $(0,0)\neq(y,z)$ and $(1,1)\neq(y,z))$, then $F_{1}$ is
distributive over $F_{2}$ ($F_{1}$ is $F_{2}$-distributive, for short).

######  Proposition 5.1.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $T^{*}$ is a pseudo-
t-norm on $X$. If $T^{*}$ is $T$-distributive, for any
$T\in\mathcal{WAO}_{1}(X)$, then $T^{*}\in\mathcal{AO}_{1}(X)$.

###### Proof.

Let $x,y\in X$ such that $x\unlhd y$. The fact that the meet operation
$(\wedge)$ is a pseudo-t-norm on $X$ implies that $T^{*}(x,z)\wedge
T^{*}(y,z)=T^{*}(x\wedge y,z)=T^{*}(x,z)$. Thus, $T^{*}$ is increasing. Now,
we prove that $T^{*}$ is associative. Since $T^{*}$ is a pseudo-t-norm on $X$
and increasing, it holds that
$T^{*}(x,T^{*}(y,z))=T^{*}(T^{*}(x,y),T^{*}(x,z))\unlhd T^{*}(T^{*}(x,y),z)$
and $T^{*}(T^{*}(x,y),z)=T^{*}(T^{*}(x,z),T^{*}(y,z))\unlhd
T^{*}(x,T^{*}(y,z))$, for any $x,y,z\in X$. Thus,
$T^{*}(x,T^{*}(y,z))=T^{*}(T^{*}(x,y),z)$, for any $x,y,z\in X$. Hence,
$T^{*}$ is associative. Since $T^{*}$ is commutative and has $1$ as a neutral
element, it follows that $T^{*}$ is a t-norm on $X$. ∎

In the following proposition, we give a $T$-distributive t-norm, for any
pseudo-t-norm $T$ on bounded trellis.

######  Proposition 5.2.

Let $(X,\unlhd,\wedge,\vee)$ be a bounded trellis. Then the smallest pseudo-t-
norm (t-norm) $T_{D}$ is $T$-distributive, for any $T\in\mathcal{WAO}_{1}(X)$.

###### Proof.

Let $T$ be an arbitrary pseudo-t-norm on $X$. We must show that the equality

$T_{D}(x,T(y,z))=T\left(T_{D}(x,y),T_{D}(x,z)\right)$

holds for every element $x,y,z$ of $X$ such that $y\neq 1$ or $z\neq 1$.
Suppose that $z\neq 1$. Then we discuss the following two possible cases:

  1. (i)

If $x=1$, then $T_{D}(x,T(y,z))=T(y,z)=T(T_{D}(x,y),T_{D}(x,z))$.

  2. (ii)

If $x\neq 1$, we have two possible cases:

    1. (i)

If $y=1$, then $T_{D}(x,T(y,z))=T_{D}(x,z)=0$ and
$T(T_{D}(x,y),T_{D}(x,z))=T(x,0)$. Proposition 3.11 guarantees that
$T(x,0)=0$, for any $x\in X$. Then the equality is holds.

    2. (ii)

If $y\neq 1$, then $T(y,z)\unlhd y\unlhd 1$ and $y\neq 1$, for any pseudo-t-
norm $T$. Thus, $T(y,z)\neq 1$. Hence, $T_{D}(x,T(y,z))=0$ and
$T\left(T_{D}(x,y),T_{D}(x,z)\right)=T(0,0)=0$.

Therefore, $T_{D}$ is $T$-distributive, for any $T\in\mathcal{WAO}_{1}(X)$. ∎

######  Proposition 5.3.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis, $T$ a pseudo-t-norm and
$T^{*}$ is a t-norm on $X$. If $T$ is $T^{*}$-distributive, then
$T\unlhd_{\mathcal{WAO}}T^{*}$ (i.e., $T$ is weaker that $T^{*}$).

###### Proof.

Let $x,y\in X$, then
$T(x,y)=T(T^{*}(x,1),y)=T^{*}(T(x,y),T(1,y))=T^{*}(T(x,y),y)$. Since $T^{*}$
is a t-norm, it holds that $T^{*}(T(x,y),y)\unlhd T^{*}(x,y)$. Thus
$T(x,y)\unlhd_{\mathcal{WAO}}T^{*}(x,y)$, for any $x,y\in X$. Hence, $T$ is
weaker that $T^{*}$. ∎

######  Proposition 5.4.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis and $T^{*}$ is a t-norm
on $X$. If $T$ is $T^{*}$-distributive, for any $T\in\mathcal{WAO}_{1}(X)$,
then $T^{*}=\wedge$. Moreover, $(X,\unlhd,\wedge,\vee)$ is a lattice.

###### Proof.

Let an arbitrary pseudo-t-norm $T$ and $T^{*}$ is a t-norm on $X$, then
Proposition 5.3 guarantees that $T\unlhd_{\mathcal{WAO}}T^{*}$. Suppose that
$T=\wedge$, then $\wedge\unlhd_{\mathcal{WAO}}T^{*}$. Since $T^{*}$ is
conjunctive, it holds that $\wedge=T^{*}$. Thus, $\wedge$ is associative.
Hence, $(X,\unlhd,\wedge,\vee)$ is a lattice. ∎

######  Proposition 5.5.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis, $T$ is a pseudo-t-norm
and $S$ is a pseudo-t-conorm on $X$. The following implications hold:

  1. (i)

If $S$ is $T$-distributive, then $T$ is idempotent;

  2. (ii)

If $T$ is $S$-distributive, then $S$ is idempotent.

###### Proof.

  1. (i)

Since $T$ is pseudo-t-norm and $S$ is pseudo-t-conorm on $X$, it follows that
$x=S(x,0)=S(x,T(0,0))=T(S(x,0),S(x,0))=T(x,x)$, for any $x\in X$. Thus, $T$ is
idempotent.

  2. (ii)

The proof is similar to that of (i).

∎

Proposition 5.5 leads to the following corollary.

######  Corollary 5.1.

Let $(X,\unlhd,\wedge,\vee,0,1)$ be a bounded trellis, $T$ is a pseudo-t-norm
and $S$ is a pseudo-t-conorm on $X$. Then it holds that:

  1. (i)

If $S$ is $T$-distributive and $T$ satisfying $T(x\wedge y,x\wedge y)\unlhd
T(x,y)$, for any $x,y\in X$, then $T=\wedge$.

  2. (ii)

If $T$ is $S$-distributive and $S$ satisfying $S(x\vee y,x\vee y)\unlhd
S(x,y)$, for any $x,y\in X$, then $S=\vee$.

##  6 Relationship among $\mathcal{WAO}_{e}(X)$ and isomorphisms on bounded
trellises

In this section, we conjugate elements of $\mathcal{WAO}_{1}(X)$ (resp.
elements of $\mathcal{WAO}_{0}(X)$) and an isomorphism map on a bounded
trellis $X$. First, we start by the following proposition.

######  Proposition 6.1.

Let $(X_{1},\unlhd_{1},\wedge_{1},\vee_{1})$,
$(X_{2},\unlhd_{2},\wedge_{2},\vee_{2})$ be two trellises and
$\rho:X_{1}\longrightarrow X_{2}$ an isomorphism. Then
$\rho(X_{1}^{tr})\subseteq X_{2}^{tr}$ .

###### Proof.

Let $x,y\in X_{2}$ and $z\in\rho(X_{1}^{tr})$ such that
$x\unlhd_{2}y\unlhd_{2}z$. Then there exist $x^{\prime},y^{\prime}\in X_{1}$
and $z^{\prime}\in X_{1}^{tr}$ such that
$\rho(x^{\prime})\unlhd_{2}\rho(y^{\prime})\unlhd_{2}\rho(z^{\prime})$. From
the increasingness of $\rho^{-1}$, it holds that
$x^{\prime}\unlhd_{1}y^{\prime}\unlhd_{1}z^{\prime}$. The fact that
$z^{\prime}\in X_{1}^{tr}$ implies that $x^{\prime}\unlhd_{1}z^{\prime}$,
i.e., $x^{\prime}\wedge_{1}z^{\prime}=x^{\prime}$. Since $\rho$ is
homomorphism; it follows that
$\rho(x^{\prime})\wedge_{2}\rho(z^{\prime})=\rho(x^{\prime}\wedge_{1}z^{\prime})=\rho(x^{\prime})$.
Hence, $\rho(x^{\prime})\unlhd_{2}\rho(z^{\prime})$, i.e., $x\unlhd_{2}z$.
Thus, $z\in X^{tr}$. Therefore, $\rho(X_{1}^{tr})\subseteq X_{2}^{tr}$ . ∎

######  Proposition 6.2.

Let $(X_{1},\unlhd_{1},\wedge_{1},\vee_{1})$,
$(X_{2},\unlhd_{2},\wedge_{2},\vee_{2})$ be two trellises and
$\rho:X_{1}\longrightarrow X_{2}$ an isomorphism. Then $[x,y,z]\in
X_{1}^{\wedge_{1}}$ (resp. $[x,y,z]\in X_{1}^{\vee_{1}}$) if and only if
$[\rho(x),\rho(y),\rho(z)]\in X_{2}^{\wedge_{2}}$ (resp.
$[\rho(x),\rho(y),\rho(z)]\in X_{2}^{\vee_{2}}$).

###### Proof.

Let $x,y,z\in X_{1}$ such that $[x,y,z]\in X_{1}^{\wedge_{1}}$. Since $\rho$
is an isomorphism, then

$\displaystyle\rho(x)\wedge_{2}(\rho(y)\wedge_{2}\rho(z))$
$\displaystyle=\rho(x)\wedge_{2}\rho(y\wedge_{1}z))$
$\displaystyle=\rho(x\wedge_{1}(y\wedge_{1}z))$
$\displaystyle=\rho((x\wedge_{1}y)\wedge_{1}z)$
$\displaystyle=\rho(x\wedge_{1}y)\wedge_{2}\rho(z))$
$\displaystyle=(\rho(x)\wedge_{2}\rho(y))\wedge_{2}\rho(z)\,.$

Therefore, $[\rho(x),\rho(y),\rho(z)]\in X_{2}^{\wedge_{2}}$. In a similar
way, we prove that $[x,y,z]\in X_{1}^{\vee_{1}}$ if and only if
$[\rho(x),\rho(y),\rho(z)]\in X_{2}^{\vee_{2}}$ . ∎

######  Proposition 6.3.

$(X_{1},\unlhd_{1},\wedge_{1},\vee_{1},0_{1},1_{1})$,
$(X_{2},\unlhd_{2},\wedge_{2},\vee_{2},0_{2},1_{2})$ be two bounded trellises,
$T\in\mathcal{WAO}_{1}(X_{2})$ and $\rho:X_{1}\longrightarrow X_{2}$ an
isomorphism. Then the binary operation $T^{\rho}$ defined by:

$T^{\rho}(x,y)=\rho^{-1}(T(\rho(x),\rho(y)))$, for any $x,y\in X_{1}$ ,

is an element of $\mathcal{WAO}_{1}(X_{1})$.

###### Proof.

One easily verifies that $T^{\rho}$ is commutative and satisfies the boundary
condition. Now, let $(x,y)\in X_{1}\times X_{1}^{tr}$ such that
$x\unlhd_{1}y$. Proposition 6.1 assures that $\rho(y)\in X_{2}^{tr}$. Since
$T$ is weakly-increasing, it holds that
$T(\rho(x),\rho(z))\unlhd_{2}T(\rho(y),\rho(z))$, for any $z\in X_{1}$. The
fact that $\rho^{-1}$ is increasing on $X_{2}$ implies that
$\rho^{-1}(T(\rho(x),\rho(z)))\unlhd_{1}\rho^{-1}(T(\rho(y),\rho(z)))$, for
any $z\in X_{1}$, i.e., $T^{\rho}(x,z)\unlhd_{1}T^{\rho}(y,z)$, for any $z\in
X_{1}$. Hence, $T^{\rho}$ is weakly-increasing on $X_{1}$. Next, we prove that
$T^{\rho}$ is weakly-associative. Let $x,y,z\in X_{1}$ such that $[x,y,z]\in
X_{1}^{\wedge_{1}}$. Proposition 6.2 assures that
$(\rho(x),\rho(y),\rho(z))\in X_{2}^{\wedge_{2}}$. Thus

$\displaystyle T^{\rho}(T^{\rho}(x,y),z)$
$\displaystyle=\rho^{-1}(T(\rho(T^{\rho}(x,y)),\rho(z)))$
$\displaystyle=\rho^{-1}(T(\rho(\rho^{-1}(T(\rho(x),\rho(y)))),\rho(z)))$
$\displaystyle=\rho^{-1}(T(T(\rho(x),\rho(y)),\rho(z)))$
$\displaystyle=\rho^{-1}(T(\rho(x),T(\rho(y),\rho(z))))$
$\displaystyle=\rho^{-1}(T(\rho(x),\rho(\rho^{-1}(T(\rho(y),\rho(z))))))$
$\displaystyle=\rho^{-1}(T(\rho(x),\rho(T^{\rho}(y,z))))$
$\displaystyle=T^{\rho}(x,T^{\rho}(y,z))\,.$

Hence, $T^{\rho}$ is weakly-associative on $X_{1}$. Therefore,
$T^{\rho}\in\mathcal{WAO}_{1}(X_{1})$ . ∎

Notice that in a bounded trellis $(X,\unlhd,\wedge,\vee,0,1)$, the identity
map $Id_{X}$ of $X$ (i.e., $Id_{X}(x)=x$, for any $x\in X$) is an isomorphism
(automorphism). Then $T^{Id_{X}}=T$, for any $T\in\mathcal{WAO}_{1}(X)$.

Dually, we have the following result for the elements of
$\mathcal{WAO}_{0}(X)$.

######  Proposition 6.4.

$(X_{1},\unlhd_{1},\wedge_{1},\vee_{1},0_{1},1_{1})$,
$(X_{2},\unlhd_{2},\wedge_{2},\vee_{2},0_{2},1_{2})$ be two bounded trellises,
$S\in\mathcal{WAO}_{0}(X_{2})$ and $\rho:X_{1}\longrightarrow X_{2}$ an
isomorphism. Then the binary operation $S^{\rho}$ defined by:

$S^{\rho}(x,y)=\rho^{-1}(S(\rho(x),\rho(y)))$, for any $x,y\in X_{1}$ ,

is an element of $\mathcal{WAO}_{0}(X_{1})$.

##  7 Conclusion

In this paper, we have studied the notion of pseudo-triangular norms on a
bounded trellis and provided some examples. Further, we have provided some new
class of pseudo-triangular norms and some characterisation. We intend that
this study open the door of different applications of trellis structure in
various areas using pseudo-triangular norms.

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