#  New structures for uninorms on bounded lattices††thanks: Supported by
National Natural Science Foundation of China (No.11871097)

Zhen-Yu Xiu1111 E-mail address<EMAIL_ADDRESS><EMAIL_ADDRESS>Yu-Xiu
Jiang2222Corresponding author. E-mail address<EMAIL_ADDRESS>
_1,2 College of Applied Mathematics, Chengdu University of Information
Technology,_
_ Chengdu 610000, China _

> Abstract In this article, we study new methods for constructing uninorms on
> bounded lattices. First, we present new methods for constructing uninorms on
> bounded lattices under the additional constraints and prove that some of
> these constraints are sufficient and necessary for the uninorms. Second, we
> show that the additional constraints on $t$-norms ($t$-conorms) and
> $t$-subnorms ($t$-subconorms) of some uninorms are exactly sufficient and
> necessary. At last, we give new constructions of uninorms on arbitrary
> bounded lattices by interation based on $t$-conorms and $t$-conorms.
>
> _Keywords_ : bounded lattices; $t$-norms; $t$-subnorms; uninorms

##  1 Introduction

The uninorms on the unit interval $[0,1]$, as a generalization of $t$-norms
and $t$-conorms, are introduced by Yager and Rybalov [24] and then applied to
various fieleds [12, 13, 14, 22], such as fuzzy logic, fuzzy set theory,
expert systems, neural networks and so on.

Recently, because of “For general systems, where we cannot always expect real
(or comparable) data, this extension of the underlying career together with
its structure is significant…” [7], the researchers widely study uninorms on
the bounded lattices instead of the unit interval $[0,1]$. About the methods
for the construction of uninorms, they mainly focus on t-norms (t-conorms) [1,
3, 4, 5, 6, 8, 9, 11, 19, 23], $t$-subnorms ( $t$-subconorms) [15, 18, 25],
closure operators ( interior operators) [7, 16, 21, 26] and additive
generators [17].

For the extension, as shown in many existing results, yielding uninorms on the
bounded lattice always needs some additional constraints on the lattice,
$t$-norms (resp. $t$-conorms), $t$-subnorms (resp. $t$-subconorms) or closure
operators (resp. interior operators) to guarantee that the methods work well.
However, we may ask the following question: are these additional constraints
necessary?

As we see, a lot of constructions of uninorms were obtained by the existing
methods. However, we may still ask are there more methods to construct
uninorms on bounded lattices, different from the existing methods?

In this paper, on one hand, we present new methods for constructing uninorms
on the bounded lattices under some additional constraints and show that some
constraints are necessary, that is, these constraints are sufficient and
necessary for the uninorms. Moreover, some sufficient conditions of some
uninorms are modified to be sufficient and necessary. On the other hand, we
present new methods for constructing uninorms on arbitrary bounded lattices by
the iterative method using $t$-norms and $t$-conorms, which are different from
the existing methods. This method is used to construct uninorms on arbitrary
bounded lattices for the first time.

The rest of this paper is organized as follows. In section $2$, we recall some
preliminaries. In section $3$, we construct uninorms based on the sufficient
and necessary conditions of the $t$-subnorm ($t$-subconorm) and the sufficient
and necessary constraints of bounded lattices, respectively. In section $4$,
we give the modifications of some uninorms. In Section $5$, we present a new
method for constructing uninorms on arbitrary bounded lattices by the
iterative method using $t$-norms and $t$-conorms. In Section $6$, some
conclusions are made.

##  2 Preliminaries

In this section, we recall some preliminary details concerning bounded lattice
and results related to aggregation functions on bounded lattices.

######  Definition 2.1 ([2]).

_A lattice $(L,\leq)$ is bounded if it has top and bottom elements, which are
written as $1$ and $0$, respectively, that is, there exit two elements $1,0\in
L$ such that $0\leq x\leq 1$ for all $x\in L$._

_Throughout this article, unless stated otherwise, we denote $L$ as a bounded
lattice with the top and bottom elements $1$ and $0$, respectively._

######  Definition 2.2 ([2]).

_Let $L$ be a bounded lattice, $a,b\in L$ with $a\leq b$. A subinterval
$[a,b]$ of $L$ is defined as_

$[a,b]=\\{x\in L:a\leq x\leq b\\}.$

_Similarly, we can define $[a,b[=\\{x\in L:a\leq x<b\\},]a,b]=\\{x\in
L:a<x\leq b\\}$ and $]a,b[=\\{x\in L:a<x<b\\}$. Denote
$D(e)=I_{e}\times\\{1\\}\cup\\{1\\}\times
I_{e}\cup[0,e]\times\\{1\\}\cup\\{1\\}\times[0,e]\cup[e,1]\times\\{e\\}\cup\\{e\\}\times[e,1]$,
$D(e)^{\prime}=I_{e}\times\\{1\\}\cup\\{1\\}\times
I_{e}\cup]0,e[\times\\{1\\}\cup\\{1\\}\times]0,e[\cup]e,1]\times\\{e\\}\cup\\{e\\}\times]e,1]$,
$E(e)=I_{e}\times\\{0\\}\cup\\{0\\}\times
I_{e}\cup[e,1]\times\\{0\\}\cup\\{0\\}\times[e,1]\cup[0,e]\times\\{e\\}\cup\\{e\\}\times[0,e]$,
$E(e)^{\prime}=I_{e}\times\\{0\\}\cup\\{0\\}\times
I_{e}\cup]e,1[\times\\{0\\}\cup\\{0\\}\times]e,1[\cup[0,e[\times\\{e\\}\cup\\{e\\}\times[0,e[$.
If $a$ and $b$ are incomparable, then we use the notation $a\parallel b$. For
$e\in L\setminus\\{0,1\\}$, we denote the set of all incomparable elements
with $e$ by $I_{e}$, that is, $I_{e}=\\{x\in L\mid x\parallel e\\}$._

######  Definition 2.3 ([20]).

_Let $(L,\leq,0,1)$ be a bounded lattice._
 _(i) An operation $T:L^{2}\rightarrow L$ is called a $t$-norm on $L$ if it is
commutative, associative, and increasing with respect to both variables, and
it has the neutral element $1\in L$, that is, $T(1,x)=x$ for all $x\in L$. The
greatest $t$-norm $T_{\wedge}$ on $L$ is given as $T_{\wedge}(x,y)=x\wedge
y$;_
 _(ii) An operation $S:L^{2}\rightarrow L$ is called a $t$-conorm on $L$ if it
is commutative, associative, and increasing with respect to both variables,
and it has the neutral element $0\in L$, that is, $S(0,x)=x$ for all $x\in L$.
The smallest t-conorm $S_{\vee}$ on $L$ is given as $S_{\vee}(x,y)=x\vee y$._

######  Definition 2.4 ([19]).

_Let $(L,\leq,0,1)$ be a bounded lattice. An operation $U:L^{2}\rightarrow L$
is called a uninorm on $L$ (a uninorm if $L$ is fixed) if it is commutative,
associative, and increasing with respect to both variables, and it has the
neutral element $e\in L$, that is, $U(e,x)=x$ for all $x\in L$._

######  Proposition 2.1 ([19]).

Let $(L,\leq,0,1)$ be a bounded lattice and $U$ be a uninorm on $L$ with a
neutral element $e\in L\setminus\\{0,1\\}$. Then we have the following:
(i) $T_{e}=U\mid[0,e]^{2}:[0,e]^{2}\rightarrow[0,e]$ is a $t$-norm on $[0,e]$;
(ii) $S_{e}=U\mid[e,1]^{2}:[e,1]^{2}\rightarrow[e,1]$ is a $t$-conorm on
$[e,1]$.

_$T_{e}$ and $S_{e}$ given in proposition 2.1 are called the underlying
$t$-norm and $t$-conorm of a uninorm $U$ on a bounded lattice $L$ with the
neutral element $e$, respectively. Throughout this study, we denote $T_{e}$ as
the underlying $t$-norm and $S_{e}$ as the underlying $t$-conorm of a given
uninorm $U$ on $L$._

######  Proposition 2.2 ([18]).

_Let $S$ be a nonempty set and $A_{1},A_{2},\ldots,A_{n}$ be subsets of $S$.
Let $H$ be a commutative binary operation on $S$. Then $H$ is associative on
$\bigcup\limits_{i\in I}A_{i}$ if the following statements hold:_
 _(i) $H$ is alternating associative on $(A_{i},A_{j},A_{k})$ for every
combination $\\{i,j,k\\}$ of $size$ $3$ chosen from $\\{1,2,\ldots,n\\}$;_
 _(ii) $H$ is alternating associative on $(A_{i},A_{i},A_{j})$ for every
combination $\\{i,j\\}$ of $size$ $2$ chosen from $\\{1,2,\ldots,n\\}$;_
 _(iii) $H$ is alternating associative on $(A_{i},A_{j},A_{j})$ for every
combination $\\{i,j,j\\}$ of $size$ $2$ chosen from $\\{1,2,\ldots,n\\}$;_
 _(iv) $H$ is alternating associative on $(A_{i},A_{i},A_{i})$ for every
$i\in\\{1,2,\ldots,n\\}$._

######  Definition 2.5 ([20]).

_Let $(L,\leq,0,1)$ be a bounded lattice._
 _(i) An operation $F:L^{2}\rightarrow L$ is called a $t$-subnorm on $L$ if it
is commutative, associative, increasing with respect to both variables, and
$F(x,y)\leq x\wedge y$ for all $x,y\in L$._
 _(ii) An operation $R:L^{2}\rightarrow L$ is called a $t$-subconorm on $L$ if
it is commutative, associative, increasing with respect to both variables, and
$R(x,y)\geq x\vee y$ for all $x,y\in L$._

######  Lemma 2.1 ([5]).

Let $L$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(1) If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then $z\vee t\in
I_{e}$ or $z\vee t=1$ for all $z,t\in I_{e}$.
(2) If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then $z\wedge t\in
I_{e}$ or $z\wedge t=0$ for all $z,t\in I_{e}$.

######  Remark 2.1.

By lemma 2.1, we can get the following results.
(1) If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then $x\vee y\in
I_{e}$ or $x\vee y=1$ for all ${x,y}\in I_{e}$.
(2) If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then $x\wedge y\in
I_{e}$ or $x\wedge y=0$ for all ${x,y}\in I_{e}$.
(3) If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then $R(x,y)\in
I_{e}$ or $R(x,y)=1$ for all ${x,y}\in I_{e}$.
(4) If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then $F(x,y)\in
I_{e}$ or $F(x,y)=0$ for all ${x,y}\in I_{e}$.

######  Theorem 2.1 ([10]).

_Let $e\in L\setminus\\{0,1\\}$, $T_{e}$ is a t-norm on $[0,e]$ and $S_{e}$ is
a t-norm on $[e,1]$. Then the function $U_{(T,e)}:L^{2}\rightarrow L$ and
$U_{(S,e)}:L^{2}\rightarrow L$ defined as follows _

_$U_{(T,e)}(x,y)=\begin{cases}T_{e}(x,y) &\mbox{if }(x,y)\in[0,e]^{2},\\\
x&\mbox{if }(x,y)\in I_{e}\times[0,e],\\\ y&\mbox{if }(x,y)\in[0,e]\times
I_{e},\\\ x\vee y\vee e&\mbox{}otherwise,\\\ \end{cases}$
and_

_$U_{(S,e)}(x,y)=\begin{cases}S_{e}(x,y) &\mbox{if }(x,y)\in[e,1]^{2},\\\
x&\mbox{if }(x,y)\in I_{e}\times[e,1],\\\ y&\mbox{if }(x,y)\in[e,1]\times
I_{e},\\\ x\wedge y\wedge e&\mbox{}otherwise.\\\ \end{cases}$_

######  Theorem 2.2 ([5]).

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(i) Let $T_{e}$ be a $t$-norm on $[0,e]$ such that $T_{e}(x,y)>0$ for all
$x,y>0$ and $S_{e}$ be a $t$-conorm on $[e,1]$ such that $S_{e}(x,y)<1$ for
all $x,y<1$. If $y\parallel x$ for all $x\in I_{e}$ and $y\in[e,1[$, then the
function $U_{1}^{e}:L^{2}\rightarrow L$ is a uninorm on $L$ with the neutral
element $e$, where

$U_{1}^{e}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if }(x,y)\in
I_{e}\times[e,1[\cup I_{e}\times]0,e[,\\\ y&\mbox{if }(x,y)\in[e,1[\times
I_{e}\cup]0,e[\times I_{e},\\\ x\vee y&\mbox{if }(x,y)\in I_{e}^{2}\cup
I_{e}\times\\{1\\}\cup\\{1\\}\times
I_{e}\cup]0,e[\times\\{1\\}\cup\\{1\\}\times]0,e[,\\\ x\wedge y&\mbox{
}otherwise.\\\ \end{cases}$
(ii) Let $T_{e}$ be a $t$-norm on $[0,e]$ such that $T_{e}(x,y)>0$ for all
$x,y>0$ and $S_{e}$ be a $t$-conorm on $[e,1]$ such that $S_{e}(x,y)<1$ for
all $x,y<1$. If $y\parallel x$ for all $x\in I_{e}$ and $y\in]0,e]$, then the
function $U_{2}^{e}:L^{2}\rightarrow L$ is a uninorm on $L$ with the neutral
element $e$, where

$U_{2}^{e}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if }(x,y)\in
I_{e}\times]e,1[\cup I_{e}\times]0,e],\\\ y&\mbox{if }(x,y)\in]e,1[\times
I_{e}\cup]0,e]\times I_{e},\\\ x\wedge y&\mbox{if }(x,y)\in I_{e}^{2}\cup
I_{e}\times\\{0\\}\cup\\{0\\}\times
I_{e}\cup]e,1[\times\\{0\\}\cup\\{0\\}\times]e,1[,\\\ x\vee y&\mbox{
}otherwise.\\\ \end{cases}$

######  Theorem 2.3 ([15]).

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(i) Let $T_{e}$ be a $t$-norm on $[0,e]$ and $R$ is a $t$-subconorm on $L$
such that $R(x,y)<1$ for all $x,y<1$. If $x\parallel y$ for all $x\in I_{e}$
and $y\in[e,1[$, then the function $U_{R}^{e}:L^{2}\rightarrow L$ is a uninorm
on $L$ with the neutral element $e$, where

$U_{R}^{e}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
x&\mbox{if }(x,y)\in I_{e}\times[0,e]\cup I_{e}\times[e,1[,\\\ y&\mbox{if
}(x,y)\in[0,e]\times I_{e}\cup[e,1[\times I_{e},\\\ R(x,y)&\mbox{if }(x,y)\in
I_{e}^{2}\cup]e,1]^{2},\\\ x\vee y&\mbox{if }(x,y)\in D(e),\\\ x\wedge
y&\mbox{ }otherwise.\\\ \end{cases}$
(ii) Let $S_{e}$ be a $t$-conorm on $[e,1]$ and $F$ is a $t$-subnorm on $L$
such that $F(x,y)>0$ for all $x,y>0$. If $x\parallel y$ for all $x\in I_{e}$
and $y\in]0,e]$, then the function $U_{F}^{e}:L^{2}\rightarrow L$ is a uninorm
on $L$ with the neutral element $e$, where

$U_{F}^{e}(x,y)=\begin{cases}S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\
x&\mbox{if }(x,y)\in I_{e}\times]0,e]\cup I_{e}\times[e,1],\\\ y&\mbox{if
}(x,y)\in]0,e]\times I_{e}\cup[e,1]\times I_{e},\\\ F(x,y)&\mbox{if }(x,y)\in
I_{e}^{2}\cup[0,e[^{2},\\\ x\wedge y&\mbox{if }(x,y)\in E(e),\\\ x\vee
y&\mbox{ }otherwise.\\\ \end{cases}$

######  Theorem 2.4 ([25]).

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(i) Let $T_{e}$ be a $t$-norm on $[0,e]$ and $R$ be a $t$-subconorm on
$L\setminus[0,e]$, then the function $U_{R}:L^{2}\rightarrow L$ defined by

$U_{R}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\ x&\mbox{if
}(x,y)\in[0,e[\times(L\setminus[0,e]),\\\ y&\mbox{if
}(x,y)\in(L\setminus[0,e])\times[0,e[,\\\ y&\mbox{if
}(x,y)\in\\{e\\}\times(L\setminus[0,e]),\\\ x&\mbox{if
}(x,y)\in(L\setminus[0,e])\times\\{e\\},\\\ R(x,y)&\mbox{if
}(x,y)\in(L\setminus[0,e])^{2},\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $y<x$ for all $x\in
I_{e}$ and $y\in]0,e[$.
(ii) Let $S_{e}$ be a $t$-conorm on $[e,1]$ and $F$ be a $t$-subnorm on
$L\setminus[e,1]$, then the function $U_{F}:L^{2}\rightarrow L$ defined by

$U_{F}(x,y)=\begin{cases}S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if
}(x,y)\in]e,1]\times(L\setminus[e,1]),\\\ y&\mbox{if
}(x,y)\in(L\setminus[e,1])\times]e,1],\\\ y&\mbox{if
}(x,y)\in\\{e\\}\times(L\setminus[e,1]),\\\ x&\mbox{if
}(x,y)\in(L\setminus[e,1])\times\\{e\\},\\\ F(x,y)&\mbox{if
}(x,y)\in(L\setminus[e,1])^{2},\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $x<y$ for all $x\in
I_{e}$ and $y\in]e,1[$.

######  Theorem 2.5 ([7]).

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(i) Let $T_{e}$ be a $t$-norm on $[0,e]$, then the function
$U_{\wedge}:L^{2}\rightarrow L$ defined by

$U_{\wedge}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
x\wedge y&\mbox{if }(x,y)\in[0,e[\times I_{e}\cup
I_{e}\times[0,e[\cup[0,e[\times[e,1]\cup[e,1]\times[0,e[,\\\ y&\mbox{if
}(x,y)\in\\{e\\}\times I_{e},\\\ x&\mbox{if }(x,y)\in I_{e}\times\\{e\\},\\\
x\vee y&otherwise,\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $x>y$ for all $x\in
I_{e}$ and $y\in[0,e[$.
(ii) Let $S_{e}$ be a $t$-conorm on $[e,1]$, then the function
$U_{\vee}:L^{2}\rightarrow L$ defined by

$U_{\vee}(x,y)=\begin{cases}S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x\vee
y&\mbox{if }(x,y)\in]e,1]\times I_{e}\cup
I_{e}\times]e,1]\cup[0,e[\times]e,1]\cup]e,1]\times[0,e[,\\\ y&\mbox{if
}(x,y)\in\\{e\\}\times I_{e},\\\ x&\mbox{if }(x,y)\in I_{e}\times\\{e\\},\\\
x\wedge y&otherwise,\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $x<y$ for all $x\in
I_{e}$ and $y\in]e,1]$.

######  Theorem 2.6 ([10]).

_Assume that $T:L^{2}\rightarrow L$ and $S:L^{2}\rightarrow L$ are a t-norm
and a t-conorm, respectively. Let $e\in L$ be an element distinct from both
$0_{L}$ and $1_{L}$. Let us denote by $U_{d}:L^{2}\rightarrow L$ and
$U_{c}:L^{2}\rightarrow L$ the following:_

_$U_{d}(x,y)=\begin{cases}T(x,y) &\mbox{if }(x,y)\in[0,e]^{2},\\\
S(x,y)&\mbox{if }(x,y)\in(e,1]^{2},\\\ x&\mbox{if
}(x,y)\in(I_{e}\cup(e,1])\times[0,e],\\\ y&\mbox{if
}(x,y)\in[0,e]\times(I_{e}\cup(e,1]),\\\ S(x\vee a,y\vee
a)&\mbox{}otherwise,\\\ \end{cases}$
and_

_$U_{c}(x,y)=\begin{cases}T(x,y) &\mbox{if }(x,y)\in[0,e)^{2},\\\
S(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if
}(x,y)\in(I_{e}\cup[0,e))\times[e,1],\\\ y&\mbox{if
}(x,y)\in[e,1]\times(I_{e}\cup[0,e)),\\\ T(x\wedge a,y\wedge
a)&\mbox{}otherwise.\\\ \end{cases}$
_

##  3  New construction of uninorms on bounded lattices

In this section, we present new construction of uninorms on some appropriate
bounded lattices $L$, which are constructed by means of $t$-conorms
($t$-norms) and $t$-subconorms ($t$-subnorms). Moreover, the sufficient and
necessary conditions are provided for these uninorms.

######  Theorem 3.1.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$,
$T_{e}$ be a $t$-norm on $[0,e]$ and $S_{e}$ be a $t$-conorm on $[e,1]$.
$(i)$ If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then the
function $U_{1}:L^{2}\rightarrow L$ defined by

$U_{1}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if }(x,y)\in
I_{e}\times[e,1[\cup I_{e}\times[0,e[,\\\ y&\mbox{if }(x,y)\in[e,1[\times
I_{e}\cup[0,e[\times I_{e},\\\ x\vee y&\mbox{if }(x,y)\in{I_{e}}^{2}\cup
I_{e}\times\\{1\\}\cup\\{1\\}\times
I_{e}\cup[0,e[\times\\{1\\}\cup\\{1\\}\times[0,e[,\\\ x\wedge
y&\mbox{}otherwise,\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $S_{e}(x,y)<1$ for all
$x,y<1$.
$(ii)$ If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then the
function $U_{2}:L^{2}\rightarrow L$ defined by

$U_{2}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if }(x,y)\in
I_{e}\times]e,1]\cup I_{e}\times]0,e],\\\ y&\mbox{if }(x,y)\in]e,1]\times
I_{e}\cup]0,e]\times I_{e},\\\ x\wedge y&\mbox{if }(x,y)\in{I_{e}}^{2}\cup
I_{e}\times\\{0\\}\cup\\{0\\}\times
I_{e}\cup]e,1]\times\\{0\\}\cup\\{0\\}\times]e,1],\\\ x\vee
y&\mbox{}otherwise,\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $T_{e}(x,y)>0$ for all
$x,y>0$.

Proof. Let $(L,\leq,0,1)$ be a bounded lattice with $e\in
L\setminus\\{0,1\\}$. If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$,
then we give the proof of the fact that $U_{1}$ is a uninorm iff
$S_{e}(x,y)<1$ for all $x,y<1$. The same result for $U_{2}$ can be obtained by
similar arguments.

Necessity. Let $U_{1}$ be a uninorm on $L$ with the identity $e$. Then we
prove that $S_{e}(x,y)<1$ for all $x,y<1$. Assume that there are some elements
$x\in]e,1[$ and $y\in]e,1[$ such that $S_{e}(x,y)=1$. If $z\in I_{e}$, then
$U_{1}(x,U_{1}(y,z))=U_{1}(x,z)=z$ and
$U_{1}(U_{1}(x,y),z)=U_{1}(S_{e}(x,y),z)=U_{1}(1,z)=1$. Since $S_{e}(x,y)=1$,
the associativity property of $U_{1}$ is violated. Then $U_{1}$ is not a
uninorm on $L$, which is a contradiction. Hence, $S_{e}(x,y)<1$ for all
$x,y<1$.

Sufficiency. To prove the sufficiency, we compare $U_{1}$ with $U_{1}^{e}$ in
Theorem 2.2 and find that besides the condition that $T_{e}(x,y)>0$ for all
$x,y>0$, the values of $U_{1}$ and $U_{1}^{e}$ are different on
$I_{e}\times\\{0\\}\cup\\{1\\}\times\\{0\\}$, $\\{0\\}\times
I_{e}\cup\\{0\\}\times\\{1\\}$ and $]0,e[\times]0,e[$. So, combined with
Theorem 2.2, the proof of the sufficiency can be reduced to the above areas.
The commutativity of $U_{1}$ and the fact that $e$ is a neutral element are
evident. Hence, we prove only the monotonicity and the associativity of
$U_{1}$.

I. Monotonicity: We prove that if $x\leq y$, then for all $z\in L$,
$U_{1}(x,z)\leq U_{1}(y,z)$.

If one of $x$, $y$, $z$ is equal to $1$, then it is clear that $U_{1}(x,z)\leq
U_{1}(y,z)$. There is no $1$ in any of the following cases.

1\. Let $x\in[0,e[$.

1.1. $y\in[0,e[$,

1.1.1. $z\in I_{e}$,

$U_{1}(x,z)=z=U_{1}(y,z)$

1.2. $y\in[e,1[$,

1.2.1. $z\in I_{e}$,

$U_{1}(x,z)=z=U_{1}(y,z)$

1.3. $y\in I_{e}$,

1.3.1. $z\in[0,e[$,

$U_{1}(x,z)=T_{e}(x,z)\leq y=U_{1}(y,z)$

1.3.2. $z\in[e,1[$,

$U_{1}(x,z)=x\leq y=U_{1}(y,z)$

1.3.3. $z\in I_{e}$,

$U_{1}(x,z)=z\leq y\vee z=U_{1}(y,z)$

2\. Let $x\in I_{e}$.

2.1. $y\in I_{e}$,

2.1.1. $z\in[0,e[$,

$U_{1}(x,z)=x\leq y=U_{1}(y,z)$

II. Associativity: We need to prove that
$U_{1}(x,U_{1}(y,z))=U_{1}(U_{1}(x,y),z)$ for all $x,y,z\in L$.

If one of $x,y,z$ is equal to $1$, then it is clear that
$U_{1}(x,U_{1}(y,z))=1=U_{1}(U_{1}(x,y),z)$. So there is no $1$ in any of the
following cases.

1\. Let $x\in[0,e[$.

1.1. $y\in[0,e[$,

1.1.1. $z\in I_{e}$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,z)=z=U_{1}(T_{e}(x,y),z)=U_{1}(U_{1}(x,y),z)$

1.2. $y\in[e,1[$,

1.2.1. $z\in I_{e}$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,z)=z=U_{1}(x,z)U_{1}(U_{1}(x,y),z)$

1.3. $y\in I_{e}$,

1.3.1. $z\in[0,e[$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y)=y=U_{1}(y,z)=U_{1}(U_{1}(x,y),z)$

1.3.2. $z\in[e,1[$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y)=y=U_{1}(y,z)=U_{1}(U_{1}(x,y),z)$

1.3.3. $z\in I_{e}$,

1.3.3.1. $y\vee z\in I_{e}$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y\vee z)=y\vee z=U_{1}(y,z)=U_{1}(U_{1}(x,y),z)$

1.3.3.2. $y\vee z=1$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y\vee z)=1=y\vee
z=U_{1}(y,z)=U_{1}(U_{1}(x,y),z)$

2\. Let $x\in[e,1[$.

2.1. $y\in[0,e[$,

2.1.1. $z\in I_{e}$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,z)=z=U_{1}(y,z)=U_{1}(U_{1}(x,y),z)$

2.2. $y\in I_{e}$,

2.2.1. $z\in[0,e[$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y)=y=U_{1}(y,z)=U_{1}(U_{1}(x,y),z)$

3\. Let $x\in I_{e}$.

3.1. $y\in[0,e[$,

3.1.1. $z\in[0,e[$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,T_{e}(y,z))=x=U_{1}(x,z)=U_{1}(U_{1}(x,y),z)$

3.1.2. $z\in[e,1[$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y)=x=U_{1}(x,z)=U_{1}(U_{1}(x,y),z)$

3.1.3. $z\in I_{e}$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,z)=x\vee z=U_{1}(x,z)=U_{1}(U_{1}(x,y),z)$

3.2. $y\in[e,1[$,

3.2.1. $z\in[0,e[$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,z)=x=U_{1}(x,z)=U_{1}(U_{1}(x,y),z)$

3.3. $y\in I_{e}$,

3.3.1. $z\in[0,e[$,

3.3.1.1. $x\vee y\in I_{e}$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y)=x\vee y=U_{1}(x\vee y,z)=U_{1}(U_{1}(x,y),z)$

3.3.1.2. $x\vee y=1$,

$U_{1}(x,U_{1}(y,z))=U_{1}(x,y)=x\vee y=1=U_{1}(x\vee
y,z)=U_{1}(U_{1}(x,y),z)$

Hence, $U_{1}$ is a uninorm on $L$ with a neutral element $e$.

It is worth pointing out that if the condition that $x\parallel y$ for all
$x\in I_{e}$ and $y\in[e,1[$ in Theorem 3.1 is not satisfied, then $U_{1}$ may
not be a uninorm on $L_{1}$. Similarly, if the condition that $x\parallel y$
for all $x\in I_{e}$ and $y\in]0,e]$ in Theorem 3.1 is not satisfied, then
$U_{2}$ may not be a uninorm on $L_{1}$ . The following example will show the
above facts.

######  Example 3.1.

Consider the bounded lattice $L_{1}=\\{0,a,e,c,f,g,1\\}$ depicted by the Hasse
diagram in Fig.$1$, where $a\in[0,e[$, $g\in]e,1]$, $c\in I_{e}$, $f\in
I_{e}$, $g>c$, $g>f$,$a<c$ and $a<f$. Take $T_{e}(x,y)=T_{\wedge}(x,y)$ on
$[0,e]$ and $S_{e}(x,y)=S_{\vee}(x,y)$ on $[e,1]$. The function $U_{1}$ on
$L_{1}$, shown in Table 1, is not a uninorm on $L_{1}$ with the neutral
element $e$. In fact, the associativity is not satisfied, since
$U_{1}(a,U_{1}(c,f))=U_{1}(a,g)=a$ and $U_{1}(U_{1}(a,c),f)=U_{1}(c,f)=g$.
Similarly, the function $U_{2}$ on $L_{1}$, shown in Table 2, is not a uninorm
on $L_{1}$ with the neutral element $e$. In fact, the associativity is not
satisfied, since $U_{2}(g,U_{2}(c,f))=U_{2}(g,a)=g$ and
$U_{2}(U_{2}(g,c),f)=U_{2}(c,f)=a$.

$0$$a$$c$$e$$f$$g$$1$_Fig.1. The lattice $L_{1}$_

Table 1: The function $U_{1}$ on $L_{1}$ given in Fig.1.
$U_{1}$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $c$ | $f$ | $0$ | $1$
$a$ | $0$ | $a$ | $a$ | $c$ | $f$ | $a$ | $1$
$e$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
$c$ | $c$ | $c$ | $c$ | $c$ | $g$ | $c$ | $1$
$f$ | $f$ | $f$ | $f$ | $g$ | $f$ | $f$ | $1$
$g$ | $0$ | $a$ | $g$ | $c$ | $f$ | $g$ | $1$
$1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
Table 2: The function $U_{2}$ on $L_{1}$ given in Fig.1.
$U_{2}$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $a$ | $a$ | $c$ | $f$ | $g$ | $1$
$e$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
$c$ | $0$ | $c$ | $c$ | $c$ | $a$ | $c$ | $c$
$f$ | $0$ | $f$ | $f$ | $a$ | $f$ | $f$ | $f$
$g$ | $0$ | $g$ | $g$ | $c$ | $f$ | $g$ | $1$
$1$ | $0$ | $1$ | $1$ | $c$ | $f$ | $1$ | $1$

######  Theorem 3.2.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
$(i)$ Let $T_{e}$ be a $t$-norm on $[0,e]$ and $R$ be a $t$-subconorm on $L$.
If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then the function
$U_{3}:L^{2}\rightarrow L$ defined by

$U_{3}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\ x&\mbox{if
}(x,y)\in I_{e}\times]0,e]\cup I_{e}\times]e,1[,\\\ y&\mbox{if
}(x,y)\in]0,e]\times I_{e}\cup]e,1[\times I_{e},\\\ R(x,y)&\mbox{if }(x,y)\in
I_{e}^{2}\cup]e,1]^{2},\\\ x\vee y&\mbox{if }(x,y)\in D(e)^{\prime},\\\
x\wedge y&\mbox{o}therwise,\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $T_{e}(x,y)>0$ for all
$x,y>0$ and $R(x,y)<1$ for all $x,y<1$.
$(ii)$ Let $F$ be a $t$-subnorm on $L$ and $S_{e}$ be a $t$-conorm on $[e,1]$.
If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then the function
$U_{4}:L^{2}\rightarrow L$ defined by

$U_{4}(x,y)=\begin{cases}S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if
}(x,y)\in I_{e}\times[e,1[\cup I_{e}\times]0,e[,\\\ y&\mbox{if
}(x,y)\in[e,1[\times I_{e}\cup]0,e[\times I_{e},\\\ F(x,y)&\mbox{if }(x,y)\in
I_{e}^{2}\cup[0,e[^{2},\\\ x\wedge y&\mbox{if }(x,y)\in E(e)^{\prime},\\\
x\vee y&\mbox{o}therwise,\end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $F(x,y)>0$ for all
$x,y>0$ and $S_{e}(x,y)<1$ for all $x,y<1$.

Proof. Let $(L,\leq,0,1)$ be a bounded lattice with $e\in
L\setminus\\{0,1\\}$. If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$,
then we give the proof of the fact that $U_{3}$ is a uninorm iff
$T_{e}(x,y)>0$ for all $x,y>0$ and $R(x,y)<1$ for all $x,y<1$. The same result
for $U_{4}$ can be obtained by similar arguments.

Necessity. Let the function $U_{3}$ be a uninorm on $L$ with the identity $e$.
Then we will prove that $T_{e}(x,y)>0$ for all $x,y>0$ and $R(x,y)<1$ for all
$x,y<1$. Assume that there are some elements $x\in]0,e[$ and $y\in]0,e[$ such
that $T_{e}(x,y)=0$. If $z\in I_{e}$, then we obtain
$U_{3}(x,U_{3}(y,z))=U_{3}(x,z)=z$ and
$U_{3}(U_{3}(x,y),z)=U_{3}(T_{e}(x,y),z)=U_{3}(0,z)=0$. Since $T_{e}(x,y)=0$,
the associativity property is violated. Assume that there are some elements
$x\in]e,1[$ and $y\in]e,1[$ such that $R(x,y)=1$. If $z\in I_{e}$, then we
obtain $U_{3}(x,U_{3}(y,z))=U_{3}(x,z)=z$ and
$U_{3}(U_{3}(x,y),z)=U_{3}(R(x,y),z)=U_{3}(1,z)=1$. Since $R(x,y)=1$, the
associativity property is violated. Hence, $U_{3}$ is not a uninorm on $L$
which is a contradiction. Therefore, the conditions that $T_{e}(x,y)>0$ for
all $x,y>0$ and $R(x,y)<1$ for all $x,y<1$ are necessary.

Sufficiency. To prove the sufficiency, we compare $U_{3}$ with $U_{R}^{e}$ in
Theorem 2.3 and find that besides the condition that $T_{e}(x,y)>0$ for all
$x,y>0$, the values of $U_{3}$ and $U_{R}^{e}$ are different on
$I_{e}\times\\{0\\}\cup\\{1\\}\times\\{0\\}$, $\\{0\\}\times
I_{e}\cup\\{0\\}\times\\{1\\}$ and $]0,e[\times]0,e[$. So, combined with
Theorem 2.3, the proof of the sufficiency can be reduced to the above areas.

The commutativity of $U_{3}$ and the fact that $e$ is a neutral element are
evident. Hence, we prove only the monotonicity and the associativity of
$U_{3}$.

I. Monotonicity: We prove that if $x\leq y$, then for all $z\in L$,
$U_{3}(x,z)\leq U_{3}(y,z)$.

1.$x=0$ or $y=0$ or $z=0$

$U_{3}(x,z)=0\leq U_{3}(y,z)$

II. Associativity: We need to prove that
$U_{3}(x,U_{3}(y,z))=U_{3}(U_{3}(x,y),z)$ for all $x,y,z\in L$.

If one of $x,y,z$ is equal to $0$, then it is clear that
$U_{3}(x,U_{3}(y,z))=U_{3}(U_{3}(x,y),z)$. Hence, $U_{3}$ is a uninorm on $L$
with a neutral element $e$.

It is worth pointing out that if the condition that $x\parallel y$ for all
$x\in I_{e}$ and $y\in[e,1[$ in Theorem 3.2 is not satisfied, then $U_{3}$ may
not be a uninorm on $L_{1}$. Similarly, if the condition that $x\parallel y$
for all $x\in I_{e}$ and $y\in]0,e]$ in Theorem 3.2 is not satisfied, then
$U_{4}$ may not be a uninorm on $L_{1}$.

######  Example 3.2.

Consider the bounded lattice $L_{1}=\\{0,a,e,c,f,g,1\\}$ depicted by the Hasse
diagram in Fig.$1$.

(1) If we take $T_{e}(x,y)=T_{\wedge}(x,y)$ on $[0,e]$ and $R(x,y)=x\vee y$,
then the function $U_{3}$ on $L_{1}$, shown in Table 3, is not a uninorm on
$L_{1}$ with the neutral element $e$. In fact, the associativity is not
satisfied, since $U_{3}(a,U_{3}(c,f))=U_{3}(a,g)=a$ and
$U_{3}(U_{3}(a,c),f)=U_{3}(c,f)=g$.

(2) If we take $S_{e}(x,y)=S_{\vee}(x,y)$ on $[e,1]$ and $F(x,y)=x\wedge y$,
then the function $U_{4}$ on $L_{1}$, shown in Table 4, is not a uninorm on
$L_{1}$ with the neutral element $e$. In fact, the associativity is not
satisfied, since $U_{4}(g,U_{4}(c,f))=U_{4}(g,a)=g$ and
$U_{4}(U_{4}(g,c),f)=U_{4}(c,f)=a$.

Table 3: The function $U_{3}$ on $L_{1}$ given in Fig.1.
$U_{3}$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $a$ | $a$ | $c$ | $f$ | $a$ | $1$
$e$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
$c$ | $0$ | $c$ | $c$ | $c$ | $g$ | $c$ | $1$
$f$ | $0$ | $f$ | $f$ | $g$ | $f$ | $f$ | $1$
$g$ | $0$ | $a$ | $g$ | $c$ | $f$ | $g$ | $1$
$1$ | $0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$

Table 4: The function $U_{4}$ on $L_{1}$ given in Fig.1.
$U_{4}$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $1$
$a$ | $0$ | $a$ | $a$ | $c$ | $f$ | $g$ | $1$
$e$ | $0$ | $a$ | $e$ | $c$ | $f$ | $g$ | $1$
$c$ | $0$ | $c$ | $c$ | $c$ | $a$ | $c$ | $1$
$f$ | $0$ | $f$ | $f$ | $a$ | $f$ | $f$ | $1$
$g$ | $0$ | $g$ | $g$ | $c$ | $f$ | $g$ | $1$
$1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$

######  Theorem 3.3.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
$(i)$ Let $T_{e}$ be a $t$-norm on $[0,e]$, and $R$ be a $t$-subconorm on $L$.
If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then the function
$U_{5}:L^{2}\rightarrow L$ defined by

$U_{5}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\ x\wedge
y&\mbox{if }(x,y)\in[0,e[\times I_{e}\cup
I_{e}\times[0,e[\cup[0,e[\times]e,1]\cup]e,1]\times[0,e[,\\\ y&\mbox{if
}(x,y)\in\\{e\\}\times I_{e}\cup\\{e\\}\times]e,1],\\\ x&\mbox{if }(x,y)\in
I_{e}\times\\{e\\}\cup]e,1]\times\\{e\\},\\\ R(x,y)&\mbox{if
}(x,y)\in{I_{e}}^{2},\\\ 1&\mbox{if }(x,y)\in I_{e}\times]e,1]\cup]e,1]\times
I_{e}\cup]e,1]^{2},\end{cases}$
is a uninorm on $L$ with the neutral element $e$ iff $x>y$ for all $x\in
I_{e}$ and $y\in[0,e[$.
$(ii)$ Let $S_{e}$ be a $t$-conorm on $[e,1]$, and $F$ be a $t$-subnorm on
$L$. If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then the function
$U_{6}:L_{2}\rightarrow L$ defined by

$U_{6}(x,y)=\begin{cases}S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x\vee
y&\mbox{if }(x,y)\in]e,1]\times I_{e}\cup
I_{e}\times]e,1]\cup[0,e[\times]e,1]\cup]e,1]\times[0,e[,\\\ y&\mbox{if
}(x,y)\in\\{e\\}\times I_{e}\cup\\{e\\}\times[0,e[,\\\ x&\mbox{if }(x,y)\in
I_{e}\times\\{e\\}\cup[0,e[\times\\{e\\},\\\ F(x,y)&\mbox{if
}(x,y)\in{I_{e}}^{2},\\\ 0&\mbox{if }(x,y)\in I_{e}\times[0,e[\cup[0,e[\times
I_{e}\cup[0,e[^{2},\end{cases}$
is a uninorm on $L$ with the neutral element $e$ iff $x<y$ for all $x\in
I_{e}$ and $y\in]e,1]$.

Proof. Let $(L,\leq,0,1)$ be a bounded lattice with $e\in
L\setminus\\{0,1\\}$. If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$,
then we give the proof of the fact that $U_{5}$ is a uninorm iff $x>y$ for all
$x\in I_{e}$ and $y\in[0,e[$. The same result for $U_{6}$ can be obtained
using similar arguments.

Necessity. Let the function $U_{5}$ be a uninorm on $L$ with the identity $e$.
Then we will prove that $x>y$ for all $x\in I_{e}$ and $y\in[0,e[$. Assume
that there are some elements $x\in I_{e}$ and $y\in[0,e[$ such that
$x\parallel y$. If $z\in]e,1]$, then $U_{5}(y,U_{5}(x,z))=U_{5}(y,1)=y$ and
$U_{5}(U_{5}(y,x),z)=U_{5}(x\wedge y,z)=x\wedge y\wedge z=x\wedge y$. Since
$x\parallel y$, the associativity property is violated. Then $U_{5}$ is not a
uninorm on $L$ which is a contradiction. Hence, $x>y$ for all $x\in I_{e}$ and
$y\in[0,e[$.

Sufficiency. Let $x>y$ for all $x\in I_{e}$ and $y\in[0,e[$, with $x\parallel
y$ for all $x\in I_{e}$ and $y\in[e,1[$. The commutativity of $U_{5}$ and the
fact that $e$ is a neutral element are evident. Hence, we prove only the
monotonicity and the associativity of $U_{5}$.

I. Monotonicity: We prove that if $x\leq y$, then for all $z\in L$,
$U_{5}(x,z)\leq U_{5}(y,z)$. The proof is split into all possible cases. If
one of $x$, $y$, $z$ is equal to $e$, then it is clear that
$U_{5}(x,z)=z=U_{5}(y,z)$

1\. Let $z=e$.

$U_{5}(x,z)=x\leq y=U_{5}(y,z)$

2\. Let $x=e$.

2.1. $y=e$,

$U_{5}(x,z)=z=U_{5}(y,z)$

2.2. $y\in]e,1]$,

2.2.1. $z\in[0,e[$,

$U_{5}(x,z)=z=U_{5}(y,z)$

2.2.2. $z\in]e,1]$,

$U_{5}(x,z)=z\leq 1=U_{5}(y,z)$

2.2.3. $z\in I_{e}$,

$U_{5}(x,z)=z\leq 1=U_{5}(y,z)$

3\. Let $y=e$.

3.1. $x\in[0,e[$,

3.1.1. $z\in[0,e[$,

$U_{5}(x,z)=T_{e}(x,z)\leq z=U_{5}(y,z)$

3.1.2. $z\in]e,1]$,

$U_{5}(x,z)=x\leq z=U_{5}(y,z)$

3.1.3. $z\in I_{e}$,

$U_{5}(x,z)=x\wedge z=x\leq z=U_{5}(y,z)$

There is no $e$ in any of the following cases.

1\. Let $x\in[0,e[$.

1.1. $y\in[0,e[$,

1.1.1. $z\in[0,e[$,

$U_{5}(x,z)=T_{e}(x,z)\leq T_{e}(y,z)=U_{5}(y,z)$

1.1.2. $z\in]e,1]$,

$U_{5}(x,z)=x\wedge z=x\leq y=y\wedge z=U_{5}(y,z)$

1.1.3. $z\in I_{e}$,

$U_{5}(x,z)=x\wedge z=x\leq y=y\wedge z=U_{5}(y,z)$

1.2. $y\in]e,1]$,

1.2.1. $z\in[0,e[$,

$U_{5}(x,z)=T_{e}(x,z)\leq z=y\wedge z=U_{5}(y,z)$

1.2.2. $z\in]e,1]$,

$U_{5}(x,z)=x\wedge z=x\leq 1=U_{5}(y,z)$

1.2.3. $z\in I_{e}$,

$U_{5}(x,z)=x\wedge z=x\leq 1=U_{5}(y,z)$

1.3. $y\in I_{e}$,

1.3.1. $z\in[0,e[$,

$U_{5}(x,z)=T_{e}(x,z)\leq z=y\wedge z=U_{5}(y,z)$

1.3.2. $z\in]e,1]$,

$U_{5}(x,z)=x\wedge z=x\leq 1=U_{5}(y,z)$

1.3.3. $z\in I_{e}$,

$U_{5}(x,z)=x\wedge z=x\leq R(y,z)=U_{5}(y,z)$

2\. Let $x\in]e,1]$.

2.1. $y\in]e,1]$,

2.1.1. $z\in[0,e[$,

$U_{5}(x,z)=x\wedge z=z=y\wedge z=U_{5}(y,z)$

2.1.2. $z\in]e,1]$,

$U_{5}(x,z)=1=U_{5}(y,z)$

2.1.3. $z\in I_{e}$,

$U_{5}(x,z)=1=U_{5}(y,z)$

3\. Let $x\in I_{e}$.

3.1. $y\in I_{e}$,

3.1.1. $z\in[0,e[$,

$U_{5}(x,z)=x\wedge z=z=y\wedge z=U_{5}(y,z)$

3.1.2. $z\in]e,1]$,

$U_{5}(x,z)=1=U_{5}(y,z)$

3.1.3. $z\in I_{e}$,

$U_{5}(x,z)=R(x,z)\leq R(y,z)=U_{5}(y,z)$

II. Associativity: We need to prove that
$U_{5}(x,U_{5}(y,z))=U_{5}(U_{5}(x,y),z)$ for all $x,y,z\in L$. The proof is
split into all possible cases. If one of $x,y,z$ is equal to $e$, then it is
clear that $U_{5}(x,U_{5}(y,z))=U_{5}(U_{5}(x,y),z)$. So there is no $e$ in
any of the following cases. By Proposition 2.2, we only need to consider the
following cases:

1\. Let $x\in[0,e[$.

1.1. $y\in[0,e[$,

1.1.1. $z\in[0,e[$,

$U_{5}(x,U_{5}(y,z))=T_{e}(x,T_{e}(y,z))=T_{e}(T_{e}(x,y),z)=U_{5}(U_{5}(x,y),z)$

1.1.2. $z\in]e,1]$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,y\wedge z)=T_{e}(x,y)=U_{5}(T_{e}(x,y),z)=$

$U_{5}(U_{5}(x,y),z)$

1.1.3. $z\in I_{e}$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,y\wedge z)=T_{e}(x,y)=U_{5}(T_{e}(x,y),z)=$

$U_{5}(U_{5}(x,y),z)$

1.2. $y\in]e,1]$,

1.2.1. $z\in]e,1]$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,1)=x=U_{5}(x,z)=U_{5}(U_{5}(x,y),z)$

1.2.2. $z\in I_{e}$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,1)=x,U_{5}(U_{5}(x,y),z)=U_{5}(x,z)=x$

and $U_{5}(y,U_{5}(x,z))=U_{5}(y,x)=x$

1.3. $y\in I_{e}$,

1.3.1. $z\in I_{e}$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,R(y,z))=x=U_{5}(x,z)=U_{5}(U_{5}(x,y),z)$

2\. Let $x\in]e,1]$.

2.1. $y\in]e,1]$,

2.1.1. $z\in]e,1]$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,1)=1=U_{5}(1,z)=U_{5}(U_{5}(x,y),z)$

2.1.2. $z\in I_{e}$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,1)=1=U_{5}(1,z)=U_{5}(U_{5}(x,y),z)$

2.2. $y\in I_{e}$,

2.2.1. $z\in I_{e}$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,R(y,z))=1=U_{5}(1,z)=U_{5}(U_{5}(x,y),z)$

3\. Let $x\in I_{e}$.

3.1. $y\in I_{e}$,

3.1.1. $z\in I_{e}$,

3.1.1.1. $R(x,y)=1,R(y,z)=1$,

$U_{5}(x,U_{5}(y,z))=U_{5}(x,R(y,z))=1=U_{5}(R(x,y),z)=U_{5}(U_{5}(x,y),z)$

3.1.1.2. $R(x,y)=1,R(y,z)\in I_{e}$,

$\displaystyle U_{5}(x,U_{5}(y,z))$ $\displaystyle=$ $\displaystyle
U_{5}(x,R(y,z))=R(x,R(y,z))$ $\displaystyle=$ $\displaystyle
R(R(x,y),z)=R(1,z)=1$ $\displaystyle=$ $\displaystyle
U_{5}(1,z)=U_{5}(R(x,y),z)$ $\displaystyle=$ $\displaystyle
U_{5}(U_{5}(x,y),z)(\mathrm{by\ Remark\ \ref{Re2.8}(3)})$

3.1.1.3. $R(x,y)\in I_{e},R(y,z)=1$,

$\displaystyle U_{5}(x,U_{5}(y,z))$ $\displaystyle=$ $\displaystyle
U_{5}(x,R(y,z))=U_{5}(x,1)$ $\displaystyle=$ $\displaystyle
1=R(x,1)=R(x,R(y,z))$ $\displaystyle=$ $\displaystyle
R(R(x,y),z)=U_{5}(R(x,y),z)$ $\displaystyle=$ $\displaystyle
U_{5}(U_{5}(x,y),z)(\mathrm{by\ Remark\ \ref{Re2.8}(3)})$

3.1.1.4. $R(x,y)\in I_{e},R(y,z)\in I_{e}$,

$U_{5}(x,U_{5}(y,z))=R(x,R(y,z))=R(R(x,y),z)=U_{5}(U_{5}(x,y),z)$

Hence, we obtain that $U_{5}(x,U_{5}(y,z))=U_{5}(U_{5}(x,y),z)$ for all
$x,y,z\in L$ by Proposition 2.2. Therefore, $U_{5}$ is a uninorm on $L$ with
the neutral element $e$.

It is worth pointing out that if the condition that $x\parallel y$ for all
$x\in I_{e}$ and $y\in[e,1[$ in Theorem 3.3 is not satisfied, then $U_{5}$ may
not be a uninorm on $L_{2}$ with the neutral element $e$. Similarly, if the
condition that $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$ in Theorem
3.3 is not satisfied, then $U_{6}$ may not be a uninorm on $L_{2}$ with the
neutral element $e$. The following example will show the above facts.

######  Example 3.3.

Consider the bounded lattice $L_{2}=\\{0,a,e,c,d,f,g,1\\}$ depicted by the
Hasse diagram in Fig.$2$, where $a\in[0,e[$, $g\in]e,1]$, $c\in I_{e}$, $d\in
I_{e}$, $f\in I_{e}$, $g>c$, $g>d$, $g>f$, $a<c$, $a<d$ and $a<f$. If we take
$T_{e}(x,y)=T_{\wedge}(x,y)$ on $[0,e]$ and $R(x,y)=x\vee y$, then the
function $U_{5}$ on $L_{2}$, shown in Table 5, is not a uninorm on $L_{2}$
with the neutral element $e$. In fact, the associativity is not satisfied,
since $U_{5}(f,U_{5}(c,d))=U_{5}(f,d)=g$ and
$U_{5}(U_{5}(f,c),d)=U_{5}(g,d)=1$. Similarly, if we take
$S_{e}(x,y)=S_{\vee}(x,y)$ on $[e,1]$ and $F(x,y)=x\wedge y$, then the
function $U_{6}$ on $L_{2}$, shown in Table 6, is not a uninorm on $L_{2}$
with the neutral element $e$. In fact, the associativity is not satisfied,
since $U_{6}(f,U_{6}(c,d))=U_{6}(f,c)=a$ and
$U_{6}(U_{6}(f,c),d)=U_{6}(a,d)=0$.

$0$$a$$c$$d$$e$$f$$g$$1$_Fig.2. The lattice $L_{2}$_

Table 5: The function $U_{5}$ on $L_{2}$ given in Fig.2.
$U_{5}$ | $0$ | $a$ | $e$ | $c$ | $d$ | $f$ | $g$ | $1$
---|---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $a$ | $a$ | $a$ | $a$ | $a$ | $a$ | $a$
$e$ | $0$ | $a$ | $e$ | $c$ | $d$ | $f$ | $g$ | $1$
$c$ | $0$ | $a$ | $c$ | $c$ | $d$ | $g$ | $1$ | $1$
$d$ | $0$ | $a$ | $d$ | $d$ | $d$ | $g$ | $1$ | $1$
$f$ | $0$ | $a$ | $f$ | $g$ | $g$ | $f$ | $1$ | $1$
$g$ | $0$ | $a$ | $g$ | $1$ | $1$ | $1$ | $1$ | $1$
$1$ | $0$ | $a$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
Table 6: The function $U_{6}$ on $L_{2}$ given in Fig.2.
$U_{6}$ | $0$ | $a$ | $e$ | $c$ | $d$ | $f$ | $g$ | $1$
---|---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $g$ | $1$
$a$ | $0$ | $0$ | $a$ | $0$ | $0$ | $0$ | $g$ | $1$
$e$ | $0$ | $a$ | $e$ | $c$ | $d$ | $f$ | $g$ | $1$
$c$ | $0$ | $0$ | $c$ | $c$ | $c$ | $a$ | $g$ | $1$
$d$ | $0$ | $0$ | $d$ | $c$ | $d$ | $a$ | $g$ | $1$
$f$ | $0$ | $0$ | $f$ | $a$ | $a$ | $f$ | $g$ | $1$
$g$ | $g$ | $g$ | $g$ | $g$ | $g$ | $g$ | $g$ | $1$
$1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$

######  Corollary 3.1.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$,
$x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, $x>y$ for all $x\in
I_{e}$ and $y\in[0,e[$. Suppose that $U_{5}$ is the uninorm on $L$ defined as
in Theorem 3.3, $U_{R}$ is the uninorm in Theorem 2.4 and $U_{\wedge}$ is the
uninorm in Theorem 2.5. Then $U_{5}\geq U_{R}$ and $U_{5}\geq U_{\wedge}$.

######  Corollary 3.2.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$,
$x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, $x<y$ for all $x\in
I_{e}$ and $y\in]e,1]$. Suppose that $U_{6}$ is the uninorm on $L$ defined as
in Theorem 3.3, $U_{F}$ is the uninorm in Theorem 2.4 and $U_{\vee}$ is the
uninorm in Theorem 2.5. Then $U_{6}\leq U_{F}$ and $U_{6}\leq U_{\vee}$.

##  4 Modifications of some uninorms

In this section, we mainly modify some uninorms, in which some constraints are
actually necessary and sufficient.

######  Theorem 4.1.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(i) Let $T_{e}$ be a $t$-norm on $[0,e]$ and $S_{e}$ be a $t$-conorm on
$[e,1]$. If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then the
function $U_{1}^{e}:L^{2}\rightarrow L$ defined by

$U_{1}^{e}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if }(x,y)\in
I_{e}\times[e,1[\cup I_{e}\times]0,e[,\\\ y&\mbox{if }(x,y)\in[e,1[\times
I_{e}\cup]0,e[\times I_{e},\\\ x\vee y&\mbox{if }(x,y)\in I_{e}^{2}\cup
I_{e}\times\\{1\\}\cup\\{1\\}\times
I_{e}\cup]0,e[\times\\{1\\}\cup\\{1\\}\times]0,e[,\\\ x\wedge y&\mbox{
}otherwise,\\\ \end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $T_{e}(x,y)>0$ for all
$x,y>0$ and $S_{e}(x,y)<1$ for all $x,y<1$.
(ii) Let $T_{e}$ be a $t$-norm on $[0,e]$ and $S_{e}$ be a $t$-conorm on
$[e,1]$. If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then the
function $U_{2}^{e}:L^{2}\rightarrow L$ defined by

$U_{2}^{e}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\ x&\mbox{if }(x,y)\in
I_{e}\times]e,1[\cup I_{e}\times]0,e],\\\ y&\mbox{if }(x,y)\in]e,1[\times
I_{e}\cup]0,e]\times I_{e},\\\ x\wedge y&\mbox{if }(x,y)\in I_{e}^{2}\cup
I_{e}\times\\{0\\}\cup\\{0\\}\times
I_{e}\cup]e,1[\times\\{0\\}\cup\\{0\\}\times]e,1[,\\\ x\vee y&\mbox{
}otherwise,\\\ \end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $T_{e}(x,y)>0$ for all
$x,y>0$ and $S_{e}(x,y)<1$ for all $x,y<1$.

Proof. If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then we give
the proof of the fact that $U_{1}^{e}$ is a uninorm iff $T_{e}(x,y)>0$ for all
$x,y>0$ and $S_{e}(x,y)<1$ for all $x,y<1$. The same result for $U_{2}^{e}$
can be obtained by similar arguments.

Necessity. Let $U_{1}^{e}$ be a uninorm on $L$ with the identity $e$. Next we
will prove that $T_{e}(x,y)>0$ for all $x,y>0$ and $S_{e}(x,y)<1$ for all
$x,y<1$. Assume that there exist $x\in]0,e[$ and $y\in]0,e[$ such that
$T_{e}(x,y)=0$. Take $z\in I_{e}$. Then
$U_{1}^{e}(x,U_{1}^{e}(y,z))=U_{1}^{e}(x,z)=z$ and, since $T_{e}(x,y)=0$,
$U_{1}^{e}(U_{1}^{e}(x,y),z)=U_{1}^{e}(T_{e}(x,y),z)=U_{1}^{e}(0,z)=0$. Hence,
the associativity property is violated. Assume that there exist $x\in]e,1[$
and $y\in]e,1[$ such that $S_{e}(x,y)=1$. Take $z\in I_{e}$. Then
$U_{1}^{e}(x,U_{1}^{e}(y,z))=U_{1}^{e}(x,z)=z$ and, since $S_{e}(x,y)=1$,
$U_{1}^{e}(U_{1}^{e}(x,y),z)=U_{1}^{e}(S_{e}(x,y),z)=U_{1}^{e}(1,z)=1$. Hence,
the associativity property is violated. So, $U_{1}^{e}$ is not a uninorm on
$L$ which is a contradiction. Therefore, the conditions that $T_{e}(x,y)>0$
for all $x,y>0$ and $S_{e}(x,y)<1$ for all $x,y<1$ are necessary.

Sufficiency. This proof is the same as Theorem 2.2.

######  Remark 4.1.

In Theorem 2.2, the conditions that $T_{e}$ is a $t$-norm on $[0,e]$ such that
$T_{e}(x,y)>0$ for all $x,y>0$ and $S_{e}$ is a $t$-conorm on $[e,1]$ such
that $S_{e}(x,y)<1$ for all $x,y<1$ are sufficient for $U_{1}^{e}$ and
$U_{2}^{e}$. In fact, these conditions are also necessary for $U_{1}^{e}$ and
$U_{2}^{e}$ in Theorem 4.1.

######  Remark 4.2.

Let $U_{1}$ and $U_{2}$ be the uninorms on $L$ defined as in Theorem 3.1, and
let $U_{1}^{e}$ and $U_{2}^{e}$ be the uninorms on $L$ defined as in Theorem
4.1.

(1) It is easy to see that the values of $U_{1}$ and $U_{1}^{e}$ are different
on $I_{e}\times\\{0\\}\cup\\{1\\}\times\\{0\\}$ and $\\{0\\}\times
I_{e}\cup\\{0\\}\times\\{1\\}$. In fact, $U_{1}=x$ and $U_{1}^{e}=0$ when
$(x,y)\in I_{e}\times\\{0\\}\cup\\{1\\}\times\\{0\\}$; $U_{1}=y$,
$U_{1}^{e}=0$ when $(x,y)\in\\{0\\}\times I_{e}\cup\\{0\\}\times\\{1\\}$. The
values of $U_{1}$ could be $0$ on $]0,e[\times]0,e[$; however, the values of
$U_{1}^{e}$ are not equal to $0$ on $]0,e[\times]0,e[$.

(2) The additional constraints of $U_{1}$ and $U_{1}^{e}$ are different, that
is, the condition that $T_{e}(x,y)>0$ for all $x,y>0$ is not needed for
$U_{1}$.

According to (1) and (2), by changing the values of $U_{1}^{e}$, we can obtain
a new method to construct the nuinorm $U_{1}$ and then additional constraints
are different for $U_{1}^{e}$ and $U_{1}$.

Similarly, the relationship between $U_{2}$ and $U_{2}^{e}$ is same as that of
$U_{1}$ and $U_{1}^{e}$.

######  Theorem 4.2.

Let $(L,\leq,0,1)$ be a bounded lattice with $e\in L\setminus\\{0,1\\}$.
(i) Let $T_{e}$ be a $t$-norm on $[0,e]$ and $R$ be a $t$-subconorm on $L$. If
$x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then the function
$U_{R}^{e}:L^{2}\rightarrow L$ defined by

$U_{R}^{e}(x,y)=\begin{cases}T_{e}(x,y)&\mbox{if }(x,y)\in[0,e]^{2},\\\
x&\mbox{if }(x,y)\in I_{e}\times[0,e]\cup I_{e}\times[e,1[,\\\ y&\mbox{if
}(x,y)\in[0,e]\times I_{e}\cup[e,1[\times I_{e},\\\ R(x,y)&\mbox{if }(x,y)\in
I_{e}^{2}\cup]e,1]^{2},\\\ x\vee y&\mbox{if }(x,y)\in D(e),\\\ x\wedge
y&\mbox{ }otherwise,\\\ \end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $R(x,y)<1$ for all
$x,y<1$.
(ii) Let $S_{e}$ be a $t$-conorm on $[e,1]$ and $F$ be a $t$-subnorm on $L$.
If $x\parallel y$ for all $x\in I_{e}$ and $y\in]0,e]$, then the function
$U_{F}^{e}:L^{2}\rightarrow L$ defined by

$U_{F}^{e}(x,y)=\begin{cases}S_{e}(x,y)&\mbox{if }(x,y)\in[e,1]^{2},\\\
x&\mbox{if }(x,y)\in I_{e}\times]0,e]\cup I_{e}\times[e,1],\\\ y&\mbox{if
}(x,y)\in]0,e]\times I_{e}\cup[e,1]\times I_{e},\\\ F(x,y)&\mbox{if }(x,y)\in
I_{e}^{2}\cup[0,e[^{2},\\\ x\wedge y&\mbox{if }(x,y)\in E(e),\\\ x\vee
y&\mbox{ }otherwise,\\\ \end{cases}$
is a uninorm on $L$ with $e\in L\setminus\\{0,1\\}$ iff $F(x,y)>0$ for all
$x,y>0$.

Proof. If $x\parallel y$ for all $x\in I_{e}$ and $y\in[e,1[$, then we give
the proof of the fact that $U_{R}^{e}$ is a uninorm iff $R(x,y)<1$ for all
$x,y<1$. The same result for $U_{F}^{e}$ can be obtained using similar
arguments.

Necessity. Let $U_{R}^{e}$ be a uninorm on $L$ with the identity $e$. Then we
will prove that $R(x,y)<1$ for all $x,y<1$. Assume that there exist
$x\in]e,1[$ and $y\in]e,1[$ such that $R(x,y)=1$. Take $z\in I_{e}$. Then
$U_{R}^{e}(x,U_{R}^{e}(y,z))=U_{R}^{e}(x,z)=z$ and, since $R(x,y)=1$,
$U_{R}^{e}(U_{R}^{e}(x,y),z)=U_{R}^{e}(R(x,y),z)=U_{R}^{e}(1,z)=1$. Hence, the
associativity property of $U_{R}^{e}$ is violated. So, $U_{R}^{e}$ is not a
uninorm on $L$ which is a contradiction. Therefore, the condition that
$R(x,y)<1$ for all $x,y<1$ is necessary.

Sufficiency. This proof is the same as Theorem 2.3.

######  Remark 4.3.

In Theorem 2.3, the conditions that $R$ is a $t$-subconorm on $L$ such that
$R(x,y)<1$ for all $x,y<1$ and $F$ is a $t$-subnorm on $[0,e]$ such that
$F(x,y)>0$ for all $x,y>0$ are sufficient for $U_{R}^{e}$ and $U_{F}^{e}$,
respectively. In fact, the conditions are also necessary for $U_{F}^{e}$ and
$U_{R}^{e}$ in Theorem 4.2, respectively.

######  Remark 4.4.

Let $U_{3}$ and $U_{4}$ be the uninorms on $L$ defined as in Theorem 3.2, and
let $U_{R}^{e}$ and $U_{F}^{e}$ be the uninorms on $L$ defined as in Theorem
4.2.

(1) It is easy to see that the values of $U_{3}$ and $U_{R}^{e}$ are different
on $I_{e}\times\\{0\\}\cup\\{1\\}\times\\{0\\}$ and $\\{0\\}\times
I_{e}\cup\\{0\\}\times\\{1\\}$. In fact, $U_{3}=0$ and $U_{R}^{e}=x$ when
$(x,y)\in I_{e}\times\\{0\\}\cup\\{1\\}\times\\{0\\}$; $U_{3}=0$ and
$U_{R}^{e}=y$ when $(x,y)\in\\{0\\}\times I_{e}\cup\\{0\\}\times\\{1\\}$. The
values of $U_{3}$ are not equal to $0$ on $]0,e[\times]0,e[$; however, the
values of $U_{R}^{e}$ could be $0$ on $]0,e[\times]0,e[$.

(2) The additional constraints of $U_{3}$ and $U_{R}^{e}$ are different, that
is, the condition that $T_{e}(x,y)>0$ for all $x,y>0$ is needed for $U_{3}$.

According to (1) and (2), by changing the values of $U_{R}^{e}$, we can obtain
a new method to construct the nuinorm $U_{3}$ and then additional constraints
are different for $U_{R}^{e}$ and $U_{3}$ .

Similarly, the relationship between $U_{4}$ and $U_{F}^{e}$ is same as that of
$U_{3}$ and $U_{R}^{e}$.

##  5  New constructions of uninorms by interation

In this section, we present new constructions of uninorms on bounded lattices
by interation based on $t$-conorms and $t$-conorms.

######  Theorem 5.1.

Let $(L,\leq,0,1)$ be a bounded lattice.
(i) Let $\\{a_{0},a_{1},a_{2},\ldots,a_{n}\\}$ be a finite chain in $L$ with
$0_{L}=a_{0}<a_{1}<a_{2}<\ldots<a_{n}=1_{L}$ on $(L,\leq,0,1)$, $T_{e}$ be a
$t$-norm on $[0,a_{1}]$ and $S_{i}$ be a $t$-conorm on $[a_{i-1},a_{i}]$ of
$L$ for $i\in\\{2,3,\ldots,n\\}$. Then for $i\in\\{2,3,\ldots,n\\}$, the
function $U_{i}^{S}:[0,a_{i}]^{2}\rightarrow[0,a_{i}]$ defined as follows

$U_{i}^{S}(x,y)=\begin{cases}U_{i-1}^{S}(x,y)&\mbox{if
}(x,y)\in[0,a_{i-1}]^{2},\\\ S_{i}(x,y)&\mbox{if
}(x,y)\in]a_{i-1},a_{i}]^{2},\\\ x\vee y&\mbox{if
}(x,y)\in[0,e]\times(I_{a_{i-1}}^{e}\cup]a_{i-1},a_{i}])\\\ &\mbox{ }\ \ \ \ \
\ \ \ \ \ \ \cup(I_{a_{i-1}}^{e}\cup]a_{i-1},a_{i}])\times[0,e],\\\ x&\mbox{if
}(x,y)\in I_{e,a_{i-1}}\times[0,e],\\\ y&\mbox{if }(x,y)\in[0,e]\times
I_{e,a_{i-1}},\\\ S_{i}(x\vee a_{i-1},y\vee a_{i-1})&\mbox{ }otherwise,\\\
\end{cases}$
is a uninorm on $[0,a_{i}]$, where $U_{1}^{S}=T_{e}$.
(ii) Let $\\{b_{0},b_{1},b_{2},\ldots,b_{n}\\}$ be a finite chain in $L$ with
$0_{L}=b_{n}\leq b_{n-1}\leq b_{n-2}\leq\ldots\leq b_{0}=1$ on $(L,\leq,0,1)$,
$S_{e}$ be a t-conorm on $[b_{1},1]$ and $T_{i}$ be a t-norm on
$[b_{i},b_{i-1}]$ for $i\in\\{2,3,\ldots,n\\}$. Then for
$i\in\\{2,3,\ldots,n\\}$, the function
$U_{i}^{T}:[b_{i},1]\rightarrow[b_{i},1]$ defined as follows

$U_{i}^{T}(x,y)=\begin{cases}U_{i-1}^{T}&\mbox{if }(x,y)\in[b_{i-1},1]^{2},\\\
T_{i}(x,y)&\mbox{if }(x,y)\in[b_{i},b_{i-1}[^{2},\\\ x\wedge y&\mbox{if
}(x,y)\in[e,1]\times(I_{b_{i-1}}^{e}\cup[b_{i},b_{i-1}[)\\\ &\mbox{ }\ \ \ \ \
\ \ \ \ \ \ \cup(I_{b_{i-1}}^{e}\cup[a_{i},b_{i-1}[)\times[e,1],\\\ x&\mbox{if
}(x,y)\in I_{e,b_{i-1}}\times[e,1],\\\ y&\mbox{if }(x,y)\in[e,1]\times
I_{e,b_{i-1}},\\\ T_{i}(x\wedge b_{i-1},y\wedge b_{i-1})&\mbox{ }otherwise,\\\
\end{cases}$
is a uninorm on $[b_{i},1]$, where $U_{1}^{T}=S_{e}$.

Proof. Let $(L,\leq,0,1)$ be a bounded lattice with
$a_{0},a_{1},a_{2},\ldots,a_{n}\in L$ such that $0_{L}=a_{0}\leq a_{1}\leq
a_{2}\leq\ldots\leq a_{n}=1$. Then we give the proof of the fact that
$U_{i}^{S}$ is a uninorm. The same result for $U_{i}^{T}$ can be obtained
using similar arguments.

If we take $i=2$ in the definition of $U_{i}^{S}$, then we can see that
$U_{2}^{S}$ is exactly the uninorm $U_{d}$ in Theorem 2.6.

Assume that $U_{k}^{S}$ is a uninorm. Then we prove that $U_{k+1}^{S}$ is a
uninorm, i.e.

$U_{k+1}^{S}(x,y)=\begin{cases}U_{k}^{S}(x,y)&\mbox{if
}(x,y)\in[0,a_{k}]^{2},\\\ S_{k+1}(x,y)&\mbox{if
}(x,y)\in]a_{k},a_{k+1}]^{2},\\\ x\vee y&\mbox{if
}(x,y)\in[0,e]\times(I_{a_{k}}^{e}\cup]a_{k},a_{k+1}])\\\ &\mbox{ }\ \ \ \ \ \
\ \ \ \ \ \cup(I_{a_{k}}^{e}\cup]a_{k},a_{k+1}])\times[0,e],\\\ x&\mbox{if
}(x,y)\in I_{e,a_{k}}\times[0,e],\\\ y&\mbox{if }(x,y)\in[0,e]\times
I_{e,a_{k}},\\\ S_{k+1}(x\vee a_{k},y\vee a_{k})&\mbox{ }otherwise.\\\
\end{cases}$

The commutativity of $U_{k+1}^{S}$ and the fact that $e$ is a neutral element
are clearly ture. Hence, we prove only the monotonicity and the associativity
of $U_{k+1}^{S}$.

I. Monotonicity: We prove that if $x\leq y$, then $U_{k+1}^{S}(x,z)\leq
U_{k+1}^{S}(y,z)$ for all $z\in L$. The proof is split into all possible
cases:

1\. Let $x\in[0,e]$.

1.1. $y\in[0,e]$,

1.1.1. $z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$,

Since the monotonicity property of $U_{k}^{S}(x,y)$, the monotonicity is true.

1.1.2. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}\cup]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=z=U_{k+1}^{S}(y,z)$

1.2. $y\in I_{e}^{a_{k}}$,

1.2.1. $z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$,

Since the monotonicity property of $U_{k}^{S}(x,y)$, the monotonicity is true.

1.2.2. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=z<z\vee a_{k}=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

1.2.3. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=z=S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

1.3. $y\in]e,a_{k}]$,

1.3.1. $z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$,

Since the monotonicity property of $U_{k}^{S}(x,y)$, the monotonicity is true.

1.3.2. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=z<z\vee a_{k}=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

1.3.3. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=z=S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

1.4. $y\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

1.4.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq x<y=U_{k+1}^{S}(y,z)$

1.4.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq a_{k}<y\vee a_{k}=S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

1.4.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=z<z\vee a_{k}\leq S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

1.4.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=z\leq S_{k+1}(y\vee a_{k},z)=U_{k+1}^{S}(y,z)$

1.5. $y\in]a_{k},a_{k+1}]$,

1.5.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)<y=U_{k+1}^{S}(y,z)$

1.5.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=U_{K}^{S}(x,z)<y=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

1.5.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=z<z\vee a_{k}\leq S_{k+1}(y,z\vee a_{k})=U_{k+1}^{S}(y,z)$

1.5.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=z\leq S_{k+1}(y,z)=U_{k+1}^{S}(y,z)$

2\. Let $x\in I_{e}^{a_{k}}$.

2.1. $y\in I_{e}^{a_{k}}$,

2.1.1. $z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$,

Since the monotonicity property of $U_{k}^{S}(x,y)$, the monotonicity is true.

2.1.2. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}=S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

2.1.3. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

2.2. $y\in]e,a_{k}]$,

2.2.1. $z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$,

Since the monotonicity property of $U_{k}^{S}(x,y)$, the monotonicity is true.

2.2.2. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}=S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

2.2.3. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

2.3. $y\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

2.3.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq x<y=U_{k+1}^{S}(y,z)$

2.3.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq a_{k}<y\vee a_{k}=S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

2.3.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}\leq
S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

2.3.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=z\leq S_{k+1}(y\vee a_{k},z)=U_{k+1}^{S}(y,z)$

2.4. $y\in]a_{k},a_{k+1}]$,

2.4.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)<y=U_{k+1}^{S}(y,z)$

2.4.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)<y=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

2.4.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}\leq
S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

2.4.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\leq
S_{k+1}(y,z)=U_{k+1}^{S}(y,z)$

3\. Let $x\in]e,a_{k}]$.

3.1. $y\in]e,a_{k}]$,

3.1.1. $z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$,

Since the monotonicity property of $U_{k}^{S}(x,y)$, the monotonicity is true.

3.1.2. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}=S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

3.1.3. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

3.2. $y\in I_{a_{k}}^{e}$,

3.2.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq x<y=U_{k+1}^{S}(y,z)$

3.2.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq a_{k}<y\vee a_{k}=S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

3.2.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}\leq
S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

3.2.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\leq S_{k+1}(y\vee
a_{k},z)=U_{k+1}^{S}(y,z)$

3.3. $y\in]a_{k},a_{k+1}]$,

3.3.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq a_{k}<y=U_{k+1}^{S}(y,z)$

3.3.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=U_{k}^{S}(x,z)\leq a_{k}<y=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)$

3.3.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}\leq
S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

3.4.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\leq
S_{k+1}(y,z)=U_{k+1}^{S}(y,z)$

4\. Let $x\in I_{a_{k}}^{e}$.

4.1. $y\in I_{a_{k}}^{e}$,

4.1.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=x\leq y=U_{k+1}^{S}(y,z)$

4.1.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]\cup I_{a_{k}}^{e}\cup
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$,

Since the monotonicity property of $S_{k+1}$, the monotonicity is true.

4.2. $y\in]a_{k},a_{k+1}]$,

4.2.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=x<y=U_{k+1}^{S}(y,z)$

4.2.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=x\vee a_{k}\leq
y=S_{k+1}(y\vee a_{k},z\vee a_{k})$

$=U_{k+1}^{S}(y,z)$

4.2.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})\leq S_{k+1}(y,z\vee
a_{k})=U_{k+1}^{S}(y,z)$

4.2.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=S_{k+1}(x\vee a_{k},z)\leq
S_{k+1}(y,z)=U_{k+1}^{S}(y,z)$

5\. Let $x\in I_{e,a_{k}}$.

5.1. $y\in I_{e,a_{k}}$,

5.1.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=x\leq y=U_{k+1}^{S}(y,z)$

5.1.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]\cup I_{a_{k}}^{e}\cup
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$,

Since the monotonicity property of $S_{k+1}$, the monotonicity is true.

5.2. $y\in I_{a_{k}}^{e}$,

5.2.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=x<y=U_{k+1}^{S}(y,z)$

5.2.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]\cup I_{a_{k}}^{e}\cup
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$,

Since the monotonicity property of $S_{k+1}$, the monotonicity is true.

5.3. $y\in]a_{k},a_{k+1}]$,

5.3.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=x<y=U_{k+1}^{S}(y,z)$

5.3.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=x\vee a_{k}\leq
y=S_{k+1}(y\vee a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

5.3.3. $z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})\leq S_{k+1}(y\vee
a_{k},z\vee a_{k})=U_{k+1}^{S}(y,z)$

5.3.4. $z\in]a_{k},a_{k+1}]$,

$U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=S_{k+1}(x\vee a_{k},z)\leq
S_{k+1}(y,z)=U_{k+1}^{S}(y,z)$

6\. Let $x\in]a_{k},a_{k+1}]$.

6.1. $y\in]a_{k},a_{k+1}]$,

6.1.1. $z\in[0,e]$,

$U_{k+1}^{S}(x,z)=x\leq y=U_{k+1}^{S}(y,z)$

6.1.2. $z\in I_{e}^{a_{k}}\cup]e,a_{k}]\cup I_{a_{k}}^{e}\cup
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$,

Since the monotonicity property of $S_{k+1}$, the monotonicity is true.

Combining the above cases, we obtain that $U_{k+1}^{S}(x,z)\leq
U_{k+1}^{S}(y,z)$ holds for all $x,y,z\in L$ such that $x\leq y$. Therefore,
$U_{k+1}^{S}$ is monotonic.

II. Associativity: It can be shown that
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)$ for all
$x,y,z\in L$. By Proposition 2.2, we only need to consider the following
cases:

1\. If $x,y,z\in[0,e]\cup I_{e}^{a_{k}}\cup]e,a_{k}]$, then since $U_{k}^{S}$
is associative, $U_{k+1}^{S}$ is associative.

2\. If $x,y,z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since
$S_{k+1}$ is associative, $U_{k+1}^{S}$ is associative.

3\. If $x,y\in[0,e],z\in I_{e}^{a_{k}}\cup]e,a_{k}]$, then since $U_{k}^{S}$
is associative, $U_{k+1}^{S}$ is associative.

4\. If $x,y\in[0,e],z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}\cup]a_{k},a_{k+1}]$,
then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,z)=z=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z).$
Thus $U_{k+1}^{S}$ is associative.

5\. If $x,y\in I_{e}^{a_{k}},z\in]e,a_{k}]$, then since $U_{k}^{S}$ is
associative, $U_{k+1}^{S}$ is associative.

6\. If $x,y\in I_{e}^{a_{k}},z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z\vee a_{k})=z\vee a_{k}=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee
a_{k})=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z).$ Thus
$U_{k+1}^{S}$ is associative.

7\. If $x,y\in I_{e}^{a_{k}},z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z$ and
$U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee a_{k})=z$. Thus $U_{k+1}^{S}$ is associative.

8\. If $x,y\in]e,a_{k}],z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z\vee a_{k})=z\vee a_{k}$ and
$U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=S_{k+1}(U_{k}^{S}(x,y)\\\
\vee a_{k},z\vee a_{k})=z\vee a_{k}$. Thus $U_{k+1}^{S}$ is associative.

9\. If $x,y\in]e,a_{k}],z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z)=S_{k+1}(x\vee a_{k},z\vee a_{k})=z$ and
$U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee a_{k})=z$. Thus $U_{k+1}^{S}$ is associative.

10\. If $x,y\in I_{a_{k}}^{e},z\in I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since
$S_{k+1}$ is associative, $U_{k+1}^{S}$ is associative.

11\. If $x,y\in I_{e,a_{k}},z\in]a_{k},a_{k+1}]$, then since $S_{k+1}$ is
associative, we have $U_{k+1}^{S}$ is associative.

12\. If $x\in[0,e],y,z\in I_{e}^{a_{k}}\cup]e,a_{k}]$, then since $U_{k}^{S}$
is associative, $U_{k+1}^{S}$ is associative.

13\. If $x\in[0,e],y,z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=S_{k+1}(y\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(y,z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z).$ Thus $U_{k+1}^{S}$
is associative.

14\. If $x\in[0,e],y,z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y,z))=S_{k+1}(y,z)=U_{k+1}^{S}(y,z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z).$
Thus $U_{k+1}^{S}$ is associative.

15\. If $x\in I_{e}^{a_{k}},y,z\in]e,a_{k}]$, then since $U_{k}^{S}$ is
associative, $U_{k+1}^{S}$ is associative.

16\. If $x\in I_{e}^{a_{k}},y,z\in I_{a_{k}}^{e}\cup
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since $S_{k+1}$ is associative,
$U_{k+1}^{S}$ is associative.

17\. If $x\in]e,a_{k}],y,z\in I_{a_{k}}^{e}\cup
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since $S_{k+1}$ is associative,
$U_{k+1}^{S}$ is associative.

18\. If $x\in I_{a_{k}}^{e},y,z\in I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since
$S_{k+1}$ is associative, $U_{k+1}^{S}$ is associative.

19\. If $x\in I_{e,a_{k}},y,z\in]a_{k},a_{k+1}]$, then then since $S_{k+1}$ is
associative, $U_{k+1}^{S}$ is associative.

20\. If $x\in[0,e],y\in I_{e}^{a_{k}},z\in]e,a_{k}]$, then since $U_{k}^{S}$
is associative, $U_{k+1}^{S}$ is associative.

21\. If $x\in[0,e],y\in I_{e}^{a_{k}},z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,
then $U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(a_{k},z\vee
a_{k}))=z\vee a_{k}=S_{k+1}(a_{k},z\vee a_{k})=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee
a_{k})=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(a_{k},z\vee
a_{k})=z\vee a_{k}$. Thus $U_{k+1}^{S}$ is associative.

22\. If $x\in[0,e],y\in I_{e}^{a_{k}},z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z)=z,U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee a_{k})=z$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee a_{k},z\vee
a_{k})=z$. Thus $U_{k+1}^{S}$ is associative.

23\. If $x\in[0,e],y\in]e,a_{k}],z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=S_{k+1}(y\vee a_{k},z\vee a_{k})=z\vee
a_{k},U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee a_{k})=z\vee a_{k}$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee a_{k},z\vee
a_{k})=z\vee a_{k}$. Thus $U_{k+1}^{S}$ is associative.

24\. If $x\in[0,e],y\in]e,a_{k}],z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z)=z,U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee a_{k})=z$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee a_{k},z\vee
a_{k})=z$. Thus $U_{k+1}^{S}$ is associative.

25\. If $x\in[0,e],y\in I_{a_{k}}^{e},z\in I_{e,a_{k}}$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=S_{k+1}(y\vee a_{k},z\vee
a_{k}),U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee
a_{k},z\vee a_{k})$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee a_{k},z\vee
a_{k})$. Thus $U_{k+1}^{S}$ is associative.

26\. If $x\in[0,e],y\in I_{a_{k}}^{e},z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=S_{k+1}(y\vee
a_{k},z),U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee
a_{k},z)$ and $U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee
a_{k},z)$. Thus $U_{k+1}^{S}$ is associative.

27\. If $x\in[0,e],y\in I_{e,a_{k}},z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=S_{k+1}(y\vee
a_{k},z),U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee
a_{k},z)$ and $U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee
a_{k},z)$. Thus $U_{k+1}^{S}$ is associative.

28\. If $x\in I_{e}^{a_{k}},y\in]e,a_{k}],z\in I_{a_{k}}^{e}\cup I_{e,a_{k}}$,
then $U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z\vee a_{k})=z\vee a_{k}=S_{k+1}(U_{k}^{S}(x,y)\vee
a_{k},z\vee
a_{k})=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,S_{k+1}(x\vee a_{k},z\vee
a_{k}))=S_{k+1}(x\vee a_{k},z\vee a_{k})=z\vee a_{k}$. Thus $U_{k+1}^{S}$ is
associative.

29\. If $x\in I_{e}^{a_{k}},y\in]e,a_{k}],z\in]a_{k},a_{k+1}]$, then
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(x,S_{k+1}(y\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(x,z\vee a_{k})=S_{k+1}(x\vee a_{k},z\vee
a_{k})=z=S_{k+1}(U_{k}^{S}(x,y)\vee a_{k},z\vee
a_{k})=U_{k+1}^{S}(U_{k}^{S}(x,y),z)=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)$ and
$U_{k+1}^{S}(y,U_{k+1}^{S}(x,z))=U_{k+1}^{S}(y,S_{k+1}(x\vee a_{k},z\vee
a_{k}))=U_{k+1}^{S}(y,z)=S_{k+1}(y\vee a_{k},z\vee a_{k})=z$. Thus
$U_{k+1}^{S}$ is associative.

30\. If $x\in I_{e}^{a_{k}},y\in I_{a_{k}}^{e},z\in
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since $S_{k+1}$ is associative,
$U_{k+1}^{S}$ is associative.

31\. If $x\in I_{e}^{a_{k}},y\in I_{e,a_{k}},z\in]a_{k},a_{k+1}]$, then since
$S_{k+1}$ is associative, $U_{k+1}^{S}$ is associative.

32\. If $x\in]e,a_{k}],y\in I_{a_{k}}^{e},z\in
I_{e,a_{k}}\cup]a_{k},a_{k+1}]$, then since $S_{k+1}$ is associative,
$U_{k+1}^{S}$ is associative.

33\. If $x\in]e,a_{k}],y\in I_{e,a_{k}},z\in]a_{k},a_{k+1}]$, then since
$S_{k+1}$ is associative, $U_{k+1}^{S}$ is associative.

34\. If $x\in I_{a_{k}}^{e},y\in I_{e,a_{k}},z\in]a_{k},a_{k+1}]$, then since
$S_{k+1}$ is associative, $U_{k+1}^{S}$ is associative.

Form $1$ to $34$, we obtain that
$U_{k+1}^{S}(x,U_{k+1}^{S}(y,z))=U_{k+1}^{S}(U_{k+1}^{S}(x,y),z)$ for all
$x,y,z\in L$ by Proposition 2.2. Therefore, $U_{k+1}^{S}$ is a uninorm on $L$
with the neutral element $e$.

Thus, $U_{i}^{S}$ is a uninorm on $L$ with the neutral element $e$ for
$i\in\\{2,3,\ldots,n\\}$.

######  Corollary 5.1.

Let $(L,\leq,0,1)$ be a bounded lattice.
(i) If we take $S_{i}=S_{\vee}$ on $[a_{i-1},a_{i}]$ of $L$ for
$i\in\\{2,3,\ldots,n\\}$ in Theorem 5.1, then we obtain the following uninorm
on $L$.

$U_{i}^{\vee}(x,y)=\begin{cases}U_{i-1}^{\vee}(x,y)&\mbox{if
}(x,y)\in[0,a_{i-1}]^{2},\\\ x\vee y&\mbox{if
}(x,y)\in]a_{i-1},a_{i}]^{2}\cup[0,e]\times(I_{a_{i-1}}^{e}\cup]a_{i-1},a_{i}])\\\
&\mbox{ }\ \ \ \ \ \ \ \ \ \ \
\cup(I_{a_{i-1}}^{e}\cup]a_{i-1},a_{i}])\times[0,e],\\\ x&\mbox{if }(x,y)\in
I_{e,a_{i-1}}\times[0,e],\\\ y&\mbox{if }(x,y)\in[0,e]\times I_{e,a_{i-1}},\\\
x\vee y\vee a_{i-1}&\mbox{ }otherwise.\\\ \end{cases}$

(ii) If we take $T_{i}=T_{\wedge}$ on $[b_{i},b_{i-1}]$ of $L$ for
$i\in\\{2,3,\ldots,n\\}$ in Theorem 5.1, then we obtain the following uninorm
on $L$.

$U_{i}^{\wedge}(x,y)=\begin{cases}U_{i-1}^{\wedge}(x,y)&\mbox{if
}(x,y)\in[b_{i-1},1]^{2},\\\ x\wedge y&\mbox{if
}(x,y)\in[b_{i},b_{i-1})^{2}\cup[e,1]\times(I_{b_{i-1}}^{e}\cup[b_{i},b_{i-1}[)\\\
&\mbox{ }\ \ \ \ \ \ \ \ \ \ \
\cup(I_{b_{i-1}}^{e}\cup[b_{i},b_{i-1}[)\times[e,1],\\\ x&\mbox{if }(x,y)\in
I_{e,b_{i-1}}\times[e,1],\\\ y&\mbox{if }(x,y)\in[e,1]\times I_{e,b_{i-1}},\\\
x\wedge y\wedge b_{i-1}&\mbox{ }otherwise.\\\ \end{cases}$

######  Corollary 5.2.

Let $(L,\leq,0,1)$ be a bounded lattice.
(i) If we take $i=2$ in Corollary 5.1, then we obtain the uninorm $U_{(T,e)}$
in Theorem 2.1.
(ii) If we take $i=2$ in Corollary 5.1, then we obtain the uninorm $U_{(S,e)}$
in Theorem 2.1.

##  6 Conclusion

In this article, we still investigate the constructions of uninorms on bounded
lattices.

On one hand, we investigate the constructions of uninorms on some appropriate
bounded lattices with $e\in L\setminus\\{0,1\\}$, based on some additional
constraints. In this progress, we mainly focus on the rationality of these
constraints and then show that some of these constraints are sufficient and
necessary for the construction of uninorms. Meanwhile, we study the
constraints of some uninorms and find that some constraints are also
necessary. In the future research, when considering the new construction of
uninorms, if the constraints are needed, we expect they are sufficient and
necessary.

On the other hand, we consider new methods of constructing of uninorms, that
is, the iterative method using $t$-norms and $t$-conorms. In the future
research, we will give more structures of uninorms by this method.

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