# Some Examples of Weak Uninorms

Authors: Jiachao Wu; Yayan Yuan
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

It is proved that, except for the uninorms and the nullnorms, there are no continuous weak uninorms who have no more than one nontrivial idempotent element. And some examples of discontinuous weak uninorms are shown. All of these examples are notn-uninorms, thus not uninorms or nullnorms.


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## Body

## 1. Introduction

A mapping from[0,1]2to[0,1]is called an AMC operator [1, 2], if it is associative, monotone nondecreasing in both variables, and commutative. The most famous AMC operator in fuzzy mathematics is thet-norms [3–5] and thet-conorms. In recent years, the weak forms of thet-norms interest the logisticians, such as the uninorms [6, 7], the nullnorms (t-operators) [8, 9], the n-uninorms [10, 11], and the weak uninorms [12, 13].It is easy to find that the weak uninorms are the most general class; that is, all thet-norms, thet-conorms, the (n-)uninorms, and the nullnorms are weak uninorms. Conversely, it is not valid; that is, a weak uninorm could be none of the others [12].As we all know, for a weak uninormR, its idempotent elements are the pointsxsubject to R(x,x)=x. The elements 0 and 1 are the trivial idempotent elements of all the weak uninorms. All the common examples of the nontrivial weak uninorms are with infinite idempotents. Then, the following problem arises.Problem 1 (see [14]).
Is there a nontrivial weak uninorm with no more than one nontrivial idempotent elements?This problem can be divided into two parts: continuous weak uninorms and discontinuous ones. In this paper, we will give answers to this problem separately.The content will be arranged as follows: in Section2, some basic definitions will be given, and it will be proved that there is no nontrivial continuous weak uninorms with none or one nontrivial idempotent element. In Section 3, some examples of weak uninorms with nontrivial idempotent elements are given. These examples give positive answers to the problem above. Section 4 also shows examples of weak uninorms, which have one or more idempotent elements. Section 5 gives a conclusion of this paper.


## 2. Continuous Weak Uninorms

Definition 2 (see [3, 4]).
An AMC operatorTis at-norm, if∀x∈[0,1],T(x,1)=x.
An AMC operatorSis at-conorm, if for anyx∈[0,1],S(x,0)=x.Definition 3 (see [6, 7]).
An AMC operatorUis said to be uninorm if there exists some elemente∈[0,1]called the neutral element such that for all x∈[0,1],U(x,e)=x.Clearly, ife=1thenUis at-norm and ife=0thenUis at-conorm [7].Definition 4.
An AMC operatorVis called a nullnorm [8] (t-operator [9]) if there exists an absorbing elementλ∈[0,1]; that is,V(x,λ)=λ, such thatfor all x∈[0,λ],V(x,0)=xand for all x∈[λ,1], V(x,1)=x.Obviously, ifλ=0, then the nullnorm is at-norm; and ifλ=1, then it is at-conorm.Definition 5 (see [10, 11]).
{
e
1
,
e
2
,
…
,
e
n
}
z
1
,
z
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,
…
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z
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1is called ann-neutral element of an AMC operatorUn, if for0=z0<z1<⋯<zn=1andei∈[zi-1,zi], we haveUn(ei,x)=x,∀x∈[zi-1,zi].
An AMC operator is called ann-uninorm, if it has ann-neutral element{e1,e2,…,en}z1,z2,…,zn-1.Obviously, each uninorm and nullnorm is ann-uninorm. And the converse is not valid. Examples could be found in [10].Definition 6 (see [12, 13]).
An AMC operatorRis named a weak uninorm if for anyx∈[0,1], there exists some elementux∈[0,1]such thatR(x,ux)=x.Ifuxin particular is a fixed valueefor allx∈[0,1], theneis called the neutral element ofR. And in this case,Rcomes to be uninorm [7]. If there is some elementλ∈[0,1], subject to for all x∈[0,λ],ux=0, and for all x∈[λ,1], ux=1, it is a nullnorm [8].One can easily see thatn-uninorms are weak uninorms. However, a weak uninorm may not be ann-uninorm and thus neither a nullnorm nor a uninorm. Examples are in [12, 13].As a result, the problem in the introduction arises. And now, let us give an answer to it: there are no nontrivial continuous weak uninorms, but there exist discontinuous ones.Theorem 7 (Theorem  7 in [15]).
IfRis an AMC operator on[0,1] that satisfies the following conditions: (i)
its idempotent elements are just 0 and 1;(ii)
there exists some elementx∈(0,1)withR(x,x)>x;(iii)
there exists some elementy∈(0,1)withR(y,y)<y;
thenRis not continuous, and(d,d)is a discontinuous point, withdthe demarcation point.From this theorem, we have the following theorem.Theorem 8.
LetRbe a continuous weak uninorm with no nontrivial idempotents. Then, we have the following results: (1)
if there exists somex∈(0,1) subject to R(x,x)<x, thenRis at-norm;(2)
if there exists somex∈(0,1) subject to R(x,x)>x, thenRis at-conorm.Before the proof, let us show the following lemma firstly.Lemma 9.
LetRbe a continuous AMC operator with no idempotent elements except 0 and 1. (1)
Iffor all x∈(0,1),R(x,x)<x, then for anyx,y∈(0,1), there exists some natural numbern, subject toyRn<x.(2)
If for allx∈(0,1),R(x,x)>x, then for anyx,y∈(0,1), there exists some natural numbern, subject toyRn>x.Proof.
(1) Since for allx∈(0,1),R(x,x)<x, for anyn,
(1)xRn+1=R(xR2,xRn-1)≤R(x,xRn-1)=xRn.
Thus, the limitlimn→∞xRnexists, denoted bylimn→∞xRn=z. From the continuity ofR, we could know that
(2)zR2=(limn→∞xRn)2=limn→∞xR2n=z.
This means that zis an idempotent element ofR. Becausez<x,z=0, that is, for anyx∈(0,1), the limit ofxRnis 0. Thus, for any x,y∈(0,1), there exists some natural numbern, subject toyRn<x.
(2) It is similarly.Now, let us show the proof of Theorem8.Proof.
Since a weak uninorm is an AMC operator. From Theorem7, we could know that the squares of the elements in(0,1)are either all strictly smaller than themselves or all strictly bigger than themselves; that is,
(3)either∀x∈[0,1],R(x,x)<x,or∀x∈(0,1),R(x,x)>x.(1)
for all x∈(0,1),R(x,x)<x. Let’s show that Ris at-norm. From the definition of weak uninorms, for anyx∈(0,1), there is somey∈[0,1], subject toR(x,y)=x. Ify≠1, from Lemma 9, there exists somen, subject toyRn<x. Thus,x=R(x,y)=⋯=R(x,yRn)  ≤R(x,x)<x, contradiction. Therefore,y=1; that is, for anyx∈(0,1),R(x,1)=x. From the monotonicity ofR, we could know that R(0,1)=0and from the idempotence of 1, we haveR(1,1)=1. As a result,
(4)∀x∈[0,1],R(x,1)=x,that is,Ris at-norm.(2)
Similarly.This theorem shows that there are no continuous weak uninorms with no nontrivial idempotents, except thet-norms and thet-conorms. For weak uninorms with just one nontrivial idempotent element, we have a similar result.Theorem 10.
There are no continuous weak uninorms with just one idempotent elementλ∈(0,1), except the uninorms and the nullnorms.Proof.
Suppose thatRis a continuous weak uninorm, with just one nontrivial idempotent elementλ. Let’s show that it is either a uninorm or a nullnorm.
LetR′(x,y)=R(λx,λy)/λ, andR′′(x,y)=R((1-λ)x+λ,(1-λ)y+λ)/(1-λ), for all x,y∈[0,1]. Obviously,R′andR′′are continuous AMC operators. Let’s show that they are weak uninorms, that is, for each elementx,uxexists.
Sinceλ, 0, and 1 are idempotent elements andRis continuous, from Theorem  2.8 in [14] (or Definition  3.1 in [9]),R′andR′′are continuous weak uninorms.
Obviously, they have no nontrivial idempotent elements. From Theorem8, they aret-norms ort-conorms. (1)
IfR′is at-norm andR′′is at-conorm, thenλis the neutral element ofR; that is,Ris a uninorm.(2)
IfR′is at-conorm andR′′is at-norm, thenλis an absorbing elementλ∈[0,1], and for all x∈[0,λ],V(x,0)=xand for all x∈[λ,1],V(x,1)=x; that is,Ris a nullnorm.(3)
If both of them aret-norms, let us show that R is at-norm, a special uninorm. It just needs to show for all x∈[0,λ],
(5)R(x,1)=R(R(x,λ),1)=R(x,R(λ,1))=R(x,λ)=x,which could get thatR′andR′′aret-norms.(4)
Similarly, if both of them aret-conorms,Ris also at-conorm, a special uninorm.


## 3. Weak Uninorms with No Nontrivial Idempotent Elements

In this section, we will give some examples of weak uninorms, which have no nontrivial idempotent elements. And none of them is ann-uninorm; that is, all the examples in this section are nontrivial weak uninorms.Example 11.
The unit interval[0,1]is divided into infinitely many sections as(1-(1/2n-1),1-(1/2n)], withn=1,2,…and 1. Letan=1-(1/2n-1); define a mappingR1as follows:
(6)R1(x,y)={an+2n-1(x-an)(y-an),if(x,y)∈(an,an+1]2,1,ifx=1,ory=1,min{x,y},otherwise.
Then,R1is a weak uninorm with no nontrivial idempotent elements; that is, its idempotent elements are just 0 and 1. But it is not ann-uninorm and thus neither a uninorm nor a nullnorm.Actually, for(x,y)∈(an,an+1]2,R1(x,y)=an+((1/2)TP(2n(x-an),2n(y-an))/2n). For convenience, it is shortly rewritten as
(7)R1(x,y)={12TP,if(x,y)∈(an,an+1]2,1,ifx=1,ory=1,min{x,y},otherwise.
In this formula, there is a symbol(1/2)TP. It means an injection of the operator(1/2)TP, that is, for all (x,y)∈(an,an+1]2,R1(x,y)=an+((1/2)TP(2n(x-an),2n(y-an))/2n), instead of R1(x,y)=  (1/2)TP (x,y). Similar for the following examples.Proof.
Obviously,R1is monotone and commutative. Let’s show it is associative.
For anyx,y,z∈[0,1], if one of them is 0 or 1, then it is trivial.
If there is somen, subject tox,y,z∈(an,an+1], from the associativity of0.5TP,R1is associative.
If there are somem<n, subject tox,y∈(an,an+1]andz∈(am,am+1], then we haveR1(x,y)>an≥am+1≥z,R1(x,z)=zandR1(y,z)=z. Therefore,
(8)R1(R1(x,y),z)=z=R1(x,z)=R1(x,R1(y,z)).
If there are somem<n, subject tox∈(an,an+1]andy,z∈(am,am+1], thenR1(x,y)=y, R1(x,z)=zandR1(y,z)<am+1≤an≤x. Thus,R1(x,R1(y,z))=R1(y,z). Therefore,
(9)R1(R1(x,y),z)=R1(y,z)=R1(x,R1(y,z)).
The last case is thatx∈(an,a{n+1}], y∈(am,a{m+1}] and z∈(al,a{l+1}], with different l, m, n. In this case, the result is always equal to the smallest element; that is, the associative law is valid. Now, the associativity ofR1has already been proved.
Next, let us show thatR1is a weak uninorm with no idempotent elements, except 0 and 1.
For anyx∈(0,1), there is somen, subject tox∈(an,an+1]. Lety=an+1+(1/2n+3), thenR1(x,y)=x. Together withR1(0,0)=0andR1(1,1)=1,R1is a weak uninorm.
Since(1/2)TP(x,x)<x, for all x∈(0,1], we haveR1(y,y)<y, for all y∈(0,1).
As a result,R1is a weak uninorm with idempotent elements no more than 0 and 1.
It is obvious that it is not ann-uninorm, thus, neither a uninorm nor a nullnorm.Note that it is not difficult to find that([0,1),R1|[0,1)2), in whichR1|[0,1)2is the restriction of the weak uninormR1on the square[0,1)2, is the ordinal sum [16] of the semigroups(((1-(1/2n-1),1-(1/2n)],TP))n∈N. ButR1itself is not. Actually, similar to the proof of Example 11, we can obtain the following property.Theorem 12.
LetRbe an AMC operator on[0,1]withR(1,1)=1. If([0,1),R|[0,1)2)is the ordinal sum of the semigroups((Xα,Tα))α∈A, in whichAis an infinite set, eachXαis in the form(a,b], and eachTαis Archimedean, thenRis a weak uninorm with no idempotent elements, except 0 and 1.Next, let us construct some more examples of weak uninorms. In these examples, if the ordinal sums are replaced as in this theorem, then they are still weak uninorms with no idempotent elements.Example 13.
The unit interval[0,1]is divided into infinitely many sections as[1/2n-1,1/2n), withn=1,2,…and 0. Then, the following definedR2is a weak uninorm with no non trivial idempotent elements:
(10)R2(x,y)={0.5+0.5SP,if(x,y)∈[12n,12n-1)2,0,ifx=0,ory=0,max{x,y},otherwise;
that is, for(x,y)∈[1/2n,1/2n-1)2,R2(x,y)=(1/2n)+(0.5+0.5SP(2n(x-(1/2n)),2n(y-(1/2n)))/2n).R
2is the dual ofR1. Thus, it is a weak uninorm.Example 14.
For some givenn0, define a mappingR3as follows:
(11)R3(x,y)={min{x,y},ifx=1,y≤an0ory=1,x≤an0,R1(x,y),otherwise,
in whichan0=1-(1/2n0). Then, R3is a weak uninorm with idempotent elements 0 and 1 only.
Note that, in this example, ifan0≠1-(1/2n0), the associativity will not be valid; that is,R3will no longer be a weak uninorm.Example 15.
InR1, ifTPis replaced byTL, denoted as R1′, then it will no longer be a weak uninorm. Since the associativity is not valid,
(12)R1′(0.2,R1′(0.6,0.6))=R1′(0.2,0.5)<0.2,R1′(R1′(0.2,0.6),0.6)=R1′(0.2,0.6)=0.2.This example shows thatTPcould not be replaced by any Archimedeant-norm.Example 16.
The following is a weak uninorm with only trivial idempotent elements:(13)R4(x,y)={R1,if(x,y)∈[0,0.5)2,1,if(x,y)∈[0.5,1]2,max{x,y},otherwise.
The demarcation point ofR4is 0.5.Example 17.
LetR5be defined by
(14)R5(x,y)={R1,if(x,y)∈[0,0.5)2,min{x,y},if(x,y)∈[0.5,1]×[0,0.25]∪[0,0.25]×[0.5,1],1,if(x,y)∈[0.5,1]2,max{x,y},otherwise.
Then, R5is a weak uninorm with nontrivial idempotent elements. See Figure 1.Figure 1
Graphical representation ofR5.


## 4. Examples of Weak Uninorms with One or More Nontrivial Idempotent Elements

Example 18.
The following definedR6andR7are weak uninorms, with just one nontrivial idempotent element 0.5:
(15)R6(x,y)={R1,if(x,y)∈[0,0.5)2,R1,if(x,y)∈[0.5,1)2,1,ifx=1,ory=1,0.5,otherwise,R7(x,y)={R1,if(x,y)∈[0,0.5)2,TL,if(x,y)∈(0.5,1]2,0.5,otherwise.Example 19.
Define a mappingR8by
(16)R8(x,y)={TP,if(x,y)∈[0,0.5)2,R1,if(x,y)∈(0.5,1]2,min{x,y},otherwise.
Then, R8is a weak uninorm with idempotent elements0,0.5, and 1.Examples18 and 19 are constructed byR1. The next example is not in this case.Example 20.
Define mappingsR9andR10(see Figure 2) as follows:
(17)R9(x,y)={TP,if(x,y)∈[0.6,1]2,SP,if(x,y)∈[0.3,0.6]2min{x,y},ifx<0.3,y=1;orx=1,y<0.3,0.6,if(x,y)∈[0.6,1]×[0.3,0.6]∪[0.3,0.6]×[0.6,1],0,otherwise,(18)R10(x,y)={TP,if(x,y)∈[0.6,1]2,R2,if(x,y)∈(0.3,0.6]2min{x,y},ifx≤0.3,y=1;orx=1,y≤0.3,0.6,if(x,y)∈[0.6,1]×[0.3,0.6]∪[0.3,0.6]×[0.6,1],0,otherwise.
ThenR9is a weak uninorm with two nontrivial idempotent elements 0.3 and 0.6;R10is a weak uninorm with just one nontrivial idempotent element 0.6.Figure 2
Graphical representation ofR9(orR10).


## 5.  Conclusion

In this paper, it is proved that there are no nontrivial continuous weak uninorms with none or one idempotent element. Moreover, some nontrivial examples of weak uninorms are given. These examples are with no more than two nontrivial idempotent elements, which is a positive answer to the question in [14].



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