eduardo silva palmeira - federal university of rio grande ...€¦ · eduardo silva palmeira on...
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UNIVERSIDADE FEDERAL DO RIO GRANDE DO NORTE
CENTRO DE CIENCIAS EXATAS E DA TERRA
DEPTO. DE INFORMATICA E MATEMATICA APLICADA
POS-GRADUACAO EM SISTEMAS E COMPUTACAO
On Extension of Fuzzy Connectives
Eduardo Silva Palmeira
Natal/RN
february 2013
Eduardo Silva Palmeira
On Extension of FuzzyConnectives
Thesis presented to the Graduate Program in
Systems and Computation of Department of In-
formatics and Applied Mathematics of Federal
University of Rio Grande do Norte in a fulfill-
ment of the Requirements for the degree of Phi-
losophy Doctor .
Advisor:Prof. Dr. Benjamın Rene Callejas Bedregal
Natal/RN
2013
I dedicate this thesis to my family and specially to my wife and my
daughter who always encouraged and accompanied my studies
giving me all needed support.
Acknowledgements
I have a lot of thanks to do, because a thesis is not built in days or
weeks, it is the result of a lengthy effort and a lot of persistence and
perseverance, where we deal every day with frustrations and achieve-
ments in the struggle for the consolidation of ideas and conjectures
that constitute this work today.
First of all I thank God who gave me the gift of life and that, in good
and bad times, was always by my side watching and illuminating my
path. For the infinite blessings He give me every day, praise and glory
be given in His name.
I thank my parents Pedro and Telma, who directly or indirectly have
always believed in me, giving me full support and that, even in the
face of life’s difficulties, never failed to give me everything they could
so that I tread my path to greatness in life. Also to my brothers
Cristina and Danillo and my nephews Hendrick and Enzo for all the
warmth.
A special thank to the woman who gave me a family, my wife Stephanie
throughout companionship and dedication shown throughout these
four years of study doing me dreams her dreams. Thanks for always
being willing to listen to my bullshits, my complaints, for believing
and supporting my projects, for supporting my boredom when I was
exhausted. Anyway, for being an angel in my life and always be watch-
ing with great affection and love for me and our family. Maybe there
is no a word deep enough to express all my gratitude and what I’m
feeling right now writing these words. So I just say: I love you, all
this here is not mine, it is ours!
I could not forget to thank my daughter Lunna, my sweet, my smile.
Thanks for having me often waited at the door to give me a hug
overpowering covered with the purest feeling of love, able to renew
me and make me forget everything bad that might exist in the world.
For my master, professor Benjamın Bedregal, for believing in me from
the beginning and for putting all possible credibility in my work. Cer-
tainly this work would not be the same, if it not had the orientation
of a person so qualified and competent, able to demonstrate a humil-
ity so great as to always treat their students with equal professional.
Moreover, I would like to thank you for your friendship and partner-
ship, and for the beers and casual conversations. I leave this course
happy for having won much more than a research partner for doing a
great friend.
Also, for GIARA research group from Pamplona-Spain, in the figure
of the professor Humberto Bustince, thank you very much for giving
me the opportunity to be researching with them for six months and
the experience that they provided me with a personal and professional
enrichment of great value.
Finally, thank you Natal, this beautiful city where I could enjoy won-
derful days and made friends that I leave here, but that I will take
forever in my heart and my memory. In particular, I would like to
thank Isaac (and Lisa), Aluisio (and Bruninha), Fagner (and Fabi-
ana), Marcelo (and Lene), Anchieta, Guga, Samir, Rogerio, Macilon,
Liliane, among others, perchance, I may have forgotten, for all friend-
ship demonstrated. I could not also forget to thank the teachers and
other employees of the LoLITA group of UFRN for the support gen-
erated during this period. Also, I would like to thank for UESC and
CAPES for the financial support.
.
Stay Hungry, Stay Foolish
(Steve Jobs)
Abstract
Let M be a sublattice of lattice L and K be a fuzzy operator (eg. a
t-norm, a t-conorm, a fuzzy fuzzy negation, etc.) on M . So, how can
we extend this operator from M to L preserving the most possible
number of its properties?
This is a very known and interesting problem and a suitable solu-
tion for this is not so trivial. In this framework we present in this
thesis two different methods for extending t-norms, t-conorms, fuzzy
negations and implications from a sublattice to a lattice consider-
ing a generalized notion of sublattices, namely (r, s)-sublattice. In
one method, named extension method via retraction, we extend fuzzy
connectives trying to obtain the smallest possible extension while the
other method aims to make an extension able to preserve the largest
number of properties of extended connective using a special extension
operator (e-operator, for short). Also, we present some results involv-
ing extension, automorphisms, De Morgan triples and t-subnorms.
Moreover, leaving a bit the scope of fuzzy logic, we investigate the
behavior of extension methods proposed in this thesis for some func-
tions related to study of images. We do the extension of Fodor and
Roubens’ equivalence functions as well as the extension of a particu-
lar case of them named restricted equivalence functions. Also in this
framework, we extend restricted dissimilarity functions and normal
Ee,N -functions.
List of Figures
2.1 Hasse diagrams of lattices L and M . . . . . . . . . . . . . . . . . 13
2.2 The relaxed idea of sublattice . . . . . . . . . . . . . . . . . . . . 16
2.3 Hasse diagrams of lattices M , L1, L2 and L3 . . . . . . . . . . . . 18
2.4 Hasse diagrams of lattices M and L . . . . . . . . . . . . . . . . . 19
2.5 Hasse diagrams of lattices M and L . . . . . . . . . . . . . . . . . 26
4.1 Diagram of interval constructor . . . . . . . . . . . . . . . . . . . 81
4.2 Diagram of intuitive idea . . . . . . . . . . . . . . . . . . . . . . . 81
4.3 Hasse diagrams of lattices M and L . . . . . . . . . . . . . . . . . 83
vii
LIST OF FIGURES
viii
List of Tables
2.1 Tables of retractions r1, r2 and r3. . . . . . . . . . . . . . . . . . . 18
2.2 Tables of pseudo-inverses s1, s2 and s3. . . . . . . . . . . . . . . . 18
2.3 A function I on L . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.1 Implication on M . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
3.2 Table of Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . 77
3.3 Properties preserved by TE and SE . . . . . . . . . . . . . . . . . 77
3.4 Properties preserved by IE . . . . . . . . . . . . . . . . . . . . . . 78
3.5 Properties preserved by NE . . . . . . . . . . . . . . . . . . . . . 78
4.1 Table of Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . 104
4.2 Properties preserved by TE� and SE� . . . . . . . . . . . . . . . . . 104
4.3 Properties preserved by NE� . . . . . . . . . . . . . . . . . . . . . 104
5.1 Restricted equivalence function on lattice M . . . . . . . . . . . . 108
5.2 Restricted equivalence function on lattice L . . . . . . . . . . . . 113
5.3 The function IREF on lattice M . . . . . . . . . . . . . . . . . . . 115
6.1 Table of Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . 139
6.2 Comparing results of TE and TE� . . . . . . . . . . . . . . . . . . 139
6.3 Comparing results of SE and SE� . . . . . . . . . . . . . . . . . . . 139
6.4 Comparing results of IE and IE� . . . . . . . . . . . . . . . . . . . 139
6.5 Comparing results of NE and NE� . . . . . . . . . . . . . . . . . . 139
ix
LIST OF TABLES
x
Contents
List of Figures vii
List of Tables ix
Contents xi
1 Introduction 1
1.1 On Extension Problem . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Fuzzy Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Restricted Equivalence Functions . . . . . . . . . . . . . . . . . . 5
1.4 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Structure of Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 Preliminaries 9
2.1 Lattice Theory: Definitions and Constructions . . . . . . . . . . . 9
2.1.1 Bounded Lattices and Homomorphisms . . . . . . . . . . . 10
2.1.2 Automorphisms and Conjugates . . . . . . . . . . . . . . . 13
2.1.3 Retractions and Sublattices . . . . . . . . . . . . . . . . . 15
2.1.4 Pseudo-quasi-metrics and Continuity . . . . . . . . . . . . 21
2.2 Fuzzy Connectives . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2.1 T-norms and T-conorms . . . . . . . . . . . . . . . . . . . 23
2.2.2 Fuzzy Negations . . . . . . . . . . . . . . . . . . . . . . . . 25
2.2.3 Fuzzy Implications . . . . . . . . . . . . . . . . . . . . . . 31
2.2.4 T-subnorms . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.2.5 De Morgan Triples . . . . . . . . . . . . . . . . . . . . . . 37
xi
CONTENTS
3 Extension Method via Retractions 41
3.1 T-norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.2 T-conorms and Fuzzy Negations . . . . . . . . . . . . . . . . . . . 46
3.2.1 Negations Obtained from Fuzzy Connectives . . . . . . . . 51
3.3 Fuzzy Implications . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.3.1 (S,N)-implications . . . . . . . . . . . . . . . . . . . . . . 62
3.3.2 R-implications . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.4 De Morgan Triples . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.5 Extension and Automorphisms . . . . . . . . . . . . . . . . . . . . 69
3.6 T-subnorms and T-subconorms . . . . . . . . . . . . . . . . . . . 74
3.7 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
4 Extension Method via e-operators 79
4.1 Toward e-operators . . . . . . . . . . . . . . . . . . . . . . . . . . 80
4.2 T-norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.3 T-conorms and Fuzzy Negations . . . . . . . . . . . . . . . . . . . 89
4.4 De Morgan Triples . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.5 Extension and Automorphisms . . . . . . . . . . . . . . . . . . . . 92
4.6 On Extension of n-dimensional T-norms . . . . . . . . . . . . . . 97
4.7 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5 On Restricted Equivalence Functions 105
5.1 Restricted Equivalence Functions on L . . . . . . . . . . . . . . . 106
5.1.1 Restricted Equivalence Functions and Negations . . . . . . 109
5.2 Characterization Theorem for L-REF . . . . . . . . . . . . . . . . 111
5.3 Restricted Dissimilarity Functions . . . . . . . . . . . . . . . . . . 116
5.4 Normal Ee,N -functions on L . . . . . . . . . . . . . . . . . . . . . 119
5.5 REF on L([0, 1]): Definition and characterization . . . . . . . . . 121
5.6 Extension of REF via Retractions . . . . . . . . . . . . . . . . . . 124
5.7 Extension of REF via e-operators . . . . . . . . . . . . . . . . . . 127
5.7.1 Extension of Natural Negation of REF . . . . . . . . . . . 129
5.8 Extension of Restricted Dissimilarity Function . . . . . . . . . . . 130
5.9 Extension of Normal Ee,N -functions . . . . . . . . . . . . . . . . . 133
xii
CONTENTS
6 Remarks and Further Works 137
References 143
Index 151
xiii
CONTENTS
xiv
Chapter 1
Introduction
This thesis is devoted to discuss about the problem of extending fuzzy connec-
tives from sublattices to a greater one in such way to provide a suitable extension
of them able to preserve the largest number of their properties.
Besides being a naturally interesting and challenging mathematical problem
(Hestenes [1941]; Uspenskii [1966]; Whitney [1934]), the problem of extending
functions or operators arises in many situations in various branches of computer
science such as image processing, mathematical morphology, computational se-
mantics, object-oriented programming, etc. For instance, there is a very known
operator in object-oriented programming named inheritance and behavioral sub-
typing that works as an extension operator through which one can extend a
subclass to a superclass, leveraging their behaviors (methods) and variables (at-
tributes). In mathematical morphology, erosion and dilation operators play a
similar role switching images between lattices.
In fuzzy mathematics (particularly in fuzzy logic) most of operators are func-
tions in usual sense and hence it can be considered the extension problem for
them. Within this framework, we present here two extension methods based in
a generalized notion of sublattice defined using retractions which have different
behavior in preserving properties of fuzzy operators as one can see in Chapters
3 and 4. Also, for having an applied point of view we lead with extension of
restricted equivalence functions defined on bounded lattices.
1
1. INTRODUCTION
1.1 On Extension Problem
The problem of extending functions is very know and studied by many re-
searchers in many branches of knowledge. Particularly, in exact sciences, the
problem of extending functions is widely studied in some areas such as mathe-
matics, physics, computer science, etc (see Hans-Peter and Shapiro [1997]; Hori-
uchi and Murakami [1993]; Murota and Shioura [2000]). In a general way, we can
describe this problem as follows: how to extend a given function from a subset to
entire domain preserving its main properties, that is, if L is an arbitrary set and
supposing f is a function defined and possessing a property P on a nonempty
subset M of L how can f be extended to L in order to preserve property P for
the elements of L\M? In other words, which is the best choice to define f(x) for
elements x ∈ L\M?
The answer for this is: It depends! This is very simple if we want only to
construct a new function that has L as its domain. In this case, it is enough
to define f(x) = a for a suitable and fixed a belonging to L (i.e., define f as
a constant function for the elements belonging to L\M). However, this task
becomes more complex if we want that extension of f preserve its properties. For
instance, an important theorem of analysis states that a continuous function f
defined on a bounded closed set M can be extended to the whole space preserving
its continuity Apostol [1974]; Bartle and Sherbert [1982]; Lima [1982].
1.2 Fuzzy Logic
Emerging from the important work “Fuzzy Sets Theory” proposed by Zadeh
[1965], fuzzy logic is a generalization of classical logic which typically considers
for membership degrees values in the unit interval [0,1] instead of only {0, 1}, but
in modern fuzzy logic, lattices are used to range these degrees Goguen [1967];
Gratzer [2011]; Trillas [1979].
Nowadays, fuzzy logic has awakened the interest and curiosity of several re-
searchers in a variety of scientific areas due to its broad scope and the fact that
it provides a good framework for constructing models which better approach to
reality Bedregal and Figueira [2008]; Buckley [2005]; Chalco-Cano et al. [2011];
2
Pardo and de la Fuente [2010]; Takagi [2009]; Zadrozny and Kacprzyk [2009]. In
literature, one may find a substantial amount of works that introduce a fuzzy
version of classic concepts as well as some others that deal with the improvement
of techniques and tools utilizing fuzzy mathematics Bona [2006]; Gratzer [2009];
Hajek [1998]; Jacas and Recasens [1994]; Liu [2011]; Mitra and Pal [2005]; Siler
and Buckley [2005]; Takano [2002].
Talking about applications, fuzzy logic has been widely used to generalize and
define important theories and tools in control theory, artificial intelligence, etc.
According to Bojadziev and Bojadziev [1996]
“Fuzzy sets and fuzzy logic are powerful mathematical tools for mod-
eling and controlling uncertain systems in industry, humanity, and
nature; they are facilitators for approximate reasoning in decision
making in the absence of complete and precise information. Their
role is significant when applied to complex phenomena not easily de-
scribed by traditional mathematics.”
As for every logic, an appropriate definition of connectives and rules is essen-
tial. In fuzzy logic, conjunctions are interpreted by triangular norms (t-norms)
and disjunctions are generalized by triangular conorms (t-conorms). Such as in
classical case, a t-conorm S can be constructed from a given t-norm T , and a
fuzzy negation N , by
N(S(x, y)) = T (N(x), N(y)) (1.1)
In this case, t-conorm S is said to be N -dual of T (by dually principe).
Another concept which plays an important role in fuzzy logic are De Morgan
triples (see Calvo [1992]; da Costa et al. [2011]; Garcıa and Valverde [1989];
Klement et al. [2000]), a fuzzy version of De Morgan’s law. More specifically, a
triple 〈T, S,N〉 where T is a t-norm, S is a t-conorm and N is a negation, may
be called a De Morgan triple if 〈T, S,N〉 satisfies both the equality (1.1) and
N(T (x, y)) = S(N(x), N(y)) (1.2)
In other words, given a t-norm T , 〈T, S,N〉 is a De Morgan triple if and only if
N is a strong negation and S is N -dual to T .
3
1. INTRODUCTION
In particular, notice that t-norms, t-conorms, fuzzy negations and implications
are functions as well as most fuzzy other operators so, the problem of extending
functions can be considered for them. In this context, one of the main works is
due to Saminger-Platz et al. [2008], in which the authors propose a way to extend
a given t-norm T from a complete sublattice (in the sense of the usual concept of
sublattice) M to a bounded lattice L. However, their extension method is very
drastic due to the action of function x∗ = supM{y | y 6L x, y ∈ M ∪ {0L, 1L}}(since it works as a “collapsing function”, see (3.1)). This is motivated by the
fact that they wish to propose the least possible extension of a t-norm T .
In this sense, seeking to have a more general way to extend fuzzy connectives
we would like to propose a method which could be more flexible than one provided
by Saminger-Platz et al. [2008]. To do so, we consider a generalized version of
the concept of sublattice in which M is not necessarily a subset of L, but can be
seen as a copy embedded in L.
Specifically speaking, we propose the following concept of sublattice: let M
and L be two bounded lattices. It is said that M is a (r, s)-sublattice of L if
there is a retraction r : L −→ M with pseudo inverse s : M −→ L such that
r ◦ s = idM ; i.e. M is a (r, s)-sublattice of L if M is a retract of L (see Definition
2.10). A natural example of sublattice in this sense can be given by taking a
bounded lattice K as M and the interval lattice K constructed from it as L (see
Example 4.1).
Based on notion of (r, s)-sublattice, our early research has led us to provide a
way to extend t-norms, t-conorms and fuzzy negations (named extension method
via retractions, see Palmeira and Bedregal [2012]), in which we try to achieve two
goals: (1) generalize the extension of t-norms presented in Saminger-Platz et al.
[2008]; (2) in order to preserve the largest possible number of properties of these
fuzzy connectives which are invariant by retractions. However, despite the first
goal has been achieved, the extension proposed in Palmeira and Bedregal [2012]
does not completely fulfill (2) since some important properties related to t-norms
and t-conorms are not preserved such as continuity, for example. For extension
of fuzzy negation this method does not preserve involution property and hence it
does not preserve strong negations (see Remark 3.2).
Thus, in order to find a way to solve problems presented by extension method
4
via retractions proposed in Palmeira and Bedregal [2012]; Palmeira et al. [2009] we
have turned our attention to investigate extensions that are more effective in pre-
serving properties of fuzzy connectives even this new extensions do not fulfill goal
(1). The inspiration for this comes from the following important fact: the interval
constructor (see Bedregal and Takahashi [2006]; Bedregal et al. [2006b]) provides
a natural way to identify a lattice F with F via retractions. So, we developed
in Palmeira et al. [2012b] a new method to extend t-norms (t-conorms and fuzzy
negations) using a special mapping named e-operator as in Definition 4.1 (this
method is named extension method via e-operator). To verify that this method is
more efficient than extension method via retractions in preserving properties we
investigated in Palmeira et al. [2012b] same issues as in Palmeira and Bedregal
[2012] and results were quite satisfactory. Every problem presented by extension
method via retractions were overcome. Another advantage of extension method
via e-operator becomes evident when we are extending t-norms (and t-conorms)
since some constraints in hypotheses of Theorem 3.1 are not necessary anymore,
namely M could not be a lower (r, s)-sublattice of L (see Theorem 4.1).
1.3 Restricted Equivalence Functions
Basically our studies are theoretical, however determine a wide range of pos-
sible applications. One of our main interest is applying our extension methods
for operators used in several branches of knowledge that lead with images. Our
motivation for choosing images comes from the fact that many works discuss
about images defined on lattices nowadays, and it is natural to generalize con-
cepts related to image processing for lattices in order to obtain a much more
general framework than [0,1]. In particular bounded lattices are of interest since
intensities in images can be considered as taking values in such lattices.
In this framework, after six months working together with GIARA group
(spanish acronym Research Group of Artificial Intelligence and Approximate
Reasoning) under supervision of professor H. Bustince and collaboration with
other group’s members we have worked on restricted equivalence functions, a
very known and powerful tool that provide a good measure for making a global
comparison of images, in which we discuss about the following issues:
5
1. INTRODUCTION
• To provide a formalization of the concept of restricted equivalence functions
on bounded lattices;
• To present a characterization theorem of these functions via implications;
• Apply our two extension methods for extending restricted equivalence func-
tions and test which one has a better behavior.
Results of this research period were submitted for specialized journals on fuzzy
sets (see, Palmeira et al. [submitted 2013a,s]) where we made a formalization of
definition of restricted equivalence functions on bounded lattices and its exten-
sions.
1.4 Objectives
The main objectives of this thesis is developing a consistent formalization of
methods for extending lattice-valued fuzzy connectives and other fuzzy operators
which preserve the largest possible number of their properties. Moreover, we
discuss about the efficiency of these methods for extending restricted equivalence
functions.
Hence, seeking to achieve expected results we turn our attention to investigate
the following specific goals:
• Propose a suitable definition of the generalized notion of sublattices ((r, s)-
sublattices) relaxing some classical constraints of this concept using retrac-
tions;
• Define a way to extend fuzzy connectives (t-norms, t-conorms, fuzzy nega-
tions and implications) from a (r, s)-sublattice to a lattice which is able to
preserves their properties. In this case, we develop two different methods:
(1) extension method via retraction seeking to generalize the method pro-
posed by Saminger-Platz et al. [2008] and (2) a method efficient in preserv-
ing properties of extended connectives (extension method via e-operator);
• Apply both extension method for other fuzzy operators namely t-subnorms,
De Morgan triples, etc;
6
• Establish the relation between extension and conjugate of a fuzzy connective
and other operators;
• Define restricted equivalence functions on bounded lattices and study some
related properties and operators. Also, apply our extension methods for
extending it;
• Develop a study about restricted dissimilarity functions and normal Ee,N -
functions defined on bounded lattices and its extensions.
1.5 Structure of Thesis
This thesis is composed of six chapters which are divided in sections in which
we discuss about, lattices, extension, fuzzy connectives, restricted equivalence
functions and other operators. We begin describing the research problem of this
thesis and making a discussion about state of art on our research area in Chapter
1 (Introduction). Other chapters are organized as follows:
• In Chapter 2 we make a specific formalization of main concepts used along
this work such as lattices, automorphisms, retractions, sublattices, continu-
ity and fuzzy connectives which constitute our theoretical bases;
• Chapter 3 is devoted to present our extension method via retractions. We
have chosen t-norms, t-conorms, fuzzy negations, fuzzy implications, De
Morgan triples and t-subnorms for applying this method and to test it
efficiency in preserving properties. For a better organization, we divided
this chapter in six section in order to study these issues separately. We also
discuss about the relation between extension and conjugate in Section 3.5;
• Also within the framework of extension, we propose in Chapter 4 the ex-
tension method via e-operators. In section 4.1 it is introduced a notion of
e-operator and some results are proved. The next four sections are dedi-
cated to discuss about similar issues like in Chapter 3. Section 4.6 presents
a study on extension of n-dimensional t-norm;
7
1. INTRODUCTION
• Restricted equivalence functions are addressed in Chapter 5. In section
5.1, we present the definition of L-REF and then we characterize them by
implications in Section 5.2. Section 5.3 and 5.4 are dedicated to introduce
the notion of restricted dissimilarity functions and normal Ee,N -functions
on bounded lattices and to prove some results. Section 5.5 discusses about
restricted equivalence functions on L([0, 1]) and presents a generalization
of the REF characterization theorem given in Bustince et al. [2006]. In
Sections 5.6 and 5.7 we apply our extension method for L-REF;
• Finally, in Chapter 6 we write about our main contributions and publica-
tions beyond to present ideas and conjectures for further works.
8
Chapter 2
Preliminaries
This chapter is devoted to present and discuss about main concepts and results
we are leading in this work which constitute the framework of our studies.
We start in Section 2.1 making a formalization of lattices and its morphisms.
In which follows we introduce the key-concept of (r, s)-sublattices, a generalized
way to define the notion of sublattice, some useful examples and related defini-
tions. Continuity for lattices is considered at the end of this section.
In Section 2.2 we turn our attention to recall the usual notions of lattice-
valued fuzzy connectives such as t-norms, t-conorms, fuzzy negations and fuzzy
implications. Also in this framework we discuss about t-subnorms and De Morgan
triples.
It is important to point out here that most of definitions and results present
is this chapter are from the literature, but some of them were proposed by us.
So, to highlight this fact definitions and results that came from literature we cite
its respective authors.
2.1 Lattice Theory: Definitions and Construc-
tions
As a basis for our developments a clever and consistent formalization on lattice
theory and related concepts is necessary since all of operators we are considering
in this work are lattice-valued. It is important to say that throughout this section
9
2. PRELIMINARIES
L represents a bounded lattice.
Some elementary concepts will not be shown here, however for a further read-
ing about such concepts we recommend Bedregal et al. [2007]; Birkhoff [1973];
Burris and Sankappanavar [2005]; Chen and Pham [2001]; de Cooman and Kerre
[1994]; Klement and Mesiar [2005]; Klement et al. [2000]; Klir and Yuan [1995];
Lowen [1996].
2.1.1 Bounded Lattices and Homomorphisms
It is very known that the concept of lattices has two approaches, namely: an
algebraic and an order-theoretical approaches.
Definition 2.1 Birkhoff [1973] Let L be a nonempty set and 6L be a partial
order on it. We define 〈L,6L〉 as an ord-lattice if for all a, b ∈ L the set {a, b}has a supremum and an infimum. If there are two elements 0L and 1L in L such
that 0L 6L x (bottom) and x 6L 1L (top) for each x ∈ L, then 〈L,6L, 0L, 1L〉 is
called a bounded lattice.
Definition 2.2 Birkhoff [1973] Let L be a nonempty set. If ∧L and ∨L are two
binary operations on L, then 〈L,∧L,∨L〉 is an alg-lattice provided that for each
x, y, z ∈ L, the following properties stand:
1. x ∧L y = y ∧L x and x ∨L y = y ∨L x (commutativity);
2. (x∧L y)∧L z = x∧L (y∧L z) and (x∨L y)∨L z = x∨L (y∧L z) (associativity);
3. x ∧L (x ∨L y) = x and x ∨L (x ∧L y) = x (absorption law).
If in L there are elements 0L and 1L such that, for all x ∈ L, x∨L0L = x (bottom)
and x ∧L 1L = x (top), then 〈L,∧L,∨L, 0L, 1L〉 is a bounded lattice .
A very interesting fact is that Definitions 2.1 and 2.2 are equivalent. Indeed,
it is known from the literature that given an alg-lattice L, the relation
x 6L y if and only if x ∧L y = x (2.1)
10
defines a partial order on L and hence L can be seen as an ord-lattice and vice-
versa. This allows us to use both definitions indiscriminately. We consider in
this whole work the algebraic notion of this concept (Definition 2.2) taking into
account that from this structure we can always define a partial order on L what
is very important for us to compare elements. The reason for choosing it rises up
from the lattice homomorphism as we discuss in rest of this subsection.
Remark 2.1 From now on when we say that L is a bounded lattice it means that
L has a structure as in Definition 2.2. Otherwise, an appropriate distinction will
be made.
Definition 2.3 Birkhoff [1973] A lattice L is called a complete lattice if every
subset of it has a supremum and an infimum element. Notice that every complete
lattice is bounded.
Example 2.1 The set [0, 1] endowed with the operations defined by x ∧ y =
min{x, y} and x∨ y = max{x, y} for all x, y ∈ [0, 1] is a complete bounded lattice
in the sense of Definitions 2.2 and 2.3 which has 0 as the bottom and 1 as the
top element.
Example 2.2 For all x, y ∈ [0, 1] it is possible to define the interval set L([0, 1]) =
{[x, y] ; 0 6 x 6 y 6 1} . This set equipped with the operations
[x, y] ∧L [w, z] = [x ∧ w, y ∧ z] and [x, y] ∨L [w, z] = [x ∨ w, y ∨ z].
with a ∧ b = min(a, b) and a ∨ b = max(a, b), is a complete bounded lattice (in
the sense of Definition 2.2) which has [0, 0] and [1, 1] as a bottom and a top
respectively. It is easy to see that this lattice is also obtained by considered in
L([0, 1]) the partial order [a, b] 62 [c, d] if and only if a 6 c and b 6 d.
Remark 2.2 When 6L is a partial order on L and there are two elements x
and y belonging to L such that neither x 6L y nor y 6L x, these elements are
said to be incomparable and we denote this by x ‖ y . Otherwise we say they are
comparable (notation: x ¨ y) .
11
2. PRELIMINARIES
Definition 2.4 Davey and Priestley [2002] Let (L,6L, 0L, 1L) and (M,6M , 0M , 1M)
be bounded lattices. A mapping f : L −→M is said to be a lattice ord-homomorphism
if, for all x, y ∈ L, it follows that
1. If x 6L y then f(x) 6M f(y);
2. f(0L) = 0M and f(1L) = 1M .
Definition 2.5 Davey and Priestley [2002] Let (L,∧L,∨L, 0L, 1L) and (M,∧M ,∨M , 0M , 1M) be bounded lattices. A mapping f : L −→ M is said to be a lattice
alg-homomorphism if, for all x, y ∈ L, we have
1. f(x ∧L y) = f(x) ∧M f(y);
2. f(x ∨L y) = f(x) ∨M f(y);
3. f(0L) = 0M and f(1L) = 1M .
Definition 2.6 Hungerford [2000] A given lattice homomorphism f on L is
called:
1. A monomorphism if it is injective;
2. An epimorphism if f is surjective;
3. An isomorphism when f is bijective. An automorphism is an isomorphism
from a lattice to itself.
Proposition 2.1 Every alg-homomorphism is an ord-homomorphism.
Proof: Let f : L −→ M be an alg-homomorphism. Since x 6L y if only if
x∧L y = x, therefore f(x) = f(x∧L y) = f(x)∧M f(y) and hence f(x) 6M f(y).
�
12
However, in general, the reciprocal of Proposition 2.1 does not hold. If f :
L −→ M is an ord-homomorphism, since x ∧L y 6L x and x ∧L y 6L y, so
f(x∧L y) 6M f(x) and f(x∧L y) 6M f(y). Thus, f(x∧L y) 6M f(x)∧M f(y) =
inf{f(x), f(y)}, however it is possible for f(x ∧L y) 6= inf{f(x), f(y)} to occur.
For example, consider the lattices L and M , as depicted in Hasse diagram shown
in Figure 2.1. Nevertheless, the map f : L −→ M defined by f(0L) = 0M ,
f(1L) = 1M , f(x) = u and f(y) = v, preserves infimum and supremum elements
and, hence, is an ord-homomorphism, though it is not an alg-homomorphism as
∧ operation is not preserved.
L◦
◦ ◦
◦
1L
x y
0L@@@
���
���
@@@
M◦
◦ ◦
◦
◦
1M
u v
w
0M
@@@
���
���
@@@
Figure 2.1: Hasse diagrams of lattices L and M
In other words the Proposition 2.1 says that every alg-homomorphism is order-
preserving. Due to this fact we have chosen to use the algebraic approach to
lattice homomorphisms (to see the other way to define lattice homomorphisms
we recommend Birkhoff [1973]; Davey and Priestley [2002]). From now on, alg-
homomorphisms will be called just homomorphisms for simplicity.
2.1.2 Automorphisms and Conjugates
Proposition 2.2 Let L be a bounded lattice. Then a function f : L −→ L is an
automorphism if and only if
1. f is bijective and
2. x 6L y if and only if f(x) 6L f(y).
13
2. PRELIMINARIES
Proof: If f is an automorphism, from Proposition 2.1 we just need to prove that
x 6L y when f(x) 6L f(y). Suppose that f(x) 6L f(y), then f(x)∧Lf(y) = f(x)
and so, as f is a lattice homomorphism, f(x ∧L y) = f(x). Thus, since f is
bijective, x ∧L y = x and therefore x 6L y.
Conversely, suppose that either 1. or 2. is not satisfied. If 1. is violated,
i.e., if f is not bijective, then it is not an automorphism by definition. If 2. is
violated, then we have two possible cases:
(i) either there are x, y ∈ L such that x 6L y but not f(x) 6L f(y). However,
then f(x)∧L f(y) <L f(x) = f(x∧L y), and thus f is not a homomorphism, and
hence neither an automorphism;
(ii) or there are x, y ∈ L such that f(x) 6L f(y) but not x 6L y. Put z = x∧L y.
Then z <L x, and f(x ∧L y) = f(z) 6= f(x) = f(x) ∧L f(y) due to the fact that
f is bijective; again this mean that f is not a homomorphism, and hence not an
automorphism.
�
Remark 2.3 We denote the set of all automorphisms on a bounded lattice L by
Aut(L) . This set endowed with the composition operation is a group that has
as neutral element the identity function idL. In algebra, an important tool is
the action of the groups on sets Burris and Sankappanavar [2005]; Hungerford
[2000]. In our case, the action of automorphism group transforms lattice functions
in other lattice functions.
Definition 2.7 Bustince et al. [2003] Let L be a bounded lattice and Ln = L ×· · · × L be a cartesian product of L n-times. Given a function f : Ln → L, the
action of an automorphism ρ over f results in the function fρ : Ln → L defined
as
fρ(x1, . . . , xn) = ρ−1(f(ρ(x1), . . . , ρ(xn))) (2.2)
In this case, fρ is called a conjugate of f .
Notice that if f : Ln → L is a conjugate of g : Ln → L and g is a conjugate
of h : Ln → L then f is a conjugate of h (automorphisms are closed under
14
composition) and if f is a conjugate of g, then g is also a conjugate of f since the
inverse of an automorphism is also an automorphism. Thus, an automorphism
action on set of n-ary functions on L (LLn) determines an equivalence relation on
LLn.
Let f : Ln → L be a conjugate of g : Ln → L. If f(x1, . . . , xn) 6Lg(x1, . . . , xn) for every (x1, . . . , xn) ∈ Ln then we denote it by f 6 g.
2.1.3 Retractions and Sublattices
Recall that the classical notion of sublattice is given as following.
Definition 2.8 Birkhoff [1973] An ordinary sublattice of a lattice L is a subset
M of L such that x, y ∈M imply x ∧L y ∈M and x ∨L y ∈M .
Nevertheless, we would like to work in a more flexible framework of sublattice
where M does not need to be a subset of L. Focusing on this idea, we have the
interest to define a generalized notion of sublattice using retractions.
Definition 2.9 Burris and Sankappanavar [2005] A homomorphism r of a lattice
L onto a lattice M is said to be a retraction if there exists a homomorphism s of
M into L which satisfies r ◦ s = idM . A lattice M is called a retract of a lattice
L if there is a retraction r, of L onto M .
Remark 2.4 In the literature, homomorphism s presented in Definition 2.9 has
no specific name. But here, this function play an important hole in our studies
and by this reason we give to it a special name, viz. a pseudo-inverse of retraction
r.
Notice that, if a lattice M is a retract of lattice L, then we have an identifica-
tion of M with a subset K = s(M) of L in which it is carried on some properties
of M to K, including its lattice structure via retraction r (see Figure 2.2). In
this case, K works as a algebraic copy of M (i.e. they are isomorphic) embedded
into L since r is an isomorphism when restricted to K.
Definition 2.10 Let L and M be arbitrary bounded lattices. We say that M is
a (r, s)-sublattice of L if M is a retract of L (i.e. M is a sublattice of L up to
15
2. PRELIMINARIES
Figure 2.2: The relaxed idea of sublattice
isomorphisms). In other words, M is a (r, s)-sublattice of L if there is a retraction
r of L onto M with pseudo-inverse s : M → L.
Proposition 2.3 The following statements hold:
1. M is a (r, s)-sublattice of L if and only if for all x, y ∈ M we have that
r(s(x) ∧L s(y)) ∈M and r(s(x) ∨L s(y)) ∈M ;
2. If M is an ordinary sublattice of L (in the sense of Definition 2.8) then
M is a (r, s)-sublattice of L where s is the identity function and r(x) =
sup{z ∈ L | z 6 x};
3. If M is a (r, s)-sublattice of L then s(M) equipped with restriction of oper-
ation of L is an ordinary sublattice of L.
Proof: Straight.
�
Based on the above proposition we can list at least four main advantages in
working with (r, s)-sublattice notion instead of the classical one:
• Generalize the ordinary notion: Every sublattice in the classical sense
(Definition 2.8) is also a (r, s)-sublattice (Definition 2.10). Indeed, it is
enough to consider s as the inclusion of M into L and as r the mapping
which keeps unchanged M , sends x to supM if the maximum of x and
supM is equal to x, sends x to inf M if the minimum of x and inf M is x,
and send x to x in any other case;
16
• Algebraic invariance: Since M is isomorphic with a subset s(M) of L
every property invariant under homomorphisms that holds in M works on
s(M) as well;
• Flexible identification: In Definition 2.10, a pseudo-inverse s of a re-
traction r can not be unique. This means that if there exist more than
one pseudo-inverse for the same retraction it is possible to identify M with
a subset of L in different ways1 what give us the possibility to chose the
best one for our proposes. But it must be clear that when we say that M
is a (r, s)-sublattice of L we are considering the existence of at least one
pseudo-inverse s and fixing it. However, no matter which pseudo-inverse is
taken, every result presented here remains working;
• Subclasses of sublattices: Considering (r, s)-sublattice notion allows us
to define some subclasses of this concept (see Definition 2.11) what it is not
possible for ordinary sublattices.
Remark 2.5 Throughout this thesis, it is used the concept of (r, s)-sublattice as
in Definition 2.10 instead of retract. Whenever the usual definition of sublattice
is used and it is not clear from the context, this sublattice will be called ordinary
sublattice.
Definition 2.11 Every retraction r : L −→ M (with pseudo-inverse s) which
satisfies s ◦ r 6 idL2 (idL 6 s ◦ r) is called a lower (an upper) retraction . In this
case, M is a lower (an upper) retract of L.
Example 2.3 One can easily see in Figure 2.3 that M is a lower retract (but it
is not an upper retract) of L1, M is an upper retract (but is not a lower retract)
of L2 and M is a retract (but is neither an upper nor a lower retract) of L3. In
fact, the unique possible retractions are ri : Li −→ M with i ∈ {1, 2, 3} defined
as in the Table 2.1. Their pseudo-inverses si : M → Li with i ∈ {1, 2, 3} are
respectively given as in the Table 2.2.
1For each pair (r, s) we have a different identification between M and its image by s.2If f and g are functions on a lattice L it is said that f 6 g if and only if f(x) 6L g(x) for
all x ∈ L.
17
2. PRELIMINARIES
M
◦
◦ ◦
◦
d
b c
a@@@
���
���
@@@
L1
◦
◦ ◦
◦
◦
5
3 4
2
1
@@@
���
���
@@@
L2
◦
◦
◦ ◦
◦
6
5
3 4
2@
@@
���
���
@@@
L3
◦
◦
◦ ◦
◦
◦
6
5
3 4
2
1
@@@
���
���
@@@
Figure 2.3: Hasse diagrams of lattices M , L1, L2 and L3
x r1
1 a2 a3 b4 c5 d
x r2
2 a3 b4 c5 d6 d
x r3
1 a2 a3 b4 c5 d6 d
Table 2.1: Tables of retractions r1, r2 and r3.
Example 2.4 Let M and L be bounded lattices as shown in Figure 2.4. A map-
ping r : L −→ M given by r(x) = sup{z ∈ M | s(z) 6L x} is a lower retraction
whose pseudo-inverse is a mapping s : M −→ L defined by s(1M) = 1L, s(a) = v,
s(b) = x, s(c) = y, s(d) = z and s(0M) = 0L. Therefore, it follows that M is a
(r, s)-sublattice of L in the sense of Definition 2.10.
Remark 2.6 Note that, given a lower retraction it is possible sometimes to define
an upper retraction with the same pseudo-inverse. For instance, let L and M be
x s1 s2 s3
a 1 2 1b 3 3 3c 4 4 4d 5 6 6
Table 2.2: Tables of pseudo-inverses s1, s2 and s3.
18
M◦
◦
◦ ◦
◦
◦
1
a
b c
d
0
@@@
���
���
@@@
L◦
◦
1
t
u◦◦
◦ ◦
◦
◦
v
x y
z
0
@@@
���
���
@@@
@@@
���
���
Figure 2.4: Hasse diagrams of lattices M and L
lattices as shown in Figure 2.4. If r is a lower retraction with pseudo-inverse
s as defined in the Example 2.4, then the function r′ given by r′(x) = inf{z ∈M | s(z) >L x} is an upper retraction since idL 6 s ◦ r′. It is easy to check that
s is also a pseudo-inverse of r′.
It is worth noting that if M is a (r, s)-sublattice of L then there is a retraction
r from L onto M , but it is not required to r to be a lower or an upper retrac-
tion. Nevertheless, as shown in the above remark, there may be more than one
retraction from L onto M with the same pseudo-inverse. This is a very useful
particularity of Definition 2.10 and we would like to highlight it in a definition.
Definition 2.12 Let M be a (r1, s)-sublattice of L. We say that
1. M is a lower (r1, s)-sublattice of L if r1 is a lower retraction. Notation:
M < L with respect to (r1, s);
2. M is an upper (r1, s)-sublattice of L whenever r1 is an upper retraction.
Notation: M > L with respect to (r1, s);
3. If r1 is a lower retraction and there is an upper retraction r2 : L −→M such
that its pseudo-inverse is also s, then M is called a full (r1, r2, s)-sublattice
of L. Notation: M E L with respect to (r1, r2, s).
Remark 2.7 Let L be a complete lattice. We denote the case when M is a
complete and lower (respectively upper) (r, s)-sublattice of L by M lL (by M mL) with respect to (r, s).
19
2. PRELIMINARIES
An immediate consequence of the definition of lower (upper) retraction is that,
if M E L then it follows that s ◦ r1 6 idL 6 s ◦ r2.
Proposition 2.4 Let K, M and L be bounded lattices. If KEMEL then KEL.
Proof: We shall prove that there exists a lower and an upper retraction r and
r′ from L onto K, respectively, both with a pseudo-inverse s such that s ◦ r 6idL 6 s ◦ r′.
Supposing K E M with respect to (r1, r2, s1) and M E L with respect to
(r3, r4, s2), it follows that
r1 ◦ s1 = r2 ◦ s1 = idK and s1 ◦ r1 6 idM 6 s1 ◦ r2
and
r3 ◦ s2 = r4 ◦ s2 = idM and s2 ◦ r3 6 idL 6 s2 ◦ r4
Thus, letting r = r1 ◦ r3, r′ = r2 ◦ r4 and s = s2 ◦ s1 then, for all x ∈ K, we have
r ◦ s(x) = r1(r3(s2(s1(x)))) = r1(s1(x)) = x = idK(x)
r′ ◦ s(x) = r2(r4(s2(s1(x)))) = r2(s1(x)) = x = idK(x)
and hence
s ◦ r = s2 ◦ (s1 ◦ r1) ◦ r3 = s2 ◦ r3 6 idL 6 s2 ◦ r4 6 s2 ◦ (s1 ◦ r2) ◦ r4 = s ◦ r′
Since r, r′ and s are homomorphisms (composition of homomorphisms is also a
homomorphism) then K E L with respect to (r, r′, s).
�
Proposition 2.5 Let M lL with respect to (r, s). For every nonempty set A ⊆M it holds that s(supA) = sup s(A). In other words, a pseudo-inverse of a lower
retract preserves supremum element.
Proof: Recall that s(A) = {s(t) ∈ L | t ∈ A}. Putting a = supA and
a′ = sup s(A) we shall prove that s(a) = a′.
20
On one hand, for all t ∈ s(A) there is a k ∈ A such that t = s(k). Since
k 6M a for each k ∈ A, then t = s(k) 6L s(a) by monotonicity of s, i.e. s(a) is
an upper bound of s(A) and hence a′ 6L s(a) since a′ = sup s(A).
On the other hand, if we take an arbitrary element k ∈ A it follows that
s(k) 6L a′ that implies k = r(s(k)) 6M r(a′) for all k ∈ A which means that
r(a′) is an upper bound of A. Thus by properties of supremum we have that
a 6M r(a′) and hence s(a) 6L s(r(a′)) 6L a′ since r is a lower retraction.
�
Analogously, we can prove the following.
Proposition 2.6 Let M mL with respect to (r, s). For every nonempty set A ⊆M it holds that s(inf A) = inf s(A), i.e. a pseudo-inverse of an upper retract
preserves infimum element.
Remark 2.8 It is important to point out that in Proposition 2.5 (Proposition
2.6) the hypothesis that r should be a lower retraction (an upper retraction) is
just used to prove inequality s(supA) 6L sup s(A) (inf s(A) 6L s(inf A)). It
means that sup s(A) 6L s(supA) (s(inf A) 6L inf s(A)) always holds no matter
which kind of retraction r is (i.e., lower, upper or neither).
2.1.4 Pseudo-quasi-metrics and Continuity
Intuitively, a distance among two values is a non-negative real value and so a
distance function on a set X is a function d : X × X −→ R+, where R+ is the
set of non-negative real numbers. In mathematics, several conditions have been
considered for distance functions, but the more used ones are those for metric
distances. However, in order to consider a reasonable general notion of distance,
we will consider the notion of pseudo-quasi-metric distance.
Definition 2.13 Kim [1968] Let X be a set. A function dX : X ×X −→ R+, is
a pseudo-quasi-metrics 1 if for each x, y ∈ X it satisfies the following conditions:
1In other papers, such as in Kasahara [1968], pseudo-quasi-metrics are called pre-metrics .But, the name of pre-metric have been used for a different notion of distance such as in Yaacovet al. [2011].
21
2. PRELIMINARIES
1. dX(x, x) = 0 and
2. dX(x, z) 6 dX(x, y) + dX(y, z)
In mathematics, an important requirement for functions is its continuity,
which intuitively means that small changes in the input result in small changes
in the output of function. Therefore, it is necessary to consider some way of
measuring changes, which can be made by using some distance notion.
Definition 2.14 Smyth [1992] Let dX and dY pseudo-quasi-metrics on sets X
and Y , respectively. A function f : X −→ Y is continuous w.r.t. dX and dY , or
just (dX , dY )-continuous , if for each x ∈ X and ε > 0 there exists δ > 0 such
that for any y ∈ X, if dX(x, y) 6 δ then dY (f(x), f(y)) 6 ε.
Proposition 2.7 Let X be a set and d be a pseudo-quasi-metric on X. Then
d2X : X ×X −→ R+ defined by d2X((x1, x2), (y1, y2)) =√dX(x1, y1)2 + dX(x2, y2)2
is also a pseudo-quasi-metric.
Proof: Straightforward.
�
Proposition 2.8 Let dX and dY be pseudo-quasi-metrics on the sets X and Y ,
respectively. If f : X −→ Y is (dX , dY )-continuous then f 2 : X −→ Y ×Y defined
by f 2(x) = (f(x), f(x)) is (dX , d2Y )-continuous.
Proof: Straightforward.
�
Proposition 2.9 Let dX , dY and dZ be pseudo-quasi-metrics on X, Y and Z, re-
spectively. If g : X −→ Y and f : Y −→ Z are (dX , dY ) and (dY , dZ)-continuous
then f ◦ g is (dX , dZ)-continuous.
Proof: Straightforward.
�
22
2.2 Fuzzy Connectives
Typically fuzzy logic considers for membership degrees values the unit interval
[0,1]. This logic is very rich due to its connectives (t-norm, t-conorm, fuzzy
negation and implication) can be defined in various different ways and have many
different properties since the unique consensus about how to define them is that
truth functions of these connectives have to behave classically on extremal values
0 and 1. For instance, there are several paper in the literature presenting different
definitions for fuzzy implications and variances of it.
Modern fuzzy logic consider lattices to range its degrees of truth for having
a much more general framework. In this section, we discuss about these fuzzy
connectives on bounded lattices taking into account definitions given in Baczynski
and Jayaram [2008]; Bedregal et al. [2006b, 2013]. Moreover, we discuss about
De Morgan triples and t-subnorms.
2.2.1 T-norms and T-conorms
It presented here a short formalization for the notion of t-norm and t-conorm on
bounded lattices. Moreover, some results are demonstrated as well.
Definition 2.15 Bedregal et al. [2006b] Let L be a bounded lattice. A binary
operation T : L× L −→ L is a t-norm if, for all x, y, z ∈ L, it satisfies:
(T1) T (x, y) = T (y, x) (commutativity);
(T2) T (x, T (y, z)) = T (T (x, y), z) (associativity);
(T3) If x 6L y then T (x, z) 6L T (y, z), ∀ z ∈ L (monotonicity);
(T4) T (x, 1L) = x (boundary condition).
Definition 2.16 Klement and Mesiar [2005] Let L be a bounded lattice and dL
a pseudo-quasi-metric on L. A t-norm T on L is called
1. Archimedean if for any x, y ∈ L\{0L, 1L}, there exists n ∈ N such that
T n(x) 6L y, where for any m ∈ N
T 0(x) = 1L and Tm+1(x) = T (Tm(x), x); (2.3)
23
2. PRELIMINARIES
2. dL-nilpotent if it is (d2L, dL)-continuous and each x ∈ L\{0L, 1L} is a nilpo-
tent element of T , i.e. there exists n ∈ N such that T n(x) = 0L;
3. idempotent if for each x ∈ L, T (x, x) = x; and
4. positive if T (x, y) = 0L if and only if either x = 0L or y = 0L;
5. An element a ∈ L is called a zero divisor if there exists a b ∈ L such that
T (a, b) = 0L.
Example 2.5 Let L be a bounded lattice. Thus, the function T : L × L → L
defined by T (x, y) = x ∧L y is a t-norm that generalize classical fuzzy t-norm of
minimum, i.e. TM : [0, 1] × [0, 1] → [0, 1] such that TM(x, y) = min{x, y} for all
x, y ∈ [0, 1].
Dually, it is possible to define the concept of t-conorms.
Definition 2.17 Bedregal et al. [2006b] Let L be a bounded lattice. A binary
operation S : L× L −→ L is said be a t-conorm if, for all x, y, z ∈ L, we have:
(S1) S(x, y) = S(y, x) (commutativity);
(S2) S(x, S(y, z)) = S(S(x, y), z) (associativity);
(S3) If x 6 y then S(x, z) 6 S(y, z), ∀ z ∈ L (monotonicity);
(S4) S(x, 0L) = x (boundary condition).
Remark 2.9 Notice that T (x, y) 6L x (or T (x, y) 6L y) and x 6L S(x, y) (or
y 6L S(x, y)) for all x, y ∈ L. In fact, T (x, y) 6L x∧y 6L x and x 6L x∨L y 6LS(x, y).
Example 2.6 Given an arbitrary bounded lattice L, the function S given by
S(x, y) = x ∨L y for all x, y ∈ L is a t-conorm on L that generalize the clas-
sical fuzzy t-conorm of maximum, i.e. SM(x, y) = max{x, y} for all x, y ∈ [0, 1].
Proposition 2.10 Let ρ be an automorphism on L. A t-conorm S : L×L −→ L
satisfies
S(x, y) = 1L if and only if x = 1L or y = 1L (2.4)
if and only if Sρ satisfies also it. A t-conorm satisfying (2.4) is called positive.
24
Proof: We have that Sρ(x, y) = 1L if and only if ρ−1(S(ρ(x), ρ(y))) = 1L if
and only if S(ρ(x), ρ(y)) = 1L if and only if ρ(x) = 1L or ρ(y) = 1L (by (2.4)) if
and only if x = 1L or y = 1L.
�
Similarly, it can be proved the following.
Proposition 2.11 Let ρ be an automorphism on L. A t-norm T : L× L −→ L
is positive if and only if T ρ satisfies also it.
2.2.2 Fuzzy Negations
A natural extension of notion of fuzzy negations can be done by considering
arbitrary bounded lattices as possible sets of truth values.
Definition 2.18 Bedregal et al. [2013] A mapping N : L→ L is a fuzzy negation
on L if the following properties are satisfied for each x, y ∈ L:
(N1) N(0L) = 1L and N(1L) = 0L and
(N2) If x 6L y then N(y) 6L N(x).
Moreover, a fuzzy negation N is strong if it also satisfies the involution property,
i.e.
(N3) N(N(x)) = x for each x ∈ L.
N is strict if satisfies the property:
(N4) N(x) <L N(y) whenever y <L x.
and it is called frontier if it satisfies property:
(N5) N(x) ∈ {0L, 1L} if and only if x = 0L or x = 1L.
Example 2.7 Bedregal et al. [2013] If L is an arbitrary bounded lattice, then the
functions N⊥, N> : L→ L defined by
N⊥(x) =
{1L, if x = 0L;
0L, otherwise.
and
N>(x) =
{0L, if x = 1L;
1L, otherwise.
25
2. PRELIMINARIES
for each x ∈ L are fuzzy negations on L.
Remark 2.10 An element e ∈ L is an equilibrium point of a fuzzy negation N
on L if N(e) = e. But differently from usual case and interval-valued case (see
Bedregal [2010]), strong fuzzy negations on lattices can not have an equilibrium
point.
Example 2.8 Let L and M be bounded lattices as shown in the Figure 2.5. The
function N1 : M → M defined by N1(0M) = 1M , N1(x) = y, N1(y) = x and
N1(1M) = 0M is a strong M-negation. Nevertheless, N1 has no equilibrium point.
Now, consider a function N2 : L→ L given by N2(0L) = 1L, N2(a) = e, N2(e) =
a, N2(1L) = 0L and N2(u) = u for each u ∈ {b, c, d}. In this case, N2 is a strong
fuzzy negation with three equilibrium points, namely b, c and d.
M
◦
◦ ◦
◦
1M
x y
0M@@@
���
���
@@@
L◦
◦
◦ ◦ ◦
◦
◦
1L
e
b c d
a
0L
@@@
���
���
@@@
Figure 2.5: Hasse diagrams of lattices M and L
Proposition 2.12 Let N be a strong fuzzy negation on L. Then
1. N is strict;
2. If N(x) 6L N(y) then y 6L x; and
3. N is bijective.
Proof: If y <L x then by (N2), N(x) 6L N(y). Supposing N(x) = N(y) then
N(N(x)) = N(N(y)) and so x = y, which is a contradiction with the premise.
Therefore, N(x) <L N(y).
If N(x) 6L N(y) then, by (N2), N(N(y)) 6L N(N(x)) and so y 6L x.
26
Since N is strict then N is trivially injective. Moreover, for any y ∈ L we
have N(N(y)) = y and hence N also is surjective. Thus, N is bijective.
�
From this proposition it follows that for each strong fuzzy negation we have
that x ‖ y if and only if N(x) ‖ N(y). Notice that although every lattice admits
a negation N , it is not true that every lattice admits an involutive negation.
Example 2.9 Let L be any lattice such that there exists x0 ∈ L with x0 6= 0L, 1L.
Then the mapping:
N(x) =
0L if x = 1L;
1L if x = 0L;
x0 otherwise.
is a frontier negation. Notice that this example proves that for every lattice L
with at least three elements it is possible to define a frontier negation.
Example 2.10 Consider lattice L0 obtained from lattice L in Figure 2.5 by omit-
ting the point a. Then, it does not exist a strong negation for this lattice. Indeed, if
N is such negation, then we should have that 0L < N(e) < N(b), N(c), N(d) < 1L
which is not possible due to the injectivity of a strong negation. It is important
to point out that the fact we can not define a strong negation over a lattices in
general, as shown in this example, is not due to the partial order of the lattice.
It is easy to see that it is not possible to define a strong negation on the linear
ordered lattice L = {0, 1}∪ [2, 3] as well, considering the usual linear order of real
numbers 6 on L.
Proposition 2.13 Let N : L → L be a function, ρ be an automorphism on L
and i ∈ {1, 2, 3, 4, 5}. N satisfies (Ni) if and only if Nρ satisfy (Ni). Moreover,
e is an equilibrium point of N if and only if ρ−1(e) is an equilibrium point of Nρ.
Proof: Suppose N satisfies (Ni) with i ∈ {1, 2, 3, 4}, then
(N1) Nρ(0L) = ρ−1(N(ρ(0L))) = ρ−1(N(0L)) = ρ−1(1L) = 1L. Analogously it
can be proved that Nρ(1L) = 0L;
(N2) If x 6L y then ρ(x) 6L ρ(y) and hence N(ρ(y)) 6L N(ρ(x)). Therefore, by
27
2. PRELIMINARIES
isotony of ρ−1, ρ−1(N(ρ(y)) 6L ρ−1(N(ρ(x)));
(N3) Nρ(Nρ(x)) = ρ−1(N(ρ(ρ−1(N(ρ(x)))))) = ρ−1(N(N(ρ(x)))) = ρ−1(ρ(x)) =
x;
(N4) Straight;
(N5) If Nρ(x) ∈ {0L, 1L} then, by Equation (2.2) and considering that ρ−1(0L) =
0L and ρ−1(1L) = 1L we have N(ρ(x)) ∈ {0L, 1L}. Thus, by (N5), ρ(x) ∈ {0L, 1L}which implies that x = 0L or x = 1L;
If N(e) = e then N(ρ(ρ−1(e))) = e and hence Nρ(ρ−1(e)) = ρ−1(N(ρ(ρ−1(e)))) =
ρ−1(e).
Reciprocal is straightforward from the previous item and the fact that for any
function f : L→ L, (fρ)ρ−1
= f .
�
Corollary 2.1 Let N : L → L be a function and ρ be an automorphism on L.
N is a (strong, frontier) fuzzy negation if and only if Nρ is a (strong, frontier)
fuzzy negation.
Proof: Straightforward from Proposition 2.13.
�
Klement et al. [2000]; Klir and Yuan [1995] observed that it is possible to
obtain, in a canonical way, a fuzzy negation NT from a t-norm T . This negation
is called natural negation of T or negation induced by T . In the most general
case, where we consider a t-norm on a bounded lattice L, it is not always possible
to obtain a natural negation, because the construction of NT is based in the
supremum of, possibly, infinite sets (this concept was generalized for lattices by
Bedregal et al. [2012a]).
Proposition 2.14 Let L be a complete lattice and T be a t-norm on L. Then
the function NT : L→ L defined by
NT (x) = sup{z ∈ L | T (x, z) = 0L} (2.5)
is an fuzzy negation .
28
Proof: According to Definition 2.18 we shall prove that NT satisfies (N1) and
(N2). Hence
(N1) NT (1L) = sup{z ∈ L | T (1L, z) = 0L} = sup{0L} = 0L and NT (0L) =
sup{z ∈ L | T (0L, z) = 1L} = supL = 1L;
(N2) If x 6L y then for any z ∈ L, T (x, z) 6L T (y, z) and therefore, if T (y, z) =
0L then T (x, z) = 0L. So, {z ∈ L | T (y, z) = 0L} ⊆ {z ∈ L | T (x, z) = 0L}.Hence, NT (y) = sup{z ∈ L | T (y, z) = 0L} 6L sup{z ∈ L | T (x, z) = 0L} =
NT (x).
�
Theorem 2.1 Let T be a t-norm on L. If T is positive then NT = N⊥.
Proof: If x 6= 0L and z ∈ L then, by (T4), T (x, z) = 0L if and only if z = 0L.
So, by Equation (2.5), NT (x) = sup{0L} = 0L. Therefore, NT = N⊥.
�
Theorem 2.2 Let T be a t-norm on L. If NT is a frontier negation then each
x ∈ L\{0L} is a zero divisor of T .
Proof: If x 6= 1L, then, as NT is frontier, NT (x) 6= 0L and so sup{z ∈L | T (x, z) = 0L} 6= 0L, that is, {z ∈ L | T (x, z) = 0L} 6= {0L}. Thus, since
T (x, 0L) = 0L, {0L} ⊂ {z ∈ L | T (x, z) = 0L}. Therefore, there exists z ∈ L\{0L}such that T (x, z) = 0L. Hence, x is a zero divisor of T .
�
Theorem 2.3 Let T be a t-norm on L and ρ be an automorphism on L. Then
NρT = NT ρ.
Proof: Let x ∈ L, then
NρT (x) = ρ−1(NT (ρ(x)))
= ρ−1(sup{z∈L | T (ρ(x), z)=0L})= ρ−1(sup{z∈L | T ρ(x, ρ−1(z))=0L})= sup{ρ−1(z) ∈ L | T ρ(x, ρ−1(z))=0L} (∗)= sup{z∈L | T ρ(x, z) = 0L}= NT ρ(x)
29
2. PRELIMINARIES
(∗) Notice that every monotone homomorphism preserves supremum element.
�
It is also possible to define this kind of negation from a given t-conorm S on
a complete bounded lattice L, as one can see in the following proposition.
Proposition 2.15 Let L be a complete lattice and S be a t-conorm on L. Then
the function NS : L→ L defined by
NS(x) = inf{z ∈ L | S(x, z) = 1L} (2.6)
is an L-negation called natural negation of S .
Proof:
(N1)
NS(1L) = inf{z ∈ L | S(1L, z) = 1L} = inf L = 0L and NS(0L) = inf{z ∈L | S(0L, z) = 1L} = inf{1L} = 1L.
(N2)
If x 6L y then for any z ∈ L we have that S(x, z) 6L S(y, z) and hence
if S(x, z) = 1L then S(y, z) = 1L. Thus, NS(y) = inf{z ∈ L | S(y, z) =
1L} 6L inf{z ∈ L | S(x, z) = 1L} = NS(x).
�
Proposition 2.16 Bedregal et al. [2013] Let S be a t-conorm on a complete
bounded lattice L. If S is positive then NS = NL>
Proposition 2.17 Bedregal et al. [2013] Let S be a t-conorm on L and ρ be an
automorphism on L. Then (NS)ρ = NSρ.
30
2.2.3 Fuzzy Implications
In the literature several notions for fuzzy implications have been considered
(see for example Baczynski [2004]; Bustince et al. [2003]; Fodor and Roubens
[1994]; Mas et al. [2007]; Yager [1983]). But, the unique consensus is that a fuzzy
implication should have the same behavior than classical implication when the
crisp case is considered Fodor and Roubens [1994], i.e. for values 0 and 1. Here
we consider the notion given by Bedregal et al. [2013].
Definition 2.19 Bedregal et al. [2013] Let L be a bounded lattice. A function
I : L2 → L is called a fuzzy implication if it satisfies the following properties:
(FPA) I(y, z) 6L I(x, z) whenever x 6L y (First place antitonicity);
(SPI) I(x, y) 6L I(x, z) whenever y 6L z (Second place isotonicity);
(CC1) I(0L, 0L) = 1L (Corner condition 1);
(CC2) I(1L, 1L) = 1L(Corner condition 2);
(CC3) I(1L, 0L) = 0L (Corner condition 3)
The set of implications on L will be denoted by IL.
Example 2.11 Let L be a bounded lattice. Thus, functions I⊥, I> : L × L → L
given by
I⊥(x, y) =
{1L, if x = 0L or y = 1L;
0L, otherwise.
and
I>(x, y) =
{0L, if x = 1L and y = 0L;
1L, otherwise.
for all x, y ∈ L are fuzzy implications.
For all x, y, z ∈ L, we define the following properties of I:
(WFPA) if x 6L y, x ¨ z and y ¨ z, then I(y, z) 6L I(x, z) (weak first place
antitonicity);
(WSPI) if y 6L z, x ¨ y and x ¨ z, then I(x, y) 6L I(x, z) (weak second place
isotonicity);
(RB) I(x, 1L) = 1L (right boundary conditions);
31
2. PRELIMINARIES
(LB) I(0L, y) = 1L (left boundary condition);
(CC4) I(0L, 1L) = 1L(corner condition 4);
(NP) I(1L, y) = y for each y ∈ L (left neutrality principle);
(EP) I(x, I(y, z)) = I(y, I(x, z)) for all x, y, z ∈ L (exchange principle);
(IP) I(x, x) = 1L for each x ∈ L (identity principle);
(OP) I(x, y) = 1L if and only if x 6L y (ordering property);
(LOP) if x 6L y then I(x, y) = 1L (left ordering property);
(WLOP) if x 6L y or x ‖L y then I(x, y) = 1L (weak left ordering property);
(IBL) I(x, I(x, y)) = I(x, y) for all x, y, z ∈ L (iterative Boolean law);
(CP) I(x, y) = I(N(y), N(x)) being N a strong L-negation (contrapositivity
property);
(L-CP) I(N(x), y) = I(N(y), x) (left contraposition law);
(R-CP) I(x,N(y)) = I(y,N(x)) (right contraposition law);
(P) I(x, y) = 0L if and only if x = 1L and y = 0L (positivety) .
It is easy to verify that (SPI) and (CC1) imply (LB). Moreover, (FPA) and
(CC2) imply (LB) and consequently (CC4).
Remark 2.11 Notice that if L is a totally ordered set then (WFPA), (WSPI)
and (WLOP) are equivalent to (FPA), (SPI) and (LOP), respectively.
Example 2.12 Let L be a bounded lattice (see Figure 2.5) and N2 the strong
fuzzy negation on L in Example 2.8. The function I : L2 → L given by
I(x, y) =
1L, if x 6L y;
N2(x), if y = 0L and x 6= 0L;
y, if x = 1L;
e, otherwise.
(2.7)
satisfies the properties (FPA), (OP ), (CP ) (with respect to N2) and (P ). It is a
easy to see that it holds by Table 2.3.
Lemma 2.1 If a function I : L2 → L satisfy (FPA) and (CP) for some strong
fuzzy negation N , then I also satisfies (SPI).
32
I 0L a b c d e 1L
0L 1L 1L 1L 1L 1L 1L 1La e 1L 1L 1L 1L 1L 1Lb b e 1L e e 1L 1Lc c e e 1L e 1L 1Ld d e e e 1L 1L 1Le a e e e e 1L 1L
1L 0L a b c d e 1L
Table 2.3: A function I on L
Proof: If y 6L z then N(z) 6L N(y) and so by (FPA), I(N(y), N(x)) 6LI(N(z), N(x)) for each x ∈ L. Therefore, by (CP), I(N(N(x)), N(N(y))) 6LI(N(N(x)), N(N(z))). Hence, I(x, y) 6L I(x, z) whereas N is strong.
�
Proposition 2.18 Let ρ be an automorphism on L, I : L×L→ L be a function
and P ∈ {(FPA), (SPI), (CC1), (CC2), (CC4), (LB), (RB)}. I satisfies P if
and only if Iρ also satisfies P .
Proof: See Proposition 10 in Bedregal et al. [2013].
�
A special type of fuzzy implication that we would like to study is the (S,N)-
implication, that is, an implication defined from a t-conorm S and a fuzzy nega-
tion N .
Definition 2.20 Baczynski and Jayaram [2008] Let S be a t-conorm on L and
N be a fuzzy negation on L. The function IS,N : L× L −→ L given by
IS,N(x, y) = S(N(x), y) (2.8)
for all x, y ∈ L is called a (S,N)-implication. If N is strong then I is called
a strong implication or S-implication. In this case, S and N are said to be
generators of I.
33
2. PRELIMINARIES
Proposition 2.19 Bedregal et al. [2013] Let I : L× L→ L be a function and ρ
an automorphism on L. Thus, I is an (S,N)-implication on L generated from S
and N if and only if Iρ is an (S,N)-implication on L generated from Sρ and Nρ.
In other words, (IS,N)ρ = ISρ,Nρ.
Another special type of implication we would like to consider in this work is
R-implication. Taking into account that there exists an isomorphism between
classical two-valued logic and classical set theory, it is possible to see that if K
and G are subsets of a set X then the identity
Kc ∪G = (K\G)c =⋃{P ⊆ X | K ∩ P ⊆ G}
holds, where Kc is the complement of set K (see Baczynski and Jayaram [2008]).
The R-implications (residual implications) are generalizations of the this iden-
tity in fuzzy logic.
Definition 2.21 Baczynski and Jayaram [2008] Let L be a complete bounded
lattice. A function I : L × L → L is called an R-implication if there exists a
t-norm T such that for all x, y ∈ L we have
I(x, y) = sup{t ∈ L | T (x, t) 6L y} (2.9)
We denote this implication generated from a t-norm T by IT .
To finish this section we present the notion of negations defined from implica-
tions. There exists a natural way to define this particular class of fuzzy negations
on [0, 1] based on the fact that a propositional formula p is logically equivalent
(in classic logic) to p →⊥ where ⊥ denotes the absurd (see Lemma 1.4.14, pg
18 in Baczynski [2004]). Bedregal et al. [2013] have generalized this concept for
bounded lattices as in the following proposition.
Proposition 2.20 Let L be a bounded lattice. If a function I : L × L → L
satisfies (FPA), (CC1) and (CC3) then the function NI : L→ L defined for each
x ∈ L by
NI(x) = I(x, 0L) (2.10)
is a fuzzy negation on L called the natural negation of I .
34
Next lemma provides necessary conditions for NI to be a fuzzy negation and
also establishes some properties.
Lemma 2.2 Bedregal et al. [2013] Let L be a bounded lattice. If a function
I : L× L→ L satisfies (EP) and (OP) then
1. NI is a fuzzy negation;
2. x 6L NI(NI(x)) for each x ∈ L;
3. NI ◦NI ◦NI = NI .
2.2.4 T-subnorms
As a generalization of the concept of t-norms, it was presented by Klement
et al. [2000] the notion of triangular subnorms as follows.
Definition 2.22 A function F : [0, 1]2 −→ [0, 1] that satisfies, for all x, y, z ∈[0, 1], the properties
1. F (x, y) = F (y, x)
2. F (x, F (y, z)) = F (F (x, y), z)
3. F (x, z) 6 F (y, z) whenever x 6 y
4. F (x, y) 6 min(x, y)
is called a t-subnorm.
It is clear that every t-norm is a t-subnorm. However, the reciprocal of this
affirmation is not true, in general. For instance, the function F : [0, 1]2 −→ [0, 1],
defined by F (x, y) = 0, is a t-subnorm, but not a t-norm. However, it is always
possible to construct a t-norm from a t-subnorm as is shown in the following
proposition.
35
2. PRELIMINARIES
Proposition 2.21 If F is a t-subnorm then the function T : [0, 1]2 −→ [0, 1],
given by
T (x, y) =
{F (x, y), if(x, y) ∈ [0, 1[2
min(x, y), otherwise
is a t-norm.
Proof: See Klement et al. [2000].
�
Naturally, the concept of t-subnorm may be generalized for bounded lattices.
Definition 2.23 Let L be a bounded lattice. A function F : L×L −→ L is called
a t-subnorm on L if it satisfies the following properties:
1. F (x, y) = F (y, x)
2. F (x, F (y, z)) = F (F (x, y), z)
3. F (x, z) 6L F (y, z) whenever x 6L y
4. F (x, y) 6L x ∧L y
Proposition 2.22 If F is a t-subnorm on bounded lattice L, then T defined by
T (x, y) =
{F (x, y), if(x, y) ∈ (L\{1L})2
x ∧L y, otherwise
is a t-norm on L.
Proof: Since F is a t-subnorm and x∧L y is a t-norm, then T is commutative,
associative and monotone. Thus, we shall only prove that T satisfies the boundary
condition, i.e., T (x, 1L) = x for each x ∈ L.
But, for each x ∈ L we have x = x ∧L 1L = T (x, 1L) since x 6 1L for all
x ∈ L. Therefore, T is a t-norm on L.
36
�
Note that it is possible to define a t-subconorm R dually from the definition of
t-subnorm just by replacing the property F (x, y) 6L x∧L y for x∨L y 6L R(x, y).
Precisely, we have
Definition 2.24 Let L be a bounded lattice. A function R : L×L −→ L is called
a t-subconorm on L if it satisfies the following properties
1. R(x, y) = R(y, x)
2. R(x,R(y, z)) = R(R(x, y), z)
3. R(x, z) 6L R(y, z) whenever x 6L y
4. x ∨L y 6L R(x, y)
Of course, a dual proof of Proposition 2.22 can be given to prove the following
proposition.
Proposition 2.23 If R is a t-subconorm on bounded lattice L, then S defined by
S(x, y) =
{R(x, y), if(x, y) ∈ (L\{0L})2
x ∨L y, otherwise
is a t-conorm on L.
2.2.5 De Morgan Triples
De Morgan’s laws represent, among other things, a way to relate disjunctive
and conjunctive operators via negations. Both in set theory and formal logics
these laws are very important tools to simplify equations, formal proofs and
other tasks. Specifically, these laws are given by
α ∧ β ≡ ¬(¬α ∨ ¬β) and α ∨ β ≡ ¬(¬α ∧ ¬β) (2.11)
37
2. PRELIMINARIES
or
¬(α ∧ β) ≡ ¬α ∨ ¬β and ¬(α ∨ β) ≡ ¬α ∧ ¬β (2.12)
In fuzzy logic, De Morgan’s laws can be naturally generalized using t-norms, t-
conorms and negations as operators. Nevertheless, due to the several possibilities
that exist to define these operators in fuzzy logic, De Morgan’s laws can be
generalized in different ways as seen in Calvo [1992]; da Costa et al. [2011]; Garcıa
and Valverde [1989]; Kolesarosa and Mesiar [2010]; Palmeira and Bedregal [2012].
Here, we choose a version that we believe is the most general one because it does
not impose any constraints for operators involved and it is more fateful with
Equations (2.11) and (2.12).
Definition 2.25 Calvo [1992] Let T be a t-norm, S a t-conorm and N a fuzzy
negation, all defined on the same bounded lattice L. We say that 〈T, S,N〉 is a
De Morgan triple if, for all x, y ∈ L, we have
1. N(T (x, y)) = S(N(x), N(y));
2. N(S(x, y)) = T (N(x), N(y)).
Remark 2.12 Naturally, every time we are talking about De Morgan triple 〈T, S,N〉its operators are considered defined on the same lattice. In order to highlight the
lattice concerned, we shall simply say that 〈T, S,N〉 is a De Morgan triple on L
when T , S and N are a t-norm, a t-conorm and a fuzzy negation all defined on
L.
It is important to point out that there are some fuzzy negations which are not
involutive and hence for some t-norms, t-conorms and fuzzy negations only one
of the items of Definition 2.25 holds true, as can be seen in the example below.
Example 2.13 Let L be the bounded lattice shown in Figure 2.4. It is easy to
check that the function N : L −→ L, given by setting N(0L) = 1L, N(z) = v,
N(x) = x, N(y) = N(u) = y, N(v) = N(t) = z and N(1L) = 0L, is a fuzzy
38
negation. Thus, considering the t-norm T and the t-conorm S defined by
T 0 z x y v u t 1
0 0 0 0 0 0 0 0 0
z 0 z z z z z z z
x 0 z x z x z x x
y 0 z z y y y y y
v 0 z x y v u v v
u 0 z z y u y y u
t 0 z x y v y v t
1 0 z x y v u t 1
S 0 z x y v u t 1
0 0 z x y v u t 1
z z z x y v 1 1 1
x x x x v v 1 1 1
y y y v y v 1 1 1
v v v v v v 1 1 1
u u 1 1 1 1 1 1 1
t t 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
then it follows that
N ◦ T 0 z x y v u t 1
0 1 1 1 1 1 1 1 1
z 1 v v v v v v v
x 1 v x v x v x x
y 1 v v y y y y y
v 1 v x y z y z z
u 1 v v y y y y y
t 1 v x y z y z z
1 1 v x y z y z 0
S ◦ (N ×N) 0L z x y v u t 1L
0 1 1 1 1 1 1 1 1
z 1 v v v v v v v
x 1 v x v x v x x
y 1 v v y y y y y
v 1 v x y z y z z
u 1 v v y y y y y
t 1 v x y z y z z
1 1 v x y z y z 0
Therefore N(T (a, b)) = S(N(a), N(b)) for all a, b ∈ L. However, on the other
hand, we see that N ◦ S 6= T ◦ (N ×N) as evidenced on the following tables:
N ◦ S 0 z x y v u t 1
0 1 v x y z y z 0
z v v x y z 0 0 0
x x x x z z 0 0 0
y y y z y z 0 0 0
v z z z z z 0 0 0
u y 0 0 0 0 0 0 0
t z 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0
T ◦ (N ×N) 0 z x y v u t 1
0 1 v x y z y z 0
z v v x y z y z 0
x x x x z z z z 0
y y y z y z y z 0
v z z z z z z z 0
u y y z z z y z 0
t z z z z z z z 0
1 0 0 0 0 0 0 0 0
39
2. PRELIMINARIES
Thus, taking into account such a possibility, we define a relaxed notion of De
Morgan triples.
Definition 2.26 Let T be a t-norm, S be a t-conorm and N be a fuzzy negation,
all of which are defined on bounded lattice L. A triple 〈T, S,N〉 is a De Morgan
T -semitriple (De Morgan S-semitriple) if, for all x, y ∈ L we have N(T (x, y)) =
S(N(x), N(y)) (N(S(x, y)) = T (N(x), N(y))).
It is worth noting that, in the case that N is a strong negation, the notions
of T -semitriple and S-semitriple coincide.
Notice also that it is possible to define De Morgan triples for t-subnorms,
t-subconorms and fuzzy negations as follows.
Definition 2.27 Let F be a t-subnorm on L, R be a t-subconorm on L and N
be a fuzzy negation. A triple 〈F,R,N〉 is called a De Morgan triple if, for all
x, y ∈ L, it satisfies
1. N(F (x, y)) = R(N(x), N(y));
2. N(R(x, y)) = F (N(x), N(y)).
Definition 2.28 Let F be a t-subnorm on L, R be a t-subconorm on L and
N be a fuzzy negation. A triple 〈F,R,N〉 is called a De Morgan F -semitriple
(R-semitriple) if, for all x, y ∈ L, it holds that N(F (x, y)) = R(N(x), N(y))
(N(R(x, y)) = F (N(x), N(y))).
40
Chapter 3
Extension Method via
Retractions
As we have seen in Chapter 1, the extension problem is a well investigated
subject in several different branches of knowledge. Here we are interested in
developing a general method to solve the problem of extending fuzzy operator
from a sublattice in order to preserve the largest number of its properties.
Our main goal in this chapter is presenting an extension method for fuzzy
connectives. Without loss of generality let’s introduce the extension problem for
t-norms. Consider an ordinary sublattice M of a bounded lattice L (i.e. M ⊆ L)
and T a t-norm on M . Since a t-norm is particularly a function it is natural to
think to extend T from M to L in order to obtain a new t-norm TE on L in
such way that this extension should preserve some properties of T by means TE
should have same properties of T . This problem is not trivial even when M is an
ordinary sublattice of L, as one can see in Example 3.1 below.
Example 3.1 Let M be an ordinary sublattice of L and TM a t-norm on M .
Define T : L× L −→ L such that
T (x, y) =
TM(x, y), if (x, y) ∈M ×M ;
x, if y = 1L;
y, if x = 1L;
0L, otherwise.
41
3. EXTENSION METHOD VIA RETRACTIONS
Thus, we have T |M×M = TM , i.e., T is an extension of TM . However, T is
not a t-norm as the order-preserving property of T fails (item (4) of Definition
2.15). This is due to the fact that we set T (x, y) = 0 for elements such that
(x, y) /∈M ×M , x 6= 1L and y 6= 1L.
Recent investigations points to a solution to this problem. In Saminger-Platz
et al. [2008] is shown a way to make this kind of extension of t-norms using the
ordinary definition of sublattice. Precisely,
WLTM (x, y) =
{x ∧L y, if 1L ∈ {x, y};
TM∪{0L,1L}(x∗, y∗), otherwise.(3.1)
where M is a (complete) ordinary sublattice of L, TM is a t-norm on M , x∗ =
supM{z | z 6L x, z ∈M ∪ {0L, 1L}} and
TM∪{0L,1L}(x, y) =
x ∧L y, if 1L ∈ {x, y};
0L, if 0L ∈ {x, y};TM(x, y), otherwise.
(3.2)
Notice that operator x∗ acts as an collapsing function flattening elements
belonging to L\M on M .
However, to have a more flexible environment where the collapsing function
can be defined in different ways, we would like to extend t-norms considering
our generalized notion of sublattice (see Definition 2.10). In fact, if M is a
(r, s)-sublattice of L (in this case, there are a retraction r : L −→ M and a
pseudo-inverse s : M −→ L such that r ◦ s = idM) and T a t-norm on M , how
can we define a function TE on L that works on s(M) as T works on M and that
is able to “carry on” to TE the largest possible number of properties that T has.
For instance, if T is Archimedean (see Definition 2.16) how to define a t-norm
TE such way that it is also Archimedean?
In this chapter it is presented a extension method via retractions that gener-
alize the Saminger-Platz’s extension method and its efficiency in preserving prop-
erties is demonstrated. Throughout this chapter L always represents a bounded
lattice.
42
3.1 T-norms
Notice that the idea behind the extension of t-norms proposed by Saminger-
Platz is to take the minimum values of those elements that do not belong to M .
We approach this problem in a similar way seeking to consider a more general
environment provided by the concept of (r, s)-sublattice.
Theorem 3.1 Let M < L with respect to (r, s). If T is a t-norm on M then,
the function TE : L× L −→ L defined by
TE(x, y) =
{x ∧L y, if 1L ∈ {x, y}
s(T (r(x), r(y))), otherwise.(3.3)
is a t-norm that extends T from M to L.
Proof: Firstly note that, if 0L ∈ {x, y} then TE(x, y) = 0L, so TE satisfies
the Equation (3.2). Moreover, it is clear that TE extends T from M to L, is
commutative and has neutral element 1L. Thus, it remains to prove that TE is
monotone and associative.
• Monotonicity
Let x, y, z ∈ L where x 6L y. In this case, supposing 1L /∈ {x, y, z},we have that r(x) 6M r(y) and hence TE(x, z) = s(T (r(x), r(z))) 6Ls(T (r(y), r(z))) = TE(y, z). Moreover, we have the following property:
(i) If x = 1L then y = 1L then TE(x, z) = TE(y, z).
(ii) Consider y = 1L and 1L /∈ {x, z}. Due to T (x, y) 6L y for all x, y ∈ Lthen
TE(x, z) = s(T (r(x), r(z))) 6L s(r(z)) 6L z = 1L ∧ z = TE(y, z)
(iii) Suppose z = 1L. Thus, TE(x, z) = x∧Lz = x 6L y = y∧Lz = TE(y, z).
• Associativity
43
3. EXTENSION METHOD VIA RETRACTIONS
First, suppose that 1L /∈ {x, y, z}. Then:
TE(TE(x, y), z) = TE(s(T (r(x), r(y))), z)
= s(T (r(s(T (r(x), r(y)))), r(z)))
= s(T (T (r(x), r(y)), r(z)))
= s(T (r(x), T (r(y), r(z))))
= s(T (r(x), r(s(T (r(y), r(z))))))
= TE(x, s(T (r(y), r(z))))
= TE(x, TE(y, z))
(i) If x = 1L then:
TE(TE(x, y), z) = TE(TE(1L, y), z) = TE(y, z) = TE(x, TE(y, z))
(ii) If y = 1L then:
TE(TE(x, y), z) = TE(TE(x, 1L), z) = TE(x, z) = TE(x, TE(y, z))
(iii) If z = 1L then:
TE(TE(x, y), z) = TE(TE(x, y), 1L) = TE(x, y) = TE(x, TE(y, z))
�
Notice that TE is a t-norm on L satisfying the identity r(TE(x, y)) = T (r(x), r(y)),
for all x, y ∈ L. In other words, TE is defined in such way that the following di-
agram commutes:
L× L r×r //
TE
��
M ×M
T
��L
r //M
Remark 3.1 If M is a complete sublattice of L in the sense of Definition 2.8
(i.e. M is a subset of L), r(x) = sup{z | s(z) 6L x} for each x ∈ L and s
44
is an inclusion map, then TE = WLTM . In other words, the extension proposed
by Saminger-Platz as shown in Equation (3.1) is an intense of our extension.
Another interesting fact is that T can be extended in several ways since pseudo-
inverse s is not unique. For each possible pseudo-inverse, TE is a different ex-
tension of T from M to L.
Corollary 3.1 Let M < L with respect to (r, s) and T be a t-norm on M . If
r : L −→M is such that r(x) = 0M if and only if x = 0L and TE is the extension
of T from M to L, as defined in Equation (3.3), then
1. if T is Archimedean then TE is Archimedean;
2. if T has a nilpotent element then TE has a nilpotent element;
3. if T has an idempotent element then TE has an idempotent element.
4. if a is a zero divisor of T then s(a) is a zero divisor of TE.
Proof:
1. Let (x, y) ∈ (L\{0L, 1L})2. Since r(x) = 0M if and only if x = 0L then
(r(x), r(y)) ∈ (M\{0M , 1M})2. Thus, by hypothesis T is Archimedean, and
hence, there is an n ∈ N with T (r(x), . . . , r(x))︸ ︷︷ ︸n−times
6L r(y). Therefore,
TE(x, . . . , x) = s(T (r(x), . . . , r(x))) 6L s(r(y)) 6L y
2. Let x ∈M\{0M , 1M} be a nilpotent element of T . Then s(x) ∈ L\{0L, 1L}is a nilpotent element of TE. In fact, let k be an integer such that
T (x, . . . , T (x, x))︸ ︷︷ ︸k−times
= 0M
Hence, since r ◦ s = IdM , it follows that
TE(s(x), . . . , TE(s(x), s(x)) · · ·) = s(T (r(s(x)), . . . , r(s(T (r(s(x)), r(s(x)))))))
= s(T (x, . . . , T (x, x))) = s(0M) = 0L
45
3. EXTENSION METHOD VIA RETRACTIONS
3. Suppose that a ∈ M is an idempotent element of T , i.e., T (a, a) = a.
Thus, since TE(s(a), s(a)) = s(a) ∧L s(a) = s(a) when s(a) = 1L and,
TE(s(a), s(a)) = s(T (r(s(a)), r(s(a)))) = s(T (a, a)) = s(a) otherwise, then
s(a) is an idempotent element of TE.
4. If a ∈ M\{0M , 1M} is a zero divisor of T then there is a b ∈ M\{0M , 1M}such that T (a, b) = 0M . Thus
TE(s(a), s(b)) = s(T (r(s(a)), r(s(b)))) = s(T (a, b)) = s(0M) = 0L
�
Example 3.2 However, extension method via retractions presented in Theorem
3.1 fails in preserving some properties of a given t-norm T . Indeed, if M < L
with respect to (r, s) and T is a t-norm on M , except for the trivial cases (i.e.
when r is not an isomorphism), even T satisfies cancellation law
T (x, y) = T (x, z) ⇒ x = 0M or y = z (3.4)
then its extension TE does not satisfies it. Notice that, if a, b, c ∈ L\{1L}without loss of generality and supposing TE(a, b) = TE(a, c) then we have that
s(T (r(a), r(b))) = s(T (r(a), r(c))) implies T (r(a), r(b)) = T (r(a), r(c)) since s is
injective and hence r(a) = 0M and r(b) = r(c). But, we can not conclude that
a = 0L or b = c.
3.2 T-conorms and Fuzzy Negations
It is also possible to apply the method of extending t-norms for t-conorms and
fuzzy negations under similar conditions as one can see in Propositions 3.1 and
3.3 below.
Proposition 3.1 Let M > L with respect to (r, s). If S is a t-conorm on M
46
then SE : L× L −→ L defined by
SE(x, y) =
{x ∨L y, if 0L ∈ {x, y}
s(S(r(x), r(y))), otherwise.(3.5)
is a t-conorm which extends S from M to L.
Proof: As in the prove of Theorem 3.1 it is easy to see that SE is commutative
and monotone. We shall prove properties (S2) and (S3) of Definition 2.17.
• Associativity
Let x1, x2, x3 ∈ L. The case x1 = x2 = x3 = 0L is straightforward from
definition.
Suppose that 0L /∈ {x1, x2, x3}. Thus,
SE(SE(x1, x2), x3) = SE(s(S(r(x1), r(x2))), x3)
= s(S(r(s(S(r(x1), r(x2)))), r(x3)))
= s(S(S(r(x1), r(x2)), r(x3)))
= s(S(r(x1), S(r(x2), r(x3))))
= s(S(r(x1), r(s(S(r(x2), r(x3))))))
= SE(x1, s(S(r(x2), r(x3))))
= SE(x1, SE(x2, x3))
If x1 = 0L and 0L /∈ {x2, x3} we have
SE(SE(x1, x2), x3) = SE(x2, x3) = SE(x1, SE(x2, x3))
If x2 = 0L and 0L /∈ {x1, x3} then
SE(SE(x1, x2), x3) = SE(x1, x3) = SE(x1, SE(x2, x3))
If x3 = 0L and 0L /∈ {x1, x2} we have
SE(SE(x1, x2), x3) = SE(x1, x2) = SE(x1, SE(x2, x3))
47
3. EXTENSION METHOD VIA RETRACTIONS
Finally, if only xi 6= 0L, where i ∈ {1, 2, 3}, then
SE(SE(x1, x2), x3) = xi = SE(x1, SE(x2, x3))
• Monotonicity
Let x1, x2 and x3 be elements of L such that x1 6L x2. Thus, we have the
following cases:
(i) Firstly, suppose that 0L /∈ {x1, x2, x3}. As such,
SE(x1, x3) = s(S(r(x1), r(x3))) 6L s(S(r(x2), r(x3))) = SE(x2, x3)
(ii) Now, if x1 = 0L and 0L /∈ {x2, x3}, then SE(x1, x3) = x3 6L s(r(x3)) 6Ls(S(r(x2), r(x3))) = SE(x2, x3);
(iii) In the case that x3 = 0L and 0L /∈ {x1, x2} we have: SE(x1, x3) =
x1 ∨L 0L = x1 6L x2 = x2 ∨L 0L = SE(x2, x3);
(iv) If x1 = x2 = x3 = 0L then SE(x1, x3) = 0L = SE(x2, x3).
�
Proposition 3.2 Let M > L with respect to (r, s). Suppose that S is a t-conorm
on M and r is such that r(x) = 1M (r(x) = 0M) if and only if x = 1L (x = 0L).
Thus
1. if S has a nilpotent element then SE has a nilpotent element;
2. if S has an idempotent element then SE has an idempotent element;
3. if a is a zero divisor of S then s(a) is a zero divisor of SE;
4. if S is positive then SE is positive.
Proof:
48
1. Let x ∈M\{0M , 1M} be a nilpotent element of S. Then s(x) ∈ L\{0L, 1L}is a nilpotent element of SE. In fact, let k be an integer such that
S (x, . . . , S(x, x))︸ ︷︷ ︸k−times
= 0M
Hence, since r ◦ s = IdM , it follows that
SE(s(x), . . . , s(x)) = SE(s(x), . . . , SE(s(x), s(x)) · · ·)= s(S(r(s(x)), . . . , r(s(S(r(s(x)), r(s(x)))))))
= s(S(x, . . . , S(x, x))) = s(0M) = 0L
2. Suppose that a ∈M is an idempotent element of S, i.e., S(a, a) = a. Thus,
since SE(s(a), s(a)) = s(a) ∨L s(a) = s(a) when s(a) = 0L and
SE(s(a), s(a)) = s(S(r(s(a)), r(s(a)))) = s(S(a, a)) = s(a)
otherwise. Then s(a) is an idempotent element of SE.
3. If a ∈ M\{0M , 1M} is a zero divisor of S then there is a b ∈ M\{0M , 1M}such that S(a, b) = 0M . Thus
SE(s(a), s(b)) = s(S(r(s(a)), r(s(b)))) = s(S(a, b)) = s(0M) = 0L
4. By (3.5) if SE(x, y) = 1L and y = 0L (or x = 0L) then 1L = SE(x, 0L) =
x ∨L 0L = x (or 1L = SE(0L, y) = 0L ∨L y = y), i.e. SE(x, y) = 1L implies
x = 1L or y = 1L. Now, suppose that SE(x, y) = 1L and 0L /∈ {x, y}. Thus
s(S(r(x), r(y))) = 1L and hence S(r(x), r(y)) = 1M by injetivety of s. By
hypotesis S is positive, then r(x) = 1M or r(y) = 1M what allow us to
conclude that x = 1L or y = 1L.
Reciprocally, if x = 1L then SE(1L, y) = 1L whereas y = 0L and
SE(1L, y) = s(S(r(1L), r(y))) = s(S(1M , r(y))) = s(1M) = 1L
if y 6= 0L. Analogously, one can prove that if y = 1L then SE(x, y) = 1L.
49
3. EXTENSION METHOD VIA RETRACTIONS
�
Proposition 3.3 Let M be a (r, s)-sublattice of L and N : M −→M be a fuzzy
negation. Then NE(x) = s(N(r(x))) for each x ∈ L is a fuzzy negation that
extends N from M to L.
Proof: Let x, y ∈ L such that x 6L y. As r is monotone, we have r(x) 6L r(y).
In this case, N(r(y)) 6M N(r(x)) since N is a fuzzy negation. Consequently, con-
sidering the monotonicity of s, we can conclude that s(N(r(y))) 6L s(N(r(x))),
that is, NE(y) 6L NE(x).
Moreover,
NE(0L) = s(N(r(0L))) = s(N(0M)) = s(1M) = 1L
NE(1L) = s(N(r(1L))) = s(N(1M)) = s(0M) = 0L
Therefore, according to Definition 2.18, it is clear that NE is a fuzzy negation on
L which extends N from M to L.
�
It is worth noting that the hypotheses of Proposition 3.3 require only that r
can be a retraction (it does not need to be neither a lower nor an upper retraction)
and hence if r is a lower retraction or an upper retraction the result remains valid.
This fact allows us to extend fuzzy negations in a more flexible way than t-norms
and t-conorms.
Remark 3.2 Note that the extension of fuzzy negation does not preserve invo-
lution property, i.e. in general NE is not a strong negation even if N is. For
instance, let M = [0, 1] and L = [0, 2]. Considering the retraction r : L → M
such that r(x) = x if x ∈ [0, 1] and else r(x) = 1 which has as a pseudo-inverse
function s : M → L given by s(x) = x for all x ∈ [0, 1) and s(1) = 2, then if N is
an arbitrary strong negation on M we have that NE(NE(4\3)) = 2 which shows
that NE does not satisfies the involution property.
Proposition 3.4 Let M be a (r, s)-sublattice of L and N : M −→ M a strong
negation. Then e ∈ M is an equilibrium point of N if and only if s(e) is an
equilibrium point of NE.
50
Proof: Let e ∈ M be an equilibrium point of N , i.e. N(e) = e. Hence by
Proposition 3.3 we have that NE(s(e)) = s(N(r(s(e)))) = s(N(e)) = s(e) what
allow us to say that s(e) is a equilibrium point of NE.
Conversely, if s(e) is an equilibrium point ofNE thus sinceN(r(x)) = r(NE(x))
for each x ∈ L it follows that N(e) = N(r(s(e))) = r(NE(s(e))) = r(s(e)) = e.
�
3.2.1 Negations Obtained from Fuzzy Connectives
In this subsection we discuss about a particular class of fuzzy negations, those
that one can obtain from either a t-norm or a t-conorm or a fuzzy implication
as introduced in Section 2.2.2. Also, for such negations we establish the relation
between its extensions and the negation obtained from extension of each fuzzy
connective (for instance, see Palmeira et al. [2012a]).
Theorem 3.2 Let MlL with respect to (r, s). Moreover, suppose that r(x) = 0M
if and only if x = 0L. If T is a t-norm on M , then (NT )E 6 NTE .
Proof: By Theorem 3.3 and Proposition 2.14 it follows that
NTE(x) = sup{y ∈ L | TE(x, y) = 0L} (3.6)
and
(NT )E(x) = s(NT (r(x))) = s(sup{z ∈M | T (r(x), z) = 0M}) (3.7)
Take an arbitrary x ∈ L and let be A = {y ∈ L | TE(x, y) = 0L}. We will
prove that (NT )E(x) ∈ A and hence it is possible to conclude that (NT )E(x) 6LNTE(x) since NTE(x) = supA.
Thus, we shall prove that TE(x, (NT )E(x)) = 0L. In fact, if x = 1L it fol-
lows that (NT )E(x) = 0L then by Theorem 3.1 we have TE(x, (NT )E(x)) = 0L.
On the other hand, if (NT )E(x) = 1L then 1L = s(NT (r(x))) = s(sup{z ∈M | T (r(x), z) = 0M}) and hence sup{z ∈ M | T (r(x), z) = 0M} = 1M since s
is injective. Thus, {z ∈ M | T (r(x), z) = 0M} = M that implies r(x) = 0M , i.e.
x = 0L. Then it can be concluded that TE(x, (NT )E(x)) = 0L.
51
3. EXTENSION METHOD VIA RETRACTIONS
Otherwise, when x 6= 1L and (NT )E(x) 6= 1L we have that
TE(x, (NT )E(x)) = s(T (r(x), r((NT )E(x))))
= s(T (r(x), r(s(NT (r(x))))))
= s(T (r(x), NT (r(x))))
Note thatNT (r(x)) = sup{z ∈M | T (r(x), z) = 0M} then T (r(x), NT (r(x))) =
0M . Thus s(T (r(x), NT (r(x)))) = 0L and hence TE(x, (NT )E(x)) = 0L by (5).
Therefore, (NT )E(x) 6 NTE(x) for all x ∈ L, i.e. (NT )E 6 NTE .
�
Proposition 3.5 Let M mL with respect to (r, s). If S is a t-conorm on M then
(NS)E(x) = s(NS(r(x))) = s(inf{z ∈M | S(r(x), z) = 1M}) is a negation on L.
Proof: Straightforward from Proposition 3.3.
�
Theorem 3.3 Let M m L with respect to (r, s). If S is a t-conorm on M then
NSE 6 (NS)E.
Proof: Firstly, take x ∈ L such that x 6= 0L. In this case we have
(NS)E(x) = s(NS(r(x))) = s(inf{z ∈M | S(r(x), z) = 1M}) (3.8)
and(NSE)(x) = inf{t ∈ L | SE(x, t) = 1L}
= inf{t ∈ L | s(S(r(x), r(t))) = 1L}= inf{t ∈ L | S(r(x), r(t)) = 1M}
(3.9)
It is important to say here that in Equation (3.9) above we can hide the case
where SE(x, t) = x ∨L t = 1L without loss of generality because if t = 0L in this
case we have that SE(x, t) = 1L if and only if x = 1L and it is easy to see that
(NS)E(1L) = (NSE)(1L).
Thus, being
A = {z ∈M | S(r(x), z) = 1M} and B = {t ∈ L | S(r(x), r(t)) = 1M}
52
thus for all z ∈ A, by (3.8), we have that S(r(x), r(s(z))) = S(r(x), z) = 1M which
means that s(z) ∈ B for each z ∈ A, i.e. s(A) ⊆ B and hence inf B 6 inf s(A)1. But s(inf A) = inf s(A) by Proposition 2.6 what allow us to conclude that
inf B 6 s(inf A). Therefore, it follows that NSE 6 (SN)E.
Finally, to complete the proof we shall consider the case when x = 0L. Thus
(NS)E(0L) = s(NS(r(1L)))
= s(NS(1M))
= s(inf{z ∈M | S(1M , z) = 1M})= s(inf{1M}) = s(1M) = 1L
and
(NSE)(0L) = inf{t ∈ L | TE(0L, t) = 1L} = inf{1L} = 1L
�
Note that, considering the proof of Theorem 3.3 above, for each t ∈ B we
have that S(r(x), r(t)) = 1M which means that r(t) ∈ A and hence s(r(t)) ∈ s(A)
what does not allows us to conclude that t ∈ s(A) since it holds only t 6L s(r(t)).
However, if we suppose that for each x ∈ L the set Ax = {z ∈ M | S(r(x), z) =
1M} is an ideal2 of M for all t ∈ L such that s(r(t)) ∈ s(Ax) then we have that
t ∈ s(Ax) since t 6L s(r(t)). Therefore for all x ∈ L it follows that B ⊆ s(Ax)
and hence s(inf Ax) = inf s(Ax) 6 B which implies (NS)E 6 NSE . In this case,
considering Theorem 3.3 we have the following corollary.
Corollary 3.2 Let M m L with respect to (r, s) and S be a t-conorm on M . If
for each x ∈ L the set Ax = {z ∈ M | S(r(x), z) = 1M} is an ideal of M then
NSE = (NS)E.
Proposition 3.6 If M > L with respect to (r, s) then SL> 6 (SM> )E.
Proof: By definition we have that
SL>(x, y) =
{x ∨L y, if x = 0L or y = 0L;
1L, otherwise.
1note that if C ⊆ D then inf D 6 inf C, see Lima [1982]2A nonempty subset K of a bounded lattice L is called an ideal if x ∈ K and y 6L x then
y ∈ K.
53
3. EXTENSION METHOD VIA RETRACTIONS
and
(SM> )E(x, y) =
{x ∨L y, if x = 0L or y = 0L;
s(SM> (r(x), r(y))), otherwise.
Note that
0M /∈ {r(x), r(y)} ⇒ s(SM> (r(x), r(y))) = s(1M) = 1L (3.10)
On the other hand
r(x) = 0M or r(y) = 0M ⇒ SM> (r(x), r(y)) = r(x) ∨M r(y) (3.11)
and hence
x ∨L y 6 s(r(x)) ∨M s(r(y)) = s(r(x) ∨M r(y)) = s(SM> (r(x), r(y))) (3.12)
Therefore, by Equations (3.10), (3.11) and (3.12) we can conclude that SL> 6
(SM> )E.
�
Proposition 3.7 Let M m L with respect to (r, s). Suppose r(x) = 1M if and
only if x = 1L, then it follows that NL> = (NM
> )E. Moreover, if S is a positive
t-conorm on M then NSE = (NM> )E.
Proof: Note that
NL>(x) =
{0L, if x = 1L;
1L, otherwise.
and (NM> )E(x) = s(NM
> (r(x))) where
NM> (r(x)) =
{0M , if r(x) = 1M ;
1M , otherwise.
But, by hipothesis r(x) = 1M if and only x = 1L and hence
NM> (r(x)) =
{0M , if x = 1L;
1M , otherwise.
54
Therefore, (NM> )E(x) = 0L if x = 1L and (NM
> )E(x) = 1L else, i.e. (NM> )E = NL
>.
Moreover, if S is a positive t-conorm on M then by Proposition 3.2 SE is a
positive t-conorm on L which implies that NSE = NL> = (NM
> )E by Proposition
2.16 .
�
3.3 Fuzzy Implications
In the following theorem it is presented a way to extend fuzzy implications
as an application of the method of extending fuzzy operators as introduced in
Theorem 3.1.
Now, we apply extension method via retractions for fuzzy implications and
discuss about related results. Also within the framework of implications two
special classes of this fuzzy connective are considered, namely (S,N)-implications
and R-implications.
Theorem 3.4 Let M be a (r, s)-sublattice of L. If I is an implication on M then
function IE : L× L −→ L given by
IE(x, y) = s(I(r(x), r(y))) (3.13)
for all x, y ∈ L, is an implication on L. In this case, IE is called the extension
of I from M to L.
Proof: Let x, y, z ∈ L. We shall prove that IE satisfies the axioms of Definition
2.19. Thus,
1. (FPA) Suppose x 6L y. Thus, by monotonicity of r, we have r(x) 6L r(y)
and hence by (FPA) and the isotonicity of s it follows that
IE(y, z) = s(I(r(y), r(z)))
6L s(I(r(x), r(z)))
= IE(x, z)
55
3. EXTENSION METHOD VIA RETRACTIONS
2. (SPI) Now, let y 6L z. Again, by monotonicity of r we have r(y) 6L r(z).
ThenIE(x, y) = s(I(r(x), r(y)))
6L s(I(r(x), r(z)))
= IE(x, z)
3. Moreover, we have that
– (CC1) IE(0L, 0L) = s(I(r(0L), r(0L))) = s(I(0M , 0M)) = s(1M) = 1L;
– (CC2) IE(1L, 1L) = s(I(r(1L), r(1L))) = s(I(1M , 1M)) = s(1M) = 1L;
– (CC3) IE(1L, 0L) = s(I(r(1L), r(0L))) = s(I(1M , 0M)) = s(0M) = 0L;
Therefore, by (1), (2) and (3) it can be concluded that IE is an implication on L.
�
It is worth noting that as for fuzzy negations we do not need to impose an
additional property for the retract r to extend implications. Note that for t-
conorms, r must be an upper retraction (for t-norms r must be a lower retraction
(see Palmeira and Bedregal [2012])).
Proposition 3.8 Under the same conditions as in Theorem 3.4, if I is an im-
plication on M satisfying some of properties (LB), (RB), (CC4), (EP), (IP),
(IBL), (CP), (L-CP) and (R-CP) then IE is an implication on L which satisfies
the same properties.
Proof: If I is an implication onM then, by Theorem 3.4, IE(x, y) = s(I(r(x), r(y))
for all x, y ∈ L. Thus:
(LB)
By hypothesis I(0M , x) = 1M for all x ∈M . Then IE(0L, y) = s(I(r(0L), r(y))) =
s(I(0M , r(y))) = s(1M) = 1L for all y ∈ L.
(RB)
Now, considering that I(x, 0M) = 1M for all x ∈ M , it follows that IE(a, 0L) =
s(I(r(a), r(0L))) = s(I(r(a), 0M)) = s(1M) = 1L for all a ∈ L.
(CC4)
IE(0L, 1L) = s(I(r(0L), r(1L))) = s(I(0M , 1M)) = s(1M) = 1L.
56
(EP)
IE(x, IE(y, z)) = s(I(r(x), r(IE(y, z)))) by eq. (3.13)
= s(I(r(x), r(s(I(r(y), r(z)))))) by eq. (3.13)
= s(I(r(x), I(r(y), r(z))) by def. 2.9
= s(I(r(y), I(r(x), r(z))) by (EP )
= s(I(r(y), r(s(I(r(x), r(z)))))) by def. 2.9
= IE(y, IE(x, z))
(IP)
Supposing that I(x, x) = 1M for each x ∈ M then IE(y, y) = s(I(r(y), r(y))) =
s(1M) = 1L for all y ∈ L.
(IBL)
For all x, y ∈ L, supposing I satisfies (IBL) we have that
IE(x, IE(x, y)) = s(I(r(x), r(IE(x, y)))) by eq. (3.13)
= s(I(r(x), r(s(I(r(x), r(y)))))) by eq. (3.13)
= s(I(r(x), I(r(x), r(y)))) by def. 2.9
= s(I(r(x), r(y))) by (IBL)
= IE(x, y)
(CP)
Let N be a fuzzy negation on M such that I(x, y) = I(N(y), N(x)). Thus, by
Proposition 3.3, NE is a fuzzy negation on L. Hence
IE(NE(y), NE(x)) = s(I(r(NE(y)), r(NE(x)))) by eq. (3.13)
= s(I(r(s(N(r(y)))), r(s(N(r(x)))))) by prop. 3.3
= s(I(N(r(y)), N(r(x)))) by def. 2.9
= s(I(r(x), r(y))) by (CP )
= IE(x, y)
(L-CP)
57
3. EXTENSION METHOD VIA RETRACTIONS
Now, being N a fuzzy negation for which I satisfies (L-CP), then
IE(NE(x), y) = s(I(r(NE(x)), r(y))) by eq. (3.13)
= s(I(r(s(N(r(x)))), r(y))) by prop. 3.3
= s(I(N(r(x)), r(y))) by def. 2.9
= s(I(N(r(y)), r(x))) by (L− CP )
= s(I(r(s(N(r(y)))), r(x))) by def. 2.9
= IE(NE(y), x)
(R-CP) Analogously to previous item.
�
Corollary 3.3 Let M be a (r, s)-sublattice of L, ρ be an automorphism on M
and I : M ×M →M be a function. If I satisfies P ∈ {(FPA), (SPI), (CC1),
(CC2), (CC4), (LB), (RB)} then (Iρ)E also satisfies P .
Proof: Straightforward from Proposition 3.8 and Proposition 2.18.
�
Example 3.3 We show here that the method of extending implications fails in
preserving properties (NP) and (OP). Let M and L be the bounded lattices shown
in Figure 2.4. It is clear that the function I given by
I(x, y) =
{1M , if x 6M y;
y, otherwise.
is a fuzzy implication on M (this function works as a generalization of Godel
implication for an arbitrary bounded lattice). It can be easily seen in Table 3.1
that implication I satisfies the properties (NP) and (OP).
However, considering the lower retraction r(x) = sup{z ∈ M | s(z) 6L x}from L into M and its pseudo-inverse s defined by s(1M) = 1L, s(a) = v, s(b) = x,
s(c) = y, s(d) = z and s(0M) = 0L (see Example 2.4), the extension IE of fuzzy
implication I does not satisfy properties (NP) and (OP). Indeed, if we take the
pair (1L, t) ∈ L2 then we have
IE(1L, t) = s(I(r(1L), r(t)) = s(I(1M , a)) = s(a) = v 6= t
58
i.e. for IE the property (NP) does not hold. To see that IE also does not satisfy
(OP) it is enough to take the pair (u, v) ∈ L2 and consider that
IE(u, v) = s(I(r(u), r(v)) = s(I(c, a)) = s(1M) = 1L
while we have that u ‖ v. In other words, IE(m,n) = 1L does not imply that
m 6L n, for all m,n ∈ L.
I 0M d c b a 1M
0M 1M 1M 1M 1M 1M 1Md 0M 1M 1M 1M 1M 1Mc 0M d 1M b 1M 1Mb 0M d c 1M 1M 1Ma 0M d c b 1M 1M
1M 0M d c b a 1M
Table 3.1: Implication on M
However, if I is a fuzzy implication on M satisfying (NP) and (OP), the
following weak versions of these properties hold for its extension IE:
(W-NP): For each y ∈ L, if I satisfies (NP), we have that
IE(1L, y) = s(I(r(1L), r(y))) = s(I(1M , r(y))) = s(r(y)) 6L y
(L-OP): If x 6L y then IE(x, y) = 1L. Indeed, let x, y ∈ L such that x 6L y
and suppose that I satisfies (L-OP). In this case, r(x) 6M r(y) since r
is monotone. Thus, by (L-OP), I(r(x), r(y)) = 1M and hence IE(x, y) =
s(I(r(x), r(y))) = s(1M) = 1L.
Remark 3.3 It is worth noting also that property (P) is not preserved by this
method of extending fuzzy implications, in general. It is easy to see that if an
implication I satisfies (P) then we have that IE(1L, 0L) = 0L. But if IE(x, y) = 0L
it does not imply that x = 1L and y = 0L. For instance, let M and L1 be the
bounded lattices shown in Figure 2.3 and take the retraction r1 : L1 → M with
pseudo-inverse s1 : M → L1 as defined in Example 2.3. In this case the fuzzy
59
3. EXTENSION METHOD VIA RETRACTIONS
implication I> : M ×M → M given by I>(x, y) = a if x = d and y = a and
I>(x, y) = d else (see Example 2.11) satisfies (P) but for its extension (I>)E we
have that
(I>)E(5, 2) = s1(I>(r1(5), r1(2))) = s1(I>(d, a)) = s1(a) = 1
which means that there are x, y ∈ L1 (in this case x = 5 and y = 2. Notice that
0L1 = 1 and 1L1 = 5) such that (I>)E(x, y) = 0L1 but (x, y) 6= (1L1 , 0L1). The
following proposition shows a weak version of the sufficiency part of property (P).
Proposition 3.9 Let M be a (r, s)-sublattice of L and I be a fuzzy implication
on M which satisfies property (P). Then:
1. If M < L and IE(x, y) = 0L then x = 1L;
2. If M > L and IE(x, y) = 0L then y = 0L.
Proof: If M is a (r, s)-sublattice of L then there exists a retraction r : L→M
and a pseudo-inverse s : M → L such that r ◦ s = idM . Thus, supposing that
IE(x, y) = 0L then s(I(r(x), r(y)) = 0L = s(0M) for each x, y ∈ L, and hence, by
injectivity of s and (P), we have that
I(r(x), r(y)) = 0M implies r(x) = 1M and r(y) = 0M (3.14)
1. If M < L then we also have that s ◦ r 6 idL and hence 1L = s(1M) =
s(r(x)) 6L x by (3.14). Therefore, we must have x = 1L since 1L is the
supremum element of L.
2. In case that M > L it follows that idL 6 s ◦ r. Again, by (3.14) we have
y 6L s(r(y)) = s(0M) = 0L. It means that y = 0L considering that 0L is
the infimum of L.
�
In which follows we present a result about negation defined from fuzzy impli-
cations.
60
Proposition 3.10 Let M be a (r, s)-sublattice of L and I : M ×M → M be a
function satisfying (EP) and (OP). Then
1. NIE(x) := IE(x, 0L) for all x ∈ L is a fuzzy negation on L;
2. NIE = (NI)E;
3. If r is an upper retraction then x 6L NIE(NIE(x)) for each x ∈ L;
4. NIE ◦NIE ◦NIE = NIE .
Proof:
1. Let x, y ∈ L such that x 6L y. Since r is monotone then r(x) 6M r(y)
and hence I(r(y), 0M) 6M I(r(x), 0M) by (FPA) (properties (EP) and (OP)
imply (FPA) as one can see in Lemma 6 in Baczynski [2004] for 〈[0, 1],6〉.The proof for a generic lattice L is similar.). Thus
NIE(y) = s(I(r(y), r(0L))) = s(I(r(y), 0M)) 6M s(I(r(x), 0M)) = NIE(x)
Moreover,
NIE(0L) = s(I(r(0L), r(0L))) = s(I(0M , 0M)) = s(1M) = 1L
and
NIE(1L) = s(I(r(1L), r(0L))) = s(I(1M , 0M)) = s(0M) = 0L
Therefore, it can be concluded that NIE is a fuzzy negation on L.
2. For each x ∈ L we have
NIE(x) = s(I(r(x), r(0L))) = s(I(r(x), 0M)) = s(NI(r(x))) = (NI)E(x)
61
3. EXTENSION METHOD VIA RETRACTIONS
3. Note that for each x ∈ L
NIE(NIE(x)) = s(I(r(NIE(x)), 0M))
= s(I(r(s(I(r(x), 0M))), 0M))
= s(I(I(r(x), 0M), 0M))
= s(NI(NI(r(x))))
(3.15)
If r is an upper retraction then idL 6 s ◦ r. By item 2 of Lemma 2.2 we
have that r(x) 6L NI(NI(r(x))) for each x ∈ L and hence x 6L s(r(x)) 6Ls(NI(NI(r(x)))). Thus by (3.15) it follows that x 6L NIE(NIE(x)) for each
x ∈ L.
4. Recall that r ◦ s = idM and that NI ◦ NI ◦ NI = NI (by item 3, Lemma
2.2). Thus,
NIE(NIE(NIE(x))) = s(I(r(NIE(NIE(x))), 0M))
= s(I(r(s(I(r(s(I(r(x), 0M))), 0M))), 0M))
= s(I(I(I(r(x), 0M), 0M), 0M))
= s(NI(NI(NI(r(x)))))
= s(NI(r(x)))
= s(NI(r(x), 0M))
= NIE(x)
for each x ∈ L.
�
3.3.1 (S,N)-implications
Proposition 3.11 Let M > L with respect to (r, s). If S is a t-conorm on M
and N is a negation on M then a function ISE ,NE : L × L −→ L defined by
ISE ,NE(x, y) = SE(NE(x), y)) for all x, y ∈ L is an implication in the sense of
Definition 2.19.
Proof: Straightforward from Propositions 3.1 and 3.3.
�
62
Corollary 3.4 Under the same conditions as in Proposition 3.11 it follows that I
is an (S,N)-implication on M generated from S and N if and only if IE = (IS,N)E
is an (S,N)-implication on L.
It is important to point out here that in general if I is a S-implication then
IE is not a S-implication since the extension of a strong negation is not a strong
negation (see Remark 3.2).
Proposition 3.12 Let M > L with respect to (r, s). Considering S a t-conorm
and N a negation both defined on M then it follows that ISE ,NE 6 (IS,N)E.
Proof: By Proposition 3.11 and definition of ISE ,NE we have that
ISE ,NE(x, y) = SE(NE(x), y) =
{NE(x) ∨L y, 0L ∈ {NE(x), y}
s(S(r(NE(x)), r(y))), otherwise.
Since NE(x) = s(N(r(x))) and r ◦ s = idM then
ISE ,NE(x, y) =
{NE(x) ∨L y, 0L ∈ {NE(x), y}
s(S(N(r(x))), r(y))), otherwise.
On the other hand, (IS,N)E(x, y) = s(IS,N(r(x), r(y))) = s(S(N(r(x)), r(y)))
for all x, y ∈ L. Thus, it is clear that
ISE ,NE(x, y) = (IS,N)E(x, y) whenever 0L /∈ {NE(x), y}. (3.16)
Moreover, if y = 0L then
ISE ,NE(x, 0L) = NE(x) = s(N(r(x)))
= s(S(N(r(x)), 0M))
= s(IS,N(r(x), 0M))
= (IS,N)E(x, 0L)
(3.17)
Now, if NE(x) = 0L then s(N(r(x))) = 0L = s(0M) which implies N(r(x)) =
63
3. EXTENSION METHOD VIA RETRACTIONS
0M since s is injective. Hence
ISE ,NE(x, y) = y
6 s(r(y))
= s(S(0M , r(y)))
= s(S(N(r(x)), r(y)))
= s(IS,N(r(x), r(y)))
= (IS,N)E(x, y)
(3.18)
Therefore, by (3.16), (3.17) and (3.18) it can be concluded that ISE ,NE(x, y) 6L(IS,N)E(x, y) for all x, y ∈ L.
�
Corollary 3.5 Under the same conditions as in Proposition 3.12 it follows that
NISE,NE
6 N(IS,N )E .
Proof: Direct from Proposition 3.10 and Proposition 3.12.
�
3.3.2 R-implications
Proposition 3.13 Let M l L with respect to (r, s). If T is a t-norm on M and
N is a negation on M then function ITE : L × L −→ L defined by ITE(x, y) =
sup{t ∈ L | TE(x, t) 6L y} for all x, y ∈ L is an implication in the sense of
Definition 2.19.
Proof: Straightforward from Definition 2.21 and the fact that under these
hypotheses TE is a t-norm on L as proved in Theorem 3.1.
�
Theorem 3.5 Let M l L with respect to (r, s). If T is a t-norm on M then
(IT )E 6 ITE ;
64
Proof: Firstly, take x, y ∈ L such that 1L /∈ {x, y}. In this case we have
(IT )E(x, y) = s(IT (r(x), r(y))) = s(sup{z ∈M | T (r(x), z) 6M r(y)}) (3.19)
and(ITE)(x, y) = sup{t ∈ L | TE(x, t) 6L y}
= sup{t ∈ L | s(T (r(x), r(t))) 6L y}= sup{t ∈ L | T (r(x), r(t)) 6M r(y)}
(3.20)
Being
A = {z ∈M | T (r(x), z) 6M r(y)} and B = {t ∈ L | T (r(x), r(t)) 6M r(y)}
thus for all z ∈ A, by (3.19), we have that T (r(x), r(s(z))) = T (r(x), z) 6Mr(y) which means that s(z) ∈ B for each z ∈ A, i.e. s(A) ⊆ B and hence
sup s(A) 6 supB (note that if C ⊆ D then supC 6 supD, see Lima [1982]).
But s(supA) = sup s(A) by Proposition 2.5 what allows us to conclude that
s(supA) 6 supB. Therefore (IT )E 6 ITE .
Finally, to complete the proof we must consider the cases when 1L ∈ {x, y}.We have two possibilities:
(i) y = 1L:
(IT )E(x, 1L) = s(IT (r(x), r(1L)))
= s(IT (r(x), 1M))
= s(sup{z ∈M | T (r(x), z) 6M 1M})= s(supM) = s(1M) = 1L
and
(ITE)(x, 1L) = sup{t ∈ L | TE(x, t) 6L 1L} = supL = 1L
65
3. EXTENSION METHOD VIA RETRACTIONS
(ii) x = 1L:
(IT )E(1L, y) = s(IT (r(1L), r(y)))
= s(IT (1M , r(y)))
= s(sup{z ∈M | T (1M , z) 6M r(y)})= s(sup{z ∈M | z 6M r(y)})
and
(ITE)(1L, y) = sup{t ∈ L | TE(1L, t) 6L y} = sup{t ∈ L | t 6L y}
Note that if z 6M r(y) then s(z) 6L s(r(y)) 6L y and hence using the
same argumentation of the first part of the proof of this proposition we can
conclude that (IT )E(1L, y) 6L (ITE)(1L, y).
Therefore by (i) and (ii) we have that (IT )E(x, y) 6L (ITE)(x, y) whereas 1L ∈{x, y}.
�
Corollary 3.6 Let M l L with respect to (r, s). If T is a t-norm on M then
N(IT )E 6 N(ITE
). In other words, the natural negation of (IT )E is less than or
equal to the natural negation of ITE .
Proof: Straightforward from Theorem 3.5 and Definition 2.21.
�
3.4 De Morgan Triples
After describing how to extend t-norms, t-conorms and fuzzy negations a
natural question that arises is under which conditions a De Morgan triple can be
extended, i.e. if M is a (r, s)-sublattice of L and 〈T, S,N〉 is a De Morgan triple
on M is 〈TE, SE, NE〉 also a De Morgan triple in the sense of Definition 2.25?
Unfortunately using the extension method via retraction it does not hold for
no trivial cases. There are two main reasons for that:
66
1. For extending a t-norm T from M to L r should be a lower retraction
while for extending a t-conorm S r should be an upper retraction and
hence r should be an isomorphism (in this case M and L are equal up to
isomorphism) what is a trivial case.
2. On the other hand, suppose M < L with respect to (r1, s) and M > L with
respect to (r2, s). Then we can extend T using retraction r1 as shown in
Equation (3.3) and extend S using retraction r2. For extending the negation
N no matter which retraction is token since the extension of negations there
is no constraints for retraction. So, suppose we extend N using retraction
r1. Thus, under this conditions for x, y /∈ {0L, 1L} we have that
NE(TE(x, y)) = NE(s(T (r1(x), r1(y))))
= s(N(r1(s(T (r1(x), r1(y))))))
= s(N(T (r1(x), r1(y))))
= s(S(N(r1(x)), N(r1(y))))
= s(S(r2(s(N(r1(x)))), r2(s(N(r1(y))))))
= SE(s(N(r1(x))), s(N(r1(y))))
= SE(NE(x), NE(y))
but identityNE(SE(x, y)) = TE(NE(x), NE(y)) does not hold sinceNE(x) =
S(N(r1(x))) and
NE(SE(x, y)) = NE(s(S(r2(x), r2(y))))
= s(N(r1(s(S(r2(x), r2(y))))))
= s(N(S(r2(x), r2(y))))
= s(T (N(r2(x)), N(r2(y))))
= s(T (r1(s(N(r2(x)))), r1(s(N(r2(y))))))
In other words, under this conditions Identities (1) and (2) of Definition 2.25
can not be valid at the same time. Therefore, we can affirm that extension method
via retraction does not preserve De Morgan triples.
Nevertheless, we have seen in Example 2.13 there exists triple 〈T, S,N〉 that
just satisfies one of axioms (1) and (2), i.e. when it is a De Morgan T -semitriple
67
3. EXTENSION METHOD VIA RETRACTIONS
or a De Morgan S-semitriple (see Definition 2.26).
Proposition 3.14 Let M < L with respect to (r, s). If 〈T, S,N〉 is a De Morgan
T -semitriple on M then the triple 〈TE, SE, NE〉 is a De Morgan T -semitriple on
L.
Proof: Let r and r′ be lower and upper retractions, respectively, with the same
pseudo-inverse s, such that, for each x, y ∈ L\{0L, 1L}, TE(x, y) = s(T (r(x), r(y)))
and SE(x, y) = s(S(r′(x), r′(y))). In addition, suppose that NE(x) = s(N(r(x)))
for all x ∈ L.
Let x, y ∈ L. Suppose that 0L ∈ {x, y}. Without loss of generality, we can
consider x = 0L. Then
NE(TE(x, y)) = NE(TE(0L, y)) = NE(0L) = 1L = SE(NE(x), NE(y))
Analogously, the above identity may be verified when y = 0L.
Subsequently, consider 1L ∈ {x, y}. We prove only the case of x = 1L, as the
case where y = 1L is analogous. Thus,
NE(TE(x, y)) = NE(TE(1L, y)) = NE(y) = SE(0L, NE(y)) = SE(NE(x), NE(y))
Finally, suppose that 0L, 1L /∈ {x, y}. Recall that r◦s = idM and r′ ◦s = idM .
Thus, it follows that
NE(TE(x, y)) = NE(s(T (r(x), r(y))))
= s(N(r(s(T (r(x), r(y))))))
= s(N(T (r(x), r(y))))
= s(S(N(r(x)), N(r(y))))
= s(S(r′(s(N(r(x)))), r′(s(N(r(y))))))
= SE(s(N(r(x))), s(N(r(y))))
= SE(NE(x), NE(y))
Therefore, it can be concluded that NE(TE(x, y)) = SE(NE(x), NE(y)), for all
x, y ∈ L.
�
68
Proposition 3.15 Let M > L with respect to (r, s). If 〈T, S,N〉 is a De Morgan
S-semitriple on M then the triple 〈TE, SE, NE〉 is a De Morgan S-semitriple on
L.
Proof: Similar to the proof of Proposition 3.14, now takingNE(x) = s(N(r′(x)))
for all x ∈ L.
�
3.5 Extension and Automorphisms
Studying the action of an automorphism over algebraic structures is a very in-
teresting issue. In fuzzy logic it is used among other classifying fuzzy connectives
through the concept of conjugate (see Definition 2.7).
In this section we discuss about some results involving extension and conjugate
of t-norms, t-conorms, fuzzy negations and implications. We investigate their
relation also for De Morgan triples and t-subnorms.
The main issue in this framework is: For instance, let M be a (r, s)-sublattice
of L, T a t-norm on M and ρ : M →M an automorphism. Is it possible to define
an automorphism ψ on L from ρ in order to have (T ρ)E = (TE)ψ?
Of course, a natural candidate to solve this problem should be ρE, i.e. ρE(x) =
s(ρ(r(x))) for each x ∈ L. But ρE is not an automorphism in general. For
instance, let M and L1 be the bounded lattices shown in the Figure 2.3 and
r1 : L1 → M the lower retract defined in Example 2.3 which has as pseudo-
inverse the homomorphism s1 : L1 → M given by s1(a) = 1, s1(b) = 2, s1(c) = 3
and s1(d) = 5 (as in Example 2.3 also). Thus, if we define the automorphism
ρ : M → M by ρ(a) = a, ρ(b) = c, ρ(c) = b and ρ(d) = d then it is easy to see
that ρE is not an automorphism since ρE(1) = ρE(2) = 1, i.e. ρE is not injective.
However, it is possible to establish conditions under which the question raised
up above can be considered for fuzzy connectives.
Theorem 3.6 Let M < L with respect to (r, s), ρ be an automorphism on M
and T be a t-norm on M . Moreover, suppose ψ : L −→ L is an automorphism
on L such that r ◦ ψ = ρ ◦ r. Then, (T ρ)E 6 (TE)ψ.
69
3. EXTENSION METHOD VIA RETRACTIONS
Proof: By definition, we have
(T ρ)E(x, y) =
{x ∧L y, if 1L ∈ {x, y};
s(ρ−1(T (ρ(r(x)), ρ(r(y))))), otherwise.
On the other hand, considering that 1L ∈ {ψ(x), ψ(y)} if and only if 1L ∈ {x, y},then
(TE)ψ(x, y) =
{x ∧L y, if 1L ∈ {x, y};
ψ−1(s(T (r(ψ(x)), r(ψ(y))))), otherwise.
It is clear that
1L ∈ {x, y} ⇒ (T ρ)E(x, y) = (TE)ψ(x, y) (3.21)
Otherwise, considering that r ◦ ψ = ρ ◦ r then
(T ρ)E(x, y) =
{x ∧L y, if 1L ∈ {x, y};
s(ρ(T (r ◦ ψ−1 ◦ s(r(x)), r ◦ ψ−1 ◦ s(r(y))))), otherwise.
(3.22)
(TE)ψ(x, y) =
{x ∧L y, if 1L ∈ {x, y};
s ◦ ρ ◦ r(s(T (r(ψ−1(x)), r(ψ−1(y))))), otherwise.
(3.23)
As r is a lower retraction, we have r ◦ s = IdM and s ◦ r 6 IdL. Thus
s(ρ(T (r ◦ ψ−1 ◦ s(r(x)), r ◦ ψ−1 ◦ s(r(y))))) 6L s(ρ(T (r(ψ−1(x)), r(ψ−1(y)))))
(3.24)
and
s ◦ ρ ◦ r(s(T (r(ψ−1(x)), r(ψ−1(y))))) = s(ρ(T (r(ψ−1(x)), r(ψ−1(y))))) (3.25)
for all x 6= 1L and y 6= 1L. Thus, by Equations (3.22), (4.11), (4.12) and (3.25),
we can conclude that
(T ρ)E(x, y) 6L (TE)ψ(x, y) for x 6= 1L and y 6= 1L (3.26)
70
Therefore, by Equations (4.8) and (3.26), it follows that (T ρ)E 6 (TE)ψ.
�
It is worth noting that (T ρ)E = (TE)ψ, regardless of whether s ◦ r = IdL,
i.e., r is an isomorphism of bounded lattices or M is a complete sublattice of
L in ordinary sense. The first case is trivial while second one is presented in
Proposition 3.16.
Lemma 3.1 If M is a complete sublattice of bounded lattice L in ordinary sense
(i.e., M ⊆ L), ψ is an automorphism on L such that ψ|M is an automorphism
on M and r : L −→M is a function given by r(x) = supM{y ∈M | y 6L x}, then
r ◦ ψ−1 = (ψ−1)|M ◦ r.
Proof: Note that (ψ−1)|M(r(x)) = (ψ−1)|M(supM{y ∈ M | y 6L x}) and
r(ψ−1(x)) = supM{w ∈ M | w 6L ψ−1(x)}. Thus, if k1 = sup
M{y ∈ M | y 6L x}
and k2 = supM{w ∈M | w 6L ψ−1(x)}, then we shall show that (ψ−1)|M(k1) = k2.
(i) We uphold that k1 6 x. Since (ψ−1)|M preserves order, hence (ψ−1)|M(k1) 6
ψ−1(x). So, (ψ−1)|M(k1) 6 k2 (because k2 = supM{w ∈M | w 6L ψ−1(x)}).
(ii) Conversely, k2 6 ψ−1(x) and hence ψ(k2) 6 ψψ−1(x) = x (note that
ψ preserves order). Thus, ψ(k2) 6 k1 and then k2 = ψ−1ψ(k2) 6 ψ−1(k1) =
(ψ−1)|M(k1).
Therefore, by (i) and (ii), we have (ψ−1)|M(k1) = k2.
�
Proposition 3.16 Let M be a complete ordinary sublattice of L as in Definition
2.8. Consider ψ an automorphism on L and suppose that ρ = ψ|M is an auto-
morphism on M . If T is a t-norm on M then (TE)ψ = (T ρ)E. In other words,
the conjugate of the extension of T is equal to the extension of the conjugate of
T .
Proof: Note that, due to M ⊆ L by hypothesis, the pseudo-inverse s of
retraction r works as the identity function in the definition of TE, i.e., TE(x, y) =
71
3. EXTENSION METHOD VIA RETRACTIONS
T (r(x), r(y)) for all x 6= 1L and y 6= 1L. Thus, by definition we have
(T ρ)E(x, y) =
{x ∧L y, if 1L ∈ {x, y};
ρ−1(T (ρ(r(x)), ρ(r(y)))), otherwise.(3.27)
On the other hand, considering that 1L ∈ {ψ−1(x), ψ−1(y)} if and only if 1L ∈{x, y}, then
(TE)ψ(x, y) =
{x ∧L y, if 1L ∈ {x, y};
ψ−1(T (r(ψ(x)), r(ψ(y))), otherwise.(3.28)
It is clear that if 1L ∈ {x, y}, then (T ρ)E(x, y) = (TE)ψ(x, y). Otherwise,
Identities (3.27) and (3.28) are different only in arguments of T , i.e., in the
Identity (3.27) the arguments of T are ρ−1(r(x)) and ρ−1(r(y)) while in Identity
(3.28) the arguments are r(ψ−1(x)) and r(ψ−1(y)). However, in accordance with
Lemma 4.2, we have r ◦ ψ−1 = ρ−1 ◦ r. Hence, (T ρ)E = (TE)ψ.
�
Let M and L be bounded lattices. If CM and CL are the classes of all t-norms
on M and L, respectively, then proposition above establishes that the following
diagram is commutative:
CMTE //
ψ|M
��
CL
ψ
��CM
(Tψ|M )E // CL
Remark 3.4 It is clear that if M > L with respect to (r, s) and ρ and S are
respectively an automorphism and a t-conorm on M it is possible to prove that
(Sρ)E > (SE)ψ and (Sρ)E = (SE)ψ whenever M is a complete ordinary sublattice
of L.
Theorem 3.7 Let M be a (r, s)-sublattice of L, ρ be an automorphism on M .
If N and I are respectively a fuzzy negation and an implication on M , supposing
ψ : L→ L is an automorphism on L such that r ◦ψ = ρ◦ r and ψ−1 ◦ s = s◦ρ−1,then
72
1. (Iρ)E = (IE)ψ;
2. (Nρ)E = (NE)ψ.
Proof:
1. By definition, for all x, y ∈ L we have
(Iρ)E(x, y) = s(ρ−1(I(ρ(r(x)), ρ(r(y))))) (3.29)
and
(IE)ψ(x, y) = ψ−1(s(I(r(ψ(x)), r(ψ(y))))) (3.30)
Considering that ψ−1 ◦ s = s ◦ ρ−1 and r ◦ ψ = ρ ◦ r then by (3.29) and
(3.30) it follows that
(Iρ)E(x, y) = ψ−1(s(I(r(ψ(x)), r(ψ(y))))) = (IE)ψ(x, y)
2. Similar to item 1.
�
Considering what it was established in Theorems 3.14 and 3.7, another ques-
tion arises involving extensions and automorphisms: If M is a complete sublattice
of the bounded lattice L, 〈T, S,N〉 is a De Morgan T -semitriple on M and ρ an
automorphism on M , then, is 〈(T ρ)E, (Sρ)E, (Nρ)E〉 a De Morgan T -semitriple?
There is a solution to a similar question presented in Theorem 7, proved by
da Costa et al. [2011]. Specifically speaking, it is shown that, if 〈T, S,N〉 is a De
Morgan triple on [0, 1] and ρ is an automorphism on [0, 1], then 〈T ρ, Sρ, Nρ〉 is
also a De Morgan triple. This is naturally generalized to lattices and De Morgan
T -semitriples, as shown in the following proposition.
Proposition 3.17 If 〈T, S,N〉 is a De Morgan T -semitriple on M a complete
sublattice of a bounded lattice L and ρ is an automorphism on L, then we have
that 〈(TE)ρ, (SE)ρ, (NE)ρ〉 is a De Morgan T -semitriple on L.
Proof: Analogous to the proof of Theorem 7 in da Costa et al. [2011].
73
3. EXTENSION METHOD VIA RETRACTIONS
�
Observe that, according to Proposition 3.16, if M is a complete sublattice of
bounded lattice L in ordinary sense, ψ is an automorphism on L and ρ = ψ|M is
an automorphism on M , then 〈(T ρ)E, (Sρ)E, (Nρ)E〉 = 〈(TE)ψ, (SE)ψ, (NE)ψ〉.
3.6 T-subnorms and T-subconorms
Our primary purpose in this section is to prove that, if M is a complete
sublattice of bounded lattice L and F is a t-subnorm on M , then we may apply
our technique of extending t-norms to extend t-subnorms. This is what we present
in the proposition below.
Proposition 3.18 Let M < L with respect to (r, s). If F is a t-subnorm on M
then the function FE : L× L −→ L given by
FE(x, y) =
{x ∧L y, if1L ∈ {x, y}
s(F (r(x), r(y))), otherwise
is a t-subnorm on L.
Proof: Clearly, FE is commutative, associative and monotone (see the proof
of Theorem 3.1). It remains to prove that FE(x, y) 6 x ∧L y.
Suppose x, y ∈ L. If 1L ∈ {x, y}, then, by definition, FE(x, y) = x ∧L y.
Otherwise, if z = x ∧L y then
FE(x, y) = s(F (r(x), r(y)))
6L s(r(x) ∧L r(y))
= s(r(x ∧L y))
= s(r(z))
6L z = x ∧L y
�
Corollary 3.7 Let M < L with respect to (r, s). If F is a t-subnorm on M then
a t-norm T generated from the extension FE of F to L is equal to the extension
(T ′)E of the t-norm T ′ defined from F (as in Proposition 3.18).
74
Proof: In Propositions 2.22 and 3.18, we have
T ′(x, y) =
{F (x, y), if(x, y) ∈ (L\{1L})2
x ∧L y, otherwise
and
FE(x, y) =
{x ∧L y, if1L ∈ {x, y}
s(F (r(x), r(y))), otherwise
Thus, since T ′(x, y) = F (x, y) if (x, y) ∈ (L\{1L})2 and FE(x, y) = s(F (r(x), r(y)))
if 1L /∈ {x, y}, it follows that
(T ′)E(x, y) =
{x ∧L y, if1L ∈ {x, y}
s(F (r(x), r(y))), otherwise(3.31)
and
T (x, y) =
{s(F (r(x), r(y))), if(x, y) ∈ (L\{1L})2
x ∧L y, otherwise
Thus, rewriting the above identity more conveniently, we have
T (x, y) =
{s(F (r(x), r(y))), otherwise
x ∧L y, if 1L ∈ {x, y}(3.32)
Therefore, by Identities (3.31) and (3.32) it follows that (T ′)E(x, y) = T (x, y) for
all x, y ∈ L.
�
Naturally, the notion of a t-subconorm may be defined simply be replacing the
property F (x, y) 6L x ∧L y for R(x, y) >L x ∨L y in Definition 2.23, allowing us
to apply the concept of De Morgan semitriples using t-subnorms, t-subconorms
and fuzzy negations.
Definition 3.1 Let F be a t-subnorm on L, R be a t-subconorm on L and N
be a fuzzy negation. A triple 〈F,R,N〉 is called a De Morgan F -semitriple
(R-semitriple) if, for all x, y ∈ L, it holds that N(F (x, y)) = R(N(x), N(y))
(N(R(x, y)) = F (N(x), N(y))).
75
3. EXTENSION METHOD VIA RETRACTIONS
Proposition 3.19 Let M > L with respect to (r, s). If R is a t-subconorm on
M then the function RE : L× L −→ L given by
RE(x, y) =
{x ∨L y, if0L ∈ {x, y}
s(R(r(x), r(y))), otherwise
is a t-subconorm.
Proof: Analogous to the proof of Proposition 3.18.
�
Proposition 3.20 Let M < L with respect to (r, s). Suppose F is a t-subnorm
on M , R is a t-subconorm on M and N a fuzzy negation on M . If 〈F,R,N〉 is a
De Morgan F -semitriple, then 〈FE, RE, NE〉 is also a De Morgan F -semitriple.
Proof: Similar to the proof of Theorem 3.14.
�
It is worth noting that a dual version of Proposition 3.20 holds. Moreover,
assume that 〈F,R,N〉 is a De Morgan F -semitriple on M complete sublattice
of bounded lattice L. By Corollary 3.7 and Proposition 3.20, we may conclude
that the De Morgan T -semitriple 〈T, S,N〉 obtained from the extended De Mor-
gan F -semitriple 〈FE, RE, NE〉 is equal to the extended De Morgan T -semitriple
〈TEM , SEM , NE〉 obtained from the De Morgan T -semitriple 〈T ′, S ′, N〉, where T ′
(S ′) is a t-norm generated from F (R) by Proposition 2.22 (by a dual version of
Proposition 2.22).
3.7 Final Remarks
This section is devoted to present tables describing the main properties that
are preserved by extension method via retractions for each fuzzy connective stud-
ied in this thesis. We use the symbol (√
) to check those properties that are
preserved by extension method via retraction and the symbol (×) for those prop-
erties that are not preserved by this extension method. Symbol (?) means that we
76
have not discuss about it and (-) means that it does not make sense this property
for these operators. Moreover, properties (RB), (LB), (CC4), (NP), (EP), (IP),
(OP), (P), (IBL) and (CP) related to implications (see Table 3.4) are defined on
page 32. For simplicity, we used abbreviations for the following properties:
Abbreviation Property
AC Archimedean
NI Nilpotent
IDN Idempotent
ZD Zero divisor
CL Cancellation law
C Continuity
P Positive
EC Relation between extension and conjugate
RE Relation between NKE and (NK)E with K a fuzzy connective
Table 3.2: Table of Abbreviations
t-norm and t-conorm AC NI IDN ZD CL C P EC
TE√ √ √ √
× ×√
(T ρ)E 6 (TE)ψ
SE ×√ √ √
× ×√
(Sρ)E > (SE)ψ
Table 3.3: Properties preserved by TE and SE
77
3. EXTENSION METHOD VIA RETRACTIONS
fuzzy implication (RB) (LB) (CC4) (NP) (EP) (IP) (OP) (P) (IBL) (CP)
IE√ √ √
×√ √
× ×√ √
(IS,N )E ? ? ? ? ? ? ? ? ? ?
(IT )E ? ? ? ? ? ? ? ? ? ?
Table 3.4: Properties preserved by IE
fuzzy negation Strong Strict Eq. P. EC RE
NE × ×√
(Nρ)E = (NE)ψ –
(NT )E × × ? ? (NT )E 6 NTE
(NS)E × × ? ? (NS)E > NSE
(NI)E × × ? ? (NT )E = NTE
Table 3.5: Properties preserved by NE
78
Chapter 4
Extension Method via
e-operators
We have seen in Chapter 3 that extension method via retractions fulfills the
objective of extending fuzzy connectives and others related operator. Actually,
this method is very general and can be applied for extending many other kind
of fuzzy operators. However, a disadvantage of this extension method is the fact
that some properties of these operator are not preserved. For instance, it does not
preserve strong (strict) negations, De Morgan triples and other properties. (see
Sections 3.2 and 3.4). Taking this into account, we started the investigation of
another way to extend fuzzy connectives that could be more efficient in preserving
properties.
Seeking to find out this method we present in this chapter a new way of
extending t-norms, t-conorms and fuzzy negations inspired on the idea of interval
constructor (see Bedregal and Takahashi [2006]) considering the important fact
that every lattice L can be seen as a sublattice of L = {[x, y] | x, y ∈ L and x 6Ly} in the sense of Definition 2.10. To do so, we develop an essential mapping,
named e-operator, as discussed in Section 4.1. This operator plays in an essential
role for making extension.
Indeed, results have shown that the extension method via e-operator is more
robust allowing a better performance of this extension in preserving properties of
extended connectives or operators.
79
4. EXTENSION METHOD VIA E-OPERATORS
We study here similar issues as in the previous chapter in order to show
the efficiency of the extension method via e-operator in preserving properties
against extension method via retractions. Also, we discuss about extension of
n-dimensional t-norms
4.1 Toward e-operators
Let 〈L,∧L,∨L, 0L, 1L〉 be a bounded lattice. It is well known that it is possible
to construct an interval version of it. Specifically, 〈L,∧,∨, [0L, 0L], [1L, 1L]〉 is a
bounded lattice where
L = {[x, y] | x, y ∈ L and x 6L y} (4.1)
Then, considering r1([x, y]) = x, r2([x, y]) = y for each [x, y] ∈ L and s(x) = [x, x]
for all x ∈ L, we can conclude that L E L with respect to (r1, r2, s), i.e. L is a
sublattice of L in the sense of Definition 2.10.
Moreover, notice that if T is a t-norm on L, a function I[T ] : L × L −→ Lgiven by
I[T ](X, Y ) = [T (r1(x), r1(y)), T (r2(x), r2(y))] (4.2)
for all X = [x, x], Y = [y, y] ∈ L, is a t-norm on L. Operator I is called interval
constructor and it works as described in Figure 4.1 (see Bedregal and Takahashi
[2006]; Bedregal et al. [2006b]).
The interval constructor is a very suitable and useful tool for converting
lattice-valued fuzzy operator in interval ones with a power to carry on most
important properties of these operators. So, taking into account that L is a
(r1, r2, s)-sublattice of L this constructor inspire us to provide an similar opera-
tor that could be able to works as a bridge for extending fuzzy connectives from
a (r1, r2, s)-sublattice M of bounded lattice L.
To do so, considering MEL with respect to (r1, r2, s) and T a t-norm on M we
just need to define a function that plays the role of [·, ·] for interval constructor,
i.e. a function H : M ×M → L as in Figure 4.2.
Based on these ideas we developed a generic operator having the minimal
properties necessary to be a function as H.
80
L× L
r1×r1
��
r2×r2
��L× L
T
��
L× L
T
��L× L
[·,·]��L
Figure 4.1: Diagram of interval constructor
L× L
r1×r1
��
r2×r2
��M ×M
T
��
M ×M
T
��M ×M
H��L
Figure 4.2: Diagram of intuitive idea
81
4. EXTENSION METHOD VIA E-OPERATORS
Definition 4.1 Let MEL with respect to (r1, r2, s). A mapping � : M×M −→ L
is called an extension operator on M (e-operator for short) if it is isotonic and
satisfies, for each a, b ∈M and for each x ∈ L, the following conditions:
r1(a� b) = a ∧M b and r2(a� b) = a ∨M b (4.3)
r1(x)� r2(x) = x (4.4)
In other words, if M E L with respect to (r1, r2, s) (by Definition 2.12, there
are two retractions r1, r2 : L −→ M with the same pseudo-inverse s : M −→ L
such that s◦r1 6 idL 6 s◦r2), then the e-operator � describes an isotonic way to
relate retractions r1 and r2 with the meet and join operators of M , respectively,
by (4.3).
Example 4.1 Given a bounded lattice 〈L,∧L,∨L, 0L, 1L〉 and its interval version
〈L,∧,∨, [0, 0], [1, 1]〉, as described at the beginning of this section LEL with respect
to (r1, r2, s). Considering the interval operations
[x, y] ∧ [a, b] = [x ∧L a, y ∧L b] and [x, y] ∨ [a, b] = [x ∨L a, y ∨L b].
by Definition 4.1, the mapping � : L× L −→ L defined by
a� b = [a ∧L b, a ∨L b]
for each a, b ∈ L, is trivially an e-operator on L.
In what follows we present an example of e-operator for finite lattices.
Example 4.2 Let M and L be bounded lattices depicted in Figure 4.3. Consider
the functions s : M −→ L and r1, r2 : L −→M defined by:
• s(1) = a, s(2) = b, s(3) = c, s(4) = d and s(5) = e;
• r1(s(x)) = x for any x ∈ M , r1(x1) = r1(x2) = r1(x3) = r1(x4) = 1,
r1(x5) = r1(x6) = r1(x7) = 2, r1(x8) = 3 and r1(x9) = 4; and
82
• r2(s(x)) = x for any x ∈ M , r2(x1) = 2, r2(x2) = r2(x5) = 3, r2(x3) =
r2(x6) = 4 and r2(x4) = r2(x7) = r2(x8) = r2(x9) = 5.
It is not hard to see that M E L with respect to (r1, r2, s). Now define � :
M ×M −→ L by
• x� x = s(x) and x� y = y � x for each x, y ∈M ; and
• 1 � 2 = x1, 1 � 3 = x2, 1 � 4 = x3, 1 � 5 = x4, 2 � 3 = x5, 2 � 4 = x6,
2� 5 = x7, 3� 4 = x7, 3� 5 = x9 and 4� 5 = x9.
Thus, r1(3�4) = r1(x7) = 2 = 3∧M 4 and r2(3�4) = r2(x7) = 5 = 3∨M 4 and so
Condition (4.3) is satisfied for this pair of values. Analogously, r1(x7)� r2(x7) =
2� 5 = x7 and so Condition (4.4) is satisfied for x7. By similar calculations all
remaining cases can be checked and the isotonicity of � shown. Therefore, it can
be stated that � is an e-operator on M .
M
◦
◦ ◦
◦
◦
5
3 4
2
1
@@@
���
���
@@@
L◦
◦
◦ ◦
◦
◦ ◦
◦◦ ◦
◦ ◦
◦
◦
b
c d
e
x9x8
x7
x5 x6
x4x2 x3
x1
a
@@@
@@@
@@@
���
���
����
BBBB
AAAAAAAAAAA
�����
��
HHHHH
HH
���
���
���
@@@
@@@
���
Figure 4.3: Hasse diagrams of lattices M and L
Remark 4.1 It is worth noting that given two retractions r1, r2 : L −→ M with
the same pseudo-inverse s : M −→ L such that s ◦ r1 6 idL 6 s ◦ r2 then it can
be easily concluded that r1 6 r2.
The following lemma provides us useful properties concerning the mapping �defined above.
83
4. EXTENSION METHOD VIA E-OPERATORS
Lemma 4.1 Consider MEL with respect to (r1, r2, s) and let � be an e-operator
on M . Then, for all a, b ∈M and x, y ∈ L, the following properties hold:
1. a 6M b if and only if r1(a� b) = a and r2(a� b) = b;
2. For every a ∈M we have s(a) = a� a;
3.
r1(x) 6M r1(y) and r2(x) 6M r2(y)⇔ x 6L y; (4.5)
4. r1(x) = r1(y) and r2(x) = r2(y) if and only if x = y;
5. � is commutative.
Proof:
1. Straightforward from (4.3) and the fact that, if a 6M b then a ∧M b = a
and a ∨M b = b.
2. By (4.4), it follows that s(a) = r1(s(a))� r2(s(a)) = a� a.
3. Since r1 and r2 are homomorphisms, we only need to prove the right side.
If r1(x) 6M r1(y) and r2(x) 6M r2(y) then by isotonicity of �, r1(x) �r2(x) 6M r1(y)� r2(y) and hence by Equation (4.4) we have x 6L y.
4. Straightforward from the previous property.
5. By Equation (4.3), r1(a� b) = a∧M b = b∧M a = r1(b� a) and r2(a� b) =
a ∨M b = b ∨M a = r2(b� a). Thus, by the previous item, a� b = b� a.
�
Lemma 4.2 Let M E L with respect to (r1, r2, s) and � : M ×M → L be an
e-operator. Thus, for each a, b ∈M ,
1. If a� b = 0L then a = 0M and b = 0M ;
2. If a� b = 1L then a = 1M and b = 1M ;
Proof:
84
1. Supposing a� b = 0L we have that a ∧M b = r1(a� b) = r1(0L) = 0M and
a ∨M b = r2(a� b) = r1(0L) = 0M , what means that a = 0M and b = 0M .
2. Analogous to item 1.
Proposition 4.1 Consider M E L with respect to (r1, r2, s) and take a mapping
� : M ×M −→ L satisfying Equation (4.3).Then � is an e-operator if and only
if the Equation (4.5) is satisfied.
Proof: (⇒) By Lemma 4.1.
(⇐) Suppose that y 6M z. We shall prove that � is isotonic and satisfies (4.4).
By Equation (4.3), r1(x� y) = x ∧M y 6M x ∧M z = r1(x� z) and r2(x� y) =
x ∨M y 6M x ∨M z = r2(x � z). So, by Equation (4.5) we have x � y 6L x � z.
Analogously it is possible to prove that y � x 6L z � x. So � is isotonic in each
component.
Moreover, by Equation (4.3), r1(r1(x) � r2(x)) = r1(x) ∧M r2(x) = r1(x) and
r2(r1(x) � r2(x)) = r1(x) ∨M r2(x) = r2(x). So, by Equation (4.5) (item 3 of
Lemma 4.1), it follows that r1(x)� r2(x) = x for all x ∈ L.
�
4.2 T-norms
Theorem 4.1 Let M E L with respect to (r1, r2, s) and � be an e-operator on
M . Thus, given a t-norm T on M , the function TE� : L2 −→ L defined by
TE� (x, y) = T (r1(x), r1(y))� T (r2(x), r2(y)) (4.6)
is a t-norm on L.
Proof: Firstly, note that r1 6 r2 so, by monotonicity of T , it follows that
T (r1(x), r1(y)) 6M T (r2(x), r2(y)) (4.7)
Let x, y, z ∈ L. Thus, we have:
85
4. EXTENSION METHOD VIA E-OPERATORS
– (Commutativity)
TE� (x, y) = T (r1(x), r1(y))� T (r2(x), r2(y))
= T (r1(y), r1(x))� T (r2(y), r2(x))
= TE� (y, x)
– (Associativity)
TE� (x, TE� (y, z)) =
= T (r1(x), r1(T (r1(y), r1(z))�T (r2(y), r2(z))))�T (r2(x), r2(T (r1(y), r1(z))�T (r2(y), r2(z)))) by eq. (5.14)
= T (r1(x), T (r1(y), r1(z))∧MT (r2(y), r2(z)))�T (r2(x), T (r1(y), r1(z))∨MT (r2(y), r2(z))) by eq. (4.3)
= T (r1(x), T (r1(y), r1(z)))�T (r2(x), T (r2(y), r2(z))) by eq. (5.16)
= T (T (r1(x), r1(y)), r1(z))�T (T (r2(x), r2(y)), r2(z)) by assoc. of T
= T (r1(T (r1(x), r1(y))�T (r2(x), r2(y))), r1(z))�T (r2(T (r1(x), r1(y))�T (r2(x), r2(y))), r2(z)) by eq. (4.3) and (5.16)
= TE� (TE� (x, y), z) by eq. (5.14)
– (Monotonicity)
Suppose that x 6L y. Thus, r1(x) 6M r1(y) and r2(x) 6M r2(y). So, by
monotonicity of T , T (r1(x), r1(z)) 6M T (r1(y), r1(z))) and T (r2(x), r2(z)) 6MT (r2(y), r2(z))). Thus, by isotonicity of �, T (r1(x), r1(z)) � T (r2(x), r2(z)) 6LT (r1(y), r1(z)))� T (r2(y), r2(z))). Therefore, TE� (x, z) 6L TE� (y, z).
– (Boundary Condition)
By Equations (4.4) and (5.14) we have
TE� (x, 1L) = T (r1(x), r1(1L))� T (r2(x), r2(1L)) = r1(x)� r2(x) = x
�
Example 4.3 Considering the lattices and operators defined in the Example 4.1,
86
if T is a t-norm on L then, according to Theorem 4.1 it follows that
TE� ([x, y], [a, b]) = T (r1([x, y]), r1([a, b]))� T (r2([x, y]), r2([a, b]))
= [T (x, a), T (y, b)]
Thus, by Proposition 4.10 in Bedregal et al. [2006b] the extension of T from
L to L (i.e. TE� ) is a t-norm on L.
Lemma 4.3 Under the same conditions as in Theorem 4.1, for any n ∈ N,
(TE� )n(x) = T n(r1(x))� T n(r2(x)).
Proof: We will prove it by induction on n. Trivially, (TE� )0(x) = 1L =
1M � 1M = T 0(r1(x)) � T 0(r2(x)). Assume as inductive hypothesis (IH) that
(TE� )k(x) = T k(r1(x))� T k(r2(x)). Then,
(TE� )k+1(x) = TE� ((TE� )k(x), x) by eq. (2.3)
= TE� (T k(r1(x))� T k(r2(x)), x) by IH
= T (r1(Tk(r1(x))� T k(r2(x))), r1(x))�
T (r2(Tk(r1(x))� T k(r2(x))), r2(x)) by eq. (5.14)
= T (T k(r1(x)) ∧M T k(r2(x)), r1(x))�T (T k(r1(x)) ∨M T k(r2(x)), r2(x)) by eq. (4.3)
= T (T k(r1(x)), r1(x))� T (T k(r2(x)), r2(x))
= T k+1(r1(x))� T k+1(r2(x)) by eq. (2.3)
�
Proposition 4.2 Under the same conditions as in Theorem 4.1, and considering
pseudo-quasi-metrics dM and dL for M and L respectively, it holds that:
1. If T is (d2M , dM)-continuous, r1 and r2 are (dL, dM)-continuous and � is
(d2M , dL)-continuous then TE� is (d2L, dL)-continuous;
2. If T is Archimedean, r−11 (0M) = r−12 (0M) = {0L} and r−11 (1M) = r−12 (1M)
= {1L} then TE� is Archimedean;
3. If T is dM -nilpotent, r1 and r2 are (dL, dM)-continuous, � is (d2M , dL)-
continuous, r−11 (0M) = r−12 (0M) = {0L} and r−11 (1M) = r−12 (1M) = {1L}then TE� is dL-nilpotent;
87
4. EXTENSION METHOD VIA E-OPERATORS
4. If T is idempotent then TE� is idempotent; and
5. If T is positive then TE� is positive.
Proof:
1. Since TE� = �◦ (T ×T )◦ (r1×r1×r2×r2)◦Id2L2 , T is (d2M , dM)-continuous,
r1 and r2 are (dL, dM)-continuous and � is (d2M , dL)-continuous, then the
result is straightforward from Propositions 2.7, 2.8 and 2.9 and from the
fact that trivially Id2L2 is (d2L, (d2L)2)-continuous, so it can be concluded that
TE� is (d2L, dL)-continuous.
2. Note that for any x ∈M\{0M , 1M}, Tm+1(x) 6M Tm(x). In fact,
Tm+1(x) = T (Tm(x), x) 6M Tm(x)
since T (x, y) 6M x for all x, y ∈M . Therefore, it can be inferred that
Tm(x) 6M T n(x), with n 6 m and n,m ∈ N (4.8)
Suppose that T is Archimedean. Thus, for each x, y ∈ L\{0L, 1L} by hy-
pothesis r1(x), r2(x) ∈M\{0M , 1M}, and therefore there are n,m ∈ N such
that T n(r1(x)) 6M r1(y) and Tm(r2(x)) 6M r2(y). Letting n 6 m we have
(TE� )m(x) = Tm(r1(x))� Tm(r2(x)) by Lemma 4.3
6L T n(r1(x))� T n(r2(x)) by (4.8)
6L r1(y)� r2(y) = y by (4.4)
3. If T is dM -nilpotent, r1 and r2 are (dL, dM)-continuous and � is (d2M , dL)-
continuous, then by the first item TE� is (d2L, dL)-continuous and so it just
rests to prove that each x ∈ L\{0L, 1L} is a nilpotent element of TE� . Let
x ∈ L\{0L, 1L} then by hypothesis r1(x), r2(x) ∈ M\{0L, 1L} and since
T is dM -nilpotent, there exists m,n ∈ N such that Tm(r1(x)) = 0M and
T n(r2(x)) = 0M . So, by Equation (4.8), T k(r1(x)) = 0M and T k(r2(x)) =
0M for k = max(m,n). Therefore, by Lemma 4.3, (TE� )k(x) = T k(r1(x))�T k(r2(x)) = 0M � 0M = 0L.
88
4. Let x ∈ L, then
TE� (x, x) = T (r1(x), r1(x))� T (r2(x), r2(x)) by eq. (5.14)
= r1(x)� r2(x) since T is idempotent
= x by eq. (4.4)
5. If TE� (x, y) = 0L then by Equation (5.14) T (r1(x), r1(y))�T (r2(x), r2(y)) =
0L. Thus,
0M = r2(0L) = r2(T (r1(x), r1(y))� T (r2(x), r2(y)))
= T (r1(x), r1(y)) ∨M T (r2(x), r2(y)) by eq. (4.3)
= T (r2(x), r2(y))
As T is positive, or r2(x) = 0M or r2(y) = 0M . If r2(x) = 0M then since r1 6
r2, r1(x) = 0M and hence r1(x) = r1(0L) and r2(x) = r2(0L). Therefore, by
Lemma 4.1, x = 0L. Analogously, if r2(y) = 0M it is possible to prove that
y = 0L.
�
4.3 T-conorms and Fuzzy Negations
Using the same idea as in the Theorem 4.1 it is also possible to extend t-
conorms and fuzzy negations.
Proposition 4.3 Let M EL with respect to (r1, r2, s) and � be an e-operator on
M . Thus,
1. If S is a t-conorm on M then
SE�(x, y) = S(r1(x), r1(y))� S(r2(x), r2(y)) (4.9)
is a t-conorm on L.
2. If N is a fuzzy negation on M then
NE� (x) = N(r1(x))�N(r2(x)) (4.10)
89
4. EXTENSION METHOD VIA E-OPERATORS
is a fuzzy negation on L. Moreover,
(a) If N is involutive (i.e., it satisfies (N3)) then NE� is also involutive.
(b) If a is an equilibrium point of fuzzy negation N then s(a) is an equi-
librium point of NE� .
Proof:
1. Analogous to the proof of Theorem 4.1.
2. Let be x, y ∈ L such that x 6L y. Thus, by monotonicity of r1 and r2,
we have r1(x) 6M r1(y) and r2(x) 6M r2(y) implying that N(r1(y)) 6MN(r1(x)) and N(r2(y)) 6M N(r2(x)) respectively. Hence, by isotonicity of
�, we have NE� (y) = N(r1(y))�N(r2(y)) 6L N(r1(x))�N(r2(x)) = NE
� (x).
Moreover, it follows that
NE� (0L) = N(r1(0L))�N(r2(0L)) = N(0M)�N(0M) = 1M � 1M = 1L
and
NE� (1L) = N(r1(1L))�N(r2(1L)) = N(1M)�N(1M) = 0M � 0M = 0L
Therefore, it can be concluded that NE� is a fuzzy negation on L.
(a) Now, suppose that N is involutive. Thus, for each x ∈ L, we have
NE� (NE
� (x)) = NE� (N(r1(x))�N(r2(x)))
= N(r1(N(r1(x))�N(r2(x))))
�N(r2(N(r1(x))�N(r2(x))))
= N(N(r2(x)))�N(N(r1(x)))
= r2(x)� r1(x)
= r1(x)� r2(x) = x
90
(b) We shall prove that NE� (s(a)) = s(a). Thus,
NE� (s(a)) = N(r1(s(a)))�N(r2(s(a)))
= N(a)�N(a)
= a� a = s(a)
�
One advantage of using extension method via e-operator for extending fuzzy
negations instead of extension method vie retractions is reveled by item 2(a) of
Proposition 4.3 which shows that strong (strict) negations are preserved by this
method.
4.4 De Morgan Triples
Another advantage of extension method via e-operators is that using it we can
extend De Morgan triples for no trivial cases what does not hold for extension
method via retractions as we have seen in Section 3.4.
Proposition 4.4 Let M EL with respect to (r1, r2, s) and � be an e-operator on
M . If 〈T, S,N〉 is a De Morgan triple on M then 〈TE� , SE� , NE� 〉 is a De Morgan
triple.
Proof: Let us recall that NE� (x) = N(r1(x)) � N(r2(x)) and SE�(x, y) =
S(r1(x), r1(y))� S(r2(x), r2(y)) for all x, y ∈ L. Thus, we have
91
4. EXTENSION METHOD VIA E-OPERATORS
NE� (TE� (x, y)) =
= NE� (T (r1(x), r1(y))� T (r2(x), r2(y)))
= N(r1(T (r1(x), r1(y))� T (r2(x), r2(y))))
�N(r2(T (r1(x), r1(y))� T (r2(x), r2(y))))
= N(T (r1(x), r1(y)))�N(T (r2(x), r2(y))) by Lemma 4.1(1)
= S(N(r1(x)), N(r1(y)))� S(N(r2(x)), N(r2(y))) by item 1 def. 4.1.
= S(N(r2(x)), N(r2(y)))� S(N(r1(x)), N(r1(y))) by commut. of �= S(r1(N(r1(x))�N(r2(x))), r1(N(r1(y))�N(r2(y))))
�S(r2(N(r1(x))�N(r2(x))), r2(N(r1(y))�N(r2(y)))) by Lemma 4.1(1)
= S(r1(NE� (x)), r1(N
E� (y)))� S(r2(N
E� (x)), r2(N
E� (y)))
= SE�(NE� (x), NE
� (y))
Analogously, it can be verified that NE� (SE�(x, y)) = TE� (NE
� (x), NE� (y)).
�
It is clear that it is possible to extend De Morgan T -semitriples (or De Morgan
S-semitriples) and triples 〈F,R,N〉 where F is a t-subnorm, R a t-subconorm and
N a fuzzy negation. So, proof of following corollaries are obvious.
Corollary 4.1 Let M E L with respect to (r1, r2, s) and � be an e-operator on
M . If 〈T, S,N〉 is a De Morgan T -semitriple ( De Morgan S-semitriples) on M
then 〈TE� , SE� , NE� 〉 is a De Morgan T -semitriple (or De Morgan S-semitriples)
on L.
Corollary 4.2 Let M E L with respect to (r1, r2, s) and � be an e-operator on
M . If 〈F,R,N〉 is a De Morgan triple on M then 〈FE� , R
E�, N
E� 〉 is a De Morgan
triple on L.
4.5 Extension and Automorphisms
In Section 3.5 we have discussed about relations between extension (using the
method via retractions) of fuzzy connectives and its conjugates. In fact, being
M a (r, s)-sublattice of L, ρ an automorphism on M and T a t-norm (or a t-
conorm S) on M , the results have shown that we have only (T ρ)E 6 (TE)ψ
92
((Sρ)E > (SE)ψ) where ψ is a suitable automorphism on L. This fact happens due
to the extension ρE (via retractions) of automorphism ρ is not an automorphism
on L in general.
Now we investigate this issue again but using extension method via e-operators,
i.e. given M E L with respect to (r1, r2, s), a t-norm T on M and ρ an automor-
phism on M , is the t-norm (T ρ)E generated from the extension of the conjugate
of T equal to the conjugate of the extension of T? In other words, is there an
automorphism ψ on L defined from ρ such that identity (T ρ)E = (TE� )ψ holds?
The main advantage of extension method via e-operators is that now extension
of an automorphism ρ is also an automorphism as one can see in Proposition 4.5.
Definition 4.2 Let MEL with respect to (r1, r2, s) and � be an e-operator on M .
If ρ is an automorphism on M , its extension is given by ρE(x) = ρ◦r1(x)�ρ◦r2(x),
for all x ∈ L.
The following proposition establishes conditions under which ρE is an auto-
morphism on L.
Proposition 4.5 Let M EL with respect to (r1, r2, s) and � be an e-operator on
M and ρ : M −→M an automorphism. Thus,
1. ρE is an automorphism on L;
2. The inverse of ρE is given by
(ρE)−1(x) = ρ−1 ◦ r1(x)� ρ−1 ◦ r2(x)
for all x ∈ L, i.e., (ρE)−1 = (ρ−1)E.
Proof:
1. By Proposition 2.2, we shall prove that ρE is bijective and x 6L y if and only
if ρE(x) 6L ρE(y). Note that ρE is naturally surjective since ρ, r1, r2 and
� are surjective. It remains to prove that ρE is injective. If ρE(x) = ρE(y)
then (ρ ◦ r1(x)) � (ρ ◦ r2(x)) = (ρ ◦ r1(y)) � (ρ ◦ r2(y)). Thus, r1(ρ ◦r1(x))� (ρ◦r2(x)) = r1(ρ◦r1(y))� (ρ◦r2(y)) and hence, by Equation (4.3),
93
4. EXTENSION METHOD VIA E-OPERATORS
ρ ◦ r1(x) = ρ ◦ r1(y). So, r1(x) = ρ−1(ρ(r1(x))) = ρ−1(ρ(r1(y))) = r1(y).
Analogously, it is possible to prove that r2(x) = r2(y). Therefore, by item
4 of Lemma 4.1 we have that x = y, that is, ρE is bijective.
Now, if x 6L y then r1(x) 6M r1(y) and r2(x) 6M r2(y) what allows us
to conclude that ρ(r1(x)) 6M ρ(r1(y)) and ρ(r2(x)) 6M ρ(r2(y)). Since �is isotonic then (ρ ◦ r1(x)) � (ρ ◦ r2(x)) 6L (ρ ◦ r1(y)) � (ρ ◦ r2(y)) and so
ρE(x) 6L ρE(y).
On the other hand, if ρE(x) 6L ρE(y) then (ρ ◦ r1(x)) � (ρ ◦ r2(x)) 6L(ρ ◦ r1(y))� (ρ ◦ r2(y)). Thus, r1(ρ ◦ r1(x))� (ρ ◦ r2(x)) 6L r1(ρ ◦ r1(y))�(ρ ◦ r2(y)) and therefore, by Equation (4.3), ρ ◦ r1(x)) 6M ρ ◦ r1(y)). So,
r1(x) = ρ−1(ρ(r1(x))) 6M ρ−1(ρ(r1(y))) = r1(y). Analogously, it is possible
to prove that r2(x) 6M r2(y). Therefore, by item 3 of Lemma 4.1 it follows
that x 6L y.
2. Now, we will prove that (ρE)−1 : L −→ L given by (ρE)−1(x) = (ρ−1 ◦r1(x))� (ρ−1 ◦ r2(x)) for all x ∈ L, is the inverse of ρE. We shall prove that
ρE ◦ (ρE)−1 = idL and (ρE)−1 ◦ ρE = idL. But
ρE ◦ (ρE)−1(x) = (ρ ◦ r1((ρE)−1(x)))� (ρ ◦ r2((ρE)−1(x)))
= (ρ ◦ r1((ρ−1 ◦ r1(x))� (ρ−1 ◦ r2(x))))�(ρ ◦ r2((ρ−1 ◦ r1(x))� (ρ−1 ◦ r2(x))))
= ρ(ρ−1 ◦ r1(x))� ρ(ρ−1 ◦ r2(x)) by Lemma 4.1(1)
= r1(x)� r2(x)
= x by eq. (4.4)
Analogously, it can be proved that (ρE)−1 ◦ ρE = idL.
�
Lemma 4.4 Let M E L with respect to (r1, r2, s) and � be an e-operator on M .
Thus, given an automorphism ρ : M −→M , ρE satisfies the following properties:
1. For all x ∈M we have that ρE(x� x) = ρ(x)� ρ(x);
2. ρ ◦ r1 = r1 ◦ ρE and ρ−1 ◦ r1 = r1 ◦ (ρE)−1;
94
3. ρ ◦ r2 = r2 ◦ ρE and ρ−1 ◦ r2 = r2 ◦ (ρE)−1.
Proof:
1. By item 2. of Lemma 4.1 we know that x� x = s(x) for all x ∈M . Thus
ρE(x� x) = ρ(r1(x� x))� ρ(r2(x� x))
= ρ(r1(s(x)))� ρ(r2(s(x)))
= ρ(x)� ρ(x)
2. Since r1 6 r2 and ρ is an automorphism, then ρ(r1(x)) 6M ρ(r2(x)) for all
x ∈ L and hence r1(ρ(r1(x)) � ρ(r2(x))) = ρ(r1(x)) by item 1. of Lemma
4.1. Thus, r1 ◦ ρE(x) = r1(ρ(r1(x))� ρ(r2(x))) = ρ(r1(x)) for all x ∈ L.
Analogously, one can prove that ρ−1 ◦ r1 = r1 ◦ (ρE)−1.
3. Similar to item 2.
�
The following theorem gives an answer to the problem presented at the be-
ginning of this section.
Theorem 4.2 Let M E L with respect to (r1, r2, s) and � be an e-operator on
M . Then, given an automorphism ρ and a t-norm T both defined on M , it is
possible to define an automorphism ψ on L such that (T ρ)E = (TE� )ψ.
Proof: Take as ψ the function ρE as given in Definition 4.2. Note that
r1(ψ(x)) 6M r2(ψ(x)) for all x ∈ L and hence
T (r1(ψ(x)), r1(ψ(y))) 6M T (r2(ψ(x)), r2(ψ(y)))
Thus, by item 1. of Lemma 4.1 we have that
r1(T (r1(ψ(x)),r1(ψ(y)))�T (r2(ψ(x)),r2(ψ(y))))=T (r1(ψ(x)),r1(ψ(y))) (4.11)
95
4. EXTENSION METHOD VIA E-OPERATORS
and
r2(T (r1(ψ(x)),r1(ψ(y)))�T (r2(ψ(x)),r2(ψ(y))))=T (r2(ψ(x)),r2(ψ(y))) (4.12)
Therefore, for all x, y ∈ L it follows that
(TE� )ψ(x, y) =
= ψ−1(TE� (ψ(x), ψ(y)))
= ψ−1(T (r1(ψ(x)), r1(ψ(y)))� T (r2(ψ(x)), r2(ψ(y))))
= ρ−1(r1(T (r1(ψ(x)), r1(ψ(y)))� T (r2(ψ(x)), r2(ψ(y)))))�ρ−1(r2(T (r1(ψ(x)), r1(ψ(y)))� T (r2(ψ(x)), r2(ψ(y))))
= ρ−1(T (r1(ψ(x)), r1(ψ(y))))� ρ−1(T (r2(ψ(x)), r2(ψ(y)))) by (4.11) and (4.12)
= ρ−1(T (ρ(r1(x)), ρ(r1(y))))� ρ−1(T (ρ(r2(x)), ρ(r2(y)))) by Lemma 4.4
= T ρ(r1(x), r1(y))� T ρ(r2(x), r2(y))
= (T ρ)E(x, y)
�
Results similar to those proved above can be demonstrated to t-conorms and
fuzzy negations, i.e., taking a t-conorm S and a fuzzy negation N both defined
on M , under the same conditions as in Theorem 4.2 we have that (Sρ)E = (SE�)ψ
and (Nρ)E = (NE� )ψ.
Proposition 4.6 Let M and L be two bounded lattices such that M E L with
respect to (r1, r2, s) and � be an e-operator on M . If 〈T, S,N〉 is a De Morgan
triple on M and ρ is an automorphism on M , then 〈(TE� )ρE, (SE�)ρ
E, (NE
� )ρE〉 is
a De Morgan triple on L.
Proof: Analogous to the proof of Theorem 7 in da Costa et al. [2011].
�
96
4.6 On Extension of n-dimensional T-norms
The n-dimensional fuzzy sets theory has been studied as a way to generalize
the fuzzy sets theory valued to the simplex Ln([0, 1]) = {x = (x1, x2, . . . , xn) ∈[0, 1]n | x1 6 x2 6 · · · 6 xn} for a fixed n ∈ N−{0} (see Shang et al. [2010]). Thus,
it is natural to think about the fuzzy operators (t-norms, t-conorms and fuzzy
negations) on Ln([0, 1]). For n = 2, a good formalization about interval-valued
fuzzy logic is given by Deschrijver and partners in Deschrijver [2006, 2008, 2011].
Recent studies for arbitrary n has been done by Bedregal et al. [2006a] where it is
done a formalization of n-dimensional aggregation functions, particulary t-norms,
fuzzy negations and automorphisms on Ln([0, 1]).
Based on this framework, an interesting issue that arises is that generalizing
lattice-valued aggregation functions to higher dimension using a bounded lattice
L instead of [0, 1] on the definition of Ln([0, 1]).
One can naturally define a lattice version of the set Ln([0, 1]), namely
Ln(L) = {x = (x1, x2, . . . , xn) ∈ Ln | x1 6L x2 6L · · · 6L xn} (4.13)
where L is a bounded lattice.
For each x,y ∈ Ln(L) we define by
x ∧ y = (x1 ∧L y1, x2 ∧L y2, . . . , xn ∧L yn)
and
x ∨ y = (x1 ∨L y1, x2 ∨L y2, . . . , xn ∨L yn)
the meet and join operations on Ln(L), respectively.
Denote /x/ = (x, x, . . . , x) for each x ∈ L. Thus, /0L/ and /1L/ are a
bottom and a top element of Ln(L). As an easy exercise one can prove that
〈Ln(L),∧,∨, /0L/, /1L/〉 is a bounded lattice.
A partial order on Ln(L) is given by
x 6 y ⇔ xi 6L yi for each i = 1, 2, . . . , n
The remainder of this section is devoted to define t-norms on Ln(L) (called
97
4. EXTENSION METHOD VIA E-OPERATORS
n-dimensional t-norms on L) constructed from t-norms on L, and to provide
a generalization of the extension proposed in Theorem 4.1 for n-dimensional t-
norms.
Proposition 4.7 Bedregal et al. [2006a] Let T1, T2, . . . , Tn : [0, 1]× [0, 1]→ [0, 1]
be t-norms such that T1 6 T2 6 · · · 6 Tn. Then
˜T1 · · ·Tn(x,y) = (T1(x1, y1), . . . , Tn(xn, yn)) (4.14)
is an n-dimensional t-norm. In case that T1 = T2 = · · · = Tn we denote ˜T1 · · ·Tnby TT .
A similar result can be easily shown considering a bounded lattice L instead
of [0, 1] in Proposition 4.7.
We decided to work in this section just with TT n-dimensional t-norm, but
every result presented here remains valid for ˜T1 · · ·Tn.
Corollary 4.3 Let M E L with respect to (r1, r2, s), � be an e-operator on M
and T be a t-norm on M . Then TTE� : Ln(L)2 −→ Ln(L) given by
TTE� (x,y) = (TE� (x1, y1), TE� (x2, y2), . . . , T
E� (xn, yn))
is a n-dimensional t-norm on Ln(L).
Proof: Straightforward from Theorem 4.1 and Proposition 4.7.
�
Proposition 4.8 Let M be a (r, s)-sublattice of L. Then
1. Ln(M) is a (r, s)-sublattice of Ln(L);
2. If M is a lower (upper) (r1, s)-sublattice of L then Ln(M) is a lower (upper)
(r1, s)-sublattice of Ln(L);
3. If M E L with respect to (r1, r2, s) then Ln(M) E Ln(L) with respect to
(r1, r2, s) where r1, r2 and s are suitable homomorphisms defined from r1, r2
and s respectively.
98
Proof:
1. Since M is a (r, s)-sublattice of L, by Definition 2.10 there is a retraction
r : L −→M with a pseudo-inverse s : M −→ L such that r◦s = idM . Define
a n-dimensional function r : Ln(L) −→ Ln(M) given for each x ∈ Ln(L) by
r(x) = (r(x1), r(x2), . . . , r(xn)) (4.15)
We claim that r is a n-dimensional retraction such that its pseudo-inverse
is an n-dimensional function s : Ln(M) −→ Ln(L) given by
s(x) = (s(x1), s(x2), . . . , s(xn))
for each x ∈ Ln(M).
Indeed, it is clear that r and s are n-dimensional homomorphisms since r
and s are. Moreover, for each x ∈ Ln(M), we have
r ◦ s(x) = r(s(x1), s(x2), . . . , s(xn))
= (r(s(x1)), r(s(x2)), . . . , r(s(xn)))
= (x1, x2, . . . , xn) = x
Thus r ◦ s = idLn(M) and hence Ln(M) is a (r, s)-sublattice of Ln(L) by
Definition 2.10.
2. If M is a lower (r1, s)-sublattice of L then s ◦ r1 6 idL. We shall prove that
s ◦ r1 6 idLn(L). Thus, for each x ∈ Ln(L) it follows that
s ◦ r1(x) = s(r1(x1), r1(x2), . . . , r1(xn))
= (s(r1(x1)), s(r1(x2)), . . . , s(r1(xn)))
6 (x1, x2, . . . , xn)
Analogously, one can prove that Ln(M) is an upper (r2, s)-sublattice of
Ln(L) assuming that M is an upper (r2, s)-sublattice of L.
3. Suppose thatMEL. Thus, there are a lower and an upper retractions r1 and
r2 from L onto M with the same pseudo-inverse s : M −→ L. Therefore, by
99
4. EXTENSION METHOD VIA E-OPERATORS
items 1. and 2. it is easy to check that r1(x)=(r1(x1), r1(x2), . . . , r1(xn)) and
r2(x) = (r2(x1), r2(x2), . . . , r2(xn)) for each x ∈ Ln(L) are n-dimensional
lower and upper retractions with the same n-dimensional pseudo-inverse
s(x) = (s(x1), s(x2), . . . , s(xn)) for each x ∈ Ln(M) according to which it
can be inferred that Ln(M)E Ln(L) with respect to (r1, r2, s).
�
Proposition 4.9 Let M EL with respect to (r1, r2, s) and let � be an e-operator
on M . The function �n : Ln(M)× Ln(M) −→ Ln(L) given by
a�n b = (a1 � b1, a2 � b2, . . . , an � bn) (4.16)
for all a,b ∈ Ln(M) is an e-operator on Ln(M).
Proof: Straightforward from Proposition 4.8 and from the fact that � is an
e-operator on M .
�
Corollary 4.4 Let M E L with respect to (r1, r2, s) and � be an e-operator on
M . If T is a t-norm on Ln(M) then the function TE� : Ln(L)× Ln(L) −→ Ln(L)
given by
TE�(x,y) = T(r1(x), r1(y))�n T(r2(x), r2(y))
for all x,y ∈ Ln(L), is a t-norm on Ln(L).
Proof: Straightforward from Theorem 4.1 and Propositions 4.8 and 4.9.
�
Theorem 4.3 Let M E L with respect to (r1, r2, s), � be an e-operator on M
and T be a t-norm on M . Then (TT )E� = TTE� .
100
Proof: Take x,y ∈ Ln(L). Then,
(TT )E�(x,y) = TT (r1(x), r1(y))�n T(r2(x), r2(y)) by (5.14)
= TT ((r1(x1), . . ., r1(xn)),(r1(y1), . . . ,r1(yn))�n
TT ((r2(x1), . . . , r2(xn)), (r2(y1), . . . , r2(yn)) by eq.(4.15)
= (T (r1(x1), r1(y1)), . . . , T (r1(xn), r1(yn)))�n
(T (r2(x1), r2(y1)), . . . , T (r2(xn), r2(yn))) by eq. (4.14)
= (T (r1(x1), r1(y1))� T (r2(x1), r2(y1)), . . . ,
T (r1(xn), r1(yn))� T (r2(xn), r2(yn)) by eq. (4.16)
= (TE� (x1, y1), . . . , TE� (xn, yn)) by eq. (5.14)
= TTE� (x,y) by eq. (4.14)
�
Let CM be the set of all t-norms T on M (similar to CL, CLn(M) and CLn(L)).
The theorem above shows that the following diagram is commutative:
CMTE� //
TT
��
CL
TTE�
��CLn(M)
(TT )E� // CLn(L)
A generalization to higher dimension of the concepts of t-conorm and fuzzy
negation may be done as we did here for t-norms. Results similar to those in
Corollary 4.4 and Theorem 4.3 can be proved for n-dimensional t-conorms and
n-dimensional fuzzy negations as done in Section 4.3.
Proposition 4.10 Let L be a bounded lattice. For all n ∈ N\{0} we have
Lm(L) E Ln(L) with respect to some (r1, r2, s) when n = 2m. Moreover, the
mapping � : Lm(L)× Lm(L) −→ Ln(L) defined by
(x1, . . . , xm)� (y1, . . . , ym) = (x1 ∧L y1, x1 ∨L y1, . . . , xm ∧L ym, xm ∨L ym)
is an e-operator.
101
4. EXTENSION METHOD VIA E-OPERATORS
Proof: We shall present a n-dimensional lower retraction r1 : Ln(L) −→ Lm(L),
a n-dimensional upper retraction r2 : Ln(L) −→ Lm(L) and a n-dimensional
pseudo-inverse s : Lm(L) −→ Ln(L) such that s ◦ r1 6 idLn(L) 6 s ◦ r2. Let
r1(x1, . . . , xn) = (x1, x3, x5, x7, . . . , xn−1)
and
r2(x1, . . . , xn) = (x2, x4, x6, x8 . . . , xn)
for all (x1, . . . , xn) ∈ Ln(L). It is clear that r1 and r2 are homomorphisms.
Moreover, the function given by s(x1, x2, . . . , xm) = (x1, x1, x2, x2, . . . , xm, xm) is
a homomorphism such that
s ◦ r1(x1, . . . , xn) = s(x1, x3, . . . , xn−1)
= (x1, x1, x3, x3, . . . , xn−1, xn−1)
6 (x1, x2, x3, x4, . . . , xn)
andr1 ◦ s(x1, x2, . . . , xm) = r1(x1, x1, x2, x2, . . . , xm, xm)
= (x1, x2, . . . , xm)
= idLm(L)(x1, x2, . . . , xm)
Therefore, r1 is a n-dimensional lower retraction which pseudo-inverse is s.
Analogously, it can be proved that r2 is a n-dimensional upper retraction with
pseudo-inverse s.
On the other hand,
1. clearly � is isotonic;
2. r1((x1, . . . , xm)� (y1, . . . , ym)) = r1(x1 ∧L y1, x1 ∨L y1, . . . , xm ∧L ym, xm ∨Lym) = (x1 ∧L y1, . . . , xm ∧L ym) = (x1, . . . , xm) ∧Lm(L) (y1, . . . , ym);
3. r2((x1, . . . , xm)� (y1, . . . , ym)) = (x1, . . . , xm) ∨Lm(L) (y1, . . . , ym); and
4. r1(x1, . . . , xn)�r2(x1, . . . , xn) = (x1, x3, . . . , xn−1)�(x2, x4, . . . , xn) = (x1∧Lx2, x1 ∨L x2, . . . , xn−1 ∧L xn, xn−1 ∨L xn) = (x1, x2, . . . , xn).
Therefore, � is an e-operator.
102
�
Notice that there are other possibilities for m and n such that Lm(L)ELn(L).
For example, m = 5 and n = 8. In this case r1(x1, . . . , x8) = (x1, x3, x5, x7, x8),
r2(x1, . . . , x8) = (x1, x2, x4, x6, x8) and s(x1, . . . , x5) = (x1, x2, x2, x3, x3, x4, x4,
x5).
Corollary 4.5 Let T be a t-norm on Lm(L) and n = 2m. The function TE� :
Ln(L)× Ln(L) −→ Ln(L) given by
TE�(x,y) = T(r1(x), r1(y))� T(r2(x), r2(y))
for all x,y ∈ Ln(L), is a t-norm on Ln(L).
Proof: Straightforward from Theorem 4.1 and Proposition 4.10.
�
4.7 Final Remarks
Now, considering the same symbols as in Section 3.7 and the following ab-
breviations shown in Table 4.1 we present here the main properties preserved by
extension method via e-operators.
Notice that we have not studied the extension of implications and negations
obtained from fuzzy connectives using method via e-operators what we pretend
to do as soon as possible.
103
4. EXTENSION METHOD VIA E-OPERATORS
Abbreviation Property
AC Archimedean
NI Nilpotent
IDN Idempotent
ZD Zero divisor
CL Cancellation law
C Continuity
P Positive
EC Relation between extension and conjugate
RE Relation between NKE�
and (NK)E� with K a fuzzy connective
Table 4.1: Table of Abbreviations
t-norm and t-conorm AC NI IDN ZD CL C P EC
TE�√ √ √ √ √ √ √
(T ρ)E� = (TE� )ψ
SE�√ √ √ √ √ √ √
(Sρ)E� = (SE�)ψ
Table 4.2: Properties preserved by TE� and SE�
fuzzy negation Strong Strict Eq. P. EC RE
NE�
√ √ √(Nρ)E� = (NE
� )ψ –
(NT )E� ? ? ? ? ?
(NS)E� ? ? ? ? ?
(NI)E� ? ? ? ? ?
Table 4.3: Properties preserved by NE�
104
Chapter 5
On Restricted Equivalence
Functions
It is well known that fuzzy sets theory provides an interesting framework for
several areas of knowledge. Particularly, regarding to the treatment of images
(pattern recognition, image segmentation, etc.) there is a huge amount of papers
which present important results and methods considering fuzzy sets valued on
[0, 1] (see for example Bustince et al. [2007]; Chaira and Ray [2003, 2005]; der
Weken et al. [2004]; Lopez-Molina et al. [2011]).
On image processing, a problem is to provide a good measure tool for making a
global comparison of images. In this scope, Bustince et al. introduced in Bustince
et al. [2006] the notion of restricted equivalence functions on [0, 1] (for short REF)
as a particular case of equivalence functions defined in Fodor and Roubens [1994].
Also, in Bustince et al. [2006] it is presented a theorem that characterizes REF
via implications, i.e. a way to construct these kind of functions using fuzzy
implications.
It is natural to generalize concepts related to image processing for lattices
in order to obtain a much more general framework than [0, 1]. In particular
bounded lattices are of interest since intensities in images can be considered as
taking values in such lattices.
Notice that, in particular, restricted equivalence functions are fuzzy operators
so, we can apply our two methods to extend them and see which one works better.
105
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
We start discussing about how to define REF on bounded lattices in order to
provide a construction method of these functions via implications (in Section 5.5,
we study similar issues for REF on L([0, 1])). We also work on defining restricted
dissimilarity functions and normal Ee,N -functions on bounded lattice. Finally, we
extend these functions via retractions and e-operators, in Sections 5.6 and 5.7.
5.1 Restricted Equivalence Functions on L
The problem of global comparison of two images has been studied by several
researchers in image processing (see Baddeley [1992]; Bustince et al. [2006, 2007];
der Weken et al. [2004]). One of the most used tools for this global comparison
is the equivalence functions introduced in Fodor and Roubens [1994].
Definition 5.1 A function EF : [0, 1]2 → [0, 1] is called equivalence function if
the following properties hold:
1. EF (x, y) = EF (y, x) for all x, y ∈ [0, 1];
2. EF (0, 1) = EF (1, 0) = 0;
3. EF (x, x) = 1 for all x ∈ [0, 1];
4. If x 6 x′ 6 y′ 6 y then EF (x, y) 6 EF (x′, y′).
As a particular class of these kind of functions, in Bustince et al. [2006] is
defined the notion of restricted equivalence functions.
Definition 5.2 A function REF : [0, 1]2 → [0, 1] is called a restricted equivalence
function if it satisfies the following conditions:
(1) REF (x, y) = REF (y, x) for all x, y ∈ [0, 1];
(2) REF (x, y) = 1 if and only if x = y;
(3) REF (x, y) = 0 if and only if x = 1 and y = 0, or x = 0 and y = 1;
(4) REF (x, y) = REF (N(x), N(y)) for all x, y ∈ [0, 1], N being a strong negation
on [0, 1];
(5) For all x, y, z ∈ [0, 1] such that x 6 y 6 z then REF (x, z) 6 REF (x, y) and
REF (x, z) 6 REF (y, z).
106
It is clear that every restricted equivalence function is an equivalence function
in the sense of Definition 5.1. But the reciprocal of this affirmation is not true.
For instance, the function EF : [0, 1]2 → [0, 1] given by
EF (x, y) =
{0, if x = 0 and y = 1 or x = 1 and y = 0;
1, otherwise.
for all x, y ∈ [0, 1], is an equivalence function but it is not a restricted equivalence
function (see Example 2 in Bustince et al. [2006]).
Naturally, concept of equivalence functions can be generalized for bounded
lattices as follows.
Definition 5.3 Let L be a bounded lattice. A function EF : L2 → L is called an
equivalence if it satisfies the following conditions:
(F1) EF (x, y) = EF (y, x) for all x, y ∈ L;
(F2) EF (0L, 1L) = EF (1L, 0L) = 0L;
(F3) EF (x, x) = 1L for all x ∈ L;
(F4) If x 6L y 6L z then EF (x, y) 6L EF (x, z).
The main goal of this section is presenting a formalization of the concept of
restricted equivalence functions for a bounded lattice L. A first problem arises
from the fact that, in general, it does not need to exist a strong negation for a given
lattice L. On the other hand, for many applications, specially in image processing,
it is crucial that an analog of (4) in Definition 5.2 holds, since it ensures the fact
that a given property is preserved when the negative of an image instead of the
image itself is considered. For this reason, we introduce the following definition.
Definition 5.4 Let N be a frontier negation on L. A function REF : L2 → L is
called a restricted equivalence function on L with respect to N , or just an L-REF
with respect to N , if it satisfies, for all x, y, z ∈ L, the following conditions:
(L1) REF (x, y) = REF (y, x);
107
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
(L2) REF (x, y) = 1L if and only if x = y;
(L3) REF (x, y) = 0L if and only if x = 1L and y = 0L, or x = 0L and y = 1L;
(L4) REF (x, y) = REF (N(x), N(y));
(L5) if x 6L y 6L z then REF (x, z) 6L REF (x, y).
Notice for a given lattice L a frontier negation always exists, as Example 2.9
shows. On the other hand, the requirement of N being a frontier negation can
no be weakened since otherwise a contradiction between (L3) and (L4) arises.
In any cases, and with an eye on possible applications, we will mostly deal with
REF’s defined with respect to a strong negation.
Example 5.1 Let L be a lattice with at least three elements, and take x0 ∈L\{0L, 1L}. Then we can define
REF (x, y) =
0L if x = y;
1L if {x, y} = {0L, 1L};x0 otherwise.
which is a restricted equivalence function with respect to any frontier negation N .
Notice that from (L4), (L5) and (L1) it is also possible to conclude that
REF (x, z) 6L REF (y, z) whenever x 6L y 6L z.
Example 5.2 Let M be a bounded lattice (see Figure 2.5) and N1 be a strong
M-negation as in Example 2.8. Thus, a function REF : M2 → M as defined in
the Table 5.1 is a L-REF with respect to N1 in the sense of Definition 5.4.
REF 0L x y 1L
0L 1L x y 0Lx x 1L x yy y x 1L x1L 0L y x 1L
Table 5.1: Restricted equivalence function on lattice M
Notice that in this case, the mapping N(x) = REF (0L, x) defines a strong
negation on the lattice M .
108
Proposition 5.1 Let REF be an L-REF with respect to N and ψ be an auto-
morphism on L. Then ψ ◦REF also is a restricted equivalence function also with
respect to N .
Proof: Analogous to [Bustince et al., 2006, Prop. 3].
�
Proposition 5.2 Let f : L2 → L be a commutative function and REF be an
L-REF with respect to a strong L-negation N . Then the function REFf : L2 → L
defined by
REFf (x, y) =
{REF (x, y) if x ¨ y
f(x, y) otherwise(5.1)
is an L-REF with respect to N .
Proof: Straightforward.
�
5.1.1 Restricted Equivalence Functions and Negations
The remark in Example 5.2 is just a particular case of the following general
result.
Proposition 5.3 Let REF be a restricted equivalence function on the lattice L
with respect to some frontier negation N . Then, the mapping
N0(x) = REF (0L, x)
is also a frontier negation on L.
Proof: That N0 satisfies (N1) is trivial. Let’s prove that (N2) holds for it.
Indeed, for each x, y ∈ L such that x 6L y since 0L 6L x 6L y by (L5) it follows
that N0(y) = REF (0L, y) 6L REF (0L, x) = N0(x).
�
109
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
Example 5.3 Notice that sometimes a given function can be a L-REF with re-
spect to some negation but not with respect to another one. For instance, consider
the Example 5.2. In this case, following the notation of Proposition 5.3, we have
that N0(0L) = 1L; N0(x) = x, N0(y) = y and N0(1L) = 0L, which is different
from the negation N1 considered for the definition of the REF. But for this N0
we see that REF (0L, x) = x 6= y = REF (1L, x), so property (L4) does not hold.
Corollary 5.1 Let REF be a restricted equivalence function on L with respect
to some frontier negation N and let N0 be defined as in Proposition 5.3. Then,
N0 = g ◦N , with g : L→ L defined as g(x) = REF (x, 1L).
Proof: From (L4) in the definition of restricted equivalence functions, we see
that N0(x) = REF (0L, x) = REF (1L, N(x)). From the symmmetry of REF the
first part of the statement follows.
�
Corollary 5.2 Let L be a lattice and REF be a restricted equivalence function on
L with respect to a frontier negation N such that g(x) = REF (x, 1L) is injective.
Then N is unique in the sense that if, for every x, y ∈ L, the identity REF (x, y) =
REF (N ′(x), N ′(y)) holds for some other frontier negation N ′, it follows that
N = N ′.
Proof: LetN ′ be a frontier negation such thatREF (x, y) = REF (N ′(x), N ′(y))
holds for every x, y ∈ L. Then, we have that REF (0L, x) = g(N ′(x)), so, as g is
injective, N ′(x) = g−1(REF (0L, x)) for every x ∈ L. In the same way we have
also that N(x) = g−1(REF (0L, x)). Thus, the result follows.
�
Example 5.4 Let’s consider again the lattice L of Example 5.2. Take as REF
the function defined in Example 5.1 with x0 = y. This is a restricted equiva-
lence function with respect to any strong negation defined on L, so in partic-
ular, for N = N1. But, with the notation of Proposition 5.3, we have that
N0(x) = REF (0L, x) = y = REF (0L, y) = N0(y), so N0 is not injective and
hence it is not strong.
110
Theorem 5.1 Let L be a lattice and N a strong L-negation. Then the mapping
REF (x, y) =
{1 if x = y ,
N((x ∨L y) ∧L (N(x) ∨L N(y))) otherwise.
is a restricted equivalence function with respect to N .
Proof: (L1) and (L2) are obvious.
(L3) REF (0L, 1L) = N((0L∨1L)∧ (N(0L)∨N(1L)))= N(1L∧1L) = N(1L) = 0L
and the same is valid for REF (1L, 0L). On the other hand, if REF (x, y) = 0L,
as N is a frontier negation, this means that (x ∨ y) ∧ (N(x) ∨ N(y)) = 1L. So
x ∨ y = 1L and N(x) ∨ N(y) = 1L. From the first identity it follows that either
x or y or both are equal to 1L whereas from the second identity either x = 0L or
y = 0L or both are equal to 0L. So the only possibility is {x, y} = {0, 1}.(L4) is straightforward.
(L5) Take x, y, z ∈ L such that x ≤L y ≤L z. Then REF (x, y) = N(y ∧ N(x))
and REF (x, z) = N(z ∧ N(x)). Since y ∧ N(x) ≤L z ∧ N(x), it follows that
REF (x, z) ≤L REF (x, y), as we wanted to see.
�
5.2 Characterization Theorem for L-REF
We present in this section a generalized method of constructing L-REF’s based
on fuzzy implications (see Theorem 7 of Bustince et al. [2008]). We start doing a
axiomatization of the concept of implications on bounded lattices.
The following theorem is a weak version of [Bustince et al., 2006, Theorem 7]
for the framework of bounded lattices.
Theorem 5.2 Let N : L2 → L be a strong L-negation and M : L2 → L be a
function such that, for all x, y ∈ L, it holds:
(M1) M(x, y) = M(y, x);
(M2) M(x, 1L) = x;
(M3) M(x, y) = 1L if and only if x = y = 1L;
(M4) M(x, y) = 0L if and only if x = 0L or y = 0L.
111
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
If there exist a function I : L2 → L satisfying (FPA), (OP), (CP) for N and (P)
then the function REF : L2 → L defined by
REF (x, y) = M(I(x, y), I(y, x)) (5.2)
is a L-REF with respect to N .
Proof: We shall prove that the conditions (L1) − (L5) hold. It is clear that
(M1) implies (L1). The proof for the others properties is given as follow:
(L2)
Suppose that REF (x, y) = 1L. Thus, by (5.2) we have that M(I(x, y), I(y, x)) =
1L and hence I(x, y) = 1L and I(y, x) = 1L by (M3). Therefore, by (OP ) it
follows that x 6L y and y 6L x, i.e. x = y.
Reciprocally, if x = y, that is x 6L y and y 6L x then I(x, y) = I(y, x) = 1L by
(OP ). Thus, it is easy to see that REF (x, y) = 1L.
(L3)
If REF (x, y) = 0L then M(I(x, y), I(y, x)) = 0L which allow us to conclude that
either I(x, y) = 0L or I(y, x) = 0L since M satisfies (M3). Thus, by (P ) we have
that either x = 1L and y = 0L or x = 1L and y = 0L.
On the other hand, if x = 1L and y = 0L then I(x, y) = 0L by (P ). Therefore,
REF (x, y) = M(I(x, y), I(y, x)) = M(0L, I(y, x)) = 0L by (M3). Analogously it
can be proved that REF (x, y) = 0L whenever x = 0L and y = 1L.
(L4)
For all x, y ∈ L we have that
REF (N(x), N(y)) = M(I(N(x), N(y)), I(N(y), N(x))) by (5.2)
= M(I(y, x), I(x, y)) by (CP )
= M(I(x, y), I(y, x)) by (M1)
= REF (x, y)
(L5)
Given x, y, z ∈ L such that x 6L y 6L z we have that I(x, y) = 1L and I(x, z) =
1L. Thus
REF (x, y) = M(I(x, y), I(y, x)) = M(1L, I(y, x)) = I(y, x) (5.3)
112
and
REF (x, z) = M(I(x, z), I(z, x)) = M(1L, I(y, x)) = I(z, x) (5.4)
Since y 6L z then by (FPA) it follows that I(z, x) 6L I(y, x) for all x ∈ L.
Therefore, considering this fact, by Equations (5.3) and (5.4) it can be concluded
that REF (x, z) 6L REF (x, y).
�
Example 5.5 Let L be the bounded lattice shown in Figure 2.5. If I : L2 → L
is the function defined in the Example 2.12 and M : L2 → L is a function given
by M(x, y) = min{x, y} for all x, y ∈ L then, by Theorem 5.2, REF (x, y) =
M(I(x, y), I(x, y)) is a restricted equivalence function on L (see Table 5.2).
REF 0L a b c d e 1L
0L 1L e b c d a 0a e 1L e e e e ab b e 1L e e e bc c e e 1L e e cd d e e e 1L e de a e e e e 1L e
1L 0L a b c d e 1L
Table 5.2: Restricted equivalence function on lattice L
It is worth noting that the reciprocal of Theorem 5.2 does not hold, in gen-
eral. In other words, given an L-REF with respect to a strong L-negation N
and a function M : L2 → L satisfying (M1), (M2) and (M3) it is not always
possible to define a function I : L2 → L which satisfy (FPA), (OP ), (CP ), (P )
and REF (x, y) = M(I(x, y), I(y, x)). This is due to there is not a specific way
to define restricted equivalence function for pairs (x, y) ∈ L2 such that x ‖ y(incomparable elements of L). According to properties (L2) and (L3) we can just
infer that 0L <L REF (x, y) <L 1L if x ‖ y.
It means that determinate how to define a general condition for L-REF on
incomparable elements of L2 in order to make the reciprocal of the Theorem 5.2
holds constitute a very interesting open problem to be studied.
113
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
Proposition 5.4 Let REF be an L-REF related to a strong L-negation N . Then
the function IREF : L2 → L defined by
IREF (x, y) =
{1L if x 6L y
REF (x, y) otherwise(5.5)
satisfies the properties (OP), (CP) and (P).
Proof: .
(OP )
Let x, y ∈ L. If x 6L y then, by definition, it is trivial that IREF (x, y) = 1L.
Reciprocally, if IREF (x, y) = 1L then either x 6L y or REF (x, y) = 1L and
y <L x or x ‖ y. But, note that if REF (x, y) = 1L then x = y by (L2) which is
a contradiction with both y <L x and x ‖ y. Therefore, it can be concluded that
x 6L y.
(CP )
Note that if x 6L y we have N(y) 6L N(x) and hence IREF (x, y) = 1L =
IREF (N(y), N(x)) by definition of IREF . Beyond that we can have y <L x or x ‖ ywhich implies that N(x) <L N(y) and N(x) ‖ N(y) respectively by Proposition
2.12. Thus in both cases it follows that IREF (N(y), N(x)) = REF (N(y), N(x)) =
REF (N(x), N(y)) and IREF (x, y) = REF (x, y) what allow us to conclude that
IREF (N(y), N(x)) = IREF (x, y) since REF (N(x), N(y)) = REF (x, y) by (L4).
(P )
If IREF (x, y) = 0L then either REF (x, y) = 0L and y <L x or REF (x, y) = 0L
and x ‖ y. But, the second case is a contradiction by (L3). Hence we must have
REF (x, y) = 0L and y <L x and again by (L3) it follows that x = 1L and y = 0L.
Reciprocally, if x = 1L and y = 0L then IREF (x, y) = REF (x, y) = REF (1L, 0L) =
0L.
�
Remark 5.1 Notice that if REF and IREF are functions as in Proposition 5.4
and M : L2 → L is a function as in Theorem 5.2 (satisfying also M(x, x) = x
for all x ∈ L ) then the Identity (5.2) holds.
114
Indeed, let x, y ∈ L and suppose that they are comparable (i.e. x ¨ y). In this
case, if x 6L y then IREF (x, y) = 1L and IREF (y, x) = REF (y, x) = REF (x, y).
Thus
REF (x, y) = IREF (y, x) = M(1L, IREF (y, x)) = M(IREF (x, y), IREF (y, x))
Analogously, one can prove that (5.2) holds if y 6L x.
On the other hand, if x ‖ y then IREF (x, y) = REF (x, y) and IREF (y, x) =
REF (y, x) = REF (x, y). Therefore
REF (x, y) = M(REF (x, y), REF (y, x)) = M(IREF (x, y), IREF (y, x))
Example 5.6 Here we show that the reciprocal of Theorem 5.2 does not hold
in general. Actually, we would like to highlight the fact that the property (FPA)
fails for the function IREF defined in (5.5) (by Theorem 7 in Bustince et al.
[2006] it known that IREF should satisfy (FPA), (OP), (CP) and (P)). To do
so, consider the bounded lattice M (see Figure 1) and the restricted equivalence
function defined in Example 5.2. In this case, the function IREF in (5.5) is defined
as in Table 5.3. Note that y < 1M but IREF (1M , x) = y and IREF (y, x) = x which
means that IREF (1M , x) ‖ IREF (y, x) since x ‖ y.
IREF 0L x y 1L
0L 1L 1L 1L 1Lx x 1L x 1Ly y x 1L 1L1L 0L y x 1L
Table 5.3: The function IREF on lattice M
Taking into account the important fact highlighted above, we can see that the
problem of proving the reciprocal of Theorem 5.2 is reduced to seek conditions
under which the function IREF as in Proposition 5.4 satisfies (FPA). Clearly
if all the elements of the bounded lattice L are comparable then the problem is
sold, it means:
115
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
Theorem 5.3 Let L be a total bounded lattice and suppose that M : L2 → L is
a function satisfying (M1), (M2), (M3) and M(x, x) = x for all x ∈ L. Thus a
function REF : L2 → L is an L-REF for a strong L-negation N if and only if
there exist a function IREF : L2 → L satisfying (FPA), (OP), (CP) for N and
(P) in such a way that the Equation (5.2) holds.
Proof: Let REF be a L-REF with respect to a strong negation N . Define:
IREF (x, y) =
{1L if x 6L y ;
REF (x, y) otherwise.
From Proposition 5.4 and Remark 5.1 we can conclude that IREF satisfies (OP ),
(CP ), (P ) and REF (x, y) = M(IREF (x, y), IREF (y, x)) for all x, y ∈ L. Thus, it
remains to prove that property (FPA) holds.
Take y ∈ L and x, z ∈ L such that x ≤L z. We want to see that IREF (z, y) ≤LIREF (x, y). There are three possibilities.
(i) If y ≤L x ≤L z we have that IREF (z, y) = REF (z, y) and IREF (x, y) =
REF (x, y). But by (L5), REF (z, y) ≤L REF (x, y) and hence IREF (z, y) �IREF (x, y).
(ii) When x ≤L y ≤L z it follows that IREF (z, y) = REF (z, y) ≤L 1L =
IREF (x, y).
(iii) If x ≤L z ≤L y then IREF (z, y) = 1L = IREF (x, y).
The reciprocal is straightforward from Theorem 5.2.
�
5.3 Restricted Dissimilarity Functions
The task of providing suitable ways to define metric functions to measure the
similarity or dissimilarity between two images is a very studied problem. In this
sense, Bustince et al. [2008] introduce the concept of restricted dissimilarity func-
tions valued on [0, 1] based on the concept of dissimilarity proposed by Baddeley
[1992]. Moreover, in Bustince et al. [2008] is proposed a method to construct
these kind of functions from restricted equivalence functions.
116
In this section we generalize the definition of restricted dissimilarity functions
given in Bustince et al. [2008] for bounded lattices and prove some related results.
Definition 5.5 Let L be a bounded lattice. A function dL : L2 → L is called
a restricted dissimilarity function on L (for short L-RDF) if it satisfies, for all
x, y, z ∈ L, the following conditions:
(D1) dL(x, y) = dL(y, x);
(D2) dL(x, y) = 1L if and only if either x = 1L and y = 0L or x = 0L and y = 1L;
(D3) dL(x, y) = 0L if and only if x = y;
(D4) if x 6L y 6L z then dL(x, y) 6L dL(x, z) and dL(y, z) 6L dL(x, z).
Theorem 5.4 Let REF : L2 → L be an L-REF with respect to N . If N ′ a
strong L-negation (not necessarily equal to N) then, the function defined by
dL(x, y) = N(REF (x, y)) for all x, y ∈ L (5.6)
is a restricted dissimilarity function.
Proof: Analogously to the proof of Theorem 5 in Bustince et al. [2008].
�
Corollary 5.3 Under the same conditions of Theorem 5.11, it holds that
dL(x, y) = dL(N(x), N(y)) for all x, y ∈ L (5.7)
Proof: Straightforward from Theorem 5.11 and (L4).
�
Lemma 5.1 Let G : L2 → L be a function satisfying for all x, y ∈ L(G1) G(x, y) = G(y, x);
(G2) G(x, 0L) = x;
(G3) G(x, 1L) = 1L.
Thus, if N is a strong L-negation on L then the function M : L2 → L given
117
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
by M(x, y) = N(G(N(x), N(y))) satisfies (M1), (M2) and (M3). Moreover, the
following equation holds:
N(M(x, y)) = M(N(x), N(y)) (5.8)
Proof: It is easy to see that property (M1) and Equation (5.8) hold (since
N is involutive). It remains to prove (M2) and (M3). In other words, we shall
prove that 1L is a neutral element and 0L is a annihilator of M . Thus, for all
x ∈ L, we have
M(x, 1L) = N(G(N(x), N(1L))) = N(G(N(x), 0L)) = N(0L) = 1L
and
M(x, 0L) = N(G(N(x), N(0L))) = N(G(N(x), 1L)) = N(N(x)) = x
�
The following theorem is a version for bounded lattices of Theorem 6 in
Bustince et al. [2008] (only the sufficiency part).
Theorem 5.5 Let G : L2 → L be a function satisfying (G1), (G2) and (G3).
Given a function dL : L2 → L, if there exists a function I : L2 → L for
which the properties (FPA), (OP ), (CP ) and (P ) hold and such that dL(x, y) =
G(N(I(x, y)), N(I(y, x))) for all x, y ∈ L, then dL is a restricted dissimilarity
function satisfying Equation (5.7).
Proof: Given G satisfying properties (G1), (G2) and (G3) by Lemma 5.1 the
function M : L2 → L defined by M(x, y) = N(G(N(x), N(y))) for all x, y ∈ Lis such that (M1), (M2) and (M3) hold. Thus, since I is a function satisfying
(FPA), (OP ), (CP ) and (P ) then by Theorem 5.2 a function REF : L2 → L
given by REF (x, y) = M(I(x, y), I(y, x)) for all x, y ∈ L is a restricted equiva-
lence function.
118
Moreover,
REF (x, y) = M(I(x, y), I(y, x))
= N(G(N(I(x, y)), N(I(y, x))))
= N(dL(x, y))
Since N is involutive then dL(x, y) = N(REF (x, y)) and hence by Theorem 5.11
we can affirm that dL is a restricted dissimilarity function. Also, it is clear that
dL satisfies Equation (5.7).
�
5.4 Normal Ee,N-functions on L
Another concept that can naturally be generalized for bounded lattices are
normal EN -functions (see Bustince et al. [2008]) where N is a strong fuzzy nega-
tion. It is important to say that in this section we are considering only strong
negations N which have at least one equilibrium point (as it can be seen in Section
2.2.2 there are strong negations that do not have any equilibrium point).
Definition 5.6 Let e be an equilibrium point of a strong L-negation N . A func-
tion Ee,N : L→ L is called a normal Ee,N -function associated to N (L-NEF, for
short) if it satisfies the following conditions:
1. Ee,N(x) = 1L if and only if x = e;
2. Ee,N(x) = 0L if and only if x = 0L or x = 1L;
3. For all x, y ∈ L such that either e 6L x 6L y or y 6L x 6L e it follows
Ee,N(y) 6L Ee,N(x);
4. Ee,N(x) = Ee,N(N(x)) for all x ∈ L.
As done for restricted dissimilarity function we present now a way to construct
normal Ee,N -function on bounded lattices from restricted equivalence functions.
119
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
Theorem 5.6 Let N be a strong L-negation and e be an equilibrium point of
N . If REF : L2 → L is an L-REF then the function given by Ee,N(x) =
REF (x,N(x)) for all x ∈ L is a Ee,N -function.
Proof:
1. Suppose that Ee,N(x) = 1L. Thus REF (x,N(x)) = 1L and by (L2) we
have x = N(x). Then x = e. Reciprocally, if x = e then Ee,N(e) =
REF (e,N(e)) = REF (e, e) = 1L.
2. If Ee,N(x) = 0L then REF (x,N(x)) = 0L. Hence either x = 1L and
N(x) = 0L or x = 0L and N(x) = 1L, i.e. x = 1L or x = 0L. On the other
hand, if x = 0L then Ee,N(x) = REF (0L, N(0L)) = 0L and if x = 1L then
Ee,N(x) = REF (1L, N(1L)) = 0L.
3. Note that if e 6L x 6L y then N(y) 6L N(x) 6L N(e). Since N(e) = e
we have that N(y) 6L N(x) 6L x 6L y and hence REF (y,N(y)) 6LREF (x,N(x)) by (L5). Thus Ee,N(y) 6L Ee,N(x).
Suppose now that y 6L x 6L e. In this case, we have that x 6L y 6LN(x) 6L N(y) and again by (L5) it can concluded that REF (y,N(y)) 6LREF (x,N(x)), i.e. Ee,N(y) 6L Ee,N(x).
4. For all x ∈ L we have
Ee,N(N(x)) = REF (N(x), N(N(x)))
= REF (N(x), x)
= REF (x,N(x))
= Ee,N(x)
�
Corollary 5.4 If dL is a restricted dissimilarity function on L then the function
given by Ee,N(x) = N(dL(x,N(x))), for all x, y ∈ L, is a normal Ee,N -function.
Proof: Straightforward from Theorem 5.11 and 5.12.
�
120
Theorem 5.7 Let M : L2 → L a function satisfying (M1), (M2) and (M3). If
I : L2 → L satisfies (FPA), (OP ), (CP ) and (P ) then
Ee,N(x) = M(I(x,N(x)), I(N(x), x)) (5.9)
for all x ∈ L is a normal Ee,N -function.
Proof: By Theorem 5.2 we know that REF (x, y) = M(I(x, y), I(y, x)) for all
x, y ∈ L is a L-REF. Thus Ee,N(x) = REF (x,N(x)) = M(I(x,N(x)), I(N(x), x))
is a normal Ee,N -function by Theorem 5.12.
�
Corollary 5.5 Let e be an equilibrium point of the strong L-negation N . Under
the same conditions of Theorem 5.7, we have Ee,N(x) = I(x,N(x)) for all x ∈ Lsuch that e 6L x.
Proof: If e 6L x then N(x) 6L N(e) and hence N(x) 6L e 6L x since
N(e) = e. Thus by (OP ) we have that I(N(x), x) = 1L. Therefore
Ee,N(x) = M(I(x,N(x)), I(N(x), x)) = M(I(x,N(x)), 1L) = I(x,N(x))
by Theorem 5.7 and (M2).
�
5.5 REF on L([0, 1]): Definition and characteri-
zation
One of main tasks related to restricted equivalence functions is providing
a characterization of them, i.e. propose a suitable method to construct these
functions. In this sense, it is presented in Bustince et al. [2006] (see Theorem 7)
a method based on implications (for REFs on [0, 1]). In this section we introduce
a generalization of this method for restricted equivalence functions on L([0, 1]).
121
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
As discussed in Example 2.2 from [0, 1] it is possible to define a set of intervals
L([0, 1]) = {[x, y] | 0 6 x 6 y 6 1} which is endowed with the partial order
[a, b] 62 [c, d]⇔ a 6 c and b 6 d (5.10)
for all a, b, c, d ∈ [0, 1], is a bounded lattice with bottom and top elements being
[0, 0] and [1, 1] respectively. There are several works in the literature discussing
about fuzzy operators on L([0, 1]) and its generalizations (see Bedregal et al.
[2012b]; Bilgic and Turksen [1994]; Dimuro et al. [2011]; M et al. [1996]).
Though the relation 62 generates just a partial order on L([0, 1]) it is possible
to extend this partial order to a linear order (total order) . Bustince et al. in
Bustince et al. [2012] introduced some ways to make this extension.
Definition 5.7 Bustince et al. [2012] Let (L([0, 1]),�) be a poset1. The order
� is called an admissible order if
1. � is a linear order on L([0, 1]);
2. for all [a, b], [c, d] ∈ L([0, 1]), [a, b] � [c, d] whenever [a, b] 62 [c, d].
Example 5.7 It is possible to prove that the order defined by [a, b] �Lex1 [c, d] if
and only if either a < c or a = c and b 6 d is an admissible order in L([0, 1])
motivated by the lexicographical order in R2.
Definition 5.8 Jurio et al. [2009] A function REFIV : L([0, 1])2 → L([0, 1])
is called a interval valued restricted equivalence function if it satisfies for all
X, Y, Z ∈ L([0, 1]):
1. REFIV (X, Y ) = REFIV (Y,X);
2. REFIV (X, Y ) = [1, 1] if and only if X = Y ;
3. REFIV (X, Y ) = [0, 0] if and only if either X = [1, 1] and Y = [0, 0] or
X = [0, 0] and Y = [1, 1];
1A non-empty set P endowed with a partial order 6P is called a partial order set or forshort a poset.
122
4. REFIV (X, Y ) = REFIV (N(X), N(Y )) with N a frontier negation;
5. if X 6 Y 6 Z then REF (X,Z) 6 REF (X, Y ).
Definition 5.9 Baczynski and Jayaram [2008] A function IIV : L([0, 1])2 →L([0, 1]) is called a interval fuzzy implication if it satisfies for all X, Y, Z,K ∈L([0, 1]) the following conditions:
(FPA) X 6 Z implies IIV (X, Y ) ≥ IIV (Z, Y ) (first place antitonicity);
(SPI) Y 6 K implies IIV (X, Y ) 6 IIV (X,K) (second place isotonicity);
(RB) IIV (X, [1, 1]) = [1, 1] (right corner condition);
(LB) IIV ([0, 0], Y ) = [1, 1] (left corner condition);
(CC3) IIV ([1, 1], [0, 0]) = [0, 0].
Extra properties for a given fuzzy implication IIV on L([0, 1]):
(OP) IIV (X, Y ) = [1, 1] if and only if X 6 Y (the ordering property);
(CP) IIV (X, Y ) = IIV (N(Y ), N(X)) with a strong negation N (contraposition);
(P) IIV (X, Y ) = [0, 0] if and only if X = [1, 1] and Y = [0, 0] (positive).
Theorem 5.8 Let � be an admissible order in L([0, 1]) and MIV : L([0, 1])2 →L([0, 1]) be a function satisfying (M1), (M2), (M3), (M4) and MIV (X,X) = X
for each X ∈ L([0, 1]). Thus REFIV : L([0, 1])2 → L([0, 1]) is an L-REF if and
only if there exists a function IIV : L([0, 1])2 → L([0, 1]) satisfying (FPA), (OP ),
(CP ), (P ) and such that REFIV (X, Y ) = MIV (IIV (X, Y ), IIV (Y,X))
Remark 5.2 It is worth noting that considering an admissible order (linear or-
der) on L([0, 1]) is an essential hypothesis for the Theorem 5.8 holds. For in-
stance, assuming the partial order 62 on L([0, 1]) and let REF : [0, 1]→ [0, 1] be
given by REF (x, y) = 1− |x− y|, which is a restricted equivalence function with
respect to the strong negation n(x) = 1 − x for all x ∈ [0, 1] (see Bustince et al.
[2006], example 1). Then
REFIV ([a, b], [c, d]) = [min{REF (a, c), REF (b, d)},max{REF (a, c), REF (b, d)}](5.11)
123
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
is a restricted equivalence function on (L([0, 1]),62). Nevertheless, if IIV is the
interval version of IREF defined in (5.5) then IIV does not satisfy property (FPA).
Indeed, taking X = [0.4, 0.7], Y = [0.5, 0.8] and Z = [0.5, 0.6] we have that
X 62 Y but REFIV (X,Z) = [0.9, 0.9] and REFIV (Y, Z) = [0.8, 1] which are not
comparable with respect to 62.
5.6 Extension of REF via Retractions
In this section we are interested in applying the method via retractions for
extending equivalence functions motivated by the fact that images can be repre-
sented by lattices (see Palmeira et al. [submitted 2013a]). For instance, imagine
you have an equivalence function E that makes a good comparison between im-
ages A and B. If you extend both images and want to make a new comparison
you should have a function E ′ to make it. One possible solution is extending E
in such way to preserve its properties. In this sense, our extension method can
be a suitable alternative.
Theorem 5.9 Let M be a (r, s)-sublattice of L and EF : M2 → M an equiva-
lence function. Then the function
EFE(x, y) = s(EF (r(x), r(y))) (5.12)
for each pair (x, y) ∈ L2 is an equivalence function that extends EF from M to
L.
Proof: It is clear that EFE satisfies (F1) since EF is an equivalence function.
Moreover, for all x ∈ L we have
EFE(x, x) = s(EF (r(x), r(x))) = s(1M) = 1L
and
EFE(0L, 1L) = s(EF (r(0L), r(1L))) = s(EF (0M , 1L)) = s(0M) = 0L
Analogously, one can prove that EFE(1L, 0L) = 0L. Thus, (F2) and (F3) hold.
124
It remains to prove (F4). To do so, note that for all x, y, z ∈ L such that x 6Ly 6L z then r(x) 6M r(y) 6M r(z) and hence EF (r(x), r(y)) 6M EF (r(x), r(z)).
Thus
EFE(x, y) = s(EF (r(x), r(y))) 6L s(EF (r(x), r(z))) = EFE(x, z).
�
Notice that Definition 5.4 refines Definition 5.3 since it imposes some new
constraints. Beyond property (L4), Definition 5.4 requires also that the unique
elements assigned to 0L by REF are (0L, 1L) and (1L, 0L) and that REF is
evaluated as 1L only for pairs with the same value in both coordinates. Thus,
to show that we can extend restricted equivalence functions using the method
provided by Propositions 3.1 and 3.3.
If M is a (r, s)-sublattice of L , N is a fuzzy negation on M and REF :
M2 → M a restricted equivalence function with respect to N then the function
REFE : L2 → L given by REFE(x, y) = s(REF (r(x), r(y))) satisfies property
(L4)
REFE(NE(x), NE(y)) = s(REF (r(NE(x)), r(NE(y))))
= s(REF (r(s(N(r(x)))), r(s(N(r(y))))))
= s(REF (N(r(x)), N(r(y))))
= s(REF (r(x), r(y)))
= REFE(x, y)
Moreover, if r is such that r(x) = 0L if and only if x = 0L and r(x) = 1L if and only
if x = 1L then supposing REFE(x, y) = 0L we have that s(REF (r(x), r(y))) =
0L. Since s is an injective function it follows that REF (r(x), r(y)) = 0M and
hence r(x) = 1M and r(y) = 0M or r(x) = 0M and r(y) = 1M by (L3), what allow
us to conclude that x = 1L and y = 0L or x = 0L and y = 1L, i.e. REFE satisfies
property (L3) (the necessity side of (L3) is shown in Theorem 5.9). Nevertheless
it does not satisfies (L2) in general as we can see in the following example.
Example 5.8 Let M and L be the bounded lattices pictured in Figure 2.5. Func-
tion s : M → L defined by s(0M) = 0L, s(x) = b, s(y) = d and s(1M) = 1L is a
125
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
pseudo-inverse of retraction
r(t) =
0M , t = 0L;
1M , t = 1L;
y, t = d;
x, otherwise.
and hence M is a (r, s)-sublattice of L. Then, considering the REF as in Ex-
ample 5.2, its extension to L is such that REFE(e, c) = s(REF (r(e), r(c))) =
s(REF (x, x)) = s(1M) = 1L showing that REFE does not satisfies (L2).
The above discussion allows us to enunciate the following proposition (exten-
sion of a weak version of restricted equivalence functions).
Proposition 5.5 Let M be a (r, s)-sublattice of L and REF : M2 → M a re-
stricted equivalence function. Then, function REFE : L2 → L given by REFE(x, y) =
s(REF (r(x), r(y))) for all x, y ∈ L satisfies
(L1) REFE(x, y) = REFE(y, x);
(WL2) REFE(x, x) = 1L;
(L4) REFE(x, y) = REFE(NE(x), NE(y)), being N a fuzzy negation on M ;
(L5) if x 6L y 6L z then REF (x, z) 6L REF (x, y).
Moreover, if r(x) = 0M iff x = 0L and r(x) = 1M iff x = 1L then
(L3) REFE(x, y) = 0L if and only if x = 1L and y = 0L or x = 0L and y = 1L.
Definition 5.10 Let L be an arbitrary bounded lattice. A function REF : L2 →L satisfying properties (L1), (WL2), (L3), (L4) and (L5) is called a weak re-
stricted equivalence function.
126
5.7 Extension of REF via e-operators
As we have seen in previous section the extension method proposed in Palmeira
and Bedregal [2012] is not efficient enough to extend restricted equivalence func-
tions. So, let’s apply the extension method via e-operators to verify if we can use
it to extend restricted equivalence functions.
Theorem 5.10 Let M E L with respect to (r1, r2, s), � : M ×M → L be an
e-operator and REF : M2 → M a restricted equivalence function. Then the
function REFE� : L2 → L given by
REFE� (x, y) = REF (r1(x), r1(y))�REF (r2(x), r2(y)) (5.13)
for all x, y ∈ L, is a restricted equivalence function on L.
Proof: It easy to see that REFE� is commutative since � is a commutative
operator and hence (L1) holds.
(L2)
Note that by Lemma 4.1 (item 2) and (L2) we have thatREFE� (x, x) = REF (r1(x), r1(x))�
REF (r2(x), r2(x)) = 1M �1M = s(1M) = 1L. On the other hand, supposing that
REFE� (x, y) = 1L it follows that REF (r1(x), r1(y))�REF (r2(x), r2(y)) = 1L and
then by Lemma 4.2 we have thatREF (r1(x), r1(y)) = 1M andREF (r2(x), r2(y)) =
1M . Since REF is a restricted equivalence function then r1(x) = r1(y) and
r2(x) = r2(y). Therefore, by Lemma 4.1 we can conclude that x = y.
(L3)
It is clear that
REFE� (1L, 0L) = REF (r1(1L), r1(0L))�REF (r2(1L), r2(0L))
= REF (1M , 0M)�REF (1M , 0L) = 0M � 0M
= s(0M) = 0L
Similarly REFE� (0L, 1L) = 0L is proved.
Conversely supposing REFE� (x, y) = 0L we have that REF (r1(x), r1(y)) �
REF (r2(x), r2(y)) = 0L and thenREF (r1(x), r1(y)) = 0M andREF (r2(x), r2(y)) =
0M by Lemma 4.2. Hence by Lemma 4.1 we have
127
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
(i) r1(x) = 1M and r1(y) = 0M ; or
(ii) r1(x) = 0M and r1(y) = 1M ; and
(iii) r2(x) = 1M and r2(y) = 0M ; or
(iv) r2(x) = 0M and r2(y) = 1M .
Note that (i) and (iv) can not happen simultaneously since otherwise we should
have 0M = r2(x) < r1(x) = 1M what is a contradiction with the fact that
r1(x) 6M r2(x) for every x ∈ L. A similar argument works to show that (ii) and
(iii) can not happen simultaneously. It remains to analyze two other possibilities:
(i) and (iii)
In this case, by Lemma 4.1, item 4, we have that{r1(x) = 1M = r1(1L)
r2(x) = 1M = r2(1L)⇒ x = 1L
and {r1(y) = 0M = r1(0L)
r2(y) = 0M = r2(0L)⇒ y = 0L
Analogously it can be proved that (ii) and (iv) imply x = 0L and y = 1L.
(L4)
Let N be a frontier negation on M . We shall prove that identity REFE� (x, y) =
REFE� (NE
� (x), NE� (y)) holds for each x, y ∈ L. Indeed, we have that
REFE� (NE
� (x), NE� (y)) =
= REF (r1(NE(x)), r1(N
E(y)))�REF (r2(NE(x)), r2(N
E(y))) (5.14)
Since by Lemma 4.1 item 1
REF (r1(NE(x)), r1(N
E(y))) =
= REF (r1(N(r1(x))�N(r2(x))), r1(N(r1(y))�N(r2(y))))
= REF (N(r2(x)), N(r2(y)))(5.15)
128
and
REF (r2(NE(x)), r2(N
E(y))) =
= REF (r2(N(r1(x))�N(r2(x))), r2(N(r1(y))�N(r2(y))))
= REF (N(r1(x)), N(r1(y)))(5.16)
by commutativity of � and Identity (5.14) it follows that
REFE� (NE
� (x), NE� (y)) = REF (N(r1(x)), N(r1(y)))�REF (N(r2(x)), N(r2(y)))
= REF (r1(x), r1(y))�REF (r2(x), r2(y))
= REFE� (x, y)
(L5)
Let x, y, z ∈ L such that x 6L y 6L z. In this case we have r1(x) 6Mr1(y) 6M r1(z) and r2(x) 6M r2(y) 6M r2(z) and hence REF (r1(x), r1(z)) 6MREF (r1(x), r1(y)) and REF (r2(x), r2(z)) 6M REF (r2(x), r2(y)). Then, by iso-
tonicity of � we have that
REFE� (x, z) = REF (r1(x), r1(z))�REF (r2(x), r2(z))
6L REF (r1(x), r1(y))�REF (r2(x), r2(y))
= REFE� (x, y)
�
5.7.1 Extension of Natural Negation of REF
Now we turn our attention for extending a special class of fuzzy negations
constructed from restricted equivalence functions as defined in Section 5.1.1.
Notice that if M is a (r1, r2, s)-sublattice of L, given a restricted equivalence
function REF on M its extension using method via e-operators is naturally
an extension for the negation obtained from this REF , i.e. an extension of
NREF (x) = REF (0M , x) which is given by
(NREF )E�(y) = NREF (r1(y))�NREF (r2(y))
= REF (0M , r1(y))�REF (0M , r2(y))
129
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
for all y ∈ L. This is obviously a fuzzy negation on L (see Section 4).
On the other hand, we have that
NREFE�(y) = REFE
� (0L, y)
= REF (r1(0L), r1(y))�REF (r2(0L), r2(y))
= REF (0M , r1(y))�REF (0M , r2(y))
So, it follows that (NREF )E� = NREFE�.
In a similar way we can also extend the class of fuzzy negations obtained from
weak restricted equivalence functions using extension method via retractions.
Proposition 5.6 Let M be a (r, s)-sublattice of L. If REF is a restricted equiv-
alence function on M then NREFE(x) = s(REF (0L, r(x))) for each x ∈ L is a
fuzzy negation which satisfies (NREF )E = NREFE .
Proof: Straght.
�
The idea behind Proposition 5.6 is that since it is not possible to extend REF’s
using extension method via retractions, negations generated from its extension
can be extended.
5.8 Extension of Restricted Dissimilarity Func-
tion
As we have done for restricted equivalence functions, here we apply both exten-
sion methods via retractions and via e-operators to extend restricted dissimilarity
functions.
Definition 5.11 Let L be a bounded lattice. A function dL : L2 → L is called
a restricted dissimilarity function on L (for short L-RDF) if it satisfies, for all
x, y, z ∈ L, the following conditions:
(D1) dL(x, y) = dL(y, x);
(D2) dL(x, y) = 1L if and only if either x = 1L and y = 0L or x = 0L and y = 1L;
130
(D3) dL(x, y) = 0L if and only if x = y;
(D4) if x 6L y 6L z then dL(x, y) 6L dL(x, z) and dL(y, z) 6L dL(x, z).
As we have done for restricted equivalence functions, here we apply both
extension methods via retractions and via e-operators to extend restricted dis-
similarity functions.
Proposition 5.7 Let M be a (r, s)-sublattice of L and suppose r is such that
r(x) = 0M if and only if x = 0L and r(x) = 1M if and only if x = 1L. If dM :
M2 → M a restricted dissimilarity function then, dEM(x, y) = s(dM(r(x), r(y)))
for each x, y ∈ L satisfies (D1), (D2) and (D4). Moreover, dL(x, x) = 0L for
every x ∈ L.
Proof: By Proposition 5.5 it is clear that (D1) holds. Moreover, for every
x ∈ L it follows that dEM(x, x) = s(dM(r(x), r(x))) = s(0M) = 0L. So, it remains
to prove (D2) and (D4).
Supposing dEM(x, y) = 1L then we have that s(dM(r(x), r(y))) = 1L implying
dM(r(x), r(y)) = r(1L) = 1M and hence either r(x) = 1M and r(y) = 0M or
r(x) = 0M and r(y) = 1M . Since r(x) = 0M if and only if x = 0L and r(x) = 1M
if and only if x = 1L then either x = 1L and y = 0L or x = 0L and y = 1L. On the
other hand, dEM(1L, 0L) = s(dM(r(1L), r(0M))) = s(dM(1M , 0M)) = s(1M) = 1L.
By item 1 it holds that dEM(0L, 1L) = 0L. Therefore, dEM satisfies (D2).
Now, take x, y, z ∈ L such that x 6L y 6L z. In this case, we have that
r(x) 6M r(y) 6M r(z) and hence dM(x, y) 6L dM(x, z). Then
dEM(x, y) = s(dM(r(x), r(y))) 6L s(dM(r(x), r(z))) = dEM(x, z)
Analogously, one can prove that dEM(y, z) 6L dEM(x, z) what allows us to say that
(D4) holds.
�
In other words, Proposition 5.7 says that extension method via retractions is
not efficient to extend restricted dissimilarity functions, just a weak version of this
kind of function what also happened for extending restricted equivalence functions
131
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
using same method as we have seen in Section 5.6. But, using extension method
via e-operators it is possible extend restricted dissimilarity functions successfully.
Proposition 5.8 Let M E L with respect to (r1, r2, s), � : M × M → L be
an e-operator and dM : M2 → M a restricted dissimilarity function. Then the
function (dM)E� : L2 → L given by
(dM)E�(x, y) = dM(r1(x), r1(y))� dM(r2(x), r2(y)) (5.17)
for all x, y ∈ L, is a restricted dissimilarity function on L.
Proof: Similar to proof of Theorem 4.14.
�
The following theorem proposes a way to construct restricted dissimilarity
functions from restricted equivalence functions.
Theorem 5.11 Let REF : L2 → L be an L-REF with respect to N . If N ′ a
frontier L-negation (not necessarily equal to N) then, the function defined by
dL(x, y) = N(REF (x, y)) for all x, y ∈ L (5.18)
is a restricted dissimilarity function.
Corollary 5.6 Let M E L with respect to (r1, r2, s), � : M × M → L be an
e-operator and REF : M2 →M a restricted equivalence function. If N is a fuzzy
negation on M then dL : L2 → L given by dL(x, y) = NE� (REFE
�(x, y)) for all
x, y ∈ L, is restricted dissimilarity function.
Proof: Straightforward from Proposition 4.3, Theorem 5.10 and Theorem 5.11.
�
132
5.9 Extension of Normal Ee,N-functions
Proposition 5.9 Let M be a (r, s)-sublattice of L and e be an equilibrium point
of a strong negation N : M →M . If Ee,N : M →M is a a normal Ee,N -function
associated to N then the function given by (Ee,N)E(x) = s(Ee,N(r(x))) for each
x ∈ L satisfies
1. (Ee,N)E(s(e)) = 1L;
2. if r is a lower retraction and (Ee,N)E(x) = 1L then s(e) 6L x;
3. if r is an upper retraction and (Ee,N)E(x) = 1L then x 6L s(e);
4. if r(x) = 0M if and only if x = 0L and r(x) = 1M if and only if x = 1L then
(Ee,N)E(x) = 0L if and only if x = 0L or x = 1L;
5. For all x, y ∈ L such that either s(e) 6L x 6L y or y 6L x 6L s(e) it
follows (Ee,N)E(y) 6L (Ee,N)E(x);
6. (Ee,N)E(x) = (Ee,N)E(NE(x)) for all x ∈ L;
Proof: First of all, notice that if e is an equilibrium point of N then NE(s(e)) =
s(N(r(s(e)))) = s(N(e)) = s(e) what means that s(e) is an equilibrium point of
NE. Taking this into account it follows that:
1. (Ee,N)E(s(e)) = s(Ee,N(r(s(e)))) = s(Ee,N(e)) = s(1M) = 1L;
2. If (Ee,N)E(x) = 1L for a x ∈ L then s(Ee,N(r(x))) = 1L what implies
Ee,N(r(x)) = 1M and hence by Definition 5.6 we have r(x) = e. Since r is
a lower retraction s ◦ r 6 idL then s(e) = s(r(x)) 6L x.
3. Analogous to previous item, considering that idL 6 s ◦ r if r is an upper
retraction.
4. Suppose (Ee,N)E(x) = 0L for a x ∈ L. Thus, = s(Ee,N(r(x))) = 0L which
means that Ee,N(r(x)) = 0M implying r(x) = 0M or r(x) = 1M . Hence
x = 0L or x = 1L. On the other hand, it is clear that (Ee,N)E(0L) =
(Ee,N)E(1L) = 0L.
133
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
5. If s(e) 6L x 6L y then e 6M r(x) 6M r(y). In this case, we have
that (Ee,N)E(y) = s(Ee,N(r(y))) 6L s(Ee,N(r(x))) = (Ee,N)E(x) since
Ee,N(y) 6L Ee,N(x). Analogously one can prove the same thesis supposing
y 6L x 6L s(e).
6. For each x ∈ L it follows
(Ee,N)E(NE(x)) = s(Ee,N(r(NE(x))))
= s(Ee,N(r(s(N(r(x))))))
= s(Ee,N(N(r(x))))
= s(Ee,N(r(x)))
= (Ee,N)E(x)
�
Remark 5.3 Proposition 5.9 shows that again extension method via retraction
is not efficient for extending normal Ee,N -functions since property 1 of Defini-
tion 5.6 is not satisfied by (Ee,N)E, just a weakened version of it (items 2 and
3 of Proposition 5.9) as also happened for restricted equivalence functions and
restricted dissimilarity functions.
Proposition 5.10 Let M E L with respect to (r1, r2, s), � : M × M → L be
an e-operator and e be an equilibrium point of strong negation N : M → M . If
Ee,N : M → M is a normal Ee,N -function then its extension (Ee,N)E� : L → L
given by
(Ee,N)E�(x) = Ee,N(r1(x))� Ee,N(r2(x))
for each x ∈ L, is a normal Es(e),NE�
-function.
Proof: It is clear that s(e) is an equilibrium point of NE� since e is an equilib-
rium point of N . Moreover,
1. if (Ee,N)E�(x) = 1L then Ee,N(r1(x))�Ee,N(r2(x)) = 1L and by Lemma 4.2
we have Ee,N(r1(x)) = 1M and Ee,N(r2(x)) = 1M which means that r1(x) =
e and r2(x) = e. Hence s(e) = s(r1(x)) 6L x and x 6L s(r2(x)) = s(e)
134
allowing us to conclude that x = s(e). On the other hand,
(Ee,N)E�(s(e)) = Ee,N(r1(s(e)))� Ee,N(r2(s(e)))
= Ee,N(e)� Ee,N(e)
= 1M � 1M = s(1M) = 1L
Therefore, (Ee,N)E�(x) = 1L if and only if x = s(e).
2. Note that
(Ee,N)E�(0L) = Ee,N(r1(0L))� Ee,N(r2(0L))
= Ee,N(0M)� Ee,N(0M)
= 0M � 0M = s(0M) = 0L
Similarly, one can verify that (Ee,N)E�(1L) = 0L.
Now, suppose (Ee,N)E�(x) = 0L, i.e. Ee,N(r1(x))� Ee,N(r2(x)) = 0L. Thus,
Ee,N(r1(x)) = 0M and Ee,N(r2(x)) = 0M by Lemma 4.2 and hence either
r1(x) = 0M or r1(x) = 1M or r2(x) = 0M or r2(x) = 1M which implies that
x = 0L or x = 1L.
3. Supposing s(e) 6L x 6L y it is easy to see that e 6M r1(x) 6M r1(y) and
e 6M r2(x) 6M r2(y), then Ee,N(r1(y)) 6M Ee,N(r1(x)) andEe,N(r2(y)) 6MEe,N(r2(x)) and hence Ee,N(r1(y))�Ee,N(r2(y)) 6L Ee,N(r1(x))�Ee,N(r2(x))
since � is isotonic. Therefore, (Ee,N)E�(y) = Ee,N(r1(y)) � Ee,N(r2(y)) 6LEe,N(r1(x))� Ee,N(r2(x)) = (Ee,N)E�(x).
4. Finally, take x ∈ L. We know that r1(x) 6M r2(x) and hence N(r2(x)) 6MN(r1(x)) and by Lemma 4.1 we have that r1(N(r1(x)) � N(r2(x))) =
N(r2(x)) and r2(N(r1(x))�N(r2(x))) = N(r1(x)). Therefore,
(Ee,N)E�(NE� (x)) =
135
5. ON RESTRICTED EQUIVALENCE FUNCTIONS
= Ee,N(r1(NE� (x)))� Ee,N(r2(N
E� (x))
= (Ee,N(r1(N(r1(x))�N(r2(x)))))� (Ee,N(r2(N(r1(x))�N(r2(x)))))
= Ee,N(N(r2(x)))� Ee,N(N(r1(x)))
= Ee,N(N(r1(x)))� Ee,N(N(r2(x)))
= (Ee,N)E�(x)
�
Theorem 5.12 Let N be a strong L-negation and e be an equilibrium point
of N . If REF : L2 → L is an L-REF then the function given by Ee,N(x) =
REF (x,N(x)) for all x ∈ L is a Ee,N -function.
Corollary 5.7 Let M E L with respect to (r1, r2, s), � : M × M → L be an
e-operator and REF : M2 → M a restricted equivalence function. If e is an
equilibrium point of strong negation N on M then function Es(e),NE�
: L2 → L
given by Es(e),NE�
(x) = REFE�(x,NE
� (x)) for all x ∈ L, is a normal Es(e),NE�
-
function.
Proof: Straightforward from Proposition 4.3 and Theorem 5.12.
�
136
Chapter 6
Remarks and Further Works
In this chapter we make a brief discussion of the results achieved in this thesis
and the prospects for future work. In addition, we summarize the major published
papers.
We have presented here two different methods to extend fuzzy operators. As
we have seen, the first method, named extension method via retractions, make
a generalization of the method proposed by Saminger-Platz et al. [2008] consid-
ering a more general environment provided by the notion (r, s)-sublattice, i.e.
considering that we should look to lattice M as a sublattice of another lattice L
through an algebraic immersion using retractions. This objective was successfully
achieved for t-norms, t-conorms, fuzzy negations and fuzzy implications even if,
in some cases, it has been necessary to add some particular constraints (eg, for
t-norms is necessary to require to r be an upper retraction).
However, even with these limitations, the extension method via retractions
presents some advantages with relation to the method proposed by Saminger-
Platz:
• Extension method provided by Saminger-Platz is a particular case of our
method via retraction (see Remark 3.1.1);
• Since we can identify lattice M as a (r, s)-sublattice of L in more than one
different way (as we know, a pseudo-inverse s can not be unique) then it is
possible to choose the most appropriate way to make the extension of the
operator, depending on the need presented;
137
6. REMARKS AND FURTHER WORKS
• M may not be a complete sublattice of L as required by Saminger-Platz
et al. [2008];
• Allows a greater variation of possibilities of extension of operators, since in
(3.1) it is considered only a specific retraction r(x) = sup{y ∈M | y 6L x}.
Furthermore, as is expected, we would like also this extension method could
be able to preserve the main properties that the extended operator has. However,
some results shown that not all properties of fuzzy operators are preserved by this
extension method. As we can see in Example 3.2, for no trivial cases the cancel-
lation law for triangular norms (and of course for t-conorms also) is not preserved
by extension method via retractions. This happens due to retraction r can not
be injective, since if it was, then this retraction is obviously an isomorphism (i.e.
a trivial case). Moreover, we also have seen that strong (strict) negations are not
preserved by extension method via retraction as well. For a given implications I,
property (CP) fails for its extension IE, for instance.
Aware of limitations presented by extension method via retraction, we started
investigating another way to make extension of fuzzy operators which could be
more efficient in preserving properties still considering the framework provided
by the notion of (r, s)-sublattice even if the new method had no more relation
to the method proposed by Saminger-Platz et al. [2008]. So, with inspiration
stemming from the interval operator – a kind of application very efficient in
carrying on algebraic informations from a lattice L to its interval representant L– we have proposed extension method via e-operators. The results presented by
this method were qualitatively good in terms of the preservation of properties of
extended fuzzy operators, overcoming all the limitations presented by the other
method (considering the questions studied till now) to t-norms, t-conorms and
negations. Unfortunately, due to lack of time, it was not possible to present here a
study of the application of the extension method via e-operators for implications,
but we strongly believed that the good results obtained for the other operators
studied here are also repeated for implications.
The following tables make a global comparison between results obtained for
each fuzzy connective using both methods considering the tables presented at the
end of Chapthers 3 and 4.
138
Abbreviation Property
AC Archimedean
NI Nilpotent
IDN Idempotent
ZD Zero divisor
CL Cancellation law
C Continuity
P Positive
EC Relation between extension and conjugate
RE Relation between NKE and (NK)E with K a fuzzy connective√Property preserved by this extension method√
Property not preserved by this extension method
? We have no results about this property
Table 6.1: Table of Abbreviations
t-norm AC NI IDN ZD CL C P EC
TE√ √ √ √
× ×√
(T ρ)E 6 (TE)ψ
TE�√ √ √ √ √ √ √
(T ρ)E� = (TE� )ψ
Table 6.2: Comparing results of TE and TE�
t-conorm AC NI IDN ZD CL C P EC
SE√ √ √ √
× ×√
(Sρ)E > (SE)ψ
SE�√ √ √ √ √ √ √
(Sρ)E� = (SE�)ψ
Table 6.3: Comparing results of SE and SE�
fuzzy implication (RB) (LB) (CC4) (NP) (EP) (IP) (OP) (P) (IBL) (CP)
IE√ √ √
×√ √
× ×√ √
IE� ? ? ? ? ? ? ? ? ? ?
Table 6.4: Comparing results of IE and IE�
fuzzy negation Strong Strict Eq. P. EC
NE × ×√
(Nρ)E = (NE)ψ
NE�
√ √ √(Nρ)E� = (NE
� )ψ
Table 6.5: Comparing results of NE and NE�
In order to investigate the behavior of extension methods proposed in this
thesis to other fuzzy operators outside the scope of fuzzy logic, we started apply-
139
6. REMARKS AND FURTHER WORKS
ing these methods for extending functions used to compare images, as we have
seen in Chapter 5. We did an interesting study on restricted equivalence func-
tions (REF), restricted d issimilarity functions (RDF) and normal Ee,N -functions
(NEF) valued on bounded lattices. Then, we applied our extension methods for
extension these functions and results were fateful. Again, extension method via
retractions presented limitations for extending REF, RDF and NEF while the
method via e-operators confirmed its effectiveness.
It is noteworthy that even extension method via e-operators has been achieved
better and more consistent results then extension method via retractions, this
method has an value due to it robustness. As observed by Saminger-Platz et al.
[2008] about their extension method, our method proposed in (3.3) seek to present
the most drastic way to extend fuzzy operators by means this method takes the
smallest possible extension.
Beside our published or submitted paper for main important journals concern-
ing to fuzzy sets Palmeira et al. [2012b],Palmeira and Bedregal [2012],Palmeira
et al. [submitted 2013a],Palmeira et al. [submitted 2013b],Palmeira et al. [sub-
mitted 2013c] we also published some papers in national and international con-
ferences.
It is also important to highlight that at the beginning of our doctoral studies
we have also worked on fuzzy topology and fuzzy homotopy and we had the
interest of considering extension of fuzzy homotopies between fuzzy connectives,
but due to the lack of time to make this investigation we have put this researches
in stand by. However, we would like to present here our main publications on
this for being a guide for further works.
1. E.S. Palmeira, B. Bedregal e R.H.N Santiago: Homotopia Intervalar. In:
TEMA - Tendencias em Matematica Aplicada e Computacional, v 12, p.
145–156, 2011.
2. E.S. Palmeira and B. Bedregal: On F-homotopy and F-fundamental Group.
In: 30th Annual Conference of the North American Fuzzy Information Pro-
cessing Society, El Paso, TX. Annals of the NAFIPS, 2011;
Undoubtedly, due to lack of studies on extension of fuzzy operators in the
literature and considering the originality of our extension methods, we believe
140
possess on our hands a tool with great potential for producing new scientific
articles in this area of knowledge. Thus, as a first step in the direction of the
production of future works, we are interested in applying the extension method via
e-operators to other fuzzy operators we have been extend using extension method
via retractions (implications, subclasses of negations and t-subnorms, see Akella
[2007]; Durante and Sempi [2005]; Ouyang et al. [2008]). Furthermore, we would
like to investigate the behavior of both methods for extending other operators of
our interests such as n-norms, nullnorms, copulas, additive generators of t-norms,
among others. It is also worth mentioning that we are carrying out research in
partnership with professor Radko Mesiar about the extension of n-dimensional
fuzzy implications.
141
6. REMARKS AND FURTHER WORKS
142
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150
Index
REFf , 109
(S,N)-implication, 33
(dX , dY )-continuous, 22
(r, s)-sublattice, 15
lower, 19
upper, 19
(r1, r2, s)-sublattice, 19
/x/, 97
Aut(L), 14
IT , 34
IREF , 114
L-NEF, 119
L-RDF, 117
L([0, 1]), 11
Ln(L), 97
Ln([0, 1]), 97
M < L, 19
M > L, 19
M m L, 19
M l L, 19
M E L, 19
N0, 109
NI , 34
NS, 30
NT , 28
N⊥, 25
N>, 26
REF , 107
REFIV , 123
¨, 11
62, 122
x, 97
TTE� , 98
TT , 98
r, 99
s, 99
�, 82
�n, 100
‖, 11
�, 122
�Lex1, 122
˜T1 · · ·Tn, 98
dL, 117
dL-nilpotent, 24
Alg-homomorphism, 12
Action of an automorphism, 14
Admissible order, 122
Alg-lattice, 10
Archimedean, 23
Automorphism, 12
151
INDEX
Bottom element, 10
Cancellation law, 46
Characterization Theorem of REF, 111
Complete lattice, 11
Conjugate, 14
Contrapositivity property, 32
De Morgan S-semitriple, 40
De Morgan T -semitriple, 40
De Morgan triple, 38
e-operator, 82
n-dimensional, 100
Epimorphism, 12
Equilibrium point, 26
Equivalence function
on [0, 1], 106
Exchange principle, 32
Extension method via e-operator
for De Morgan triple, 91
for negation, 89
for t-conorm, 89
for t-norm, 85
Extension method via retraction
for De Morgan T -semitriple, 68
for implication, 55
for negation, 50
for t-conorm, 46
for t-norm, 43
Extension operator, 82
n-dimensional, 100
Fuzzy implication, 31
Fuzzy Negation, 25
Homomorphism, 12
Idempotent, 24
Identity principle, 32
Implication, 31
positive, 32
Interval constructor, 80
interval fuzzy implication, 123
Interval order, 122
Involution property, 25
Isomorphism, 12
Iterative Boolean law, 32
L-REF, 107
Lattice, 10
bounded, 10
Left boundary condition, 32
Left contraposition law, 32
Left neutrality principle, 32
Left ordering property, 32
Lexicographical order, 122
Linear order, 122
Lower (r, s)-sublattice, 19
Monomorphism, 12
n-dimensional t-norm, 98
Natural negation
of I, 34
of S, 30
of T , 28
of REF, 109
Negation, 25
frontier, 25
strict, 25
152
INDEX
strong, 25
Normal Ee,N -function, 119
Ord-homomorphism, 12
Ord-lattice, 10
Ordering property, 32
Ordinary sublattice, 15
Positive, 24
Pre-metrics, 21
Pseudo-inverse, 15
n-dimensional, 99
Pseudo-quasi-metrics, 21
R-implication, 34
Restricted dissimilarity function, 117
Restricted equivalence function
interval-valued, 122
lattice-valued, 107
on [0, 1], 106
Retract, 15
lower, 17
upper, 17
Retraction, 15
lower, 17
n-dimensional, 99
upper, 17
Right boundary condition, 31
Right contraposition law, 32
S-implication, 33
t-conorm, 24
positive, 24
t-norm, 23
positive, 25
t-subconorm, 37
t-subnorm, 36
Top element, 10
Total order, 122
Triangular conorm, 24
Triangular norm, 23
Upper (r, s)-sublattice, 19
Weak first place antitonicity, 31
Weak left ordering property, 32
Weak second place isotonicity, 31
153