# Chebyshev distances associated to the second members of systems of max-
product/Lukasiewicz fuzzy relational equations

Ismaïl Baaj
Univ. Artois, CNRS, CRIL, F-62300 Lens, France
<EMAIL_ADDRESS>
###### Abstract

In this article, we study the inconsistency of a system of $\max$-product
fuzzy relational equations and of a system of $\max$-Lukasiewicz fuzzy
relational equations. For a system of $\max-\min$ fuzzy relational equations
$A\Box_{\min}^{\max}x=b$ and using the $L_{\infty}$ norm, (Baaj, 2023) showed
that the Chebyshev distance $\Delta=\inf_{c\in\mathcal{C}}\|b-c\|$, where
$\mathcal{C}$ is the set of second members of consistent systems defined with
the same matrix $A$, can be computed by an explicit analytical formula
according to the components of the matrix $A$ and its second member $b$. In
this article, we give analytical formulas analogous to that of (Baaj, 2023) to
compute the Chebyshev distance associated to the second member of a system of
$\max$-product fuzzy relational equations and that associated to the second
member of a system of $\max$-Lukasiewicz fuzzy relational equations.

_Keywords_ Fuzzy set theory ; Systems of fuzzy relational equations ;
Chebyshev approximations

##  1 Introduction

Artificial Intelligence (AI) applications based on systems of fuzzy relational
equations emerged thanks to (Sanchez, 1976, 1977)’s seminal work on solving
systems of $\max-\min$ fuzzy relational equations. (Sanchez, 1976) gave
necessary and sufficient conditions for a system to be consistent i.e., to
have solutions. (Sanchez, 1977) showed that if the system is consistent, it
has a greater solution and many minimal solutions, and he then described the
complete set of solutions of the system.

However, the inconsistency of these systems remains difficult to address.
While the majority of the approaches have investigated how to determine
approximate solutions of an inconsistent system (Di Nola et al., 1989;
Gottwald, 1986; Klir and Yuan, 1994; Pedrycz, 1983, 1990a; Wangming, 1986; Wen
et al., 2022; Wu et al., 2021, 2022; Xiao et al., 2019), Pedrycz highlighted
another strategy in (Pedrycz, 1990b). Given an inconsistent system, Pedrycz
proposed to slightly modify its second member to obtain a consistent system.
Some authors proposed algorithms for his procedure (Cuninghame-Green and
Cechlárová, 1995; Li and Fang, 2010). In a recent preprint (Baaj, 2023), for
an inconsistent system of $\max-\min$ fuzzy relational equations
$A\Box_{\min}^{\max}x=b$, the author gave an explicit analytical formula to
compute, using the $L_{\infty}$ norm, the Chebyshev distance
$\Delta=\inf_{c\in\mathcal{C}}\|b-c\|$ where $\mathcal{C}$ is the set of
second members of consistent systems defined with the same matrix $A$. As a
Chebyshev approximation of the second member $b$ is a vector $c$ such that
$\|b-c\|=\Delta$ and the system $A\Box_{\min}^{\max}x=c$ is a consistent
system (Cuninghame-Green and Cechlárová, 1995; Li and Fang, 2010), the formula
of $\Delta$ led the author of (Baaj, 2023) to give the complete description of
the structure of the set of Chebyshev approximations of $b$. He then described
the approximate solutions set of the considered inconsistent system, which is
the set of solutions of consistent systems $A\Box_{\min}^{\max}x=c$, where $c$
is a Chebyshev approximation of $b$.
In our opinion, it would be relevant to study how these results can be
extended to systems of $\max-T$ fuzzy relational equations, where $T$ is a
continuous t-norm. As it is well-known, many applications are based on systems
of fuzzy relational equations with the max-product composition or the max-
Lukasiewicz composition e.g., (Di Nola et al., 1989; Hirota and Pedrycz, 1999;
Nobuhara et al., 2006).

In this article, we tackle the problem of the inconsistency of a system of
max-product fuzzy relational equations and of a system of max-Lukasiewicz
fuzzy relational equations. We extend to systems of $\max-T$ fuzzy relational
equations, where $T$ is a continuous t-norm, the definition and the properties
of the idempotent application $F$, see (11), introduced in (Baaj, 2023). This
application was initially proposed as a reformulation of (Sanchez, 1976)’s
result for checking if a system of $\max-\min$ fuzzy relational equations
defined with a fixed matrix and a given vector used as second member is a
consistent system. The idempotence and the right-continuity of the application
$F$ lets us extend a fundamental result given for $\max-\min$ composition
(Theorem 1 of (Cuninghame-Green and Cechlárová, 1995)) to $\max-T$ composition
(Theorem 1), and to obtain the greatest Chebyshev approximation of the second
member of such a system (Corollary 1).
Then, we give an explicit analytical formula to compute the Chebyshev distance
associated to the second member of a system of $\max$-product fuzzy relational
equations (Theorem 4). In the case of a system of $\max$-Lukasiewicz fuzzy
relational equations, an analogous analytical formula is also given (Theorem
5). Each Chebyshev distance is computed according to the components of the
matrix and the second member of the considered system.

The article is structured as follows. In (Section 2), we begin by reminding
the necessary background on solving a system of $\max-T$ fuzzy relational
equations where $T$ is a continuous t-norm. We extend the idempotent
application $F$ of (Baaj, 2023) to a system of $\max-T$ fuzzy relational
equations. We remind the definition of the Chebyshev distance associated to
the second member of a system of $\max-T$ fuzzy relational equations and
extend to such a system a fundamental result given for $\max-\min$ composition
(Theorem 1 of (Cuninghame-Green and Cechlárová, 1995)). In (Section 3), we
study the solving of two inequalities, which are involved in the computation
of the Chebyshev distance associated to the second member of a system of
$\max$-product fuzzy relational equations and that associated to the second
member of a system of $\max$-Lukasiewicz fuzzy relational equations. In
(Section 4), we give the explicit analytical formulas to compute the two
Chebyshev distances. Finally, we conclude with some perspectives.

Our results are illustrated with examples (note that in all examples, the
numbers are rounded to two decimal places).

##  2 Background

Throughout this section, $T$ denotes a continuous t-norm.

In what follows, we begin by reminding the definition and properties of a
continuous t-norm and its associated residual implicator. To the t-norm
defined by the usual product, we make explicit its associated residual
implicator, as well as the one associated to the t-norm of Lukasiewicz.
We then extend to systems of $\max-T$ fuzzy relational equations the
idempotent application $F$ introduced in (Baaj, 2023) for checking if a system
of $\max-\min$ fuzzy relational equations defined with a fixed matrix and a
given vector used as second member is a consistent system. Finally, we remind
the definition of the Chebyshev distance associated to the second member of a
system of $\max-T$ fuzzy relational equations and extend to such a system a
fundamental result proven in (Cuninghame-Green and Cechlárová, 1995) for
$\max-\min$ composition.

###  2.1 Notations

The following notations were given in (Baaj, 2023) for the case of a system of
$\max-\min$ fuzzy relational equations and we reuse them in this article.

By $[0,1]^{n\times m}$ we denote the set of matrices of size $(n,m)$ i.e., $n$
rows and $m$ columns, whose components are in $[0,1]$. In particular:

  * •

$[0,1]^{n\times 1}$ denotes the set of column vectors of $n$ components,

  * •

$[0,1]^{1\times m}$ denotes the set of row matrices of $m$ components.

On the set $[0,1]^{n\times m}$, we use the order relation $\leq$ defined by:

$A\leq B\quad\text{iff we have}\quad a_{ij}\leq b_{ij}\quad\text{ for all
}\quad 1\leq i\leq n,1\leq j\leq m,$

where $A=[a_{ij}]_{1\leq i\leq n,1\leq j\leq m}$ and $B=[b_{ij}]_{1\leq i\leq
n,1\leq j\leq m}$.

For $x,y,z,u,\delta\in[0,1]$, we put:

  * •

$x^{+}=\max(x,0)$,

  * •

$\overline{z}(\delta)=\min(z+\delta,1)$,

  * •

$\underline{z}(\delta)=\max(z-\delta,0)=(z-\delta)^{+}$

and we have the following equivalence in $[0,1]$:

$\mid x-y\mid\leq\delta\Longleftrightarrow\underline{x}(\delta)\leq
y\leq\overline{x}(\delta).$ (1)

For our work, to a column-vector $b=[b_{i}]_{1\leq i\leq n}$ and a number
$\delta\in[0,1]$, we associate two column-vectors:

$\underline{b}(\delta)=[(b_{i}-\delta)^{+}]_{1\leq i\leq
n}\quad\text{and}\quad\overline{b}(\delta)=[\min(b_{i}+\delta,1)]_{1\leq i\leq
n}.$ (2)

These vectors $\underline{b}(\delta)$ and $\overline{b}(\delta)$ were already
introduced e.g., in (Cuninghame-Green and Cechlárová, 1995) (with others
notations) and in (Li and Fang, 2010).
Then, from (1), we deduce for any $c=[c_{i}]_{1\leq i\leq n}\in[0,1]^{n\times
1}$:

$\|b-c\|\leq\delta\Longleftrightarrow\underline{b}(\delta)\leq
c\leq\overline{b}(\delta).$ (3)

where $\|b-c\|=\max_{1\leq i\leq n}\mid b_{i}-c_{i}\mid$.

###  2.2 T-norms and their associated residual implicators

A triangular-norm (t-norm, see (Klement et al., 2013)) is a map
$T:[0,1]\times[0,1]\mapsto[0,1]$, which satisfies:

  * $T$ is commutative: $T(x,y)=T(y,x)$,

  * $T$ is associative: $T(x,T(y,z))=T(T(x,y),z)$,

  * $T$ is increasing with respect to the second variable: $y\leq z\,\Longrightarrow\,T(x,y)\leq T(x,z)$,

  * $T$ has $1$ as neutral element: $T(x,1)=x$.

The residual implicator associated to the t-norm $T$ is the map ${\cal
I}_{T}:[0,1]\times[0,1]\rightarrow[0,1]:(x,y)\mapsto{\cal
I}_{T}(x,y)=\sup\\{z\in[0,1]\,\mid\,T(x,z)\leq y\\}$.

The main properties of the residual implicator ${\cal I}_{T}$ of a continuous
t-norm $T$ are: for all $a,b\in[0,1]$, we have:

  * •

${\cal I}_{T}(a,b)=\max\\{z\in[0,1]\,\mid\,T(a,z)\leq b\\}$. Therefore,
$T(a,{\cal I}_{T}(a,b))\leq b$.

  * •

${\cal I}_{T}$ is left-continuous and decreasing in its first argument as well
as right-continuous and increasing in its second argument.

  * •

For all $z\in[0,1]$, we have:

$T(a,z)\leq b\Longleftrightarrow z\leq{\cal I}_{T}(a,b).$

  * •

We have $\,b\leq{\cal I}_{T}(a,T(a,b))$.

The t-norm defined by the usual product is denoted by $T_{P}$. Its associated
residual implicator is the Goguen product, and we have:

$T_{P}(x,y)=x\cdot y\quad;\quad{\cal
I}_{T_{P}}(x,y)=x\underset{GG}{\longrightarrow}y=\begin{cases}1&\text{ if
}x\leq y\\\ \frac{y}{x}&\text{ if }x>y\end{cases}.$ (4)

Lukasiewicz’s t-norm is denoted by $T_{L}$ and we have:

$T_{L}(x,y)=\max(x+y-1,0)={(x+y-1)}^{+}\quad;\quad{\cal
I}_{T_{L}}(x,y)=x\underset{L}{\longrightarrow}y=\min(1-x+y,1).$ (5)

In the rest of this section, we set a continuous t-norm denoted $T$.

###  2.3 Solving systems of $\max-T$ fuzzy relational equations

Let $A=[a_{ij}]\in[0,1]^{n\times m}$ be a matrix of size $(n,m)$ and
$b=[b_{i}]\in[0,1]^{n\times 1}$ be a vector of $n$ components. The system of
$\max-T$ fuzzy relational equations associated to $(A,b)$ is of the form:

$(S):A\Box_{T}^{\max}x=b,$ (6)

where $x=[x_{j}]_{1\leq j\leq m}\in[0,1]^{m\times 1}$ is an unknown vector of
$m$ components and the operator $\Box_{T}^{\max}$ is the matrix product that
uses the continuous t-norm $T$ as the product and $\max$ as the addition.
Equivalently, the system can also be written as:

$\max_{1\leq j\leq m}T(a_{ij},x_{j})=b_{i},\,\forall i\in\\{1,2,\dots,n\\}.$

The studies on systems of fuzzy relational equations adopt two equivalent
notation conventions, which differ in their representation of the unknown part
and the second member: either as column vectors or as row vectors. The
transpose map allows us to switch from one notation to the other.

To check if the system (S), see (6), is consistent, we compute the following
vector:

$e=A^{t}\Box_{{\cal I}_{T}}^{\min}b,$ (7)

where $A^{t}$ is the transpose of $A$ and the matrix product $\Box_{{\cal
I}_{T}}^{\min}$ uses the residual implicator ${\cal I}_{T}$ (associated to
$T$) as the product and $\min$ as the addition. The vector $e$ is the
potential maximal solution of the system $(S)$.

Thanks to (Sanchez, 1976)’s seminal work on the solving of systems $\max-\min$
fuzzy relational equations, and (Pedrycz, 1982, 1985) and (Miyakoshi and
Shimbo, 1985) (using other t-norms than $\min$), we have the following
equivalence:

$A\Box_{T}^{\max}x=b\text{ is consistent}\Longleftrightarrow
A\Box_{T}^{\max}e=b.$ (8)

In the rest of this subsection, we will give (Proposition 1) and show that the
proof of its second statement implies the equivalence (8).
The set of solutions of the system $(S)$ is denoted by:

${\cal S}={\cal S}(A,b)=\\{v\in[0,1]^{m\times
1}\,\mid\,A\,\Box_{T}^{\max}v=b\\}.$ (9)

The structure of the solution set was described by (Sanchez, 1977) (with the
$\max-\min$ composition) and (Di Nola et al., 1982, 1984, 1989).

######  Example 1.

Let:

$A=\begin{bmatrix}0.31&0.49&0.76\\\ 0.34&0.9&0.15\\\ 0.94&0.47&0.05\\\
\end{bmatrix}\text{ and }b=\begin{bmatrix}0.73\\\ 0.84\\\ 0.61\end{bmatrix}.$
(10)

To check if the systems $A\Box_{T_{P}}^{\max}x=b$ and
$A\Box_{T_{L}}^{\max}x=b$ are consistent, we compute their respective
potential greatest solution:

$e_{T_{P}}=A^{t}\Box_{{\cal I}_{T_{P}}}^{\min}b=\begin{bmatrix}0.65\\\ 0.93\\\
0.96\end{bmatrix}\quad\text{ and }\quad e_{T_{L}}=A^{t}\Box_{{\cal
I}_{T_{L}}}^{\min}b=\begin{bmatrix}0.67\\\ ~0.94\\\ 0.97\end{bmatrix}.$

These two systems are consistent because:

$A\Box_{T_{P}}^{\max}e_{T_{P}}=A\Box_{T_{L}}^{\max}e_{T_{L}}=b.$

In (Baaj, 2023), the author introduced an idempotent application denoted $F$,
to check if a system of $\max-\min$ fuzzy relational equations defined with a
fixed matrix $A=[a_{ij}]_{1\leq i\leq n,1\leq j\leq m}\in[0,1]^{n\times m}$
and a given vector $c\in[0,1]^{n\times 1}$ used as second member is a
consistent system. The application is a reformulation of (Sanchez, 1976)’s
result. For systems of $\max-T$ fuzzy relational equations which uses a
continuous t-norm $T$, we extend the definition of the application $F$ as
follows:

$F:[0,1]^{n\times 1}\longrightarrow[0,1]^{n\times 1}:c=[c_{i}]\mapsto
F(c)=[F(c)_{i}]$ (11)

where:

$\forall i\in\\{1,2,\dots,n\\},\,F(c)_{i}=\max_{1\leq j\leq
m}T(a_{ij},\min_{1\leq k\leq n}{\cal I}_{T}(a_{kj},c_{k})).$ (12)

The properties of the application $F$ given in (Baaj, 2023) for a system of
$\max-\min$ fuzzy relational equations are valid for systems of $\max-T$ fuzzy
relational equations:

######  Proposition 1.

  1. 1.

$\forall c\in[0,1]^{n\times 1}$, $F(c)\leq c$.

  2. 2.

For any vector $c\in[0,1]^{n\times 1}$ we have:

$F(c)=c\Longleftrightarrow\text{ the system }A\Box_{T}^{\max}x=c\text{ is
consistent.}$

  3. 3.

$F$ is idempotent i.e., $\forall c\in[0,1]^{n\times 1},F(F(c))=F(c)$.

  4. 4.

$F$ is increasing and right-continuous.

The application $F$ being right-continuous at a point $c\in[0,1]^{n\times 1}$
means: for any sequence $(c^{(k)})$ in $[0,1]^{n\times 1}$ such that
$(c^{(k)})$ converges to $c$ when $k\rightarrow\infty$ and verifying $\forall
k,c^{(k)}\geq c$, we have:

$F(c^{(k)})\rightarrow F(c)$ when $k\rightarrow\infty$.

###### Proof.

We adapt the proofs given in (Baaj, 2023) for the $\max-\min$ composition to
the $\max-T$ composition in the following way:

The first assertion is deduced from the inequality $T(a,{\cal I}_{T}(a,b))\leq
b$.

For the second statement, if we suppose that $F(c)=c$, it means that
$e=A^{t}\Box_{{\cal I}_{T}}^{\min}c$ is a solution of the system
$A\Box_{T}^{\max}x=c$ (so it is consistent).
For the proof of the implication $\Longleftarrow$, we use the general
inequality:

$\forall x\in[0,1]^{m\times 1},\,x\leq A^{t}\Box_{{\cal
I}_{T}}^{\min}(A\Box_{T}^{\max}x),$

which is deduced from the inequality $b\leq{\cal I}_{T}(a,T(a,b))$. We also
obtain that $e=A^{t}\Box_{{\cal I}_{T}}^{\min}c$ is the greatest solution of
the system $A\Box_{T}^{\max}x=c$. Clearly, we have also proven the equivalence
(8).

Thanks to the two assertions above, the idempotence of the application $F$ is
proved as in the case of the $\max-\min$ composition.

It is easy to prove that the application $F$ is increasing. The right-
continuity of the residual implicator ${\cal I}_{T}$ in its second argument
implies the right-continuity of the application $F$.

∎

######  Example 2.

(continued) Let $c=\begin{bmatrix}0.74\\\ 0.66\\\ 0.62\end{bmatrix}$. When we
use any of the two t-norms $T_{P}$ or $T_{L}$, we have the equality $F(c)=c$.
So the systems $A\Box_{T_{P}}^{\max}x=c$ and $A\Box_{T_{L}}^{\max}x=c$ are
consistent.

###  2.4 Chebyshev distance associated to the second member of a system of
$\max-T$ fuzzy relational equations

To the matrix $A$ and the second member $b$ of the system $(S)$ of $\max-T$
fuzzy relational equations, see (6), let us associate the set of vectors
$c=[c_{i}]\in[0,1]^{n\times 1}$ such that the system $A\Box_{T}^{\max}x=c$ is
consistent:

$\mathcal{C}=\\{c=[c_{i}]\in{[0,1]}^{n\times 1}\mid A\Box_{T}^{\max}x=c\text{
is consistent}\\}.$ (13)

This set allows us to define the Chebyshev distance associated to the second
member $b$ of the system $(S)$.

######  Definition 1.

The Chebyshev distance associated to the second member $b$ of the system
$(S):A\Box_{T}^{\max}x=b$ is:

$\Delta=\Delta(A,b)=\inf_{c\in\mathcal{C}}\|b-c\|$ (14)

where:

$\|b-c\|=\max_{1\leq i\leq n}\mid b_{i}-c_{i}\mid.$

We have the following fundamental result, proven for $\max-\min$ composition
in (Cuninghame-Green and Cechlárová, 1995), that we extend for $\max-T$
composition:

######  Theorem 1.

$\Delta=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}.$ (15)

The proof of (Theorem 1) is given in three steps.

###### Proof.

We put $\Delta^{\prime}=\inf\\{\delta\in[0,1]\,|\,\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}$.

$\bullet$ Step 1. We want to prove that we have
$\Delta^{\prime}=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}$. Indeed, as we have $\underline{b}(1)=0\leq
F(\overline{b}(1))$, the set $\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}$ is non-empty. Therefore, we can find a sequence
$(\delta_{k})$ in $[0,1]$ verifying:

  1. 1.

$\forall k,\,\underline{b}(\delta_{k})\leq F(\overline{b}(\delta_{k}))$,

  2. 2.

$\forall k,\,\delta_{k}\geq\Delta^{\prime}$,

  3. 3.

$\delta_{k}\rightarrow\Delta^{\prime}$.

As $\forall k,\overline{b}(\delta_{k})\geq\overline{b}(\Delta^{\prime})$ and
$\overline{b}(\delta_{k})\rightarrow\overline{b}(\Delta^{\prime})$, the right-
continuity of the application $F$ at $\overline{b}(\Delta^{\prime})$ implies:

$F(\overline{b}(\delta_{k}))\rightarrow F(\overline{b}(\Delta^{\prime})).$

The inequalities $\underline{b}(\delta_{k})\leq F(\overline{b}(\delta_{k}))$
lead, by the passage to the limit, to:

$\underline{b}(\Delta^{\prime})\leq F(\overline{b}(\Delta^{\prime})).$

So $\Delta^{\prime}=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}.$

$\bullet$ Step 2. We want to prove that $\Delta\leq\Delta^{\prime}$.

We have:

$\underline{b}(\Delta^{\prime})\leq
F(\overline{b}(\Delta^{\prime}))\leq\overline{b}(\Delta^{\prime}).$

By (3) and by noticing that $F(\overline{b}(\Delta^{\prime}))\in{\cal C}$, we
deduce:

$\Delta\leq\|b-F(\overline{b}(\Delta^{\prime}))\|\leq\Delta^{\prime}.$

$\bullet$ Step 3. We want to prove that $\Delta^{\prime}\leq\Delta$. For this
it is sufficient to show that:

$\forall c\in{\cal C},\,\Delta^{\prime}\leq\|b-c\|.$

Indeed, for any $c\in{\cal C}$, we put $\delta=\|b-c\|$, and we have by (3):

$\underline{b}(\delta)\leq c\leq\overline{b}(\delta).$

Therefore $F(c)=c$ and the growth of $F$ lead to:

$\underline{b}(\delta)\leq c=F(c)\leq F(\overline{b}(\delta)).$

So:

$\Delta^{\prime}\leq\delta=\|b-c\|.$

We have proved that
$\Delta=\Delta^{\prime}=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}$.

∎

We define the set of Chebyshev approximations of $b$:

######  Definition 2.

The set of Chebyshev approximations of $b$ is defined using the set
$\mathcal{C}$, see (13), and the Chebyshev distance associated to $b$
(Definition 1):

${\cal C}_{b}=\\{c\in{\cal C}\,\mid\,\|b-c\|=\Delta(A,b)\\}.$ (16)

We have:

######  Corollary 1.

$F(\overline{b}(\Delta))\in{\cal C}_{b}$ is the greatest Chebyshev
approximation of the second member $b$ of the system $(S)$.

###### Proof.

Let us first show that we have $\|b-F(\overline{b}(\Delta))\|=\Delta$.
Indeed, from (Theorem 1) and (Proposition 1), we know that
$\underline{b}(\Delta)\leq F(\overline{b}(\Delta))\leq\overline{b}(\Delta)$,
so by (3), $\|b-F(\overline{b}(\Delta))\|\leq\Delta$.

As $F(\overline{b}(\Delta))\in{\cal C}$ (Proposition 1), we know by
(Definition 1):

$\Delta\leq\|b-F(\overline{b}(\Delta))\|.$

We have proven that $F(\overline{b}(\Delta))\in{\cal C}_{b}$.

To prove that $F(\overline{b}(\Delta))$ is the greatest Chebyshev
approximation of the second member $b$ of the system $(S)$ we must prove that
$c\leq F(\overline{b}(\Delta))$ for any $c\in{\cal C}_{b}$.
Let $c\in{\cal C}_{b}$. As $\Delta=\|b-c\|$, by (3) we have:

$\underline{b}(\Delta)\leq c\leq\overline{b}(\Delta).$

By the growth of $F$, we deduce:

$c=F(c)\leq F(\overline{b}(\Delta)).$

∎

As a consequence, we have:

######  Corollary 2.

$\Delta=\min_{c\in\mathcal{C}}\|b-c\|.$ $\Delta=0\Longleftrightarrow\text{ the
system $(S)$ is consistent.}$

Therefore, $\Delta=0$ is a necessary and sufficient condition for the system
$(S)$ to be consistent.

##  3 Preliminaries computations

In this section, we solve two inequalities. The first one will be involved in
the analytical formula to compute the Chebyshev distance associated to the
second member of a system of $\max-T$ fuzzy relational equations when the
t-norm is the usual product i.e., $T=T_{P}$, see (4). The second one will be
involved in the formula to compute the Chebyshev distance when we use
Lukasiewicz’s t-norm i.e., $T=T_{L}$, see (5). A similar study was carried out
in (Baaj, 2023) for $\max-\min$ composition.

###  3.1 T-norm product $T=T_{P}$

Let $x,y,z,u\in[0,1]$ be fixed. For the t-norm product $T_{P}$, let us study
the following inequality that involves the Goguen product
$\underset{GG}{\longrightarrow}$ (see (4)), for $\delta\in[0,1]$:

$\underline{x}(\delta)\leq
u\cdot(y\underset{GG}{\longrightarrow}\overline{z}(\delta))$ (17)

where
$y\underset{GG}{\longrightarrow}\overline{z}(\delta)=\begin{cases}1&\text{ if
}y-z\leq\delta\\\ \frac{z+\delta}{y}&\text{ if }y-z>\delta\end{cases}$.
We look for the smallest value of $\delta$ so that the inequality (17) is
true.
Let:

$\sigma_{GG}(u,x,y,z)=\max[(x-u)^{+},\min(\varphi(u,x,y,z),(y-z)^{+})]$ (18)

where $\varphi(u,x,y,z)=\begin{cases}\frac{(x\cdot y-u\cdot
z)^{+}}{u+y}&\text{ if }u>0\\\ x&\text{ if }u=0\end{cases}$.

######  Proposition 2.

For all $\delta\in[0,1]$, we have:

$\underline{x}(\delta)\leq
u\cdot(y\underset{GG}{\longrightarrow}\overline{z}(\delta))\Longleftrightarrow\sigma_{GG}(u,x,y,z)\leq\delta.$

###### Proof.

If $u=0$, then $\sigma_{GG}(0,x,y,z)=x$ and we immediately get the desired
equivalence:

$(x-\delta)^{+}\leq 0\Longleftrightarrow
x\leq\delta\Longleftrightarrow\sigma_{GG}(0,x,y,z)\leq\delta.$

It remains for us to study the case where $u>0$.

Suppose that $(x-\delta)^{+}\leq
u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))$.

  * •

If $y-z\leq\delta$, we have:

$(x-\delta)^{+}\leq
u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))=u\cdot 1=u.$

We know that the inequality $(x-\delta)^{+}\leq u$ is equivalent to the
inequality $(x-u)^{+}\leq\delta$ (Baaj, 2023). As
$\min(\varphi(u,x,y,z),(y-z)^{+})\leq(y-z)^{+}\leq\delta$, we obtain:

$\sigma_{GG}(u,x,y,z)=\max[(x-u)^{+},\min(\varphi(u,x,y,z),(y-z)^{+})]\leq\delta.$

  * •

If $y-z>\delta$, we have:

$(x-\delta)^{+}\leq
u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))=u\dfrac{z+\delta}{y}\leq
u$

so, $(x-u)^{+}\leq\delta$ and moreover

$\displaystyle(x-\delta)^{+}-u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))$
$\displaystyle=\max(x-\delta-\dfrac{uz+u\delta}{y},-\dfrac{uz+u\delta}{y})$
$\displaystyle=\max(\dfrac{xy-\delta
y-uz-u\delta}{y},-\dfrac{uz+u\delta}{y})\leq 0.$

The last inequality is equivalent to $xy-\delta y-uz-u\delta\leq 0$ and also:

$\dfrac{xy-uz}{u+y}\leq\delta.$

Then:

$\bullet$ If $(xy-uz)^{+}=0$, we have:

$\varphi(u,x,y,z)=\dfrac{(xy-uz)^{+}}{u+y}=0$

and also

$\sigma_{GG}(u,x,y,z)=\max[(x-u)^{+},\min(\varphi(u,x,y,z),(y-z)^{+})=(x-u)^{+}\leq\delta.$

$\bullet$ If $(xy-uz)^{+}>0$, we have:

$\varphi(u,x,y,z)=\dfrac{xy-uz}{u+y}\leq\delta$

and also

$\displaystyle\sigma_{GG}(u,x,y,z)$
$\displaystyle=\max[(x-u)^{+},\min(\varphi(u,x,y,z),(y-z)^{+})]$
$\displaystyle=\max[(x-u)^{+},\varphi(u,x,y,z)]$ $\displaystyle\leq\delta.$

Let us prove now the remaining implication.
Suppose that $\sigma_{GG}(u,x,y,z)\leq\delta$:

  * •

If $(x-\delta)^{+}=0$, we get trivially that $(x-\delta)^{+}=0\leq
u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))$.

  * •

If $y\underset{GG}{\longrightarrow}\overline{z}(\delta)=1$, we must prove that
$(x-\delta)^{+}\leq u$, which is equivalent to $(x-u)^{+}\leq\delta$. We have:

$(x-u)^{+}\leq\max[(x-u)^{+},\min(\varphi(u,x,y,z),(y-z)^{+}]=\sigma_{GG}(u,x,y,z)\leq\delta.$

It remains for us to study the case where $(x-\delta)^{+}=x-\delta>0$ and
$y\underset{GG}{\longrightarrow}\overline{z}(\delta)<1$.

From $y\underset{GG}{\longrightarrow}\overline{z}(\delta)<1$, we deduce:

$y-z>\delta\quad\text{and}\quad
y\underset{GG}{\longrightarrow}\overline{z}(\delta)=\dfrac{z+\delta}{y}.$

Then, we have:

$\displaystyle(x-\delta)^{+}-u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))$
$\displaystyle=x-\delta-\dfrac{uz+u\delta}{y}$
$\displaystyle=\dfrac{u+y}{y}(\dfrac{xy-uz}{u+y}-\delta).$

To obtain
$(x-\delta)^{+}-u(y\underset{GG}{\longrightarrow}\overline{z}(\delta))\leq 0$,
it is sufficient to prove that

$\dfrac{xy-uz}{u+y}\leq\delta.$

  * •

If $xy\leq uz$, we have $\dfrac{xy-uz}{u+y}\leq 0\leq\delta$.

  * •

If $xy>uz$ and taking into account the inequality $y-z>\delta$, we deduce:

$\delta\geq\sigma_{GG}(u,x,y,z)\geq\min(\dfrac{(xy-
uz)^{+}}{u+y},y-z)=\dfrac{xy-uz}{u+y}.$

∎

From (Proposition 2), we immediately deduce:

######  Theorem 2.

For any $x,y,z,u\in[0,1]$, we have:

$\sigma_{GG}(u,x,y,z)=\min\\{\delta\in[0,1]\,\mid\,\underline{x}(\delta)\leq
u\cdot(y\underset{GG}{\longrightarrow}\overline{z}(\delta))\\}.$ (19)

We illustrate this result:

######  Example 3.

Let $x=0.4$, $y=0.3$, $z=0.6$ and $u=0.2$. We want to obtain the smallest
value of $\delta\in[0,1]$ so that $\underline{x}(\delta)\leq
u\cdot(y\underset{GG}{\longrightarrow}\overline{z}(\delta))$ is true. We have
$u\cdot(y\underset{GG}{\longrightarrow}z)=0.2$.

We compute $\varphi(u,x,y,z)=\frac{(0.4\cdot 0.3-0.2\cdot
0.6)^{+}}{0.2+0.3}=0.$

$\displaystyle\delta$ $\displaystyle\\!\begin{aligned}
&=\sigma_{GG}(u,x,y,z)\\\ &=\max[(0.4-0.2)^{+},\min(0,(0.3-0.6)^{+})]\\\
&=0.2.\end{aligned}$

We have $\underline{x}(\delta)=x-0.2=0.2$ and
$\overline{z}(\delta)=z+0.2=0.8$. Therefore,
$u\cdot(y\underset{GG}{\longrightarrow}\overline{z}(\delta))=0.2$ and:

$\underline{x}(\delta)=u\cdot(y\underset{GG}{\longrightarrow}\overline{z}(\delta)).$

###  3.2 Lukasiewicz’s t-norm $T=T_{L}$

For Lukasiewicz’s t-norm $T_{L}$, we study the following inequality that
involves Lukasiewicz’s implication $\underset{L}{\longrightarrow}$ (see (5))
for $\delta\in[0,1]$:

$\underline{x}(\delta)\leq\max(0,\,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u),$
(20)

where
$y\underset{L}{\longrightarrow}\overline{z}(\delta)=\begin{cases}1&\text{ if
}y-z\leq\delta\\\ 1-y+z+\delta&\text{ if }y-z>\delta\end{cases}$.
Let $v=x+u-1$ and:

$\sigma_{L}(u,x,y,z)=\min(x,\max(v^{+},\frac{(v+y-z)^{+}}{2})).$ (21)

Then:

######  Proposition 3.

For all $\delta\in[0,1]$, we have:

$\underline{x}(\delta)\leq\max(0,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u)\,\Longleftrightarrow\,\sigma_{L}(u,x,y,z)\leq\delta.$
(22)

###### Proof.

We remark that for $x\leq\delta$ the two assertions of the equivalence are
verified: we have $(x-\delta)^{+}=0$ and $\sigma_{L}(u,x,y,z)\leq
x\leq\delta$. It remains for us to show the equivalence in the case where
$x>\delta$.

Set $f(u,x)=(x+u-1)^{+},\,g(u,x,y,z)=\dfrac{(x+u-1+y-z)^{+}}{2}$. Then, we
have:

$\sigma_{L}(u,x,y,z)=\min(x,\max(f(u,x),g(u,x,y,z))).$

Let us prove that
$\underline{x}(\delta)\leq\max(0,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u)\,\Longrightarrow\,\sigma_{L}(u,x,y,z)\leq\delta$.

As we suppose that $x>\delta$, we must prove:

$\max(f(u,x),g(u,x,y,z))\leq\delta.$

This last inequality will be proven in the following two steps:

  1. 1.

If $y-z\leq\delta$, then
$y\underset{L}{\longrightarrow}\overline{z}(\delta)-u=1-u$ and we have:

$x-\delta\leq\max(0,1-u)$

which we put in the form

$\displaystyle x-\delta-\max(0,1-u)$ $\displaystyle=x-\delta+\min(0,u-1)$
$\displaystyle=\min(x-\delta,x+u-1-\delta)$ $\displaystyle\leq 0.$

As $x-\delta>0$, we deduce that $x+u-1\leq\delta$ and also:

$f(u,x)=(x+u-1)^{+}\leq\delta.$

Let us show that we also have:

$g(u,x,y,z)=\dfrac{(x+u-1+y-z)^{+}}{2}\leq\delta.$

Indeed, if $g(u,x,y,z)=0$, we have trivially $g(u,x,y,z)=0\leq\delta$.

If $g(u,x,y,z)>0$, then we have:

$\displaystyle g(u,x,y,z)$ $\displaystyle=\dfrac{x+u-1+y-z}{2}$
$\displaystyle\leq\dfrac{x+u-1+\delta}{2}$
$\displaystyle=\dfrac{x+u-1}{2}+\dfrac{\delta}{2}$
$\displaystyle\leq\dfrac{(x+u-1)^{+}}{2}+\dfrac{\delta}{2}\quad\text{(we
already know that $f(u,x)=(x+u-1)^{+}\leq\delta$)}$
$\displaystyle\leq\dfrac{\delta}{2}+\dfrac{\delta}{2}=\delta.$

In summary, we have shown
$y-z\leq\delta\,\Longrightarrow\,\max(f(u,x),g(u,x,y,z))\leq\delta$.

  2. 2.

If $y-z>\delta$, then
$y\underset{L}{\longrightarrow}\overline{z}(\delta)-u=1-y+z+\delta-u$ and we
have:

$x-\delta\leq\max(0,1-y+z+\delta-u).$

We rewrite this last inequality in the form:

$\displaystyle x-\delta-\max(0,1-y+z+\delta-u)$
$\displaystyle=x-\delta+\min(0,u-1+y-z-\delta)$
$\displaystyle=\min(x-\delta,\,\,x+u-1~+y-z-2\delta)$ $\displaystyle\leq 0.$

As $x-\delta>0$, we deduce $x+u-1~+y-z\leq 2\delta$ and also:

$g(u,x,y,z)=\dfrac{(x+y-z+u-1)^{+}}{2}\leq\delta.$

It remains for us to show that $f(u,x)=(x+u-1)^{+}\leq\delta$.
In fact, if $f(u,x)>0$, then we have:

$\displaystyle\dfrac{x+u-1+y-z}{2}$ $\displaystyle>\dfrac{x+u-1+\delta}{2}$
$\displaystyle=\dfrac{x+u-1}{2}+\dfrac{\delta}{2}$
$\displaystyle=\dfrac{f(u,x)}{2}+\dfrac{\delta}{2}.$

We deduce that $x+u-1+y-z>0$, and also:

$g(u,x,y,z)=\dfrac{x+u-1+y-z}{2}>\dfrac{f(u,x)}{2}+\dfrac{\delta}{2}.$

As we have already shown that $g(u,x,y,z)\leq\delta$, we finally obtain:

$f(u,x)<2\,g(u,x,y,z)-\delta\leq 2\delta-\delta=\delta.$

In summary, we have shown
$y-z>\delta\,\Longrightarrow\,\max(f(u,x),g(u,x,y,z))\leq\delta$.

We have completed the proof of the implication $\Longrightarrow$.

Let us assume now that:

$\sigma_{L}(u,x,y,z)=\min(x,\max((x+u-1)^{+},\dfrac{(x+u-1+y-z)^{+}}{2}))\leq\delta$

and let us show that:

$(x-\delta)^{+}\leq\max(0,y\rightarrow_{L}\overline{z}(\delta)-u).$

As we suppose that $x>\delta$, the inequality $\sigma_{L}(u,x,y,z)\leq\delta$
becomes:

$\max(f(u,x),g(u,x,y,z))=\max((x+u-1)^{+},\dfrac{(x+u-1+y-z)^{+}}{2})\leq\delta.$

Let us put the inequality to be shown in the form:

$(x-\delta)^{+}-\max(0,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u)\leq
0$

and distinguish the two cases: $y-z\leq\delta$ and $y-z>\delta$.

  1. 1.

If $y-z\leq\delta$, the inequality to be shown becomes:

$\displaystyle x-\delta-\max(0,1-u)$ $\displaystyle=x-\delta+\min(0,u-1)$
$\displaystyle=\min(x-\delta,x+u-1-\delta)$ $\displaystyle\leq 0.$

As $x-\delta>0$, we must show that $x+u-1-\delta\leq 0$, i.e.,
$x+u-1\leq\delta$.
But, we have:

$x+u-1\leq(x+u-1)^{+}=f(u,x)\leq\max(f(u,x),g(u,x,y,z))\leq\delta.$

  2. 2.

If $y-z>\delta$, the inequality to be shown becomes:

$\displaystyle x-\delta-\max(0,1-y+z+\delta-u)$
$\displaystyle=x-\delta+\min(0,u-1+y-z-\delta)$
$\displaystyle=\min(x-\delta,x+u-1+y-z-2\delta)$ $\displaystyle\leq 0.$

So, we must show that $x+u-1+y-z-2\delta\leq 0$, i.e
$\dfrac{x+u-1+y-z}{2}\leq\delta$.
But, we have:

$\dfrac{x+u-1+y-z}{2}\leq\dfrac{(x+u-1+y-z)^{+}}{2}=g(u,x,y,z)\leq\max(f(u,x),g(u,x,y,z))\leq\delta.$

We have completed the proof of implication $\Longleftarrow$.

∎

From (Proposition 3), we immediately deduce:

######  Theorem 3.

For any $x,y,z,u\in[0,1]$, we have:

$\sigma_{L}(u,x,y,z)=\min\\{\delta\in[0,1]\,\mid\,\underline{x}(\delta)\leq\max(0,\,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u)\\}.$
(23)

We illustrate this result:

######  Example 4.

Let $x=0.4$, $y=0.6$, $z=0.3$ and $u=0.6$. We want to obtain the smallest
value of $\delta\in[0,1]$ so that
$\underline{x}(\delta)\leq\max(0,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u)$
is true.
We have $\max(0,y\underset{L}{\longrightarrow}z-u)=0.1$ and $v=x+u-1=0$.

$\displaystyle\delta$ $\displaystyle\\!\begin{aligned}
&=\sigma_{L}(u,x,y,z)\\\ &=\min(0.4,\max(0,\frac{(0+0.6-0.3)^{+}}{2}))\\\
&=0.15.\end{aligned}$

We have $\underline{x}(\delta)=x-0.15=0.25$ and
$\overline{z}(\delta)=z+0.15=0.45$. Therefore,
$\max(0,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u)=0.25$ and:

$\underline{x}(\delta)=\max(0,y\underset{L}{\longrightarrow}\overline{z}(\delta)-u).$

##  4 Analytical formulas to compute the Chebyshev distances associated to the
second members of the systems $A\Box_{T_{P}}^{\max}x=b$ and
$A\Box_{T_{L}}^{\max}x=b$

In this section, we compute the Chebyshev distance $\Delta$ (Definition 1)
associated to the second member of a system of $\max$-product fuzzy relational
equations $A\Box_{T_{P}}^{\max}x=b$, by an explicit analytical formula in
terms of components of the matrix $A$ and those of the second member $b$. We
also give an analytical formula for the Chebyshev distance associated to the
second member of a system of $\max$-Lukasiewicz fuzzy relational equations
$A\Box_{T_{L}}^{\max}x=b$.
We rely on the equality:
$\Delta=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\},$ see (15), which was proven for any continuous
t-norm.

######  Notation 1.

Using a continuous t-norm $T$ whose associated implication operator is ${\cal
I}_{T}$, we set for $1\leq i,k\leq n$ and $1\leq j\leq m$:

  * •

$K_{i}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq
F(\overline{b}(\delta))_{i}\\}$,

  * •

$\beta_{j}=\min_{1\leq k\leq n}\,{\cal
I}_{T}(a_{kj},\overline{b}(\delta))_{k})$,

  * •

$W_{ij}^{T}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq
T(a_{ij},\beta_{j})\\}$.

###  4.1 Analytical formula to compute the Chebyshev distance associated to
the second member $b$ of the system $A\Box_{T_{P}}^{\max}x=b$

For any $i=1,2,\dots,n$ and $j=1,2,\dots,m$, we have in this case:

$W_{ij}^{T_{P}}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq
T_{P}(a_{ij},\beta_{j})\\}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq
a_{ij}\beta_{j}\\}.$

where $\beta_{j}=\min_{1\leq k\leq n}\,{\cal
I}_{T_{P}}(a_{kj},\overline{b}(\delta)_{k})=\min_{1\leq k\leq
n}\,a_{kj}\underset{GG}{\longrightarrow}\overline{b}(\delta)_{k}$. Then, we
have:

$W_{ij}^{T_{P}}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq\min_{1\leq
k\leq
n}\,a_{ij}\,(a_{kj}\underset{GG}{\longrightarrow}\overline{b}(\delta)_{k})\\}.$
(24)

As $F(\overline{b}(\delta))_{i}=\max_{1\leq j\leq
m}\,T_{P}(a_{ij},\beta_{j})$, we still have:

$K_{i}=\bigcup_{1\leq j\leq m}\,W_{ij}^{T_{P}}.$ (25)

The Chebyshev distance $\Delta$ associated to the second member $b$ of the
system of $\max$-product fuzzy relational equations $A\Box_{T_{P}}^{\max}x=b$
is given by the following formula:

######  Theorem 4.

$\Delta=\max_{1\leq i\leq n}\,\delta_{i}$ (26)

where for $i=1,2,\dots,n$:

$\delta_{i}=\min_{1\leq j\leq m}\,\max_{1\leq k\leq
n}\,\sigma_{GG}\,(a_{ij},b_{i},a_{kj},b_{k}).$ (27)

See (18) for the definition of $\sigma_{GG}$.

###### Proof.

For any $i=1,2,\dots,n$ and $j=1,2,\dots,m$, we deduce from (24) that for
$\delta\in[0,1]$, we have:

$\delta\in W_{ij}^{T_{P}}\,\Longleftrightarrow\,\delta\geq\,\max_{1\leq k\leq
n}\,\sigma_{GG}\,(a_{ij},b_{i},a_{kj},b_{k}).$

Using (25), we get:

$\delta\in
K_{i}\,\Longleftrightarrow\,\exists\,j\in\\{1,2,\dots,m\\}\,\,\text{such
that}\,\,\delta\geq\max_{1\leq k\leq
n}\,\sigma_{GG}\,(a_{ij},b_{i},a_{kj},b_{k}).$

So, we obtain:

$\delta\in K_{i}\,\Longleftrightarrow\,\delta\geq\min_{1\leq j\leq
m}\,\max_{1\leq k\leq n}\,\sigma_{GG}\,(a_{ij},b_{i},a_{kj},b_{k}).$

As by definition $\delta\in
K_{i}\,\Longleftrightarrow\,\underline{b}(\delta)_{i}\leq
F(\overline{b}(\delta))_{i}$ and
$\Delta=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}$, we get:

$\Delta=\max_{1\leq i\leq n}\,\min_{1\leq j\leq m}\,\max_{1\leq k\leq
n}\,\sigma_{GG}\,(a_{ij},b_{i},a_{kj},b_{k}).$

∎

The following example illustrates this result:

######  Example 5.

In what follows, we rely on the following matrix and vector, which were used
as example in (Cuninghame-Green and Cechlárová, 1995; Pedrycz, 1990b):

$A=\begin{bmatrix}1&0.4&0.5&0.7\\\ 0.7&0.5&0.3&0.5\\\ 0.2&1&1&0.6\\\
0.4&0.5&0.5&0.8\end{bmatrix}\text{ and }b=\begin{bmatrix}0.4\\\ 1\\\ 0.2\\\
0\end{bmatrix}.$ (28)

For the system $A\Box_{T_{P}}^{\max}x=b$ which uses the t-norm product
$T_{P}$, we compute from (Theorem 4):

$\displaystyle\delta_{1}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq
n}\,\sigma_{GG}\,(a_{1j},b_{1},a_{kj},b_{k})=\min(0.11,0.23,0.2,0.21)=0.11,$
$\displaystyle\delta_{2}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq
n}\,\sigma_{GG}\,(a_{2j},b_{2},a_{kj},b_{k})=\min(0.42,0.6,0.72,0.62)=0.42,$
$\displaystyle\delta_{3}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq
n}\,\sigma_{GG}\,(a_{3j},b_{3},a_{kj},b_{k})=\min(0.13,0.07,0.07,0.11)=0.07,$
$\displaystyle\delta_{4}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq n}\,\sigma_{GG}\,(a_{4j},b_{4},a_{kj},b_{k})=\min(0.0,0.0,0.0,0.0)=0.0.$

Therefore, the Chebyshev distance associated to the second member $b$ of the
system $A\Box_{T_{P}}^{\max}x=b$ is
$\Delta=\max(\delta_{1},\delta_{2},\delta_{3},\delta_{4})=0.42$.

We compute $\overline{b}(\Delta)=\begin{bmatrix}0.82\\\ 1\\\ 0.62\\\
0.42\end{bmatrix}$ and $F(\overline{b}(\Delta))=\begin{bmatrix}0.82\\\ 0.57\\\
0.62\\\ 0.42\end{bmatrix}$ is the greatest Chebyshev approximation of the
second member $b$.

###  4.2 Analytical formula to compute the Chebyshev distance associated to
the second member $b$ of the system $A\Box_{T_{L}}^{\max}x=b$

For any $i=1,2,\dots,n$ and $j=1,2,\dots,m$, we have in this case:

$W_{ij}^{T_{L}}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq
T_{L}(a_{ij},\beta_{j})\\}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq\max(0,\,\,a_{ij}+\beta_{j}-1)\\}.$

where $\beta_{j}=\min_{1\leq k\leq n}\,{\cal
I}_{T_{L}}(a_{kj},\overline{b}(\delta)_{k})=\min_{1\leq k\leq
n}\,a_{kj}\underset{L}{\longrightarrow}\overline{b}(\delta)_{k}$.

Set $u_{ij}=1-a_{ij}$. Then, we have:

$\displaystyle T_{L}(a_{ij},\beta_{j})$
$\displaystyle=\max(0,\,\,\beta_{j}-u_{ij})$
$\displaystyle=\max(0,\,\,[\min_{1\leq k\leq
n}\,a_{kj}\underset{L}{\longrightarrow}\overline{b}(\delta)_{k}]-u_{ij})$
$\displaystyle=\min_{1\leq k\leq
n}\,\max(0,\,\,[a_{kj}\underset{L}{\longrightarrow}\overline{b}(\delta)_{k}]-u_{ij})$
$\displaystyle=\min_{1\leq k\leq
n}\,([a_{kj}\underset{L}{\longrightarrow}\overline{b}(\delta)_{k}]-u_{ij})^{+}.$

So, we obtain:

$W_{ij}^{T_{L}}=\\{\delta\in[0,1]\mid\underline{b}(\delta)_{i}\leq\min_{1\leq
k\leq
n}\,([a_{kj}\underset{L}{\longrightarrow}\overline{b}(\delta)_{k}]-u_{ij})^{+}\\}.$
(29)

As $F(\overline{b}(\delta))_{i}=\max_{1\leq j\leq
m}\,T_{L}(a_{ij},\beta_{j})$, we still have:

$K_{i}=\bigcup_{1\leq j\leq m}\,W_{ij}^{T_{L}}.$ (30)

The Chebyshev distance $\Delta$ associated to the second member $b$ of the
system of $\max$-Lukasiewicz fuzzy relational equations
$A\Box_{T_{L}}^{\max}x=b$ is given by the following formula:

######  Theorem 5.

$\Delta=\max_{1\leq i\leq n}\,\delta_{i}$ (31)

where for any $i=1,2,\dots n$:

$\delta_{i}=\min_{1\leq j\leq m}\,\max_{1\leq k\leq
n}\,\sigma_{L}\,(1-a_{ij},b_{i},a_{kj},b_{k}).$ (32)

See (21) for the definition of $\sigma_{L}$.

###### Proof.

For any $i=1,2,\dots,n$ and $j=1,2,\dots,m$, we deduce from (29) that for
$\delta\in[0,1]$, we have:

$\delta\in W_{ij}^{T_{L}}\,\Longleftrightarrow\,\delta\geq\,\max_{1\leq k\leq
n}\,\sigma_{LK}\,(1-a_{ij},b_{i},a_{kj},b_{k}).$

Using (30), we get:

$\delta\in
K_{i}\,\Longleftrightarrow\,\exists\,j\in\\{1,2,\dots,m\\}\,\,\text{such
that}\,\,\delta\geq\max_{1\leq k\leq
n}\,\sigma_{LK}\,(1-a_{ij},b_{i},a_{kj},b_{k}).$

So, we obtain:

$\delta\in K_{i}\,\Longleftrightarrow\,\delta\geq\min_{1\leq j\leq
m}\,\max_{1\leq k\leq n}\,\sigma_{LK}\,(1-a_{ij},b_{i},a_{kj},b_{k}).$

As by definition $\delta\in
K_{i}\,\Longleftrightarrow\,\underline{b}(\delta)_{i}\leq
F(\overline{b}(\delta))_{i}$ and
$\Delta=\min\\{\delta\in[0,1]\mid\underline{b}(\delta)\leq
F(\overline{b}(\delta))\\}$, we get:

$\Delta=\max_{1\leq i\leq n}\,\min_{1\leq j\leq m}\,\max_{1\leq k\leq
n}\,\sigma_{LK}\,(1-a_{ij},b_{i},a_{kj},b_{k}).$

∎

######  Example 6.

We rely on the matrix $A$ and the vector $b$ used in the previous example, see
(28), to construct the system $A\Box_{T_{L}}^{\max}x=b$, which uses with the
Lukasiewicz’s t-norm $T_{L}$. From (Theorem 5), we have:

$\displaystyle\delta_{1}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq n}\,\sigma_{L}\,(1-a_{1j},b_{1},a_{kj},b_{k})=\min(0,0.4,0.35,0.25)=0,$
$\displaystyle\delta_{2}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq
n}\,\sigma_{L}\,(1-a_{2j},b_{2},a_{kj},b_{k})=\min(0.45,0.65,0.75,0.65)=0.45,$
$\displaystyle\delta_{3}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq n}\,\sigma_{L}\,(1-a_{3j},b_{3},a_{kj},b_{k})=\min(0.2,0,0,0.2)=0,$
$\displaystyle\delta_{4}$ $\displaystyle=\min_{1\leq j\leq m}\,\max_{1\leq
k\leq n}\,\sigma_{L}\,(1-a_{4j},b_{4},a_{kj},b_{k})=\min(0,0,0,0)=0.$

Therefore, the Chebyshev distance associated to the second member $b$ of the
system $A\Box_{T_{L}}^{\max}x=b$ is
$\Delta=\max(\delta_{1},\delta_{2},\delta_{3},\delta_{4})=0.45$.

We compute $\overline{b}(\Delta)=\begin{bmatrix}0.85\\\ 1\\\ 0.65\\\
0.45\end{bmatrix}$ and $F(\overline{b}(\Delta))=\begin{bmatrix}0.85\\\ 0.55\\\
0.65\\\ 0.45\end{bmatrix}$ is the greatest Chebyshev approximation of the
second member $b$.

##  5 Conclusion

In this article, we gave explicit analytical formulas for computing the
Chebyshev distance associated to the second member of a system of
$\max$-product fuzzy relational equations and that associated to the second
member of a system of $\max$-Lukasiewicz fuzzy relational equations.

The results of (Baaj, 2023) may be then extended to these two systems. For
each system, we can describe the sets of Chebyshev approximations of its
second member and its approximate solutions set. Similarly to the $\max-\min$
learning paradigm introduced in (Baaj, 2023), in which the learning error is
expressed in terms of $L_{\infty}$ norm, we can tackle the development of a
$\max$-product (resp. $\max$-Lukasiewicz) learning paradigm. The formula of
(Baaj, 2023), which computes the minimum value of the learning error according
to training data and the method to construct approximate weight matrices whose
learning error is minimal can be directly extended to the case of the
$\max$-product composition or the $\max$-Lukasiewicz composition.

As applications, we have the complete solution of the problem of the
invertibility of a fuzzy relation for $\max$-Lukasiewicz composition: we know
the set of matrices $A$ which admit a pre-inverse or a post-inverse. Moreover,
for $\max-\min$ composition, if a matrix $A$ has no preinverse (resp.
postinverse), we know how to compute, using the $L_{\infty}$ norm, an
approximate preinverse (resp. postinverse) for $A$. For both cases of the
$\max-\min$ composition and the $\max$-product composition, the problem of the
invertibility of a fuzzy relation is already solved (Wu et al., 2021, 2022).

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