# Some applications of fuzzy sets in residuated lattices

Cristina Flaut, Dana Piciu, Bianca Liana Bercea

###### Abstract

In this paper, based on ideals, we investigate residuated lattices from fuzzy
set theory and lattice theory point of view.

Ideals are important concepts in the theory of algebraic structures used for
formal fuzzy logic and first, we investigate the lattice of fuzzy ideals in
residuated lattices. Then we present applications of fuzzy sets in Coding
Theory and we study connections between fuzzy sets associated to ideals and
Hadamard codes.

Keywords: Residuated lattice, fuzzy ideal, coding theory

##  1 Introduction

The notion of residuated lattice, introduced in [17] by Ward and Dilworth,
provides an algebraic framework for fuzzy logic.

Managing certain and uncertain information is a priority of artificial
intelligence, in an attempt to imitate human thinking. To make this possible,
in [18], Zadeh introduced the concept of fuzzy subset of a nonempty set.

In this paper, we study some applications of fuzzy sets in residuated
lattices.

In [15], this concept is applied to these algebras and the fuzzy ideals are
introduced. In Section 3, we investigate more properties of fuzzy ideals and
we study their lattice structure, which is a Heyting algebra.

In Section 4 we found connections between the fuzzy sets associated to ideals
in particular residuated lattices and Hadamard codes.

##  2 Preliminaries

A residuated lattice is an algebra $(L,\vee,\wedge,\odot,\rightarrow,0,1),$
with an order $\preceq$ such that

  1. $(i)$

$(L,\vee,\wedge,0,1)$ is a bounded lattice;

  2. $(ii)$

$(L,\odot,1)$ is a commutative monoid;

  3. $(iii)$

$x\odot z\preceq y$ if and only if $x\preceq z\rightarrow y,$ for $x,y,z\in L$
, see  [17].

In this paper, $L$ will be denoted a residuated lattice, unless otherwise
stated.

A _Heyting algebra_ ([1]) is a lattice $(L,\vee,\wedge)$ with $0$ such that
for every $a,b\in L,$ there exists an element $a\rightarrow b\in L$ (called
the pseudocomplement of $a$ with respect to $b)$ where $a\rightarrow
b=\sup\\{x\in L:a\wedge x\leq b\\}.$ Heyting algebras are divisible residuated
lattices.

For $x,y\in L,$ we define $x\boxplus y=x^{\ast}\rightarrow y^{\ast\ast}$ and
$x\uplus y=x^{\ast}\rightarrow y,$ where $x^{\ast}=x\rightarrow 0.$ We remark
that $\boxplus$ is associative and commutative and $\uplus$ is only
associative.

We recall some rules of calculus in residuated lattices, see [3], [16]:

  1. $(1)$

$x\rightarrow y=1$ if and only if $x\preceq y;$

  2. $(2)$

$x,y\preceq x\uplus y\preceq x\boxplus y,x\boxplus 0=x^{\ast\ast},x\boxplus
x^{\ast}=1,x\boxplus 1=1,x\boxplus y=y\boxplus x,(x\boxplus y)\boxplus
z=x\boxplus(y\boxplus z),$ $x\preceq y\Rightarrow x\boxplus z\preceq y\boxplus
z;$

  3. $(3)$

$x\boxplus y=(x^{\ast}\odot y^{\ast})^{\ast},$ $(x\boxplus
y)^{\ast\ast}=x\boxplus y=x^{\ast\ast}\boxplus y^{\ast\ast},$ for every
$x,y,z\in L.$

An ideal in residuated lattices is a generalization of the similar notion from
MV-algebras, see [6]. This concept is introduced in [15] using the operator
$\uplus$ which is not commutative. An equivalent definition is given in [3]
using $\boxplus.$ We remark that $\boxplus$ is associative and commutative and
$\uplus$ is only associative.

###### Definition 1

([3]) An _ideal_ residuated lattice $L$ is a subset $I\neq\emptyset$ of $L$__
such that:

  1. $(i_{1})$

For $x\leq i,x\in L,$ $i\in I\Longrightarrow$ $x\in I;$

  2. $(i_{2})$

$i,j\in I$ $\Longrightarrow i\boxplus j\in I.$

Let $A$ be a non-empty set. If $[0,1]$ is the real unit interval, a fuzzzy
subset of $A$ is a function $\mu:A\longrightarrow[0,1],$ see [18]. If $\mu$ is
not a constant map, then $\mu$ is a proper fuzzy subset of $A$.

Let $B\subset A$ be a non-empty subset of $A.$ The map
$\mu_{B}:A\rightarrow[0,1],$

$\mu_{B}\left(x\right)=\\{\begin{array}[]{c}1,\text{ if }x\in B\\\ 0,\text{ if
}x\notin B.\end{array}$

(the characteristic function) is a fuzzy subset.

The notion of fuzzy ideal in residuated lattices is introduced in [15] and
some characterizations are obtained.

###### Definition 2

([15]) A _fuzzy ideal of_ a residuated lattice $L$ is a fuzzy subset $\mu$ of
$L$ such that:

  1. $(fi_{1})$

$x\preceq y\Longrightarrow$ $\mu(x)\geq\mu(y);$

  2. $(fi_{2})$

$\mu(x\uplus y)\geq\min(\mu(x),\mu(y)),$ for every $x,y\in L.$

Two equivalent definitions for fuzzy ideals are given in [15]:

A _fuzzy ideal of_ $L$ is a fuzzy subset $\mu$ of $L$ such that:

  1. $(fi_{3})$

$\mu(0)\geq\mu(x),$ for every $x\in L;$

  2. $(fi_{4})$

$\mu(y)\geq\min(\mu(x),\mu((x^{\ast}\rightarrow y^{\ast})^{\ast}),$ for every
$x,y\in L\Leftrightarrow(fi_{4}^{\prime})$
$\mu(y)\geq\min(\mu(x),\mu(x^{\ast}\odot y)),$ for every $x,y\in L$.

We denote by $\mathcal{I}(L)$ the set of ideals and by $\mathcal{FI}(L)$ the
set of fuzzy ideals of the residuated lattice $L.$

Obviously, the constant functions $\mathbf{0,1}:L\rightarrow[0,1],$
$\mathbf{0}(x)=0$ and $\mathbf{1}(x)=1,$ for every $x\in L$ are fuzzy ideals
of $L.$

There are two important fuzzy subsets in a residuated lattice $L:$ For
$I\subseteq L$ and $\alpha,\beta\in[0,1]$ with $\alpha>\beta$ is defined
$\mu_{I}:L\rightarrow[0,1]$ by

$\mu_{I}(x)=\\{\begin{array}[]{c}\alpha,\text{ if }x\in I\\\ \beta,\text{ if
}x\notin I.\end{array}$

The fuzzy subset $\mu_{I}$ is a generalization of the characteristic function
of $I,$ denoted $\varphi_{I.}$ Moreover, in [15] is proved that
$I\in\mathcal{I}(L)$ iff $\mu_{I}\in\mathcal{F}\mathcal{I}(L).$

###### Lemma 3

([15]) For $\mu\in\mathcal{FI}(L),$ the following hold:

  1. $(i)$

$\mu(x)=\mu(x^{\ast\ast})$

  2. $(ii)$

$\mu(x\uplus y)=\min(\mu(x),\mu(y)),$ for every $x,y\in L.$

For $\mu_{1}$ and $\mu_{2}$ two fuzzy subsets of $L$ is define the order
relation $\mu_{1}\subset$ $\mu_{2}$ if $\mu_{1}(x)\leq$ $\mu_{2}(x),$ for
every $x\in L.$

Moreover, for a family $\\{\mu_{i}:i\in I\\}$ of fuzzy ideals of $L$ we define
$\underset{i\in I}{\cup}\mu_{i},\underset{i\in
I}{\cap}\mu_{i}:L\rightarrow[0,1]$ by

$(\underset{i\in I}{\cup}\mu_{i})(x)=\sup\\{\mu_{i}(x):i\in I\\}\text{ and
}(\underset{i\in I}{\cap}\mu_{i})(x)=\inf\\{\mu_{i}(x):i\in I\\},\text{ for
every }x\in L,\text{ see }\cite[cite]{[\@@bibref{}{ZA}{}{}]}.$

Obviously, $\underset{i\in I}{\cap}\mu_{i}\in\mathcal{FI}(L)$ but, in general
$\underset{i\in I}{\cup}\mu_{i}$ is not a fuzzy ideal of $L,$ see [14].

We recall (see [1]) that a complete lattice $(\mathcal{A},\vee,\wedge)$ is
called Brouwerian if it satisfies the identity
$a\wedge(\underset{i}{\bigvee}b_{i})=\underset{i}{\bigvee}(a\wedge b_{i}),$
whenever the arbitrary unions exists. An element $a\in\mathcal{A}$ is called
compact if $a\leq\vee X$ for some $X\subseteq\mathcal{L}$ implies $a\leq\vee
X_{1}$ for some finite $X_{1}\subseteq X.$

###### Remark 4

([1]) Let $A$ be a set of real numbers. We say that $l\in R$ is the supremum
of $A$ if:

  1. 1.

$l$ is an upper bound for $A;$

  2. 2.

$l$ is the least upper bound: for every $\ \epsilon>0$ there is
$a_{\epsilon}\in A$ such that $a_{\epsilon}>l-\epsilon,$ i.e.,
$l<a_{\epsilon}+\epsilon.$

###### Remark 5

If $a,b$ are real numbers such that $a,b\in[0,1]$ and $a>b-\epsilon,$ for
every $\epsilon>0,$ then $a\geq b.$ Indeed, if we suppose that $a<b,$ then
there is $\epsilon_{0}>0$ such that $b-a>\epsilon_{0}>0,$ which is a
contradiction with hypothesis.

##  3 The lattice of fuzzy ideals in a residuated lattice $L$

###### Lemma 6

Let $x,y,z\in L.$ Then $x^{\ast}\boxplus(y\boxplus z)=1$ iff $x\preceq
y\boxplus z.$

Proof. If $x^{\ast}\boxplus(y\boxplus z)=1,$ then
$1=x^{\ast\ast}\rightarrow(y\boxplus
z)^{\ast\ast}=x^{\ast\ast}\rightarrow(y\boxplus z),$ so $x\preceq
x^{\ast\ast}\preceq y\boxplus z.$

Conversely, $x\preceq y\boxplus z\Rightarrow$ $x^{\ast\ast}\preceq(y\boxplus
z)^{\ast\ast}\Rightarrow x^{\ast\ast}\rightarrow(y\boxplus
z)^{\ast\ast}=1\Rightarrow x^{\ast}\boxplus(y\boxplus z)=1.$

###### Lemma 7

If $\mu\in\mathcal{FI}(L),$ then $\mu(x\boxplus y)=\mu(x\uplus
y)=\min(\mu(x),\mu(y^{\ast\ast})),$ for every $x,y\in L.$

Proof. From Lemma 3, $\mu(x\boxplus y)=\mu(x\uplus
y^{\ast\ast})=\min(\mu(x),\mu(y^{\ast\ast}))=\min(\mu(x),\mu(y))=\mu(x\uplus
y).$

###### Proposition 8

Let $\mu$ be a fuzzy subset of $L.$ Then $\mu\in\mathcal{FI}(L)$ iff it
satisfies the following conditions:

  1. $(fi_{1})$

$x\preceq y\Longrightarrow$ $\mu(x)\geq\mu(y);$

  2. $(fi_{2}^{\prime})$

$\mu(x\boxplus y)\geq\min(\mu(x),\mu(y)),$ for every $x,y\in L.$

Proof. If $\mu\in\mathcal{FI}(L),$ using Lemma 7, $(fi_{2}^{\prime})$ holds.

Conversely, assume that $(fi_{1})$ and $(fi_{2}^{\prime})$ hold and let
$x,y\in L.$ Since $x\uplus y\preceq x\boxplus y$ we obtain
$\min(\mu(x),\mu(y))\leq\mu(x\boxplus y)\leq\mu(x\uplus y),$ so $(fi_{2})$
hold. Thus, $\mu\in\mathcal{FI}(L).$

###### Proposition 9

Let $\mu$ be a fuzzy subset of $L.$ The following are equivalent:

  1. $(i)$

$\mu\in\mathcal{FI}(L);$

  2. $(ii)$

For every $x,y,z\in L,$ if $(x\boxplus y)\boxplus z^{\ast}=1$ then
$\mu(z)\geq\min(\mu(x),\mu(y));$

  3. $(iii)$

For every $x,y,z\in L,$ if $z\preceq x\boxplus y$ then
$\mu(z)\geq\min(\mu(x),\mu(y)).$

Proof. $(i)\Longrightarrow(ii).$ Let $x,y,z\in L$ such that $(x\boxplus
y)\boxplus z^{\ast}=1.$ Then $1=(x\boxplus y)^{\ast}\longrightarrow z^{\ast}$
so, $(x\boxplus y)^{\ast}\preceq z^{\ast}$. Thus, using Lemma 3 and
Proposition 8 we have $\mu(z)=\mu(z^{\ast\ast})\geq\mu((x\boxplus
y)^{\ast\ast})=\mu(x\boxplus y)\geq\min(\mu(x),\mu(y)).$

$(ii)\Longrightarrow(i).$ Since $(x\boxplus x)\boxplus 0^{\ast}=1,$ by
hypothesis, we deduce $(fi_{3}).$ Also, since $[x\boxplus(x^{\ast}\odot
y)]\boxplus y^{\ast}=$ $(x\boxplus y^{\ast})\boxplus(x^{\ast}\odot
y)=(x^{\ast}\odot y)^{\ast}\boxplus(x^{\ast}\odot y)=1,$ we obtain
$(fi_{4}^{\prime}).$Thus, $\mu\in\mathcal{FI}(L).$

$(ii)\Leftrightarrow(iii).$ Using Lemma 6, $z\preceq x\boxplus y$ iff
$(x\boxplus y)\boxplus z^{\ast}=1.$

If $\mu$ is a fuzzy subset of a residuated lattice $L,$ we denote by
$\overline{\mu}$ the smallest fuzzy ideal containing $\mu.$ $\overline{\mu}$
is called the fuzzy ideal generated by $\mu$ and it is characterized in [15],
Theorem 3.19 and [14], Theorem 5.

In the following, we show a new characterization:

###### Proposition 10

Let $L$ be a residuated lattice and $\mu,\mu^{\prime}$ $:L\rightarrow[0,1]$ be
fuzzy subsets of $L$ such that

$\mu^{\prime}(x)=\sup\\{\min(\mu(x_{1}),...,\mu(x_{n})):x\preceq
x_{1}\boxplus...\boxplus x_{n},n\in N,x_{1},....,x_{n}\in L\\},\text{ for
every }x\in L.\text{ }$

Then $\mu^{\prime}=\overline{\mu}.$

Proof. First, using Proposition 9, we will prove that
$\mu^{\prime}\in\mathcal{FI}(L).$

Let $x,y,z\in L$ such that $z\preceq x\boxplus y$ and $\epsilon>0$ arbitrary.

By definition of $\mu^{\prime}$, for $x,y\in L$ there are $n,m\in N$ and
$x_{1},....,x_{n},y_{1},....,y_{m}\in L$ such that

$x\preceq x_{1}\boxplus...\boxplus x_{n}\text{ and
}\mu^{\prime}(x)<\epsilon+\min(\mu(x_{1}),...,\mu(x_{n}))\text{ }$

and

$y\preceq y_{1}\boxplus...\boxplus y_{m}\text{ and
}\mu^{\prime}(y)<\epsilon+\min(\mu(y_{1}),...,\mu(y_{m})).\text{ }$

Then $x\boxplus y\preceq x_{1}\boxplus...\boxplus x_{n}\boxplus
y_{1}\boxplus...\boxplus y_{m}$ and $\mu^{\prime}(x\boxplus
y)=\sup\\{\min(\mu(t_{1}),...,\mu(t_{k})):x\boxplus y\preceq
t_{1}\boxplus...\boxplus t_{k},k\in N,t_{1},....,t_{k}\in L\\}$
$\geq\min(\mu(x_{1}),...,\mu(x_{n}),\mu(y_{1}),...,\mu(y_{m}))$
$=\min(\min(\mu(x_{1}),...,\mu(x_{n})),\min(\mu(y_{1}),...,\mu(y_{m})))$
$>\min(\mu^{\prime}(x)-\epsilon,\mu^{\prime}(y)-\epsilon)=\min(\mu^{\prime}(x),\mu^{\prime}(y))-\epsilon.$

Since $\epsilon$ is arbitrary, using Remark 5, we deduce that
$\mu^{\prime}(x\boxplus y)\geq\min(\mu^{\prime}(x),\mu^{\prime}(y)).$

Similarly, for $x\boxplus y$ there are $p\in N$ and $s_{1},....,s_{p}\in L$
such that

$x\boxplus y\preceq s_{1}\boxplus...\boxplus s_{p}\text{ and
}\mu^{\prime}(x\boxplus y)<\epsilon+\min(\mu(s_{1}),...,\mu(s_{p})).\text{ }$

Thus, $z\preceq s_{1}\boxplus...\boxplus s_{p},$ so
$\mu^{\prime}(z)=\sup\\{\min(\mu(z_{1}),...,\mu(z_{r})):z\preceq
z_{1}\boxplus...\boxplus z_{r},r\in N,z_{1},....,z_{r}\in L\\}$
$\geq\min(\mu(s_{1}),...,\mu(s_{p}))$ $>\mu^{\prime}(x\boxplus y)-\epsilon.$

We obtain $\mu^{\prime}(z)\geq\mu^{\prime}(x\boxplus y).$ Finally, we conclude
that $\mu^{\prime}(z)\geq\min(\mu^{\prime}(x),\mu^{\prime}(y)),$ so
$\mu^{\prime}\in\mathcal{FI}(L).$

Obviously, $\mu\subset\mu^{\prime}$ since for every $x\in L,x\preceq x\boxplus
x,$ so $\mu^{\prime}(x)\geq\min(\mu(x),\mu(x))=\mu(x).$

Also, if $\mu^{{}^{\prime\prime}}\in\mathcal{FI}(L)$ such that
$\mu\subset\mu^{{}^{\prime\prime}\text{ }}$then
$\mu^{\prime}(x)=\sup\\{\min(\mu(x_{1}),...,\mu(x_{n})):x\preceq
x_{1}\boxplus...\boxplus x_{n},n\in N,x_{1},....,x_{n}\in L\\}\leq$
$\sup\\{\min(\mu^{\prime\prime}(x_{1}),...,\mu^{\prime\prime}(x_{n})):x\preceq
x_{1}\boxplus...\boxplus x_{n},n\in N,x_{1},....,x_{n}\in
L\\}\leq\mu^{\prime\prime}(x),$ for every $x\in L,$ since $x\preceq
x_{1}\boxplus...\boxplus x_{n}\Rightarrow$
$\mu^{\prime\prime}(x)\geq\mu^{\prime\prime}(x_{1}\boxplus...\boxplus
x_{n})=\min(\mu^{\prime\prime}(x_{1}),...,\mu^{\prime\prime}(x_{n})).$Thus,
$\mu^{\prime}\subset\mu^{{}^{\prime\prime}\text{ }}$ , so $\mu^{\prime}$ is
the least fuzzy ideal of $L$ containing $\mu,$ i.e.,
$\mu^{\prime}=\overline{\mu}.$

###### Theorem 11

The lattice $(\mathcal{FI}(L),\subset$ $)$ is a complete Brouwerian lattice.

Proof. If $(\mu_{i})_{i\in I}$ is a family of fuzzy ideals of $L,$ then the
infimum of this family is $\underset{i\in I}{\sqcap}\mu_{i}=\underset{i\in
I}{\cap}\mu_{i}$ and the supremum is $\underset{i\in
I}{\sqcup}\mu_{i}=\overline{\underset{i\in I}{\cup}\mu_{i}}.$

Obviously, the lattice $(\mathcal{FI}(L),\subset$ $)$ is complete.

To prove that $\mathcal{FI}(L)$ is a Brouwerian lattice we show that for every
fuzzy ideal $\mu$ and every family $(\mu_{i})_{i\in I}$ of fuzzy ideals,
$\mu\sqcap(\underset{i\in I}{\sqcup}\mu_{i})=\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i}).$ Clearly, $\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i})\subset\mu\sqcap(\underset{i\in
I}{\sqcup}\mu_{i}),$ so we prove only that $\mu\sqcap(\underset{i\in
I}{\sqcup}\mu_{i})\subset\underset{i\in I}{\sqcup}(\mu\sqcap\mu_{i}).$

For this, let $x\in L$ and $\epsilon>0$ arbitrary.

Since $(\underset{i\in I}{\sqcup}\mu_{i})(x)=\sup\\{\min((\underset{i\in
I}{\cup}\mu_{i})(z_{1}),...,(\underset{i\in I}{\cup}\mu_{i})(z_{m})):x\preceq
z_{1}\boxplus...\boxplus z_{m},m\in N,z_{1},....,z_{m}\in L\\},$ there are
$n\in N$ and $x_{1},....,x_{n}\in L$ such that

$x\preceq x_{1}\boxplus...\boxplus x_{n}\text{ and }(\underset{i\in
I}{\sqcup}\mu_{i})(x)<\epsilon+\min((\underset{i\in
I}{\cup}\mu_{i})(x_{1}),...,(\underset{i\in I}{\cup}\mu_{i})(x_{n})).$

Using the definition of $\underset{i\in I}{\cup}\mu_{i},$ for every
$k=1,...,n$ there is $i_{k}\in N$ such that

$\text{ }(\underset{i\in I}{\cup}\mu_{i})(x_{k})<\epsilon+\mu_{i_{k}}(x_{k}).$

Thus,

$\text{ }(\underset{i\in
I}{\sqcup}\mu_{i})(x)<\epsilon+\min(\epsilon+\mu_{i_{1}}(x_{1}),...,\epsilon+\mu_{i_{n}}(x_{n})).$

Then

$(\mu\sqcap(\underset{i\in
I}{\sqcup}\mu_{i}))(x)<2\epsilon+\min(\mu(x),\mu_{i_{1}}(x_{1}),...,\mu_{i_{n}}(x_{n})).$

We consider $y_{1},....,y_{n}\in L$ such that

$y_{1}^{\ast}=(y_{2}\boxplus...\boxplus y_{n})\boxplus x^{\ast}$
$y_{n}^{\ast}=(x_{1}\boxplus...\boxplus x_{n-1})\boxplus x^{\ast}$

and for every $t=2,...,n-1$

$y_{t}^{\ast}=(x_{1}\boxplus...\boxplus
x_{t-1})\boxplus(y_{t+1}\boxplus...\boxplus y_{n})\boxplus x^{\ast}.$

Obviously, for every $t=1,...,n,$ $y_{t}^{\ast}\boxplus x=1,$ so,
$y_{t}^{\ast\ast}\preceq x^{\ast\ast}$ and
$\mu(x)=\mu(x^{\ast\ast})\leq\mu(y_{t}^{\ast\ast})=\mu(y_{t}).$

Moreover, $(y_{1}\boxplus...\boxplus y_{n})\boxplus x^{\ast}=y_{1}\boxplus
y_{1}^{\ast}=1,$ so using Lemma 6, we deduce that

$x\preceq y_{1}\boxplus...\boxplus y_{n}.$

Also, by Lemma 6, since $x\preceq x_{1}\boxplus...\boxplus x_{n}$ we have that
$y_{n}^{\ast}\boxplus x_{n}=(x_{1}\boxplus...\boxplus x_{n})\boxplus
x^{\ast}=1$ and for every $t=1,...,n-1,$ $y_{t}^{\ast}\boxplus
x_{t}=[(x_{1}\boxplus...\boxplus x_{t})\boxplus$ $(y_{t+2}\boxplus...\boxplus
y_{n}\boxplus x^{\ast}]\boxplus y_{t+1}=y_{t+1}^{\ast}\boxplus y_{t+1}=1.$

So,

$y_{t}\preceq x_{t},\text{ for every }t=1,...,n.$

Thus, we deduce that

$\mu_{i_{k}}(x_{k})\leq\mu_{i_{k}}(y_{k}),\text{ for every }k=1,...,n.$

We conclude that

$\min(\mu(x),\mu_{i_{k}}(x_{k}))\leq\min(\mu(y_{k}),\mu_{i_{k}}(y_{k}))=(\mu\sqcap\mu_{i_{k}})(y_{k}),\text{
for every }k=1,...,n.$

Thus,

$(\mu\sqcap(\underset{i\in
I}{\sqcup}\mu_{i}))(x)<2\epsilon+\min((\mu\sqcap\mu_{i_{1}})(y_{1}),...(\mu\sqcap\mu_{i_{n}})(y_{n})).$

Since $(\mu\sqcap\mu_{i_{k}})(y_{k})\leq(\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i}))(y_{k}),$ for every $k=1,...,n,$ using the fact
that $x\preceq y_{1}\boxplus...\boxplus y_{n},$ we obtain

$(\mu\sqcap(\underset{i\in
I}{\sqcup}\mu_{i}))(x)<2\epsilon+\min((\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i}))(y_{1}),...,(\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i}))(y_{n}))<2\epsilon+(\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i}))(x).$

But $\epsilon$ is arbitrary, so from Remark 5,

$(\mu\sqcap(\underset{i\in I}{\sqcup}\mu_{i}))(x)\leq(\underset{i\in
I}{\sqcup}(\mu\sqcap\mu_{i}))(x).$

By [1] and Theorem 11 we deduce that:

###### Proposition 12

If $\mu_{1},$ $\mu_{2}\in\mathcal{FI}(L)$ then

  1. $(i)$

$\mu_{1}\rightsquigarrow$
$\mu_{2}=\sup\\{\mu\in\mathcal{FI}(L):\mu_{1}\sqcap\mu\subset\mu_{2}\\}=\sqcup\\{\mu\in\mathcal{FI}(L):\mu_{1}\sqcap\mu\subset\mu_{2}\\}\in\mathcal{FI}(L);$

  2. $(ii)$

If $\mu\in\mathcal{FI}(L),$ then $\mu_{1}\sqcap\mu\subset\mu_{2}$ if and only
if $\mu\sqsubset\mu_{1}\rightsquigarrow\mu_{2}.$

Moreover,

###### Corollary 13

$(\mathcal{FI}(L),\sqcap,\sqcup,\rightsquigarrow,\mathbf{0})$ is a Heyting
algebra.

##  4 Applications of fuzzy sets in Coding Theory

###  4.1  Symmetric difference of ideals in a finite commutative and unitary
ring

In this section, we will present an application of a fuzzy sets on some
special cases of residuated algebras, namely Boolean algebras.

Let $A$ be a non-empty set and $B\subset A$ be a non-empty subset of $A.$ The
map $\mu_{B}:A\rightarrow[0,1],$

$\mu_{B}\left(x\right)=\left\\{\begin{array}[]{c}{1,~{}x\in B}\\\
{0,~{}x\notin B}\end{array}\right.,$

is called the characteristic function of the set $B$.

For two nonempty sets, $A,B,$ we define the symmetric difference of the sets
$A,B,$

$A\Delta B=\left(A-B\right)\cup\left(B-A\right)=\left(A\cup
B\right)-\left(B\cap A\right)$

###### Proposition 14

We consider $A$ and $B$ two nonempty sets.

  1. $(i)$

We have $\mu_{A\Delta B}=0$ if and only if $A=B;$

  2. $(ii)$

([13], p. 215). The following relation is true

$\mu_{A\Delta B}=\mu_{A}+\mu_{B}-2\mu_{A}\mu_{B}.$

  3. $(iii)$

Let $A_{i},i\in\\{1,2,...,n\\}$ be $n$ nonempty sets. The following relation
is true

$\mu_{A_{1}\Delta A_{2}\Delta...\Delta A_{n}}=\sum\mu_{A_{i}}-2\underset{i\neq
j}{\sum}\mu_{A_{i}}\mu_{A_{j}}+2^{2}\underset{i\neq j\neq
k}{\sum}\mu_{A_{i}}\mu_{A_{j}}\mu_{A_{k}}-...+\left(-1\right)^{n-1}2^{n-1}\mu_{A_{1}}\mu_{A_{2}}...\mu_{A_{n}}.$

###### Remark 15

Let $\left(R,+,\cdot\right)$ be a unitary and a commutative ring and
$I_{1},I_{2},...,I_{s}$ be ideals in $R$.

  1. $(i)$

For $i\neq j,$ we have $I_{i}\Delta I_{j}$ is not an ideal in $R$. Indeed,
$0\notin I_{i}\Delta I_{j}$, therefore $I_{i}\Delta I_{j}$ is not an ideal in
$R$;

  2. $(ii)$

In general, $I_{1}\Delta I_{2}\Delta...\Delta I_{n}$, for $n\geq 2$, is not an
ideal in $R$. Indeed, if $n\geq 3$ and $x,y\in I_{1}\Delta
I_{2}\Delta...\Delta I_{n},$ supposing that $x\in I_{j}$ and $y\in I_{k},$ we
have that $xy\in I_{j}$ and $xy\in I_{k},$ therefore $xy\in I_{j}\cap I_{k}$.
We obtain that $\mu_{I_{1}\Delta I_{2}\Delta...\Delta
I_{n}}\left(xy\right)=\mu_{I_{j}}\left(xy\right)+\mu_{I_{k}}\left(xy\right)-2\mu_{{}_{I_{j}}}\mu_{{}_{I_{k}}}\left(xy\right)=0,$
then $xy\notin I_{1}\Delta I_{2}\Delta...\Delta I_{n}$ and $I_{1}\Delta
I_{2}\Delta...\Delta I_{n}$ is not an ideal in $R$.

###### Definition 16

If $A=\\{a_{1},a_{2},...,a_{n}\\}$ is a finite set with $n$ elements and $B\,$
is a nonempty subset of $A$, we consider the vector
$c_{B}=(c_{i})_{i\in\\{1,2,...,n\\}},$ where $c_{i}=0$ if $a_{i}\notin B$ and
$c_{i}=1$ if $a_{i}\in B$. The vector $c_{B}$ is called the codeword attached
to the set $B$. We can represent $c_{B}$ as a string
$c_{B}=c_{1}c_{2}...c_{n}$.

###  4.2 Linear codes

We consider $p$ a prime number and $\mathbf{F}_{p^{n}}$ a finite field of
characteristic $p$. $\mathbf{F}_{p^{n}}$ is a vector space over the field
$\mathbb{Z}_{p}.$ A linear code $\mathcal{C}$ of length $n$ and dimension $k$
is a vector subspace of the vector space $\mathbf{F}_{p^{n}}$. If $p=2$, we
call this code a binary linear code. The elements of $\mathcal{C}$ are called
codewords. The weight of a codeword is the number of its elements that are
nonzero and the distance between two codewords is the Hamming distance between
them, that means represents the number of elements in which they differ. The
distance $d$ of the linear code is the minimum weight of its nonzero
codewords, or equivalently, the minimum distance between distinct codewords. A
linear code of length $n$, dimension $k$, and distance $d$ is called an
$[n,k,d]$ code (or, more precisely, $[n,k,d]_{p}$ code). The rate of a code is
$\frac{k}{n},$ that means it is an amount such that for each $k$ bits of
transmitted information, the code generates $n$ bits of data, in which $n-k$
are redundant. Since $\mathcal{C}$ is a vector subspace of dimension $k$, it
is generated by bases of $k$ vectors. The elements of such a basis can be
represented as a rows of a matrix $G$, named generating matrix associated to
the code $\mathcal{C}$. This matrix is a matrix of $k\times n$ type. (see [Gu;
10]). The codes of the type $[2^{t},t,2^{\tau-1}]_{2}$, $t\geq 2$, are called
Hadamard codes. Hadamard codes are a class of error-correcting codes (see [KK;
12], p. 183). Named after french mathematician Jacques Hadamard, these codes
are used for error detection and correction when transmitting messages are
over noisy or unreliable channels. Usually, Hadamard codes are constructed by
using Hadamard matrices of Sylvester’s type, but there are Hadamard codes
using arbitrary Hadamard matrix not necessarily of the above type (see [CR;
20]). As we can see, Hadamard codes have a good distance property, but the
rate is of a low level (see [Gu; 10]).

###### Remark 17

([11], Definition 16). The generating matrix of a Hadamard code of the type
$[2^{t},t,2^{\tau-1}]_{2}$, $t\geq 2,$ has as columns all $t$-bits vectors
over $\mathbb{Z}_{2}$ (vectors of length $t$).

##  5  Connections between Boolean algebras and Hadamard codes

In the following, we present a particular case of residuated lattices, named
MV-algebras.

###### Definition 18

([4]) An abelian monoid $\left(X,\theta,\oplus\right)$ is called MV-algebra if
and only if we have an operation $"^{\prime}"$ such that:

  1. $(i)$

$(x^{\prime})^{\prime}=x;$

  2. $(ii)$

$x\oplus\theta^{\prime}=\theta^{\prime};$

  3. $(iii)$

$\left(x^{\prime}\oplus y\right)^{\prime}\oplus y=$ $\left(y^{\prime}\oplus
x\right)^{\prime}\oplus x$, for all $x,y\in X.$ We denote it by
$\left(X,\oplus,^{\prime},\theta\right).\vskip 6.0pt plus 2.0pt minus 2.0pt$

###### Definition 19

([6], Definition 4.2.1) An algebra
$\left(W,\circ,\overline{\phantom{x}},1\right)$ of type
$\left(2,1,0\right)~{}$is called a Wajsberg algebra (or W-algebra) if and only
if for every $x,y,z\in W$, we have:

  1. $(i)$

$1\circ x=x;$

  2. $(ii)$

$\left(x\circ y\right)\circ\left[\left(y\circ z\right)\circ\left(x\circ
z\right)\right]=1;$

  3. $(iii)$

$\left(x\circ y\right)\circ y=\left(y\circ x\right)\circ x;$

  4. $(iv)$

$\left(\overline{x}\circ\overline{y}\right)\circ\left(y\circ x\right)=1.\vskip
6.0pt plus 2.0pt minus 2.0pt$

###### Remark 20

([6], Lemma 4.2.2 and Theorem 4.2.5)

  1. $(i)$

If $\left(W,\circ,\overline{\phantom{x}},1\right)$ is a Wajsberg algebra,
defining the following multiplications

$x\odot y=\overline{\left(x\circ\overline{y}\right)}$

and

$x\oplus y=\overline{x}\circ y,$

for all $x,y\in W$, we obtain that
$\left(W,\oplus,\odot,\overline{\phantom{x}},0,1\right)$ is an MV-algebra.

  2. $(ii)$

If $\left(X,\oplus,\odot,^{\prime},\theta,1\right)$ is an MV-algebra, defining
on $X$ the operation

$x\circ y=x^{\prime}\oplus y,$

it results that $\left(X,\circ,^{\prime},1\right)$ is a Wajsberg algebra.

###### Definition 21

([9]) If $\left(W,\circ,\overline{\phantom{x}},1\right)$ is a Wajsberg
algebra, on $W$ we define the following binary relation

$x\leq y~{}\text{if~{}and~{}only~{}if~{}}x\circ y=1.$ (3.2.)

This relation is an order relation, called the natural order relation on $W$.

###### Definition 22

([7]) Let $\left(X,\oplus,^{\prime},\theta\right)$ be an MV-algebra. The
nonempty subset $I\subseteq X$ is called an ideal in $X$ if and only if the
following conditions are satisfied:

  1. $(i)$

$\theta\in I$, where $\theta=\overline{1};$

  2. $(ii)$

$x\in I$ and $y\leq x$ implies $y\in I;$

  3. $(iii)$

If $x,y\in I$, then $x\oplus y\in I$.

We remark that the concept of ideal in residuated lattices is a generalization
for the notion of ideal in MV-algebras.

###### Definition 23

([6], p. 13) An ideal $P$ of the MV-algebra
$\left(X,\oplus,^{\prime},\theta\right)$ is a prime ideal in $X$ if and only
if for all $x,y\in P$ we have $(x^{\prime}\oplus y)^{\prime}\in P$ or
$(y^{\prime}\oplus x)^{\prime}\in P$.

###### Definition 24

([10], p. 56) Let $\left(W,\circ,\overline{\phantom{x}},1\right)$ be a
Wajsberg algebra and let $I\subseteq W$ be a nonempty subset. $I$ is called an
ideal in $W$ if and only if the following conditions are fulfilled:

###### Definition 25

  1. $(i)$

$\theta\in I$, where $\theta=\overline{1};$

  2. $(ii)$

$x\in I$ and $y\leq x$ implies $y\in I;$

  3. $(iii)$

If $x,y\in I$, then $\overline{x}\circ y\in I$.

###### Definition 26

Let $\left(W,\circ,\overline{\phantom{x}},1\right)$ be a Wajsberg algebra and
$P\subseteq W$ be a nonempty subset. $P$ is called a prime ideal in $W$ if and
only if for all $x,y\in P$ we have $(x\circ y)^{\prime}\in P$ or $(y\circ
x)^{\prime}\in P$.

###### Definition 27

The algebra $\left(B,\vee\wedge,\partial,0,1\right)$, equipped with two binary
operations $\vee$ and $\wedge$ and a unary operation$\ \partial$, is called a
Boolean algebra if and only if $\left(B,\vee\wedge\right)~{}$is a distributive
and a complemented lattice with

$x\vee\partial x=1,$ $x\wedge\partial x=0,$

for all elements $x\in B$. The elements $0$ and $1$ are the least and the
greatest elements from the algebra $B.\vskip 6.0pt plus 2.0pt minus 2.0pt$

###### Remark 28

  1. $(i)$

Boolean algebras represent a particular case of MV-algebras. Indeed, if
$\left(B,\vee\wedge,\partial,0,1\right)$ is a Boolean algebra, then can be
easily checked that $\left(B,\vee,\partial,0\right)$ is an MV-algebra;

  2. $(ii)$

A Boolean ring $\left(B,+,\cdot\right)$ is a unitary and commutative ring such
that $x^{2}=x,$ for each $x\in B;$

  3. $(iii)$

To a Boolean algebra $\left(B,\vee\wedge,\partial,0,1\right)$ we can associate
a Boolean ring $\left(B,+,\cdot\right),$ where

$\displaystyle x+y$ $\displaystyle=$ $\displaystyle\left(x\vee
y\right)\wedge\partial\left(x\wedge y\right),$ $\displaystyle x\cdot y$
$\displaystyle=$ $\displaystyle x\wedge y,$

for all $x,y\in B.$ Conversely, if $\left(B,+,\cdot\right)$ is a Boolean ring,
we can associate a Boolean algebra $\left(B,\vee\wedge,\partial,0,1\right),$
where

$\displaystyle x\vee y$ $\displaystyle=$ $\displaystyle x+y+xy,$
$\displaystyle x\wedge y$ $\displaystyle=$ $\displaystyle xy,$
$\displaystyle\partial x$ $\displaystyle=$ $\displaystyle 1+x;$

  4. $(iv)$

Let $\left(I,+,\cdot\right)$ be an ideal in a Boolean ring
$\left(B,+,\cdot\right)$, therefore $I$ is an ideal in the Boolean algebra
$\left(B,\vee\wedge,\partial,0,1\right)$. The converse is also true.

###### Remark 29

  1. $(i)$

If $X$ is an MV-algebra and $I$ is an ideal (prime ideal) in $X$, therefore on
the Wajsberg algebra structure, obtained as in Remark 3.7. ii), we have that
the same set $I$ is an ideal (prime ideal) in $X$ as Wajsberg algebra. The
converse is also true.

  2. $(ii)$

Finite MV-algebras of order $2^{t}$ are Boolean algebras.

  3. $(iii)$

Between ideals in a Boolean algebra and ideals in the associated Boolean ring
it is a bijective correspondence, that means, if $I$ is an ideal in a Boolean
algebra, the same set $I,$ with the corresponded multiplications, is an ideal
in the associated Boolean ring. The converse is also true.

We consider $\left(R,+,\cdot\right)$ a finite, commutative, unitary ring and
$I,J$ be two ideals. Let $c_{I}$ and $c_{J}$ be the codewords attached to
these sets, as in Definition 16.

###### Proposition 30

With the above notations, we have that:

  1. $(i)$

To the set $I\Delta J$ correspond the codeword $c_{I}+c_{J}=c_{I}\oplus
c_{J}$, where $\oplus$ is the XOR-operation;

  2. $(ii)$

If $I_{1},I_{2},...,I_{q}$ are ideals in the ring $R$ and
$c_{I_{1}},c_{I_{2}},...,c_{I_{q}}$ are the attached codewords, therefore the
vectors $c_{I_{1}},c_{I_{2}},...,c_{I_{q}}$ are linearly independent vectors.

Proof. $(i).$ It is clear, by straightforward computations.

$(ii).$ Let $R$ has $n$ elements. We work on the vector space
$V=\underset{n-time}{\underbrace{\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times...\times\mathbb{Z}_{2}}}$
over the field $\mathbb{Z}\,_{2}$. We consider
$\alpha_{1}c_{I_{1}}+...\alpha_{q}c_{I_{q}}=0,$where
$\alpha_{1},...\alpha\,_{q}\in\mathbb{Z}_{2}.$ Supposing that
$\alpha_{1}=...=\alpha\,_{q}=1$, we have that
$\alpha_{1}c_{I_{1}}+...\alpha_{q}c_{I_{q}}=0$ implies that $I_{1}\Delta
I_{2}\Delta...\Delta I_{q}=\emptyset$. Without losing the generality, since
symmetric difference is associative, from here we have that $I_{1}\Delta
I_{2}\Delta...\Delta I_{q-1}=I_{q}$, which is false, since $I_{q}$ has an
ideal structure and $I_{1}\Delta I_{2}\Delta...\Delta I_{q-1}$ is not an
ideal, from Remark 15.

With the above notations, we consider a matrix $M_{C},$ with rows the
codewords associated to the ideals $I_{1},I_{2},...,I_{q}$,

$M_{C}=\left(\begin{array}[]{c}c_{I_{1}}\\\ c_{I_{2}}\\\ ...\\\
c_{I_{q}}\end{array}\right).$

Since these rows are linearly independent vectors, the matrix $M_{C}$ can be
considered as a generating matrix for a code, called the code associated to
the ideals $I_{1},I_{2},...,I_{q}$, denoted
$\mathcal{C}_{I_{1}I_{2},...I_{q}}$.

###### Theorem 31

Let $\left(B,\vee\wedge,\partial,0,1\right)$ be a finite Boolean algebra of
order $2^{n}$. The following statements are true:

  1. $(i)$

The algebra $B$ has $n$ ideals of order $2^{n-1}$;

  2. $(ii)$

The code associated to above ideals generate a Hadamard code of the type
$[2^{n},n,2^{n-1}]_{2}$, $n\geq 2$.

Proof. $(i).$ It is clear, since ideals in the Boolean algebra structure are
ideals in the associated Boolean ring and vice-versa.

$(ii).$ Let $I_{1},I_{2},...,I_{n}$ be the ideals of order $2^{n-1}$. With the
above notations, we consider a matrix $M_{C},$ with rows the codewords
associated to these ideals,

$M_{C}=\left(\begin{array}[]{c}c_{I_{1}}\\\ c_{I_{2}}\\\ ...\\\
c_{I_{n}}\end{array}\right).$

Due to the correspondence between the ideals in the Boolean algebra structure,
the ideals in the associated Boolean ring and Proposition 30, we have that the
rows of the matrix $M_{C}$ are linearly independent vectors. Since
$I_{1},I_{2},...,I_{n}$ are the ideals of order $2^{n-1}$, the associated
codewords have $2^{n-1}$ nonzero elements, therefore the Hamming distance is
$d_{H}=2^{n-1}$. From here, we have that $M_{C}$ is a generating matrix for
the code $\mathcal{C}_{I_{1}I_{2},...I_{n}}$, which is a Hadamard code of the
type $[2^{n},n,2^{n-1}]_{2}$, $n\geq 2$.

###### Remark 32

A generating matrix $M_{C}$ of a Hadamard code $\mathcal{C}$ of the type
$[2^{n},n,2^{n-1}]_{2}$, $n\geq 2$, has $2^{n-1}n$ elements equal with $1$. If
the matrix has the following form: on the row $i$ we have the first $2^{n-i}$
elements equal to $1$, the next $2^{n-i}$ elements equal to $0$, and so on,
for $i\geq 1$, we call this form the Boolean form of the generating matrix of
the Hadamard code $\mathcal{C}$ and we denote it $M_{B}$.

###### Remark 33

  1. $(i).$

If $G$, a $r\times s$ matrix over a field $K$, is a generating matrix for a
linear code $\mathcal{C}$, then any matrix which is row equivalent to $G$ is
also a generating matrix for the code $\mathcal{C}$. Two row equivalent
matrices of the same type have the same row space. The row space of a matrix
is the set of all possible linear combinations of its row vectors, that means
it is a vector subspace of the space $K^{s}$, with dimension the rank of the
matrix $G$, $rankG$. From here, we have that two matrices are row equivalent
if and only if one can be deduced to the other by a sequence of elementary row
operations.

  2. $(ii).$

If $G$ is a generating matrix for a linear code $\mathcal{C}$, then, from the
above notations, we have that $M_{C}$ and $M_{B}$ are row equivalent,
therefore these matrices generate the same Hadamard code $\mathcal{C}$ of the
type $[2^{n},n,2^{n-1}]_{2}$, $n\geq 2$.

###### Theorem 34

With the above notations, let $M_{B}$ be the Boolean form of a generating
matrix of the Hadamard code of the type $[2^{n},n,2^{n-1}]_{2}$, $n\geq 2$. We
can construct a Boolean algebra $\mathcal{B}$ of order $2^{n}$ which has $n$
ideals of order $2^{n-1},$ with associated codewords being the rows of a
matrix $M_{B}$.

Proof. We consider the set $B_{i}=\\{0_{i},1_{i}\\}$, with
$0_{i}\leq_{i}1_{i}$, $i\in\\{1,2,...,n\\}$. On $B_{i}$ we define the
following
multiplication:$\begin{tabular}[]{l|ll}$\circ_{i}$&$0_{i}$&$1_{i}$\\\
\hline\cr$0_{i}$&$1_{i}$&$1_{i}$\\\ $1_{i}$&$0_{i}$&$1_{i}$\end{tabular}.$

It is clear that $\left(B_{i},\circ_{i},^{\prime},1_{i}\right)$, where
$0_{i}^{\prime}=1_{i}$ and $1_{i}^{\prime}=0_{i}$, is a Wajsberg algebra of
order $2$. On $B_{i}$ we have the following partial order relation
$x_{i}\leq_{i}y_{i}~{}$if and only if $x_{i}\circ_{i}y_{i}=1_{i}.$

Therefore, on the Cartesian product $\mathcal{B}=B_{1}\times
B_{2}\times...\times B_{n}$ we define a component-wise multiplication, denoted
$\diamond$. From here, we have that
$\left(\mathcal{B},\diamond,^{\prime},\mathbf{1}\right)$, where
$\left(x_{1},x_{2},...,x_{n}\right)^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},...,x_{n}^{\prime}\right)$
and $\mathbf{1=(}1,1,...,1)$, is a Wajsberg algebra of order $2^{n}$. We write
and denote the elements of $\mathcal{B}$ in the lexicographic order. The
element $\left(0_{1},0_{2},...,0_{n}\right)$, denoted $\left(0,0,...,0\right)$
or $\mathbf{0}$ it is the first element in $\mathcal{B}.$ With $\mathbf{1}$ we
denote $\mathbf{(}1,1,...,1)=\left(1_{1},1_{2},...,1_{n}\right)$ which is the
last element in $\mathcal{B}$. From Definition 3.8, on $\mathcal{B}$ we have
the following partial order relation

$x\leq_{\mathcal{B}}y~{}\text{if~{}and~{}only~{}if~{}}x\diamond
y=\mathbf{1}\text{.}$

It is clear that on $\mathcal{B}$ we have that $x\leq_{\mathcal{B}}y$ if and
only if $x_{i}\leq_{i}y_{i}$, for $i\in\\{1,2,...,n\\}$. From the Wajsberg
algebra structure we obtain the $MV$-algebra structure on $\mathcal{B}$, which
is a Boolean algebra structure, with the multiplication $x\oplus
y=x^{\prime}\diamond y$ ($\oplus$ which is the component-wise XOR-sum). The
ideals of order $2^{n-1}$ in this Boolean algebra of order $2^{n}$ are
generated by the maximal elements in respect to the order relation
$\leq_{\mathcal{B}}$. These elements have $n-1$ ”nonzero” components. First
maximal element, in the lexicographic order, is
$m_{1}=\left(0,1,1,...,1\right)$. This element generates an ideal of order
$2^{n-1}$, containing all elements $x_{j}$ equal or less than $m_{1}$ in
respect to the order relation $\leq_{\mathcal{B}}$. Indeed, all these elements
$x_{j}$ are maximum $n-2$ nonzero components and $x_{ji}\leq_{i}m_{1i}$,
$i\in\\{1,2,...,n\\},j\in\\{1,2,...,2^{n-1}\\}$, with the first component
always zero. We denote with $J_{1}$ the set all elements equal or less than
$m_{1}$. It results that $J_{1}$ with the multiplication $\oplus$ is
isomorphic to the vector space $\mathbb{Z}_{2}^{n-1}$, therefore $J_{1}$ is an
ideal in $\mathcal{B}$. The codeword corresponding to this ideal is
$\left(1,1,...,1,0,0,...,0\right)$ in which the first $2^{n-1}$ positions are
equal with $1$ and the next $2^{n-1}$ are $0$ and represent the first row of
the matrix $M_{\mathcal{B}}$. The next maximal element in lexicographic order
is $m_{2}=\left(1,0,1,...,1\right),$ with zero on the second position and $1$
in the rest. This element generates an ideal $J_{2}~{}$of order $2^{n-1}$,
containing all elements $x_{j}$ equal or less than $m_{2}$ in respect to the
order relation $\leq_{\mathcal{B}}$. All these elements $x_{j}$ are maximum
$n-2$ nonzero components and $x_{ji}\leq_{i}m_{1i}$,
$i\in\\{1,2,...,n\\},j\in\\{1,2,...,2^{n-1}\\},$ with the second component
always zero. With the same reason as above, we have that $J_{2},$ with the
multiplication $\oplus,$ is isomorphic to the vector space
$\mathbb{Z}_{2}^{n-1}$, therefore $J_{2}$ is an ideal in $\mathcal{B}$. The
codeword corresponding to this ideal is
$\left(1,1,...,1,0,0,...,0,1,1,...,0,...\right)$, with the first $2^{n-2}$
positions equal with $1$, the next $2^{n-2}$ are $0$ and so on. This codeword
represent the second row of the matrix $M_{\mathcal{B}}$, etc.

###### Example 35

In [8], the authors described all Wajsberg algebras of order less or equal
with $9$. In the following, we provide some examples of codes associated to
these algebras.

Case $n=4.$ We have two types of Wajsberg algebras of order $4$. First type is
a totally ordered set which has no proper ideals and the second type is a
partially ordered Wajsberg algebra, $W=\\{0,a,b,1\\}.$ This algebra has the
multiplication given by the following table:

$\circ$ | $0$ | $a$ | $b$ | $1$
---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$
$a$ | $b$ | $1$ | $b$ | $1$
$b$ | $a$ | $a$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $1$
.

This algebra has two proper ideals $I=\\{0,a\\}$ and $J=\\{0,b\\}$. The
associated $MV$-algebra of this algebra is a Boolean algebra. We consider
$c_{I}=\left(1,1,0,0\right)$ and $c_{J}=\left(1,0,1,0\right)$ the codewords
attached to the ideals $I$ and $J$. The matrix

$M_{C}=\left(\begin{array}[]{cccc}1&1&0&0\\\ 1&0&1&0\end{array}\right)$

is the generating matrix for the Hadamard code of the type
$\left(2^{2},2,2\right)$. As in Remark 17, this matrix has as columns all
$2$-bits vectors over $\mathbb{Z}_{2}:\\{11,10,01,00\\}$.

Case $n=8$. We consider the partially ordered Wajsberg algebra,
$W=\\{0,a,b,c,d,e,f,1\\}$ with the multiplication given by the following
table:

$\begin{tabular}[]{l|llllllll}$\circ$&$0$&$a$&$b$&$c$&$d$&$e$&$f$&$1$\\\
\hline\cr$0$&$1$&$1$&$1$&$1$&$1$&$1$&$1$&$1$\\\
$a$&$f$&$1$&$f$&$1$&$f$&$1$&$f$&$1$\\\ $b$&$e$&$e$&$1$&$1$&$e$&$e$&$1$&$1$\\\
$c$&$d$&$e$&$f$&$1$&$d$&$e$&$f$&$1$\\\ $d$&$c$&$c$&$c$&$c$&$1$&$1$&$1$&$1$\\\
$e$&$b$&$c$&$b$&$c$&$f$&$1$&$f$&$1$\\\ $f$&$a$&$a$&$c$&$c$&$e$&$e$&$1$&$1$\\\
$1$&$0$&$a$&$b$&$c$&$d$&$e$&$f$&$1$\end{tabular}.$

All proper ideals are of the form $I_{1}=\\{0,a\\}$, $I_{2}=\\{0,b\\}$,
$I_{3}=\\{0,d\\}$, $I_{4}=\\{0,a,b,c\\}$, $I_{5}=\\{0,a,d,e\\}$,
$I_{6}=\\{0,b,d,f\\}$ are also prime ideals. This algebra has three ideals of
order three $I_{4},I_{5},I_{6}$. The associated $MV$-algebra of this algebra
is a Boolean algebra. We consider $c_{I_{4}}=\left(1,1,1,1,0,0,0,0\right),$
$c_{I_{5}}=\left(1,1,0,0,1,1,0,0\right),c_{I_{6}}=\left(1,0,1,0,1,0,1,0\right)$
the codewords attached to the ideals $I_{4},I_{5},I_{6}$. The matrix

$M_{C}=\left(\begin{array}[]{cccccccc}1&1&1&1&0&0&0&0\\\ 1&1&0&0&1&1&0&0\\\
1&0&1&0&1&0&1&0\end{array}\right)$

is the generating matrix for the Hadamard code $\left(2^{3},2,2^{2}\right)$.
As in Remark 17, this matrix has as columns all $3$-bits vectors over
$\mathbb{Z}_{2}$, namely $\\{111,110,101,100,011,010,001,000\\}$.

###### Remark 36

([8], case $n=9$) If a finite Wajsberg algebra has an even number of proper
ideals, we can consider their associated codewords, as above. The obtained
generating matrix generate a linear code with Hamming distance $\geq 3$.
Indeed, for $n=9$, we consider the partially ordered Wajsberg algebra,
$W=\\{0,a,b,c,d,e,f,g,1\\}$ with the multiplication given by the following
table:

$\begin{tabular}[]{l|lllllllll}$\circ$&$0$&$a$&$b$&$c$&$d$&$e$&$f$&$g$&$1$\\\
\hline\cr$0$&$1$&$1$&$1$&$1$&$1$&$1$&$1$&$1$&$1$\\\
$a$&$g$&$1$&$1$&$g$&$1$&$1$&$g$&$1$&$1$\\\
$b$&$f$&$g$&$1$&$f$&$g$&$1$&$f$&$g$&$1$\\\
$c$&$e$&$e$&$e$&$1$&$1$&$1$&$1$&$1$&$1$\\\
$d$&$d$&$e$&$e$&$g$&$1$&$1$&$g$&$1$&$1$\\\
$e$&$c$&$d$&$e$&$f$&$g$&$1$&$f$&$g$&$1$\\\
$f$&$b$&$a$&$b$&$e$&$e$&$e$&$1$&$1$&$1$\\\
$g$&$a$&$b$&$b$&$d$&$e$&$e$&$g$&$1$&$1$\\\
$1$&$0$&$a$&$b$&$c$&$d$&$e$&$f$&$g$&$1$\end{tabular}.$

All proper ideals are $I_{1}=\\{0,a,b\\}$, $I_{2}=\\{0,c,f\\}$ and are also
prime ideals. We consider $c_{I_{1}}=\left(1,1,1,0,0,0,0,0,0\right)$ and
$c_{I_{2}}=\left(1,0,0,1,0,0,1,0,0\right),$ the codewords attached to the
ideals $I_{1},I_{2}$. The matrix

$M_{C}=\left(\begin{array}[]{ccccccccc}1&1&1&0&0&0&0&0&0\\\
1&0&0&1&0&0&1&0&0\end{array}\right)$

is the generating matrix for the linear code of the form
$\left(9,2,3\right),\mathcal{C}_{I_{1}I_{2}}$. The even numbers of ideals
assure us that the rows in the generating matrix are linear independent
vectors.

##  6 Conclusions

In this paper, based on ideals, we investigate residuated lattices from fuzzy
set theory and lattice theory point of view. Also we found connections between
the fuzzy sets associated to ideals in a Boolean algebras and Hadamard codes.
As a further research, we will study other connections between fuzzy sets and
some type of algebras of logic.

## References

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Cristina Flaut

Faculty of Mathematics and Computer Science, Ovidius University,

Bd. Mamaia 124, 900527, Constanţa, România,

http://www.univ-ovidius.ro/math/

e-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>

Dana Piciu

Faculty of Science, University of Craiova,

A.I. Cuza Street, 13, 200585, Craiova, România,

http://www.math.ucv.ro/dep_mate/

e-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>

Bianca Liana Bercea

PhD student at Doctoral School of Mathematics,

Ovidius University of Constanţa, România

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