# n-fold filters in residuated lattices

Albert Kadji, Celestin Lele, Marcel Tonga  Department of Mathematics,
University of Yde 1<EMAIL_ADDRESS>Department of Mathematics,
University of Dschang<EMAIL_ADDRESS>Department of Mathematics,
University of Yde 1<EMAIL_ADDRESS>
###### Abstract.

Residuated lattices play an important role in the study of fuzzy logic based
of t-norm. In this paper, we introduced the notions of n-fold implicative
filters, n-fold positive implicative filters, n-fold boolean filters, n-fold
fantastic filters, n-fold normal filters and n-fold obstinate filters in
residuated lattices and study the relations among them.
This generalized the similar existing results in BL-algebra with the
connection of the work of Kerre and all in [14], Kondo and all in [7], [11]
and Motamed and all in [9].
At the end of this paper, we draw two diagrams; the first one describe the
relations between some type of n-fold filters in residuated lattices and the
second one describe the relations between some type of n-fold residuated
lattices.
Key words: residuated lattices, filters, n-fold filters, n-fold residuated
lattices.

2000 Mathematics Subject Classification. Primary 06D99, 08A30

##  1\. Introduction

Since Hájek introduced his Basic Fuzzy logics, (BL-logics) in short in 1998
[1], as logics of continuous t-norms, a multitude research papers related to
algebraic counterparts of BL-logics, has been published. In [2], [3],[9] and
[13] the authors defined the notion of n-fold implicative filters, n-fold
positive implicative filters, n-fold boolean filters, n-fold fantastic
filters, n-fold obstinate filters, n-fold normal filters in BL-algebras and
studied the relation among many type of n-fold filters in BL-algebra.
The aim of this paper is to extend this research to residuated lattices with
the connection of the results obtaining in [14], [11], [7].

##  2\. Preliminaries

A residuated lattice is a nonempty set $L$ with four binary operations
$\wedge,\vee,\otimes,\rightarrow$, and two constants $0,1$ satisfying:
L-1: $\mathbb{L}(L):=(L,\wedge,\vee,0,1)$ is a bounded lattice;
L-2: $(L,\otimes,1)$ is a commutative monoid;
L-3: $x\otimes y\leq z$ iff $x\leq y\rightarrow z$ (Residuation);
A MTL-algebra is a residuated lattice $L$ which satisfies the following
condition:
L-4: $(x\rightarrow y)\vee(y\rightarrow x)=1$ (Prelinearity);
A BL-algebra is a MTL-algebra $L$ which satisfies the following condition:
L-5: $x\wedge y=x\otimes(x\rightarrow y)$ (Divisibility).
A MV-algebra is a BL-algebra $L$ which satisfies the following condition:
L-6: $\overline{\overline{x}}=x$ where $\overline{x}:=x\rightarrow 0$.

In this work, unless mentioned otherwise,
$(L,\wedge,\vee,\otimes,\rightarrow,0,1)$ will be a residuated lattice, which
will often be referred by its support set $L$.

######  Proposition 2.1.

[4],[6],[7],[11]For all $x,y,z\in L$

(1) $\displaystyle x\leq y\;\text{iff}\;x\rightarrow y=1;x\otimes y\leq
x\wedge y;$ (2) $\displaystyle x\rightarrow(y\rightarrow z)=(x\otimes
y)\rightarrow z;$ (3) $\displaystyle x\rightarrow(y\rightarrow
z)=y\rightarrow(x\rightarrow z);$ (4) $\displaystyle\text{If}\;x\leq
y,\;\text{then}\;y\rightarrow z\leq x\rightarrow z\;\text{and}\;z\rightarrow
x\leq z\rightarrow y;$ (5) $\displaystyle x\leq y\rightarrow(x\otimes
y);x\otimes(x\rightarrow y)\leq y;$ (6) $\displaystyle 1\rightarrow
x=x;x\rightarrow x=1;x\rightarrow 1=1;x\leq y\rightarrow
x,x\leq\bar{\bar{x}},\bar{\bar{\bar{x}}}=\bar{x};$ (7) $\displaystyle
x\otimes\bar{x}=0;x\otimes y=0\;\text{iff}\;x\leq\bar{y};$ (8) $\displaystyle
x\leq y\;\text{implies}\;x\otimes z\leq y\otimes z,z\rightarrow x\leq
z\rightarrow y,y\rightarrow z\leq x\rightarrow z,\bar{y}\leq\bar{x};$ (9)
$\displaystyle\overline{x\otimes y}=x\rightarrow\bar{y};$ (10) $\displaystyle
x\vee y=1\;\text{implies}\;x\otimes y=x\wedge y\;\text{and}\ x^{n}\vee
y^{n}=1\;\text{for every }\;n\geq 1;$ (11) $\displaystyle x\otimes(y\vee
z)=(x\otimes y)\vee(x\otimes z);$ (12) $\displaystyle(x\vee y)\rightarrow
z=(x\rightarrow z)\wedge(y\rightarrow z);(x\rightarrow z)\vee(y\rightarrow
z)\leq(x\wedge y)\rightarrow z;$ (13) $\displaystyle(x\vee y)\otimes(x\vee
z)\leq x\vee(y\otimes z),\;\text{hence}\;(x\vee y)^{mn}\leq x^{n}\vee y^{m};$
(14) $\displaystyle x\vee y\leq((x\rightarrow y)\rightarrow
y)\wedge((y\rightarrow x)\rightarrow x);$ (15) $\displaystyle x\rightarrow
y\leq(y\rightarrow z)\rightarrow(x\rightarrow z);$ (16) $\displaystyle
y\rightarrow x\leq(z\rightarrow y)\rightarrow(z\rightarrow x);$ (17)
$\displaystyle((x\rightarrow y)\rightarrow y)\rightarrow y=x\rightarrow y.$

Besides equations (1)-(17), we will use the following results.

Fact 1
A nonempty subset $F$ of a residuated lattice
$(L,\wedge,\vee,\otimes,\rightarrow,0,1)$ is called a filter if it satisfies:
(F1): For every $x,y\in F$, $x\otimes y\in F$;
(F2): For every $x,y\in L$, if $x\leq y$ and $x\in F$, then $y\in F$.
A deductive system of a residuated lattices $L$ is a subset $F$ containing $1$
such that for all $x,y\in L$; $x\rightarrow y\in F\ \ \text{and}\ \ x\in F\ \
\text{imply}\ \ y\in F.$
It is known that in a residuated lattices, filters and deductive systems
coincide [4].
Fact 2
The following Examples will be use as a residuated lattices which are not BL-
algebra.

######  Example 2.2.

[12] Let $L=\\{0,a,b,c,d,1\\}$ be a lattice such that $0<a<c$, $0<b<c<d<1$,
$a$ and $b$ are incomparable. Define the operations $\otimes$ and
$\rightarrow$ by the two tables. Then $L$ is a residuated lattice which is not
a BL-algebra since $(a\longrightarrow b)\vee(b\longrightarrow a)=c\neq 1$.

$\otimes$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $a$ | $0$ | $a$ | $a$ | $a$
$b$ | $0$ | $0$ | $b$ | $b$ | $b$ | $b$
$c$ | $0$ | $a$ | $b$ | $c$ | $c$ | $c$
$d$ | $0$ | $a$ | $b$ | $c$ | $c$ | $d$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
$\longrightarrow$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
$a$ | $b$ | $1$ | $b$ | $1$ | $1$ | $1$
$b$ | $a$ | $a$ | $1$ | $1$ | $1$ | $1$
$c$ | $0$ | $a$ | $b$ | $1$ | $1$ | $1$
$d$ | $0$ | $a$ | $b$ | $d$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$

$F=\\{1,b,c,d\\}$; $F_{1}=\\{1,a,c,d\\}$; $F_{2}=\\{1,c,d\\}$ are proper
filters of $L$.

######  Example 2.3.

[12] Let $L=\\{0,a,b,c,d,1\\}$ be a lattice such that $0<a,b,d,c<1$, $a,b,c,d$
are pairwise incomparable. Define the operations $\otimes$ and $\rightarrow$
by the two tables. Then $L$ is a residuated lattice which is not a BL-algebra
since $a\otimes(a\longrightarrow b)=b\neq 0=a\wedge b$.

$\otimes$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $a$ | $b$ | $d$ | $d$ | $a$
$b$ | $0$ | $b$ | $b$ | $0$ | $0$ | $b$
$c$ | $0$ | $d$ | $0$ | $d$ | $d$ | $c$
$d$ | $0$ | $d$ | $0$ | $d$ | $d$ | $d$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
$\longrightarrow$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
$a$ | $a$ | $1$ | $b$ | $c$ | $c$ | $1$
$b$ | $c$ | $a$ | $1$ | $c$ | $c$ | $1$
$c$ | $b$ | $a$ | $b$ | $1$ | $a$ | $1$
$d$ | $b$ | $a$ | $b$ | $a$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$

$F=\\{1,c,d\\}$ is a proper filter of $L$.

######  Example 2.4.

[4] Let $L=\\{0,a,b,c,d,1\\}$ be a lattice such that $0<a<c<d<1$, $0<b<c<d<1$,
$a$ and $b$ are incomparable. Define the operations $\otimes$ and
$\rightarrow$ by the two tables. Then
$(L,\wedge,\vee,\otimes,\rightarrow,0,1)$ is a residuated lattices which is
not a BL-algebra since $(a\longrightarrow b)\vee(b\longrightarrow a)=c\vee
c=c\neq 1$.

$\otimes$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $0$ | $0$ | $0$ | $a$ | $a$
$b$ | $0$ | $0$ | $0$ | $0$ | $b$ | $b$
$c$ | $0$ | $0$ | $0$ | $0$ | $c$ | $c$
$d$ | $0$ | $a$ | $b$ | $c$ | $d$ | $d$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
$\longrightarrow$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
$a$ | $c$ | $1$ | $c$ | $1$ | $1$ | $1$
$b$ | $c$ | $c$ | $1$ | $1$ | $1$ | $1$
$c$ | $c$ | $c$ | $c$ | $1$ | $1$ | $1$
$d$ | $0$ | $a$ | $b$ | $c$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$

$F=\\{1,d\\}$ is a proper filter of $L$.

######  Example 2.5.

[6] Let $L=\\{0,a,b,c,1\\}$ be a lattice such that $0<c<a,b<1$, $a,b$ are
incomparable. Define the operations $\otimes$ and $\rightarrow$ by the two
tables. Then $(L,\wedge,\vee,\otimes,\rightarrow,0,1)$ is a residuated lattice
which is not BL-algebra since $a\otimes(a\longrightarrow b)=c\neq 0=a\wedge
b$.

$\otimes$ | $0$ | $c$ | $a$ | $b$ | $1$
---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$
$c$ | $0$ | $c$ | $c$ | $c$ | $c$
$a$ | $0$ | $c$ | $a$ | $c$ | $a$
$b$ | $0$ | $c$ | $c$ | $b$ | $b$
$1$ | $0$ | $c$ | $a$ | $b$ | $1$
$\longrightarrow$ | $0$ | $c$ | $a$ | $b$ | $1$
---|---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$ | $1$
$c$ | $0$ | $1$ | $1$ | $1$ | $1$
$a$ | $0$ | $b$ | $1$ | $b$ | $1$
$b$ | $0$ | $a$ | $a$ | $1$ | $1$
$1$ | $0$ | $c$ | $a$ | $b$ | $1$

$F_{1}=\\{1,a\\},F_{2}=\\{1,b\\},F_{3}=\\{1,a,b,c\\}$ are proper filters of
$L$.

######  Definition 2.6.

[11] A residuated lattice $L$ is said to be locally finite if for every $x\neq
1$, there exists an integer $n\geq 1$ such that $x^{n}:=\underbrace{x\otimes
x\cdots\otimes x}_{n\;times}=0$.

######  Definition 2.7.

[7] Let $F$ be a filter of a residuated lattice $L$. For $x,y\in L$, a
relation $\equiv_{F}$ on $L$, define by
$x\equiv_{F}y\Longleftrightarrow(x\longrightarrow y,y\longrightarrow x)\in F$,
is a congruence on $L$ and a quotient structure $L/F$ is also a residuated
lattice where : $x/F\wedge y/F=(x\wedge y)/F$; $x/F\vee/F=(x\vee y)/F$;
$x/F\otimes y/F=(x\otimes y)/F$; $x/F\longrightarrow y/F=(x\longrightarrow
y)/F$.

######  Definition 2.8.

[14] [11] [4] A Proper filter $F$ is said to be:

  * (i)

prime if it satisfies the following condition:
For all $x,y\in L$, $x\longrightarrow y\in F$ or $y\longrightarrow x\in F$.

  * (ii)

prime of the second kind if it satisfies the following condition:
For all $x,y\in L$, $x\vee y\in F$ implies $x\in F$ or $y\in F$.

  * (iii)

prime of the third kind if it satisfies the following condition:
For all $x,y\in L$, $(x\longrightarrow y)\vee(y\longrightarrow x)\in F$.

  * (iv)

boolean if it satisfies the following condition:
For all $x\in L$, $x\vee\overline{x}\in F$.

  * (v)

boolean filter in the second kind if it satisfies the following condition:
For all $x\in L$, $x\in F$ or $\overline{x}\in F$.

######  Remark 2.9.

[14][11] [6]

  * (i)

Prime filters are prime filters in the second kind. The converse is true if
$L$ is a MTL-algebra.

  * (ii)

Prime filters are prime filters in the third kind. The converse is true if $L$
is a MTL-algebra.

  * (iii)

Boolean filters in the second kind are boolean filters.

  * (iv)

Maximal filters are prime filters in the second kind.

  * (v)

If $L$ is a MTL-algebra, then maximal filters are prime filters.

We have the following results.

######  Proposition 2.10.

[6] For any filter $F$ of a residuated lattices $L$, the following conditions
are equivalent:

  * (i)

$F$ is a maximal filter of $L$.

  * (ii)

For any $x\in L$, $x\notin F$ if and only if $\overline{x^{n}}\in F$ for some
$n\geq 1$

  * (iii)

For any $x\notin F$, there is $f\in F$ and $n\geq 1$ such that $f\otimes
x^{n}=0$

Follows from Prop.2.10, we have the following lemma:

######  Lemma 2.11.

$F$ is a maximal filter of $L$ if and only if $L/F$ is a locally finite
residuated lattice.

Now, unless mentioned otherwise, $n\geq 1$ will be an integer and $F\subseteq
L$.

##  3\. SEMI MAXIMAL FILTER IN RESIDUATED LATTICES

######  Definition 3.1.

[5] Let $F$ be a proper filter of $L$. The intersection of all maximal filters
of $L$ which contain $F$ is called the radical of $F$ and it is denoted by
Rad($F$).

######  Definition 3.2.

A proper filter $F$ of $L$ is said to be a semi maximal filter of $L$ if
Rad($F$)= $F$.

The following example shows that the notion of semi maximal filters in
residuated lattices exist and semi maximal filter may not be maximal filter.

######  Example 3.3.

Let $L$ be a residuated lattice from Example 2.2. It is easy to check that
Rad($\\{1,c,d\\}$)= $\\{1,c,d\\}$. Hence $\\{1,c,d\\}$ is a semi maximal
filter of $L$.
But $\\{1,c,d\\}\subseteq\\{1,a,c,d\\}$ and $\\{1,c,a,d\\}$ is a filter of
$L$, hence $\\{1,c,d\\}$ is a semi maximal filter which is not a maximal
filter of $L$.

######  Remark 3.4.

It is clear that maximal filters are semi maximal filters.

##  4\. N-FOLD IMPLICATIVE FILTER IN RESIDUATED LATTICES

######  Definition 4.1.

An n-fold implicative residuated lattice $L$ is a residuated lattices which
satifies the following condition:
$x^{n+1}=x^{n}$ for all $x,y\in L$.

The following examples shows that n-fold implicative residuated lattices exist
and that residuated lattice is not in general n-fold implicative residuated
lattice.

######  Example 4.2.

Let $L$ be a residuated lattice from Example 2.4. We have:

  * (i)

$a^{1+1}=0\neq a^{1}$ so $L$ is not an 1-fold implicative residuated lattice.

  * (ii)

For all $n\geq 2$, $x^{n+1}=x^{n}$ for all $x,y\in L$. So $L$ is an n-fold
implicative residuated lattice for all $n\geq 2$.

######  Definition 4.3.

$F$ is an n-fold implicative filter if it satisfies the following conditions:

  * (i)

$1\in F$

  * (ii)

For all $x,y,z\in L$, if $x^{n}\longrightarrow(y\longrightarrow z)\in F$ and
$x^{n}\longrightarrow y\in F$, then $x^{n}\longrightarrow z\in F$.

In particular 1-fold implicative filters are implicative filters.[7]

######  Example 4.4.

Let $n\geq 1$ and $L$ be a residuated lattice from Example 2.5. Simple
computations proves that $F_{1}=\\{1,a\\},F_{2}=\\{1,b\\},F_{3}=\\{1,a,b,c\\}$
are n-fold implicative filters.

The following lemma gives a characterization of n-fold implicative filters.

######  Lemma 4.5.

Let $a\in L$. Let $F$ be a filter of $L$. Then $L_{a}=\\{b\in
L:a^{n}\longrightarrow b\in F\\}$ is a filter of $L$ if and only if $F$ is an
n-fold implicative filter of $L$.

###### Proof.

Let $F$ be an n-fold implicative filter of $L$. Since $a^{n}\longrightarrow
1=1\in F$, we have $1\in L_{a}$. Let $x,y\in L$ be such that
$x,x\longrightarrow y\in L_{a}$, then $a^{n}\longrightarrow x\in F$ and
$a^{n}\longrightarrow(x\longrightarrow y)\in F$. Since $F$ is an n-fold
implicative filter of $L$, by Definition 4.3, $a^{n}\longrightarrow y\in F$,
hence $y\in L_{a}$. Therefore $L_{a}$ is a filter of $L$.
Conversely suppose that $L_{a}$ is a filter of $L$ for all $a\in L$. Let
$x,y,z\in L$ be such that $x^{n}\longrightarrow(y\longrightarrow z)\in F$ and
$x^{n}\longrightarrow y\in F$. We have $y,y\longrightarrow z\in L_{x}$, by the
hypothesis $L_{x}$ is a filter of $L$, so $z\in L_{x}$ and hence
$x^{n}\longrightarrow z\in F$. ∎

The following proposition gives another characterization of n-fold implicative
filters in residuated lattices.

######  Proposition 4.6.

Let $F$ be a filter of $L$. Then for all $x\in L$, the following conditions
are equivalent:

  * (i)

$F$ is an n-fold implicative filter of $L$.

  * (ii)

$x^{n}\longrightarrow x^{2n}\in F$.

###### Proof.

$(i)\longrightarrow(ii)$: Let $x\in L$, by Prop. 2.1 we have:
$x^{n}\longrightarrow(x^{n}\longrightarrow x^{2n})=x^{2n}\longrightarrow
x^{2n}=1\in F$ and $x^{n}\longrightarrow x^{n}=1\in F$. Since $F$ is an n-fold
implicative filter of $L$, we get $x^{n}\longrightarrow x^{2n}\in F$.
$(ii)\longrightarrow(i)$: Let $x,y,z\in L$ be such that
$x^{n}\longrightarrow(y\longrightarrow z)\in F$ and $x^{n}\longrightarrow y\in
F$. By Prop. 2.1 we have the following:

  * (1)

$x^{n}\otimes[x^{n}\longrightarrow(y\longrightarrow z)]\leq y\longrightarrow
z$.

  * (2)

$x^{n}\otimes(x^{n}\longrightarrow y)\leq y$.

  * (3)

By (1) and (2) we have : $[x^{n}\otimes[x^{n}\longrightarrow(y\longrightarrow
z)]]\otimes[x^{n}\otimes(x^{n}\longrightarrow y)]\leq
y\otimes(y\longrightarrow z)\leq z$.

  * (4)

By (3) we have : $([x^{n}\longrightarrow(y\longrightarrow
z)]\otimes(x^{n}\longrightarrow y))\otimes x^{2n}\leq z$.

  * (5)

By (4), we have :$([x^{n}\longrightarrow(y\longrightarrow
z)]\otimes(x^{n}\longrightarrow y))\leq x^{2n}\longrightarrow z$.

  * (6)

Since $x^{n}\longrightarrow(y\longrightarrow z)\in F$ and
$x^{n}\longrightarrow y\in F$, by the fact that $F$ is a filter, we get
$[x^{n}\longrightarrow(y\longrightarrow z)]\otimes(x^{n}\longrightarrow y)\in
F$

  * (7)

By (5),(6) and the fact that $F$ is a filter, we get $x^{2n}\longrightarrow
z\in F$.

  * (8)

$x^{n}\longrightarrow x^{2n}\leq(x^{2n}\longrightarrow
z)\longrightarrow(x^{n}\longrightarrow z)$

  * (9)

By (7), (8) and the fact that $x^{n}\longrightarrow x^{2n}\in F$, we obtain
$x^{n}\longrightarrow z\in F$.

Hence $F$ is an n-fold implicative filter of $L$. ∎

######  Proposition 4.7.

Let $F$ be a filter of $L$. Then for all $x,y\in L$, the following conditions
are equivalent:

  * (i)

$x^{n}\longrightarrow x^{2n}\in F$.

  * (ii)

If $x^{n+1}\longrightarrow y\in F$, then $x^{n}\longrightarrow y\in F$.

###### Proof.

$(i)\longrightarrow(ii)$: Since $(i)$ holds, by Prop. 4.6 $F$ is an n-fold
implicative filter of $L$. On the other hand by Prop. 2.1 we have :
$x^{n+1}\longrightarrow y=x^{n}\longrightarrow(x\longrightarrow y)\in F$ and
$x^{n}\longrightarrow x=1\in F$, by the fact that $F$ is an n-fold implicative
filter of $L$ we get $x^{n}\longrightarrow y\in F$.
$(ii)\longrightarrow(i)$: We have :
$x^{n+1}\longrightarrow(x^{n-1}\longrightarrow x^{2n})=x^{2n}\longrightarrow
x^{2n}=1\in F$. From this and the fact that (ii) holds, we also have :
$x^{n}\longrightarrow(x^{n-1}\longrightarrow x^{2n})\in F$.
But $x^{n+1}\longrightarrow(x^{n-2}\longrightarrow
x^{2n})=x^{n}\longrightarrow(x^{n-1}\longrightarrow x^{2n})\in F$.
From this and the fact that (ii) holds, we also have :
$x^{n}\longrightarrow(x^{n-2}\longrightarrow x^{2n})\in F$.
By repeating the process n times, we get
$x^{n}\longrightarrow(x^{n-n}\longrightarrow x^{2n})=x^{n}\longrightarrow
x^{2n}\in F$. ∎

######  Proposition 4.8.

Let $F$ be a filter of $L$. Then for all $x,y,z\in L$, the following
conditions are equivalent:

  * (i)

$x^{n}\longrightarrow x^{2n}\in F$.

  * (ii)

If $x^{n}\longrightarrow(y\longrightarrow z)\in F$, then
$(x^{n}\longrightarrow y)\longrightarrow(x^{n}\longrightarrow z)\in F$.

###### Proof.

$(i)\longrightarrow(ii)$: Assume that $x^{n}\longrightarrow(y\longrightarrow
z)\in F$. By Prop. 2.1 we have the following equations:

  * (1)

$y\longrightarrow z\leq(x^{n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow z)$.

  * (2)

$x^{n}\longrightarrow(y\longrightarrow z)\leq
x^{n}\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow z)]$.

  * (3)

$x^{n}\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow
z)]=x^{n}\longrightarrow[x^{n}\longrightarrow((x^{n}\longrightarrow
y)\longrightarrow z)]=x^{2n}\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow z]$.

  * (4)

By (2) and (3) we have : $x^{n}\longrightarrow(y\longrightarrow z)\leq
x^{2n}\longrightarrow[(x^{n}\longrightarrow y)\longrightarrow z]$.

  * (5)

Since $F$ is a filter, by (4) and the fact that
$x^{n}\longrightarrow(y\longrightarrow z)\in F$, we have :
$x^{2n}\longrightarrow[(x^{n}\longrightarrow y)\longrightarrow z]\in F$

  * (6)

$x^{2n}\longrightarrow[(x^{n}\longrightarrow y)\longrightarrow
z]\leq(x^{n}\longrightarrow
x^{2n})\longrightarrow(x^{n}\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow z])=(x^{n}\longrightarrow
x^{2n})\longrightarrow((x^{n}\longrightarrow
y)\longrightarrow[(x^{n}\longrightarrow z])$.

  * (7)

Since $F$ is a filter, by (6) and the fact that
$x^{2n}\longrightarrow[(x^{n}\longrightarrow y)\longrightarrow z]\in F$, we
have : $(x^{n}\longrightarrow x^{2n})\longrightarrow((x^{n}\longrightarrow
y)\longrightarrow[(x^{n}\longrightarrow z])\in F$.

  * (8)

Since $F$ is a filter, by (7) and the fact that $x^{n}\longrightarrow
x^{2n}\in F$, we obtain $(x^{n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow z)\in F$.

$(ii)\longrightarrow(i)$: Since $x^{n}\longrightarrow(x^{n}\longrightarrow
x^{2n})=x^{2n}\longrightarrow x^{2n}=1\in F$, by (ii) we have :
$(x^{n}\longrightarrow x^{n})\longrightarrow(x^{n}\longrightarrow x^{2n})\in
F$, hence $x^{n}\longrightarrow x^{2n}\in F$. ∎

By Prop. 4.6, Prop. 4.7 and Prop. 4.8, we have the following result:

######  Proposition 4.9.

Let $F$ be a filter of $L$. Then for all $x,y,z\in L$, the following
conditions are equivalent:

  * (i)

$F$ is an n-fold implicative filter of $L$.

  * (ii)

$x^{n}\longrightarrow x^{2n}\in F$.

  * (iii)

If $x^{n+1}\longrightarrow y\in F$, then $x^{n}\longrightarrow y\in F$.

  * (iv)

If $x^{n}\longrightarrow(y\longrightarrow z)\in F$, then
$(x^{n}\longrightarrow y)\longrightarrow(x^{n}\longrightarrow z)\in F$.

######  Proposition 4.10.

If a filter $F$ is an n-fold implicative filter, then $F$ is an (n+1)-fold
implicative filter.

###### Proof.

Let $F$ be a filter. Assume that $F$ is an n-fold implicative filter. Let
$x,y\in L$ such that $x^{n+2}\longrightarrow y\in F$, by Prop. 2.1,
$x^{n+1}\longrightarrow(x\longrightarrow y)=x^{n+2}\longrightarrow y\in F$.
Since $F$ is an n-fold implicative filter, apply Prop. 4.9(iii),
we obtain $x^{n}\longrightarrow(x\longrightarrow y)\in F$. Hence
$x^{n+1}\longrightarrow y\in F$ and by Prop. 4.9, $F$ is an (n+1)-fold
implicative filter. ∎

By the following example, we show that the converse of Prop. 4.10 is not true
in general.

######  Example 4.11.

[10]

Let $L=\\{0,a,b,1\\}$ be a lattice such that $0<a<b<1$. Define the operations
$\otimes$ and $\rightarrow$ by the two tables. Then
$(L,\wedge,\vee,\otimes,\rightarrow,0,1)$ is a residuated lattice.

$\otimes$ | $0$ | $a$ | $b$ | $1$
---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $0$ | $0$ | $a$
$b$ | $0$ | $0$ | $a$ | $b$
$1$ | $0$ | $a$ | $b$ | $1$
$\longrightarrow$ | $0$ | $a$ | $b$ | $1$
---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$
$a$ | $b$ | $1$ | $1$ | $1$
$b$ | $a$ | $b$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $1$

$\\{1\\}$ is an 3-fold implicative filter but $\\{1\\}$ is not an 2-fold
implicative filter, since $b^{2}\longrightarrow a\in\\{1\\}$ but
$b^{1}\longrightarrow a=b\notin\\{1\\}$

######  Proposition 4.12.

n-fold implicative filters are filters.

###### Proof.

Suppose that $F$ is an n-fold implicative filter of $L$. Let $z,y\in L$ such
that $y,y\longrightarrow z\in F$. We have $1^{n}\longrightarrow
y,1^{n}\longrightarrow(y\longrightarrow z)\in F$, this implies
$z=1^{n}\longrightarrow z\in F$. Hence $F$ is a deductive system of $L$ and
the thesis follows from the fact1. ∎

By the following example, we show that the converse of Prop. 4.12 is not true
in general.

######  Example 4.13.

Let $L$ be a residuated lattice from Example 4.11. $\\{1\\}$ is a filter but
$\\{1\\}$ is not an 2-fold implicative filter, since $b^{2}\longrightarrow
a\in\\{1\\}$ but $b^{1}\longrightarrow a=b\notin\\{1\\}$.

Using Prop.4.9, it is easy to show the following results:

######  Corollary 4.14.

If $L$ is an n-fold implicative residuated lattice then, the concepts of
n-fold implicative filters and filters coincide.

######  Theorem 4.15.

Let $F_{1}$ and $F_{2}$ two filters of $L$ such that $F_{1}\subseteq F_{2}$.
If $F_{1}$ is an n-fold implicative filter, then $F_{2}$ is an n-fold
implicative filter.

The following theorem gives the relation between n-fold implicative residuated
lattice and n-fold implicative filter.

######  Proposition 4.16.

Let $F$ be a filter of $L$. The following conditions are equivalent:

  * (i)

$L$ is an n-fold implicative residuated lattice.

  * (ii)

Every filter of $L$ is an n-fold implicative filter of $L$.

  * (iii)

{1} is an n-fold implicative filter of $L$.

  * (iv)

$x^{n}=x^{2n}$ for all $x\in L$.

###### Proof.

$(i)\longrightarrow(ii)$ : follows from Corollary. 4.14
$(ii)\longrightarrow(iii)$ : follows from the fact that {1} is a filter of
$L$.
$(iii)\longrightarrow(iv)$ : Assume that {1} is an n-fold implicative filter
of $L$. From Prop 4.9, we have $x^{n}\longrightarrow x^{2n}=1$ for all $x\in
L$. So $x^{n}\leq x^{2n}$ for all $x\in L$. Since $x^{2n}\leq x^{n}$ for all
$x\in L$, we obtain $x^{n}=x^{2n}$ for all $x\in L$.
$(iv)\longrightarrow(i)$ : If $x^{n}=x^{2n}$ for all $x\in L$, we have
$x^{n}\longrightarrow x^{2n}=1\in\\{1\\}$ for all $x\in L$, by Prop. 4.9,
$\\{1\\}$ is an n-fold implicative filter of $L$. Since
$x^{n}\longrightarrow(x^{n}\longrightarrow x^{n+1})=1\in\\{1\\}$ and
$x^{n}\longrightarrow x^{n}=1\in\\{1\\}$, we get $x^{n}\longrightarrow
x^{n+1}\in\\{1\\}$, that is $x^{n+1}=x^{n}$ for all $x\in L$. ∎

######  Corollary 4.17.

A filter $F$ of a residuated lattice $L$ is an n-fold implicative filter if
and only if $L/F$ is an n-fold implicative residuated lattice.

###### Proof.

Let $F$ be a filter.
Suppose that $F$ is an n-fold implicative filter. By Prop. 4.9(ii), we have
$x^{n}\longrightarrow x^{2n}\in F$ for all $x\in L$, that is
$(x^{n}\longrightarrow x^{2n})/F=1/F$ for all $x\in L$. So
$(x/F)^{n}\longrightarrow(x/F)^{2n}=(x^{n}/F)\longrightarrow(x^{2n}/F)=(x^{n}\longrightarrow
x^{2n})/F=1/F$ for all $x/F\in L/F$, by Prop. 4.16(iv), $L/F$ is an n-fold
implicative residuated lattice.
Suppose conversely that $L/F$ is an n-fold implicative residuated lattice. By
Prop. 4.16(iv), we get $(x/F)^{n}=(x/F)^{2n}$ for all $x/F\in L/F$ or
equivalently $(x^{n}/F)=(x^{2n}/F)$ for all $x\in L$. That is
$(x^{n}\longrightarrow x^{2n})/F=1/F$ for all $x\in L$. Hence
$x^{n}\longrightarrow x^{2n}\in F$ for all $x\in L$, we obtain the result by
apply Prop. 4.9(ii). ∎

By (3)[4] and Corollary 4.17, we have the following result.

######  Corollary 4.18.

A filter $F$ of a residuated lattice $L$ is an 1-fold implicative filter if
and only if $L/F$ is a Heyting algebra. As a consequence, it is easy to
observe that, a residuated lattice $L$ is a Heyting algebra if and only if
$\\{1\\}$ is an 1-fold implicative filter of $L$ if and only if $L$ is an
1-fold implicative residuated lattice.

##  5\. N-FOLD POSITIVE IMPLICATIVE FILTERS OF RESIDUATED LATTICES

######  Definition 5.1.

$F$ is an n-fold positive implicative filter if it satisfies the following
conditions:

  * (i)

$1\in F$

  * (ii)

For all $x,y,z\in L$, if $x\longrightarrow((y^{n}\longrightarrow
z)\longrightarrow y)\in F$ and $x\in F$, then $y\in F$.

In particular 1-fold positive implicative filters are positive implicative
filters.[7]

######  Example 5.2.

Let $n\geq 1$. Let $L$ be a residuated lattice from Example 2.5. Simple
computations proves that $F_{3}=\\{1,a,b,c\\}$ is an n-fold positive
implicative filter.

######  Proposition 5.3.

Every n-fold positive implicative filter is a filter.

###### Proof.

Let $F$ be an n-fold positive implicative filter of $L$, it is clear that
$1\in F$. Since for any $y\in F$, $y^{n}\longrightarrow 1=1$, by setting $z=1$
in the definition of n-fold positive implicative filter, we obtain the result.
∎

The following Example shows that filters are not n-fold positive implicative
filters in general.

######  Example 5.4.

Let $L$ be a residuated lattice from Example 2.5 and $n\geq 1$.
$F_{1}=\\{1,a\\},F_{2}=\\{1,b\\}$ are filters but not n-fold positive
implicative filters since $1\longrightarrow((b^{n}\longrightarrow
0)\longrightarrow b)\in F_{1}$ and $1\in F_{1}$, but $b\notin F_{1}$;
$1\longrightarrow((a^{n}\longrightarrow 0)\longrightarrow a)\in F_{2}$ and
$1\in F_{2}$, but $a\notin F_{2}$.

The following proposition gives a characterization of n-fold positive
implicative filter for any $n\geq 1$ .

######  Proposition 5.5.

The following conditions are equivalent for any filter $F$ and any $n\geq 1$ :

  * (i)

$F$ is an n-fold positive implicative filter

  * (ii)

For all $x,y\in L$, $(x^{n}\longrightarrow y)\longrightarrow x\in F$ implies
$x\in F$.

  * (iii)

For all $x\in L$, $\overline{x^{n}}\longrightarrow x\in F$ implies $x\in F$.

###### Proof.

$(i)\longrightarrow(ii)$ : Suppose that $F$ is n-fold positive implicative
filter of $L$ and $(x^{n}\longrightarrow y)\longrightarrow x\in F$, since
$1\longrightarrow((x^{n}\longrightarrow y)\longrightarrow
x)=(x^{n}\longrightarrow y)\longrightarrow x\in F$ and $1\in F$, we apply the
fact that $F$ is n-fold positive implicative filter of $L$ and obtain the
result.
$(ii)\longrightarrow(iii)$ : We obtain the result by setting $y=0$ in the
equation $(ii)$.
$(iii)\longrightarrow(i)$ : Suppose that
$x\longrightarrow((y^{n}\longrightarrow z)\longrightarrow y)\in F$ and $x\in
F$, from the fact that $F$ is filter, we obtain $(y^{n}\longrightarrow
z)\longrightarrow y\in F$.
On the other hand, from Prop. 2.1(4), we have : $(y^{n}\longrightarrow
z)\longrightarrow y\leq(y^{n}\longrightarrow 0)\longrightarrow y$, from the
fact that $F$ is filter, we obtain $(y^{n}\longrightarrow 0)\longrightarrow
y\in F$, we apply the hypothesis and obtain $y\in F$. ∎

######  Corollary 5.6.

A proper filter $F$ is an n-fold positive implicative filter if an only if for
all $x\in L$, $x\vee\overline{x^{n}}\in F$.

###### Proof.

Assume that for all $x\in L$, $\overline{x^{n}}\longrightarrow x\in F$ and
$x\vee\overline{x^{n}}\in F$. By Prop. 5.5, we must show that $x\in F$. Since
by (14)Prop. 2.1, $x\vee\overline{x^{n}}\leq(\overline{x^{n}}\longrightarrow
x)\longrightarrow x$, we have $(\overline{x^{n}}\longrightarrow
x)\longrightarrow x\in F$. Using the fact that
$\overline{x^{n}}\longrightarrow x\in F$, we have $x\in F$.
Conversely suppose that $F$ is an n-fold positive implicative filter. Let
$x\in L$. Let $t=x\vee\overline{x^{n}}$, we must show that $t\in F$. Since
$x\leq t$, we have $x^{n}\leq t^{n}$ and then
$\overline{t^{n}}\leq\overline{x^{n}}\leq\overline{x^{n}}\vee x=t$. Hence
$\overline{t^{n}}\leq t$ or equivalently $\overline{t^{n}}\longrightarrow
t=1$. So $\overline{t^{n}}\longrightarrow t\in F$. From this and the fact that
$F$ is an n-fold positive implicative filter, by Prop. 5.5, we get that $t\in
F$. ∎

######  Definition 5.7.

$F$ is an n-fold boolean filter if it satisfies the following conditions:
$x\vee\overline{x^{n}}\in F$ for all $x\in L$. In particular 1-fold boolean
filters are boolean filters.[4].

The extension theorem of n-fold positive implicative filters is obtained from
the following result:

######  Theorem 5.8.

Let $n\geq 1$. Let $F_{1}$ and $F_{2}$ two filters of $L$ such that
$F_{1}\subseteq F_{2}$. If $F_{1}$ is an n-fold positive implicative filter,
then so is $F_{2}$.

###### Proof.

If $F_{1}$ is an n-fold positive implicative filter, then by Corollary 5.6, we
get $\overline{x^{n}}\vee x\in F_{1}$ for all $x\in L$. Since $F_{1}\subseteq
F_{2}$, we have $\overline{x^{n}}\vee x\in F_{2}$ for all $x\in L$ and by
Corollary 5.6, $F_{2}$ is an n-fold positive implicative filter. ∎

The following theorem gives the relation between n-fold positive implicative
filters and n-fold implicative filters in residuated lattices.

######  Theorem 5.9.

Every n-fold positive implicative filter of $L$ is an n-fold implicative
filter of $L$.

###### Proof.

Let $F$ be an n-fold positive implicative filter of $L$. Let $x,y\in L$ be
such that $x^{n+1}\longrightarrow y\in F$. By Prop. 2.1 we have the following:

  * (1)

$(x^{n+1}\longrightarrow y)^{n}\longrightarrow(x^{n}\longrightarrow y)$=
$[(x^{n+1}\longrightarrow y)^{n-1}\otimes(x^{n+1}\longrightarrow
y)]\longrightarrow(x^{n}\longrightarrow y)$=
$[(x^{n+1}\longrightarrow y)^{n-1}]\longrightarrow[(x^{n+1}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow y)$=
$[(x^{n+1}\longrightarrow y)^{n-1}]\longrightarrow[(x^{n+1}\longrightarrow
y)\longrightarrow[x^{n-1}\longrightarrow(x\longrightarrow y)]$=
$[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[x^{n-1}\longrightarrow[(x^{n+1}\longrightarrow
y)\longrightarrow(x\longrightarrow y)]]$=
$[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[x^{n-1}\longrightarrow[(x\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x\longrightarrow y)]]$

  * (2)

So by (1) we have :$(x^{n+1}\longrightarrow
y)^{n}\longrightarrow(x^{n}\longrightarrow y)$=
$[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[x^{n-1}\longrightarrow[(x\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x\longrightarrow y)]]$

  * (3)

We have : $(x^{n}\longrightarrow y)\longrightarrow
y\leq[(x\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x\longrightarrow y)]$

  * (4)

By (3) we have :$[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[x^{n-1}\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow y]]\leq[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[x^{n-1}\longrightarrow[[(x\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x\longrightarrow y)]]]$

  * (5)

By (4) and (2) , we have : $((x^{n+1}\longrightarrow
y)^{n-1})\longrightarrow(x^{n-1}\longrightarrow((x^{n}\longrightarrow
y)\longrightarrow y))\leq(x^{n+1}\longrightarrow
y)^{n}\longrightarrow(x^{n}\longrightarrow y)$.

  * (6)

We have: $[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[x^{n-1}\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow y]]=[(x^{n+1}\longrightarrow
y)^{n-1}]\longrightarrow[(x^{n}\longrightarrow
y)\longrightarrow[x^{n-1}\longrightarrow y]]=[(x^{n}\longrightarrow
y)]\longrightarrow[(x^{n+1}\longrightarrow
y)^{n-1}\longrightarrow[x^{n-1}\longrightarrow y]]$

  * (7)

By (6) and (5) we get : $(x^{n}\longrightarrow
y)\longrightarrow[(x^{n+1}\longrightarrow
y)^{n-1}\longrightarrow(x^{n-1}\longrightarrow y)]\leq(x^{n+1}\longrightarrow
y)^{n}\longrightarrow(x^{n}\longrightarrow y)$

  * (8)

We have : $(x^{n}\longrightarrow y)\otimes(x^{n+1}\longrightarrow
y)^{n-1}\otimes x^{n-1}=(x^{n}\longrightarrow y)\otimes(x^{n+1}\longrightarrow
y)^{n-2}\otimes(x^{n+1}\longrightarrow y)\otimes x^{n-2}\otimes
x=(x^{n}\longrightarrow y)\otimes(x^{n+1}\longrightarrow y)^{n-2}\otimes
x^{n-2}\otimes x\otimes(x^{n+1}\longrightarrow y)$

  * (9)

We also have : $x\otimes(x^{n+1}\longrightarrow
y)=x\otimes[x\longrightarrow(x^{n}\longrightarrow y)]\leq x^{n}\longrightarrow
y$

  * (10)

Then : $(x^{n}\longrightarrow y)\otimes(x^{n+1}\longrightarrow y)^{n-2}\otimes
x^{n-2}\otimes x\otimes(x^{n+1}\longrightarrow y)\leq(x^{n}\longrightarrow
y)\otimes(x^{n+1}\longrightarrow y)^{n-2}\otimes
x^{n-2}\otimes(x^{n}\longrightarrow y)$

  * (11)

So: $(x^{n}\longrightarrow y)\otimes(x^{n+1}\longrightarrow y)^{n-1}\otimes
x^{n-1}\leq(x^{n}\longrightarrow y)^{2}\otimes(x^{n+1}\longrightarrow
y)^{n-2}\otimes x^{n-2}$

  * (12)

By (11) we get: $[(x^{n}\longrightarrow y)^{2}\otimes(x^{n+1}\longrightarrow
y)^{n-2}\otimes x^{n-2}]\longrightarrow y\leq[(x^{n}\longrightarrow
y)\otimes(x^{n+1}\longrightarrow y)^{n-1}\otimes x^{n-1}]\longrightarrow y$

  * (13)

By (12), we have: $((x^{n}\longrightarrow y)^{2}\otimes(x^{n+1}\longrightarrow
y)^{n-2})\longrightarrow(x^{n-2}\longrightarrow y)\leq((x^{n}\longrightarrow
y)\otimes(x^{n+1}\longrightarrow
y)^{n-1})\longrightarrow(x^{n-1}\longrightarrow y)$

  * (14)

So : $(x^{n}\longrightarrow y)^{2}\longrightarrow((x^{n+1}\longrightarrow
y)^{n-2}\longrightarrow(x^{n-2}\longrightarrow y))\leq(x^{n}\longrightarrow
y)\longrightarrow((x^{n+1}\longrightarrow
y)^{n-1}\longrightarrow(x^{n-1}\longrightarrow y))$

  * (15)

By (14) and (7), we have : $(x^{n}\longrightarrow
y)^{2}\longrightarrow((x^{n+1}\longrightarrow
y)^{n-2}\longrightarrow(x^{n-2}\longrightarrow y))\leq(x^{n+1}\longrightarrow
y)^{n}\longrightarrow(x^{n}\longrightarrow y)$

By repeating (15) n times, we obtain: $(x^{n}\longrightarrow
y)^{n}\longrightarrow((x^{n+1}\longrightarrow
y)^{0}\longrightarrow(x^{0}\longrightarrow y))\leq(x^{n+1}\longrightarrow
y)^{n}\longrightarrow(x^{n}\longrightarrow y)$. This implies
$(x^{n}\longrightarrow y)^{n}\longrightarrow(1\longrightarrow(1\longrightarrow
y))\leq(x^{n+1}\longrightarrow y)^{n}\longrightarrow(x^{n}\longrightarrow y)$.
Hence $(x^{n}\longrightarrow y)^{n}\longrightarrow
y\leq(x^{n+1}\longrightarrow y)^{n}\longrightarrow(x^{n}\longrightarrow y)$.
Then $((x^{n}\longrightarrow y)^{n}\longrightarrow
y)\longrightarrow((x^{n+1}\longrightarrow
y)^{n}\longrightarrow(x^{n}\longrightarrow y))=1$. Hence by Prop. 2.1 we have
: $(x^{n+1}\longrightarrow y)^{n}\longrightarrow(((x^{n}\longrightarrow
y)^{n}\longrightarrow y)\longrightarrow(x^{n}\longrightarrow y))=1\in F$.
Since $(x^{n+1}\longrightarrow y)\in F$ and $F$ is a filter, we have
$(x^{n+1}\longrightarrow y)^{n}\in F$. By the fact that $F$ is a filter and
$(x^{n+1}\longrightarrow y)^{n}\longrightarrow(((x^{n}\longrightarrow
y)^{n}\longrightarrow y)\longrightarrow(x^{n}\longrightarrow y))\in F$, we
have $((x^{n}\longrightarrow y)^{n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow y)\in F$. Since $F$ is an n-fold
positive implicative filter, by Prop. 5.5 we have: $x^{n}\longrightarrow y\in
F$.
By Prop. 4.9, $F$ is an n-fold implicative filter.
∎

The following Example shows that n-fold implicative filters may not be n-fold
positive implicative filters.

######  Example 5.10.

Let $L$ be a residuated lattice from Example 2.5 and $n\geq 1$.
$F_{1}=\\{1,a\\},F_{2}=\\{1,b\\}$ are n-fold implicative filters but not
n-fold positive implicative filters since
$1\longrightarrow((b^{n}\longrightarrow 0)\longrightarrow b)\in F_{1}$ and
$1\in F_{1}$, but $b\notin F_{1}$; $1\longrightarrow((a^{n}\longrightarrow
0)\longrightarrow a)\in F_{2}$ and $1\in F_{2}$, but $a\notin F_{2}$.

######  Proposition 5.11.

Every n-fold positive implicative filter is an (n+1)-fold positive implicative
filter.

###### Proof.

Let $F$ be an n-fold positive implicative filter of $L$. Let $x\in L$ such
that $\overline{x^{n+1}}\longrightarrow x\in F$. We show that $x\in F$. Since
$\overline{x^{n+1}}\longrightarrow x\leq\overline{x^{n}}\longrightarrow x$, by
the fact that $F$ is a filter, we get $\overline{x^{n}}\longrightarrow x\in
F$. Since $F$ is an n-fold positive implicative filter of $L$, by Prop. 5.5,
we obtain $x\in F$. By Prop. 5.5, $F$ is an (n+1)-fold positive implicative
filter of $L$. ∎

By the following example, we show that the converse of Prop. 5.11 is not true
in general.

######  Example 5.12.

Let $L$ be a residuated lattice from Example 4.11.It is clear that $\\{1\\}$
is an 3-fold positive implicative filter but $\\{1\\}$ is not an 2-fold
positive implicative filter, since $\overline{b^{2}}\longrightarrow
b\in\\{1\\}$ and $b\notin\\{1\\}$.

######  Definition 5.13.

A residuated lattice $L$ is called n-fold positive implicative residuated
lattice if it satisfies $\overline{y^{n}}\longrightarrow y=y$ for each $y\in
L$.

######  Remark 5.14.

Since $\overline{x^{n}}\longrightarrow x\leq\overline{x}\longrightarrow x$ for
all $x\in L$, it is clear that 1-fold residuated lattices are n-fold
residuated lattices. It is also clear that n-fold residuated lattices are
(n+1)-fold residuated lattices since $\overline{x^{n+1}}\longrightarrow
x\leq\overline{x^{n}}\longrightarrow x$ for all $x\in L$

The following example shows that residuated lattices are not in general n-fold
positive implicative residuated lattices.

######  Example 5.15.

Let $L$ be a residuated lattice from Example 2.5 and $n\geq 2$. $L$ is not an
n-fold positive implicative residuated lattice since $(b^{n}\longrightarrow
0)\longrightarrow b=(b\longrightarrow 0)\longrightarrow b=0\longrightarrow
b=1\neq b$

It is follows from Prop.5.5 and Prop.5.3 the following proposition:

######  Proposition 5.16.

If $L$ is an n-fold positive implicative residuated lattice then the notion of
n-fold positive implicative filter and filter coincide.

######  Proposition 5.17.

The following conditions are equivalent :

  * (i)

$L$ is an n-fold positive implicative residuated lattice.

  * (ii)

Every filter $F$ of $L$ is an n-fold positive implicative filter of $L$

  * (iii)

$\\{1\\}$ is an n-fold positive implicative filter

###### Proof.

$(i)\longrightarrow(ii)$ : Follows from Prop.5.16
$(ii)\longrightarrow(iii)$ : Follows from the fact that $\\{1\\}$ is a filter
of $L$.
$(iii)\longrightarrow(i)$: Assume that $\\{1\\}$ is an n-fold positive
implicative filter. Let $x\in L$. By Corollary 5.6, for all $x\in L$ holds
$x\vee\overline{x^{n}}=1$. By (14)Prop. 2.1,
$x\vee\overline{x^{n}}\leq(\overline{x^{n}}\longrightarrow x)\longrightarrow
x$, Hence $\overline{x^{n}}\longrightarrow x\longrightarrow x=1$ or
equivalently $\overline{x^{n}}\longrightarrow x\leq x$, and by the fact that
$x\leq\overline{x^{n}}\longrightarrow x$, we have
$\overline{x^{n}}\longrightarrow x=x$. ∎

The following result which follows from Prop. 5.17 and Prop. 4.16, gives the
relation between n-fold positive implicative residuated lattices and n-fold
implicative residuated lattices.

######  Proposition 5.18.

n-fold positive implicative residuated lattices are n-fold implicative
residuated lattices.

By the following example, we show that the converse of Prop. 5.18 is not true
in general.

######  Example 5.19.

A residuated lattice $L$ from Example 2.5 is an n-fold implicative residuated
lattice but by Example 5.15, it is not an fold positive implicative residuated
lattice.

The following result which follows from Prop. 5.17 and Corollary 5.6, gives
new a characterization of n-fold positive implicative residuated lattices.

######  Corollary 5.20.

A residuated lattice $L$ is an n-fold positive implicative residuated lattice
if and only if it satisfies $\overline{y^{n}}\vee y=1$ for each $y\in L$.

######  Proposition 5.21.

If $L$ is a totaly ordered residuated lattice, then any n-fold positive
implicative filter $F$ is maximal filter of $L$ and $L/F$ is a locally finite
residuated lattice.

###### Proof.

Let $L$ be a totaly ordered residuated lattice. Assume that $F$ is n-fold
positive implicative filter and let $x\in L$ be such an element that $x\notin
F$. From Prop. 5.5 an assumption $\overline{x^{n}}\leq x$, or equivalently
$\overline{x^{n}}\ \longrightarrow x=1\in F$ leads to a contradiction $x\in
F$, so we necessarily have $x<\overline{x^{n}}$. Therefore $x^{n+1}=0\in F$
and so $\overline{x^{n+1}}=1\in F$ . The thesis follows from Prop. 2.10. ∎

At consequence of Prop. 5.21, we have the following results:

######  Corollary 5.22.

A totaly ordered residuated lattice is a locally finite if $\\{1\\}$ is an
n-fold positive implicative filter. A totaly ordered n-fold positive
implicative residuated lattice is a locally finite.

######  Proposition 5.23.

A filter $F$ of $L$ is an n-fold positive implicative filter if and only $L/F$
is an n-fold positive implicative residuated lattice.

###### Proof.

Suppose that $F$ is an n-fold positive implicative filter. Let $x\in L$ be
such that $\overline{(x/F)^{n}}\longrightarrow x/F\in\\{1/F\\}$, then
$(\overline{x^{n}}\longrightarrow x)/F=\overline{(x/F)^{n}}\longrightarrow
x/F=1/F$. So $(\overline{x^{n}}\longrightarrow x)\in F$, by the fact that $F$
is an n-fold positive implicative filter, we have $x\in F$. Hence
$x/F\in\\{1/F\\}$, by Prop. 5.5, $\\{1/F\\}$ is an n-fold positive implicative
filter of $L/F$, by Prop. 5.17, $L/F$ is an n-fold positive implicative
residuated lattice.
Suppose conversely that $L/F$ is an n-fold positive implicative residuated
lattice. Let $x\in L$ be such that $\overline{x^{n}}\longrightarrow x\in F$.
We have $(\overline{x^{n}}\longrightarrow x)/F=1/F$, this implies
$\overline{(x/F)^{n}}\longrightarrow x/F\in\\{1/F\\}$. Since $L/F$ is an
n-fold positive implicative residuated lattice, by Prop. 5.17, $\\{1/F\\}$ is
an n-fold positive implicative filter of $L/F$. Hence $x/F\in\\{1/F\\}$ or
equivalently $x\in F$. By Prop. 5.5, $F$ is an n-fold positive implicative
filter of $L$. ∎

The following examples shows that the notion of n-fold positive implicative
residuated lattices exist.

######  Example 5.24.

Let $L$ be a residuated lattice from Example 2.5 and $n\geq 2$. Since
$F_{3}=\\{1,a,b,c\\}$ is an n-fold positive implicative filter, by Prop. 5.23,
$L/F_{3}$ is an n-fold positive implicative residuated lattice.

Follows from Corollary 5.6, we have the following proposition.

######  Proposition 5.25.

In any residuated lattices, the concepts of n-fold boolean filters and n-fold
positive implicative filters coincide.

######  Definition 5.26.

$L$ is an n-fold boolean residuated lattice if it satisfies the following
conditions:
$x\vee\overline{x^{n}}=1$ for all $x\in L$. In particular 1-fold boolean
residuated lattices are boolean algebra.

It is easy to observe that:

######  Remark 5.27.

A residuated lattice $L$ is an n-fold boolean residuated lattice if and only
if $\\{1\\}$ is an n-fold boolean filter of $L$ if and only if $L$ is an
n-fold positive implicative residuated lattice.

##  6\. N-Fold Normal Filter in Residuated Lattices

In [13], the authors introduce the notion of n-fold normal filter in BL-
algebra. This motives us to introduce the notion of n-fold normal filter in
residuated lattices.

######  Proposition 6.1.

Let $n\geq 1$. The following conditions are equivalent for any filter $F$:

  * (i)

For all $x,y,z\in L$, if $z\longrightarrow((y^{n}\longrightarrow
x)\longrightarrow x)\in F$ and $z\in F$, then $(x\longrightarrow
y)\longrightarrow y\in F$.

  * (ii)

For all $x,y\in L$, $(y^{n}\longrightarrow x)\longrightarrow x\in F$ implies
$(x\longrightarrow y)\longrightarrow y\in F$.

###### Proof.

Let $F$ be a filter which satisfying (i) . Assume that $(y^{n}\longrightarrow
x)\longrightarrow x\in F$. We have $(y^{n}\longrightarrow x)\longrightarrow
x=1\longrightarrow((y^{n}\longrightarrow x)\longrightarrow x)\in F$ and $1\in
F$, by the fact that $F$ satisfying (i), we obtain $(x\longrightarrow
y)\longrightarrow y\in F$. Conversely, let
$z\longrightarrow((y^{n}\longrightarrow x)\longrightarrow x)\in F$ and $z\in
F$, Since $F$ is a filter, we have $(y^{n}\longrightarrow x)\longrightarrow
x\in F$. By hypothesis, we obtain $(x\longrightarrow y)\longrightarrow y\in
F$. ∎

######  Definition 6.2.

A filter $F$ is an n-fold normal filter if it satisfies one of the conditions
of Prop. 6.1. In particular 1-fold normal filters are normal filters.

######  Example 6.3.

Let $n\geq 1$. Let $L$ be a residuated lattice from Example 2.4. Simple
computations proves that $F_{3}=\\{1,a,b,c\\}$ is an n-fold normal filter.

The following example shows that filters may not be n-fold normal in general.

######  Example 6.4.

Let $L$ be a residuated lattice from Example 2.4. $F_{2}=\\{1,b\\}$ is not an
n-fold normal filter since $(a^{n}\longrightarrow 0)\longrightarrow
0=1\in\\{1,b\\}$ but $[(0\longrightarrow a)\longrightarrow
a]=a\notin\\{1,b\\}$.

######  Proposition 6.5.

n-fold positive implicative filters are n-fold normal filters.

###### Proof.

Assume that $F$ is an n-fold positive implicative filter an let
$(x^{n}\longrightarrow y)\longrightarrow y\in F$.
We must show that $(y\longrightarrow x)\longrightarrow x\in F$.
Since $y\leq(y\longrightarrow x)\longrightarrow x$, by (4) Prop.2.1, we obtain
$(x^{n}\longrightarrow y)\longrightarrow y\leq(x^{n}\longrightarrow
y)\longrightarrow((y\longrightarrow x)\longrightarrow x)$.
From this and the fact that $F$ is a filter, we obtain $(x^{n}\longrightarrow
y)\longrightarrow((y\longrightarrow x)\longrightarrow x)\in F$.
Since $x\leq(y\longrightarrow x)\longrightarrow x$, by (4)Prop.2.1 we have
$x^{n}\leq((y\longrightarrow x)\longrightarrow x)^{n}$, hence
$(x^{n}\longrightarrow y)\longrightarrow[(y\longrightarrow x)\longrightarrow
x]\leq([(y\longrightarrow x)\longrightarrow x]^{n}\longrightarrow
y)\longrightarrow[(y\longrightarrow x)\longrightarrow x]$.
Since $(x^{n}\longrightarrow y)\longrightarrow[(y\longrightarrow
x)\longrightarrow x]\in F$, by the fact that $F$ is a filter, we obtain
$([(y\longrightarrow x)\longrightarrow x]^{n}\longrightarrow
y)\longrightarrow[(y\longrightarrow x)\longrightarrow x]\in F$. From this and
the fact that $F$ is an n-fold positive implicative filter, we obtain the
result by apply Prop.5.5. ∎

By the following example, we show that the converse of Prop. 6.5 is not true
in general.

######  Example 6.6.

Let $L$ be a lattice from Example 4.11 $\\{1\\}$ is an 1-fold normal filter
but $\\{1\\}$ is not an 1-fold positive implicative filter.

##  7\. n-fold fantastic Filter in Residuated Lattices

######  Definition 7.1.

Let $n\geq 1$. $F$ is an n-fold fantastic filter $L$ if it satisfies the
following conditions:

  * (i)

$1\in F$

  * (ii)

For all $x,y\in L$, $y\longrightarrow x\in F$ implies $[(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x\in F$.

In particular 1-fold fantastic filters are fantastic filters.[7]

######  Example 7.2.

Let $n\geq 1$. Let $L$ be a residuated lattice from Example 2.3. It is easy to
check that $\\{1\\}$ is an n-fold fantastic filter.

The following example shows that filters may not be n-fold fantastic in
general.

######  Example 7.3.

Let $L$ be a residuated lattice from Example 2.2. $\\{1\\}$ is not an n-fold
fantastic filter since $a\longrightarrow c=1\in\\{1\\}$ but
$[(c^{n}\longrightarrow a)\longrightarrow a]\longrightarrow c=c\notin\\{1\\}$.

######  Proposition 7.4.

Let $n\geq 1$. n-fold positive implicative filters are n-fold fantastic
filters.

###### Proof.

Assume that $F$ is an n-fold positive implicative filter. Let $x,y\in L$ be
such that $y\longrightarrow x\in F$.
By Prop. 2.1, we have:
$x\leq[((x^{n}\longrightarrow y)\longrightarrow y)\longrightarrow x]$. (a)
Then by Prop. 2.1, we also have:
$x^{n}\leq[((x^{n}\longrightarrow y)\longrightarrow y)\longrightarrow x]^{n}$.
(b)
By (b) and Prop. 2.1, we get, $(x^{n}\longrightarrow
y)\geq[((x^{n}\longrightarrow y)\longrightarrow y)\longrightarrow
x]^{n}\longrightarrow y$. (c)
By Prop. 2.1, we get, $y\longrightarrow x\leq((x^{n}\longrightarrow
y)\longrightarrow y)\longrightarrow((x^{n}\longrightarrow y)\longrightarrow
x)$. (d)
We also have :
$((x^{n}\longrightarrow y)\longrightarrow
y)\longrightarrow((x^{n}\longrightarrow y)\longrightarrow
x)=((x^{n}\longrightarrow y)\longrightarrow(((x^{n}\longrightarrow
y)\longrightarrow y)\longrightarrow x)$. (e)
So, by (d)and(e), we get, $y\longrightarrow x\leq((x^{n}\longrightarrow
y)\longrightarrow(((x^{n}\longrightarrow y)\longrightarrow y)\longrightarrow
x)$. (f)
By (c) and Prop. 2.1, we get,
$[((x^{n}\longrightarrow y)\longrightarrow[((x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x]\leq[[((x^{n}\longrightarrow
y)\longrightarrow y)\longrightarrow x]^{n}\longrightarrow
y]\longrightarrow[[((x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x]]$ (g)
By (f) and (g) , we obtain,
$y\longrightarrow x\leq[[((x^{n}\longrightarrow y)\longrightarrow
y)\longrightarrow x]^{n}\longrightarrow
y]\longrightarrow[[((x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x]]$ (h)
Since $F$ is a filter (see Prop. 5.3), by (h) and the fact that
$y\longrightarrow x\in F$, we get :
$[[((x^{n}\longrightarrow y)\longrightarrow y)\longrightarrow
x]^{n}\longrightarrow y]\longrightarrow[[((x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x]]\in F$. (w)
By (w), Prop. 5.5 and the fact that $F$ is an n-fold positive implicative
filter, we obtain $[[((x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow x]]\in F$.
Hence $F$ is an n-fold fantastic filter.
∎

The following example shows that n-fold fantastic filters may not be n-fold
positive implicative in general.

######  Example 7.5.

Let $n\geq 1$. Let $L$ be a residuated lattice from Example 2.3. It is easy to
check that $\\{1\\}$ is an n-fold fantastic filter, but not n-fold positive
implicative filter since $\overline{a^{n}}\longrightarrow a\in F$ and $a\notin
F$.

######  Proposition 7.6.

Let $n\geq 1$. n-fold fantastic filters are n-fold normal filters.

###### Proof.

Assume that $F$ is an n-fold fantastic filter. Let $x,y\in L$ be such that
$(x^{n}\longrightarrow y)\longrightarrow y\in F$ and $t=(y\longrightarrow
x)\longrightarrow x$. We must show that $t\in F$.
By Prop. 2.1, we have:
$y\leq(y\longrightarrow x)\longrightarrow x$, so $(x^{n}\longrightarrow
y)\longrightarrow y\leq[(x^{n}\longrightarrow
y)\longrightarrow[(y\longrightarrow x)\longrightarrow x]]$, that is
$(x^{n}\longrightarrow y)\longrightarrow y\leq[(x^{n}\longrightarrow
y)\longrightarrow t]$. (a)
Since $(x^{n}\longrightarrow y)\longrightarrow y\in F$, by (a) an the fact
that $F$ is a filter, it follows that $(x^{n}\longrightarrow y)\longrightarrow
t\in F$. (b)
By (b) and the fact that $F$ is an n-fold fantastic filter, we get that
$[(t^{n}\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)]\longrightarrow t\in F$. (c)
By Prop. 2.1, we also have $t^{n}\longrightarrow(x^{n}\longrightarrow
y)=x^{n}\longrightarrow(t^{n}\longrightarrow y)$, so
$(t^{n}\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow
y)=(x^{n}\longrightarrow(t^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)$ and then
$[(x^{n}\longrightarrow(t^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)]\longrightarrow t\in F$. (d)
On the other hand, by Prop. 2.1, we also have $(t^{n}\longrightarrow
y)\longrightarrow y\leq(x^{n}\longrightarrow(t^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)$. (e)
Since $x\leq t$, it follows that $(x^{n}\longrightarrow y)\longrightarrow
y\leq(t^{n}\longrightarrow y)\longrightarrow y$. (f)
Since $F$ is a filter and $(x^{n}\longrightarrow y)\longrightarrow y\in F$, by
(f) we obtain $(t^{n}\longrightarrow y)\longrightarrow y\in F$. (g)
Since $F$ is a filter, by (g) and (e) it follows that
$(x^{n}\longrightarrow(t^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)\in F$. (h)
Since $F$ is a filter, by (h) and (d) it follows that $t\in F$.
Hence $F$ is an n-fold normal filter. ∎

######  Lemma 7.7.

For all $x,y\in L$, we have:
$[(x^{n}\longrightarrow x^{2n})\otimes(x^{2n}\longrightarrow y)]\leq
x^{n}\longrightarrow y$.

###### Proof.

Let $x,y\in L$, by Prop.2.1 we have the following:

  * (1)

$x^{n}\longrightarrow x^{2n}\leq[(x^{2n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow y)]$

  * (2)

By (1) we have : $[(x^{2n}\longrightarrow y)\otimes(x^{n}\longrightarrow
x^{2n})]\leq[(x^{2n}\longrightarrow y)\otimes((x^{2n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow y))]$

  * (3)

$[(x^{2n}\longrightarrow y)\otimes((x^{2n}\longrightarrow
y)\longrightarrow(x^{n}\longrightarrow y))]\leq x^{n}\longrightarrow y$

By (2) and (3) we have : $[(x^{n}\longrightarrow
x^{2n})\otimes(x^{2n}\longrightarrow y)]\leq x^{n}\longrightarrow y$. ∎

######  Proposition 7.8.

Let $n\geq 1$. Let $F$ a filter of $L$. If $F$ is n-fold fantastic and n-fold
implicative filter, then $F$ is an n-fold positive implicative filter.

###### Proof.

Let $x,y\in L$ be such that $(x^{n}\longrightarrow y)\longrightarrow x\in F$.
Assume that $F$ is both n-fold fantastic filter and n-fold implicative filter.
Since $F$ is a n-fold fantastic filter, by the fact that
$(x^{n}\longrightarrow y)\longrightarrow x\in F$, we have :
$[(x^{n}\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)]\longrightarrow x\in F$.
By Lemma 7.7 and residuation we get : $(x^{n}\longrightarrow
x^{2n})\leq(x^{2n}\longrightarrow y)\longrightarrow(x^{n}\longrightarrow y)$.
So, $(x^{n}\longrightarrow
x^{2n})\leq[x^{n}\longrightarrow(x^{n}\longrightarrow
y)]\longrightarrow(x^{n}\longrightarrow y)$. Hence, $[(x^{n}\longrightarrow
x^{2n})\longrightarrow x]\geq[[x^{n}\longrightarrow(x^{n}\longrightarrow
y)]\longrightarrow(x^{n}\longrightarrow y)]\longrightarrow x$. Since $F$ is a
filter, by the fact that $[(x^{n}\longrightarrow(x^{n}\longrightarrow
y))\longrightarrow(x^{n}\longrightarrow y)]\longrightarrow x\in F$, we have :
$(x^{n}\longrightarrow x^{2n})\longrightarrow x\in F$. Since $F$ is and n-fold
implicative filter, by Prop. 4.9, we get, $x\in F$. By Prop. 5.5, $F$ is an
n-fold positive implicative filter.

∎

Follows from Prop. 7.4, Prop. 7.8, Prop. 5.9, it is easy to show the following
theorem.

######  Theorem 7.9.

Let $n\geq 1$. Let $F$ a filter. $F$ is an n-fold positive implicative filter
of $L$ if and only if $F$ is n-fold fantastic and n-fold implicative filter of
$L$.

######  Definition 7.10.

$L$ is said to be n-fold fantastic residuated lattice if for all $x,y\in L$,
$y\longrightarrow x=[(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x$.

The following example shows that the notion of n-fold fantastic residuated
lattice exist.

######  Example 7.11.

Let $n\geq 1$. Let $L$ be a residuated lattice from Example 2.3. It is easy to
check that $L$ is an n-fold fantastic residuated lattice.

The following example shows that residuated lattices may not be n-fold
fantastic in general.

######  Example 7.12.

Let $L$ be a residuated lattice from Example 2.2. $L$ is not an n-fold
fantastic residuated lattice since $a\longrightarrow c=1\neq
c=[(c^{n}\longrightarrow a)\longrightarrow a]\longrightarrow c$.

The following proposition gives a characterization of n-fold fantastic
residuated lattice.

######  Proposition 7.13.

$L$ is an n-fold fantastic residuated lattice if and only if the inequality
$(x^{n}\longrightarrow y)\longrightarrow y\leq(y\longrightarrow
x)\longrightarrow x$ holds for all $x,y\in L$.

###### Proof.

Assume that $L$ is an n-fold fantastic residuated lattice. Let $x,y\in L$. We
have $[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow[(y\longrightarrow x)\longrightarrow x]=(y\longrightarrow
x)\longrightarrow[[(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x]$. (a)
By hypothesis $y\longrightarrow x=[[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow x]$. Hence $(y\longrightarrow
x)\longrightarrow[[(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x]=1$. (b)
By (a) and (b), we get $[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow[(y\longrightarrow x)\longrightarrow x]=1$ or equivalently
$[(x^{n}\longrightarrow y)\longrightarrow y]\leq[(y\longrightarrow
x)\longrightarrow x]$.
Suppose conversely that the inequality $(x^{n}\longrightarrow
y)\longrightarrow y\leq(y\longrightarrow x)\longrightarrow x$ holds for all
$x,y\in L$. Then $(y\longrightarrow x)\longrightarrow[[(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x]=[(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow[(y\longrightarrow x)\longrightarrow x]$.
(e)
Since $(x^{n}\longrightarrow y)\longrightarrow y\leq(y\longrightarrow
x)\longrightarrow x$, by Prop. 2.1, we get $[(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow[(x^{n}\longrightarrow y)\longrightarrow
y]\leq[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow[(y\longrightarrow x)\longrightarrow x]$, that is
$1\leq[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow[(y\longrightarrow x)\longrightarrow x]$ or equivalently
$[(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow[(y\longrightarrow
x)\longrightarrow x]=1$. (f)
By (e) and (f), its follows that $(y\longrightarrow
x)\leq[[(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow x]$. (g)
Since $y\leq(x^{n}\longrightarrow y)\longrightarrow y$, we also get by Prop.
2.1, $y\longrightarrow x\geq[[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow x]$. (h)
From (h) and (g), we obtain $y\longrightarrow x=[[(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x]$. Hence $L$ is an n-fold fantastic
residuated lattice. ∎

######  Proposition 7.14.

The following conditions are equivalent for any filter $F$:

  * (i)

$L$ is an n-fold fantastic residuated lattice.

  * (ii)

Every filter $F$ of $L$ is an n-fold fantastic filter of $L$

  * (iii)

$\\{1\\}$ is an n-fold fantastic filter of $L$.

###### Proof.

$(i)\longrightarrow(ii)$ : Follows from Definition 7.10
$(ii)\longrightarrow(iii)$ : Follows from the fact that $\\{1\\}$ is a filter
of $L$.
$(iii)\longrightarrow(i)$: Assume that $\\{1\\}$ is an n-fold fantastic
filter.
Let $x,y\in L$ and $t=(y\longrightarrow x)\longrightarrow x$. By Prop. 2.1,
$y\leq t$. So $y\longrightarrow t=1$ and by the hypothesis, we have
$[(t^{n}\longrightarrow y)\longrightarrow y]\longrightarrow t=1$,
that is $[(t^{n}\longrightarrow y)\longrightarrow y]\leq t$. (w)
On the other hand, $x\leq t$ implies $x^{n}\leq t^{n}$, hence
$[(x^{n}\longrightarrow y)\longrightarrow y]\leq(t^{n}\longrightarrow
y)\longrightarrow y$. (z)
By (z) and (w), it follows that $[(x^{n}\longrightarrow y)\longrightarrow
y]\leq t=(y\longrightarrow x)\longrightarrow x$. Hence by Prop. 7.13, $L$ is
an n-fold fantastic residuated lattice. ∎

Combine Prop.7.14, Prop.5.17, Prop.4.16 and Theorem 7.9, we have the following
result:

######  Corollary 7.15.

Let $n\geq 1$. $L$ is an n-fold positive implicative residuated lattice if and
only if $L$ is n-fold fantastic residuated lattice and n-fold implicative
residuated lattice.

The following corollary gives a characterization of n-fold fantastic filter in
residuated lattice.

######  Corollary 7.16.

Let $n\geq 1$. Let $F$ be a filter of $L$. Then $F$ is an n-fold fantastic
filter if and only $L/F$ is is an n-fold fantastic residuated lattice.

###### Proof.

Let $F$ be a filter of $L$. Assume that $F$ is an n-fold fantastic filter. We
show that $L/F$ is is an n-fold fantastic residuated lattice. Let $x,y\in L$
be such that $y/F\longrightarrow x/F\in\\{1/F\\}$, then $(y\longrightarrow
x)/F=1/F$ or equivalently $y\longrightarrow x\in F$. Since $F$ is an n-fold
fantastic filter, we get $[(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow x\in F$ or equivalently $([(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x)/F=1/F$, so $([((x/F)^{n}\longrightarrow
y/F)\longrightarrow y/F]\longrightarrow x/F)\in\\{1/F\\}$. Hence $\\{1/F\\}$
is an n-fold fantastic filter of $L/F$, therefore by Prop. 7.14, $L/F$ is is
an n-fold fantastic residuated lattice.
Conversely, assume that $L/F$ is is an n-fold fantastic residuated lattice.
Let $x,y\in L$ be such that $y\longrightarrow x\in F$ then $(y\longrightarrow
x)/F=1/F$ or equivalently $y/F\longrightarrow x/F\in\\{1/F\\}$. Since $L/F$ is
is an n-fold fantastic residuated lattice, by Prop. 7.14, $\\{1/F\\}$ is an
n-fold fantastic filter of $L/F$. From this and the fact that
$y/F\longrightarrow x/F\in\\{1/F\\}$, we have: $([((x/F)^{n}\longrightarrow
y/F)\longrightarrow y/F]\longrightarrow x/F)\in\\{1/F\\}$ or equivalently
$([(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow x)/F=1/F$, so
$[(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow x\in F$. Hence $F$
is an n-fold fantastic filter. ∎

The extension theorem of n-fold fantastic filters is obtained from the
following result:

######  Theorem 7.17.

Let $n\geq 1$. Let $F_{1}$ and $F_{2}$ two filters of $L$ such that
$F_{1}\subseteq F_{2}$. If $F_{1}$ is an n-fold fantastic filter, then so is
$F_{2}$.

###### Proof.

Let $x,y\in L$ be such that $y\longrightarrow x\in F_{2}$. Since $F_{1}$ is an
n-fold fantastic filter, by Corollary. 7.16, $L/F_{1}$ is an n-fold positive
implicative residuated lattice. So $([((x/F_{1})^{n}\longrightarrow
y/F_{1})\longrightarrow y/F_{1}]\longrightarrow
x/F_{1})=y/F_{1}\longrightarrow x/F_{1}$, so $(y\longrightarrow
x)\longrightarrow([(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x)\in F_{1}$, so $(y\longrightarrow x)\longrightarrow([(x^{n}\longrightarrow
y)\longrightarrow y]\longrightarrow x)\in F_{2}$. Since $F_{2}$ is a filter of
$L$, by the fact that $y\longrightarrow x\in F_{2}$ and $(y\longrightarrow
x)\longrightarrow([(x^{n}\longrightarrow y)\longrightarrow y]\longrightarrow
x)\in F_{2}$, we get $([(x^{n}\longrightarrow y)\longrightarrow
y]\longrightarrow x)\in F_{2}$. Hence $F_{2}$ is an n-fold fantastic filter. ∎

##  8\. n-fold obstinate Filters in Residuated Lattices

######  Definition 8.1.

A filter $F$ is an n-fold obstinate filter of $L$ if it satisfies the
following conditions for any $n\geq 1$:

  * (i)

$0\notin F$

  * (ii)

For all $x,y\in L$, $x,y\notin F$ imply $x^{n}\longrightarrow y\in F$ and
$y^{n}\longrightarrow x\in F$

In particular 1-fold obstinate filters are obstinate filters.[8]

The following proposition gives a characterization of n-fold obstinate filter
of $L$.

######  Proposition 8.2.

For any $n\geq 1$, the following conditions are equivalent for any filter $F$:

  * (i)

$F$ is an n-fold obstinate filter

  * (ii)

For all $x\in L$, if $x\notin F$ then there exist $m\geq 1$ such that
$(\overline{x^{n}})^{m}\in F$

###### Proof.

$(i)\longrightarrow(ii)$ : Suppose that $F$ is n-fold obstinate filter of $L$.
Let $x\in L$ be such that $x\notin F$. By setting $y=0$ in Definition 8.1, we
get: $x^{n}\longrightarrow 0\in F$. Hence for $m=1$, we have:
$(\overline{x^{n}})^{m}\in F$
$(ii)\longrightarrow(i)$: Conversely, let $x,y\notin F$, we show that
$x^{n}\longrightarrow y\in F$ and $y^{n}\longrightarrow x\in F$. By the
hypothesis, there are $m_{1},m_{2}\geq 1$ such that
$(\overline{x^{n}})^{m_{1}},(\overline{y^{n}})^{m_{2}}\in F$.
By Prop. 2.1(8), we have:
$(\overline{x^{n}})^{m_{1}}\leq\overline{x^{n}}\leq x^{n}\longrightarrow y$
(a)
and $(\overline{y^{n}})^{m_{2}}\leq\overline{x}^{n}\leq y^{n}\longrightarrow
x$ (b)
Since $F$ is a filter, by (a) and (b), we get $x^{n}\longrightarrow y\in F$
and $y^{n}\longrightarrow x\in F$. ∎

######  Example 8.3.

Let $L$ be a lattice from Example 2.2. $F=\\{1,b,c,d\\}$ is a proper filter of
$L$. Using Prop. 8.2, for any $n\geq 1$, it is easy to check that $F$ is an
n-fold obstinate filter of $L$.

The following example shows that any filters may not be n-fold obstinate
filter.

######  Example 8.4.

Let $L$ be a lattice from Example 2.3. $F=\\{1,c,d\\}$ is a proper filter of
$L$. For any $n\geq 1$, we have: $a,b\notin F$ but $a^{n}\longrightarrow
b=b\notin F$. Hence $F$ is not an n-fold obstinate filter of $L$.

######  Proposition 8.5.

Every n-fold obstinate filter of $L$ is a (n+1)-fold obstinate filter of $L$.

###### Proof.

Let $F$ be an n-fold obstinate filter of $L$ and $x,y\notin F$. We show that
$x^{n+1}\longrightarrow y\in F$ and $y^{n+1}\longrightarrow x\in F$. By
hypothesis, $x^{n}\longrightarrow y\in F$ and $y^{n}\longrightarrow x\in F$.
(c)
By Prop. 2.1(8), $x^{n+1}\leq x^{n}$ and $y^{n+1}\longrightarrow y^{n}$.
Then, $x^{n}\longrightarrow y\leq x^{n+1}\longrightarrow y$ and
$y^{n}\longrightarrow x\leq y^{n+1}\longrightarrow x$. (d)
By (c) and (d) and the fact that $F$ is a filter, we obtain
$x^{n+1}\longrightarrow y\in F$ and $y^{n+1}\longrightarrow x\in F$. Hence $F$
is a (n+1)-fold obstinate filter of $L$. ∎

The extension theorem of n-fold obstinate filters is obtained from the
following result and any $n\geq 1$:

######  Theorem 8.6.

Let $F_{1},F_{2}$ two filter of $L$ be such that $F_{1}\subseteq F_{2}$. If
$F_{1}$ is an n-fold obstinate filter of $L$ then so is $F_{2}$.

###### Proof.

Let $F_{1},F_{2}$ two filter of $L$ be such that $F_{1}\subseteq F_{2}$.
Assume that $F_{1}$ is an n-fold obstinate filter of $L$, and let $x\notin
F_{2}$. Since $F_{1}\subseteq F_{2}$, we have $x\notin F_{1}$. Since $F_{1}$
is an n-fold obstinate filter of $L$, by Prop. 8.2, there exist $m\geq 1$ such
that $(\overline{x^{n}})^{m}\in F_{1}$. Since $F_{1}\subseteq F_{2}$, we have
$(\overline{x^{n}})^{m}\in F_{2}$. Therefore there exist $m\geq 1$ such that
$(\overline{x^{n}})^{m}\in F_{2}$. Hence by Prop. 8.2, $F_{2}$ is an n-fold
obstinate filter of $L$. ∎

From Theorem 8.6, it is easy to shows the following result for any $n\geq 1$:

######  Corollary 8.7.

$\\{1\\}$ is an n-fold obstinate filter of $L$ if and only if every filter of
$L$ is an n-fold obstinate filter of $L$.

######  Proposition 8.8.

The following conditions are equivalent for any filter $F$ and any $n\geq 1$:

  * (i)

$F$ is an n-fold obstinate filter

  * (ii)

$F$ is a maximal and n-fold positive implicative filter

  * (iii)

$F$ is a maximal and n-fold implicative filter

###### Proof.

$(i)\longrightarrow(ii)$: Assume that $F$ is an n-fold obstinate filter. We
show that $F$ is a maximal and n-fold positive implicative filter. At first we
show that $F$ is a maximal. Let $x\notin F$, since $F$ is an n-fold obstinate
filter, by Prop. 8.2, there exist $m\geq 1$ such that
$(\overline{x^{n}})^{m}\in F$. Since
$(\overline{x^{n}})^{m}\leq\overline{x^{n}}$, by the fact that $F$ is a
filter, we get $\overline{x^{n}}\in F$, by Prop. 2.10, $F$ is a maximal
filter.
On the other hand, assume in the contrary that there exist $x\in L$ such that
$\overline{x^{n}}\longrightarrow x\in F$ and $x\notin F$. Since $F$ is an
n-fold obstinate filter, by Prop. 8.2, there exist $m\geq 1$ such that
$(\overline{x^{n}})^{m}\in F$. Since
$(\overline{x^{n}})^{m}\leq\overline{x^{n}}$, by the fact that $F$ is a
filter, we get $\overline{x^{n}}\in F$. Since $F$ is a filter, by the fact
that $\overline{x^{n}}\longrightarrow x\in F$, we obtain $x\in F$ which is a
contradiction. Hence for all $x\in L$, $\overline{x^{n}}\longrightarrow x\in
F$ implies $x\in F$.
By Prop. 5.5, $F$ is an n-fold positive implicative filter.
$(ii)\longrightarrow(iii)$: follows from Prop. 5.9
$(iii)\longrightarrow(i)$: Assume that $F$ is a maximal and n-fold implicative
filter of $L$. Let $x,yL$ be such that $x,y\notin F$. By Lemma 4.5,
$L_{x}=\\{b\in L:x^{n}\longrightarrow b\in F\\}$ is a filter of $L$.
$L_{y}=\\{b\in L:y^{n}\longrightarrow b\in F\\}$ is a filter of $L$.
Let $z\in F$, since $z\leq x^{n}\longrightarrow z$, we have
$x^{n}\longrightarrow z\in F$, hence $x^{n}\longrightarrow z\in F$, so $z\in
L_{x}$ and we obtain $F\subseteq L_{x}$. On the other hand,
$x^{n}\longrightarrow x=1\in F$ since $x^{n}\subseteq x$, hence $x\in L_{x}$.
By hypothesis, $x\notin F$, So $F\varsubsetneq L_{x}\subseteq L$. Since $F$ is
a maximal filter of $L$, we get $L_{x}=L$. Therefore $y\in L_{x}$ or
equivalently $x^{n}\longrightarrow y\in F$. Similarly, we get
$y^{n}\longrightarrow x\in F$. Hence $F$ is an n-fold obstinate filter of $L$.
∎

######  Proposition 8.9.

The following conditions are equivalent for any filter $F$ and any $n\geq 1$:

  * (i)

$F$ is an n-fold obstinate filter

  * (ii)

$F$ is a maximal and n-fold boolean filter

  * (iii)

$F$ is a prime of second kind and n-fold boolean filter

###### Proof.

$(i)\longrightarrow(ii)$: Assume that $F$ is an n-fold obstinate filter. First
observe that by Prop. 8.8, $F$ is a maximal filter. Let $x\in L$. We have two
cases: $x\in F$ or $x\notin F$
case 1: $x\in F$. Since $x\leq x\vee\overline{x^{n}}$, by the fact that $F$ is
a filter, we have $x\vee\overline{x^{n}}\in F$.
case 2: $x\notin F$. Since $F$ is an n-fold obstinate filter, by Prop. 8.2,
there exist $m\geq 1$ such that $(\overline{x^{n}})^{m}\in F$. Since
$(\overline{x^{n}})^{m}\leq\overline{x^{n}}\leq x\vee\overline{x^{n}}$, we
have $(\overline{x^{n}})^{m}\leq x\vee\overline{x^{n}}$. By the fact that $F$
is a filter, we have $x\vee\overline{x^{n}}\in F$.
Since in both the two cases $x\vee\overline{x^{n}}\in F$, it is clear that for
all $x\in L$, $x\vee\overline{x^{n}}\in F$, hence $F$ is an n-fold boolean
filter.
$(ii)\longrightarrow(iii)$: Follows in the fact that a maximal filter of $L$
is a prime filter of second kind of $L$.
$(iii)\longrightarrow(i)$: Assume that $F$ is a prime filter of the second
kind and n-fold boolean filter. Let $x\in L$ be such that $x\notin F$. $F$ is
an n-fold boolean filter, we have $x\vee\overline{x^{n}}\in F$. Since $F$ is a
prime filter of the second kind , by the fact $x\notin F$, we have
$\overline{x^{n}}^{1}\in F$. by Prop. 8.2, $F$ is an n-fold obstinate filter.
∎

Combine Prop. 8.8 and Prop. 7.4, we have the following proposition.

######  Proposition 8.10.

Any n-fold obstinate filter $F$ is an n-fold fantastic filter. The converse is
not true in general.

The following example shows that the converse of the Proposition 8.10 is not
true in general.

######  Example 8.11.

Let $L$ be a residuated lattice from Example 2.3. It is easy to check that
$\\{1\\}$ is an n-fold fantastic filter but not n-fold obstinate filter.

Follows from Prop.8.9 and Prop.5.25, we have the following result:

######  Corollary 8.12.

The following conditions are equivalent for any filter $F$ and any $n\geq 1$:

  * (i)

$F$ is an n-fold obstinate filter

  * (ii)

$F$ is a prime filter in the second kind and n-fold positive implicative
filter

######  Proposition 8.13.

for any $n\geq 1$, a filter $F$ is an n-fold obstinate filter if and only if
every filter of $L/F$ is an n-fold obstinate filter of $L/F$.

###### Proof.

Assume that $F$ is an n-fold obstinate filter. Let $x,y\in L$ be such that
$x/F,y/F\notin\\{1/F\\}$, then $x,y\notin F$. Since $F$ is an n-fold obstinate
filter, we have $x^{n}\longrightarrow y,y^{n}\longrightarrow x\notin F$. From
this we have, $(x^{n}\longrightarrow y)/F,(y^{n}\longrightarrow x)/F\notin
1/F$ or equivalently $(x/F)^{n}\longrightarrow y/F,(y/F)^{n}\longrightarrow
x/F\notin 1/F$. Hence $\\{1/F\\}$ is an n-fold obstinate filter of $L/F$ and
by Corollary , every filter of $L/F$ is an n-fold obstinate filter of $L/F$.
Conversely, let $x,y\notin F$. Then $x/F,y/F\notin\\{1/F\\}$. Since
$\\{1/F\\}$ is an n-fold obstinate filter of $L/F$, we have
$(x/F)^{n}\longrightarrow y/F,(y/F)^{n}\longrightarrow x/F\notin 1/F$ or
equivalently $(x^{n}\longrightarrow y)/F,(y^{n}\longrightarrow x)/F\notin
1/F$. So $x^{n}\longrightarrow y,y^{n}\longrightarrow x\notin F$ and $F$ is an
n-fold obstinate filter. ∎

######  Definition 8.14.

A residuated lattice $L$ is said to be an n-fold obstinate residuated it
satisfies the following condition :
For all $x,y\in L$, $x,y\neq 1$ implies $x^{n}\longrightarrow y=1$ and
$y^{n}\longrightarrow x=1$.

######  Proposition 8.15.

The following conditions are equivalent for any $n\geq 1$:

  * (i)

$L$ is an n-fold obstinate residuated lattice

  * (ii)

$\\{1\\}$ is an n-fold obstinate filter of $L$.

  * (iii)

Every filter of $L$ is n-fold obstinate

###### Proof.

$(i)\longrightarrow(ii)$: Obvious
$(ii)\longrightarrow(iii)$:Follows from Theorem 8.6
$(iii)\longrightarrow(i)$:Assume that every filter of $L$ is n-fold obstinate,
then $\\{1\\}$ is an n-fold obstinate filter of $L$ since $\\{1\\}$ is a
filter of $L$. The thesis follows by setting $F=\\{1\\}$ in Definition 8.1. ∎

The following example shows that the notion of n-fold obstinate residuated
lattice exist.

######  Example 8.16.

Let $L$ be a lattice from Example 2.2. $F$ is an n-fold obstinate filter of
$L$, then by Prop. 8.13 and Prop. 8.15, $L/F$ is an n-fold obstinate
residuated lattice, for any $n\geq 1$.

The following example shows that any residuated lattice may not be n-fold
obstinate residuated lattice.

######  Example 8.17.

Let $L$ be a lattice from Example 2.4. For any $n\geq 1$, $F=\\{1,d\\}$ is not
an n-fold obstinate filter of $L$, then by Prop. 8.13 and Prop. 8.15, $L$ is
not an n-fold obstinate residuated lattice.

Follows from Prop. 8.15, Prop. 8.9 and Prop. 5.27, we have the following
proposition.

######  Proposition 8.18.

n-fold obstinate residuated lattices are n-fold boolean residuated lattices

##  9\.  Diagram among type of n-fold filters in Residuated Lattices

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