The Lattice of L-fuzzy Filters in a Given $R_{0}$-algebra

Abstract

In the present paper, the L-fuzzy filter theory on $R_0$-algebras is further studied. Some new properties of L-fuzzy filters are given. Representation theorem of L-fuzzy filter which is generated by a fuzzy set is established. It is proved that the set consisting of all L-fuzzy filters on a given $R_{0}$-algebra, under the L-fuzzy set-inclusion order $\Subset$, forms a complete distributive lattice.

1 Introduction

To make the computers simulate beings in dealing with certainty and uncertainty in information is one important task of artificial intelligence. Logic appears in a “sacred” (resp., a “profane”) form which is dominant in proof theory (resp., model theory). The role of logic in mathematics and computer science is twofold–as a tool for applications in both areas, and a technique for laying the foundations. Nonclassical logic [1] including many-valued logic and fuzzy logic takes the advantage of classical logic to handle information with various facets of uncertainty [2], such as fuzziness and randomness. At present, nonclassical logic has become a formal and useful tool for computer science to deal with fuzzy information and uncertain information. $R_0$-algebra is an important class of non-classical fuzzy logical algebras which was introduced by Wang in [3] by providing an algebra proof of the completeness theorem of the formal deductive system $\mathcal {L}^*$. From then, $R_0$-algebras has been extensively investigated by many researchers. Among them, Jun and Liu studied the theory of filters in $R_0$-algebras in [4]. The concept of fuzzy sets is introduced firstly by Zadeh in [5]. Liu and Li in [6] proposed the concept of fuzzy filters of $R_0$-algebras and discussed some their properties by using fuzzy sets theory. As an extension of the concept of fuzzy filter, in [7] the author and Xu propose the notion of L-fuzzy filters of $R_0$-algebras in terms of the notion of L-fuzzy set in [8], where the prefix L a lattice. In this paper, we will further research the properties of L-fuzzy filters in $R_0$-algebras. The lattice structural feature of the set containing all of L-fuzzy filters in a given $R_0$-algebra is investigated. It should be noticed that when $L=[0,1]$, then [0, 1]-fuzzy sets are originally meant fuzzy sets. Since [0, 1] is a special completely distributive lattice, to investigate properties of L-fuzzy filters, sometimes we assume that the prefix L is a completely distributive lattice.

2 Preliminaries

Definition 1

(cf. [3]). Let M be an algebra of type $(\lnot , \vee , \rightarrow )$, where $\lnot$ is a unary operation, $\vee$ and $\rightarrow$ are binary operations. $(M, \lnot , \vee , \rightarrow , 1)$ is called an $R_0$-algebra if there is a partial order $\leqslant$ such that $(M, \leqslant , 1)$ is a bounded distributive lattice with the greatest element 1, $\vee$ is the supremum operation with respect to $\leqslant$, $\lnot$ is an order-reversing involution, and the following conditions hold for every $a, b, c\in {M}$:

• (M1) $\lnot {a}\rightarrow \lnot {b}=b\rightarrow {a}$;

• (M2) $1\rightarrow {a}=a, a\rightarrow {a}=1$;

• (M3) $b\rightarrow {c}\leqslant (a\rightarrow {b})\rightarrow (a\rightarrow {c})$;

• (M4) $a\rightarrow (b\rightarrow {c})=b\rightarrow (a\rightarrow {c})$;

• (M5) $a\rightarrow (b\vee {c})=(a\rightarrow {b})\vee (a\rightarrow {c}), a\rightarrow (b\wedge {c})=(a\rightarrow {b})\wedge (a\rightarrow {c})$;

• (M6) $(a\rightarrow {b})\vee ((a\rightarrow {b})\rightarrow (\lnot {a}\vee {b}))=1$.

Lemma 1

(cf. [3]). Let M be an $R_0$-algebra, $a, b, c\in {M}$. Then the following properties hold.

• (P1) $a\leqslant {b}$ if and only if $a\rightarrow {b}=1$;

• (P2) $a\leqslant {b}\rightarrow {c}$ if and only if $b\leqslant {a}\rightarrow {c}$;

• (P3) $(a\vee {b})\rightarrow {c}=(a\rightarrow {c})\wedge (b\rightarrow {c}), (a\wedge {b})\rightarrow {c}=(a\rightarrow {c})\vee (b\rightarrow {c})$;

• (P4) If $b\leqslant {c}$, then $a\rightarrow {b}\leqslant {a}\rightarrow {c}$, and if $a\leqslant {b}$, then $b\rightarrow {c}\leqslant {a}\rightarrow {c}$;

• (P5) $a\rightarrow {b}\geqslant \lnot {a}\vee {b}$ and $a\wedge \lnot {a}\leqslant {b}\vee \lnot {b}$;

• (P6) $(a\rightarrow {b})\vee (b\rightarrow {a})=1$ and $a\vee {b}=((a\rightarrow {b})\rightarrow {b})\wedge ((b\rightarrow {a})\rightarrow {a})$;

• (P7) $a\rightarrow (b\rightarrow {a})=1$ and $a\rightarrow (\lnot {a}\rightarrow {b})=1$;

• (P8) $a\rightarrow {b}\leqslant {a}\vee {c}\rightarrow {b}\vee {c}$ and $a\rightarrow {b}\leqslant {a}\wedge {c}\rightarrow {b}\wedge {c}$;

• (P9) $a\rightarrow {b}\leqslant ({a}\rightarrow {c})\vee ({c}\rightarrow {b})$.

Lemma 2

(cf. [3]). Let M be an $R_0$-algebra. Define a new operator $\otimes$ on M such that $a\otimes {b}=\lnot (a\rightarrow \lnot {b})$, for every $a, b, c\in {M}$. Then the following properties hold.

• (P10) $(M, \otimes , 1)$ is a commutative monoid with the multiplicative unit element 1;

• (P11) If $a\leqslant {b}$, then $a\otimes {c}\leqslant {b}\otimes {c}$;

• (P12) $0\otimes {a}=0$ and $a\otimes \lnot {a}=0$;

• (P13) $a\otimes {b}\leqslant {a}\wedge {b}$ and $a\otimes (a\rightarrow {b})\leqslant {b}$ and $a\leqslant {b}\rightarrow (a\otimes {b})$;

• (P14) $a\otimes {b}\rightarrow {c}=a\rightarrow (b\rightarrow {c})$ and $a\otimes ({b}\vee {c})=(a\otimes {b})\vee (a\otimes {c})$.

Let X be a non-empty set and L a lattice. A map $\mathscr {A}: X\rightarrow {L}$ is called an L-fuzzy subset on X. The set of all L-fuzzy subsets on X is denoted by $\mathcal {F}_{L}(X)$. (cf. [8]). Let $\mathscr {A}$ and $\mathscr {B}$ be two L-fuzzy subsets on X. We define $\mathscr {A}\Cap \mathscr {B}$, $\mathscr {A}\Cup \mathscr {B}$ , $\mathscr {A}\Subset \mathscr {B}$ and $\mathscr {A}=\mathscr {B}$ as follows:

• (i) $(\mathscr {A}\Cap \mathscr {B})(x)=\mathscr {A}(x)\wedge \mathscr {B}(x)$, for all $x\in {X}$;

• (ii) $(\mathscr {A}\Cup \mathscr {B})(x)=\mathscr {A}(x)\vee \mathscr {B}(x)$, for all $x\in {X}$;

• (iii) $\mathscr {A}\Subset \mathscr {B}\Longleftrightarrow \mathscr {A}(x)\leqslant \mathscr {B}(x)$, for all $x\in {X}$;

• (iv) $\mathscr {A}=\mathscr {B}\Longleftrightarrow (\mathscr {A}\Subset \mathscr {B} \text{ and } \mathscr {B}\Subset \mathscr {A})$.

3 On L-fuzzy Filters in $R_0$-algebras

In this section, we recall the definition of L-fuzzy filters and give their some new properties.

Definition 2

(cf. [7]). Let M be an $R_0$-algebra and L a lattice. An L-fuzzy subset $\mathscr {A}$ on M is said to be an L-fuzzy filter of M, if it satisfies the following conditions:

• (LF1) $\mathscr {A}(1)\geqslant \mathscr {A}(a)$ for all $a\in {M}$ ;

• (LF2) $\mathscr {A}(b)\geqslant {\mathscr {A}}(a)\wedge {\mathscr {A}}(a\rightarrow {b})$ for all $a, b\in {M}$.

The set of all L-fuzzy filters of M is denoted by $\mathbf{LFil}(M)$.

Theorem 1

Let M be an $R_0$-algebra, L a lattice and $\mathscr {A}$ an L-fuzzy subset on M. Then $\mathscr {A}\in \mathbf{LFil}(M)$ if and only if it satisfies the following conditions:

• (LF3) $a\leqslant {b}$ implies $\mathscr {A}(b)\geqslant \mathscr {A}(a)$ for all $a,b\in {M}$;

• (LF4) $\mathscr {A}(a\otimes {b})\geqslant \mathscr {A}(a)\wedge \mathscr {A}(b)$ for all $a, b\in {M}$.

Proof

Assume that $\mathscr {A}\in \mathbf{LFil}(M)$. From Theorem 6 in [7], we know that $\mathscr {A}$ satisfies the condition (LF3). Let $a, b\in {M}$, since $a\leqslant {b}\rightarrow (a\otimes {b})$, by (LF2) and (LF3)), we have that $\mathscr {A}(a\otimes {b})\geqslant \mathscr {A}(b)\wedge \mathscr {A}(b\rightarrow (a\otimes {b})) \geqslant \mathscr {A}(a)\wedge \mathscr {A}(b)$. Thus $\mathscr {A}$ also satisfies the condition (LF4). Conversely, Assume that $\mathscr {A}$ satisfies the condition (LF3) and (LF4). since $a\leqslant 1$, by (LF3) we have $\mathscr {A}(1)\geqslant \mathscr {A}(a)$. Thus $\mathscr {A}$ satisfies the condition (LF1). From $a\otimes (a\rightarrow {b})\leqslant {b}$, (LF3) and (LF4), it follows that $\mathscr {A}(b)\geqslant \mathscr {A}(a\otimes (a\rightarrow {b})) \geqslant \mathscr {A}(a)\wedge {\mathscr {A}}(a\rightarrow {b})$. Thus $\mathscr {A}$ satisfies the condition (LF2). Therefore $\mathscr {A}\in \mathbf{LFil}(M)$ by Definition 2.

Definition 3

Let M be an $R_0$-algebra, L a lattice and $\mathscr {A}$ an L-fuzzy subset on M. An L-fuzzy subset $\mathscr {A}^{\lambda }$ on M is defined as follows:

$$\begin{aligned} \mathscr {A}^{\lambda }(a)= {\left\{ \begin{array}{ll} \mathscr {A}(a),&{} a\ne 1,\\ \mathscr {A}(1)\vee \lambda , &{}a=1, \end{array}\right. } \end{aligned}$$

(1)

for all $a\in {M}$, where $\lambda \in {L}$.

Theorem 2

Let M be an $R_0$-algebra, L a lattice and $\mathscr {A}\in \mathbf{LFil}(M)$. Then $\mathscr {A}^{\lambda }\in \mathbf{LFil}(M)$ for all $\lambda \in {L}$.

Proof

Firstly, for all $a, b\in {M}$, let $a\leqslant {b}$, we consider the following two cases:

• (i) Assume that $b=1$. If $a=1$, we have that $\mathscr {A}^{\lambda }(b)=\mathscr {A}(1)\vee \lambda =\mathscr {A}^{\lambda }(a)$. If $a\ne 1$, by using $\mathscr {A}\in \mathbf{LFil}(M)$ and (LF1), we have that $\mathscr {A}^{\lambda }(b)$=$\mathscr {A}(1)\vee \lambda \geqslant \mathscr {A}(1)\geqslant \mathscr {A}(a)=\mathscr {A}^{\lambda }(a).$

• (ii) Assume that $b\ne 1$, then $a\ne 1$. It follows that $\mathscr {A}^{\lambda }(b)=\mathscr {A}(b)\geqslant \mathscr {A}(a)=\mathscr {A}^{\lambda }(a)$ from $\mathscr {A}\in \mathbf{LFil}(M)$ and (LF3).

Summarize above two cases, we conclude that $a\leqslant {b}$ implies $\mathscr {A}^{\lambda }(b)\geqslant \mathscr {A}^{\lambda }(a)$, for all $a, b\in {M}$. That is, $\mathscr {A}^{\lambda }$ satisfies (LF3).

Secondly, for all $a, b\in {M}$, we consider the following two cases:

• (i) Assume that $a\otimes {b}=1$. If $a=b=1$, it is obvious that

  $$\begin{aligned} \mathscr {A}^{\lambda }(a\otimes {b})=\mathscr {A}(1)\vee \lambda =\mathscr {A}^{\lambda }(a)\wedge \mathscr {A}^{\lambda }(b). \end{aligned}$$

If $a=1, b\ne 1$ or $a\ne 1, b=1$, then $a\otimes {b}\ne 1$, it is a contradiction.

If $a\ne 1$ and $b\ne 1$, it follows that $\mathscr {A}^{\lambda }(a)\wedge \mathscr {A}^{\lambda }(b)=\mathscr {A}(a)\wedge \mathscr {A}(b) \leqslant \mathscr {A}(a\otimes {b})=\mathscr {A}(1)\leqslant \mathscr {A}(1)\vee \lambda =\mathscr {A}^{\lambda }(a\otimes {b})$ from $\mathscr {A}\in \mathbf{LFil}(M)$, (LF4) and (1).

• (ii) Assume that $a\otimes {b}\ne 1$. If $a=b=1$, it is obvious a contradiction.

If $a=1, b\ne 1$ or $a\ne 1, b=1$, let’s assume $a=1, b\ne 1$, then $a\otimes {b}=\lnot (1\rightarrow \lnot {b})=b$, and so $\mathscr {A}^{\lambda }(a)\wedge \mathscr {A}^{\lambda }(b)\leqslant \mathscr {A}^{\lambda }(b) =\mathscr {A}(b)=\mathscr {A}(a\otimes {b})=\mathscr {A}^{\lambda }(a\otimes {b}).$

If $a\ne 1$ and $b\ne 1$, it follows that $\mathscr {A}^{\lambda }(a\otimes {b})=\mathscr {A}(a\otimes {b})\geqslant \mathscr {A}(a)\wedge \mathscr {A}(b) =\mathscr {A}^{\lambda }(a)\wedge \mathscr {A}^{\lambda }(b)$ from $\mathscr {A}\in \mathbf{LFil}(M)$ and (LF4).

Summarize above two cases, we conclude that $\mathscr {A}^{\lambda }(a\otimes {b})\geqslant \mathscr {A}^{\lambda }(a)\wedge \mathscr {A}^{\lambda }(b)$, for all $a, b\in {M}$. That is, $\mathscr {A}^{\lambda }$ satisfies (LF4).

Thus it follows that $\mathscr {A}^{\lambda }\in \mathbf{LFil}(M)$ from Theorem 1.

Definition 4

Let M be an $R_0$-algebra, L a lattice and $\mathscr {A}, \mathscr {B}$ two L-fuzzy subsets on M. Defined L-fuzzy subsets $\mathscr {A}^\mathscr {B}$ and $\mathscr {B}^\mathscr {A}$ on M as follows: for all $a\in {M}$,

$$\begin{aligned} \mathscr {A}^{\mathscr {B}}(a)= {\left\{ \begin{array}{ll} \mathscr {A}(a),&{} a\ne 1,\\ \mathscr {A}(1)\vee \mathscr {B}(1), &{}a=1, \end{array}\right. }\ \ \ \text{ and }\ \ \ \mathscr {B}^{\mathscr {A}}(a)= {\left\{ \begin{array}{ll} \mathscr {B}(a),&{} a\ne 1,\\ \mathscr {B}(1)\vee \mathscr {A}(1), &{}a=1. \end{array}\right. } \end{aligned}$$

(2)

Corollary 1

Let M be an $R_0$-algebra, L a lattice and $\mathscr {A}, \mathscr {B}$ two L-fuzzy subsets on M. If $\mathscr {A}, \mathscr {B}\in \mathbf{LFil}(M)$. Then $\mathscr {A}^{\mathscr {B}}, \mathscr {B}^{\mathscr {A}}\in \mathbf{LFil}(M)$.

Definition 5

Let M be an $R_0$-algebra, L a completely lattice and $\mathscr {A}, \mathscr {B}$ two L-fuzzy subsets on M. An L-fuzzy set $\mathscr {A}\uplus \mathscr {B}$ on M is defined as follows: for all $a, x, y\in {M}$,

$$\begin{aligned} (\mathscr {A}\uplus \mathscr {B})(a)=\bigvee \limits _{{x}\otimes {y}\leqslant {a}} \left[ \mathscr {A}(x)\wedge \mathscr {B}(y)\right] . \end{aligned}$$

(3)

Theorem 3

Let M be an $R_0$-algebra, L a completely distributive lattice and $\mathscr {A}, \mathscr {B}$ two L-fuzzy subsets on M. If $\mathscr {A}, \mathscr {B}\in \mathbf{LFil}(M)$. Then $\mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}\in \mathbf{LFil}(M)$.

Proof

Firstly, for all $a, b\in {M}$, let $a\leqslant {b}$, then $\{x\otimes {y}|x\otimes {y}\leqslant {a}\}\subseteq \{x\otimes {y}|x\otimes {y}\leqslant {b}\}$, and so

$$\begin{aligned} \left( \mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}\right) (b) =&\bigvee \limits _{x\otimes {y}\leqslant {b}}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge {\mathscr {B}}^{\mathscr {A}}(y)\right] \\ \geqslant&\bigvee \limits _{x\otimes {y}\leqslant {a}}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge {\mathscr {B}}^{\mathscr {A}}(y)\right] =\left( \mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}\right) (a). \end{aligned}$$

Hence $\mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}$ satisfies (LF3). Secondly, for all $a, b\in {M}$, we have that

and so $\mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}$ also satisfies (LF4). Hence $\mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}\in \mathbf{LFil}(M)$ by Theorem 1.

4 Generated L-fuzzy Filter by an L-fuzzy Subset

In this section, we give the notion of generated L-fuzzy filter by an L-fuzzy subset and establish its representation theorem.

Definition 6

Let M be an $R_0$-algebra, L a lattice and $\mathscr {A}$ an L-fuzzy subset on M. An L-fuzzy filter $\mathscr {B}$ of M is called the generated L-fuzzy filter by $\mathscr {A}$, denoted $\langle {\mathscr {A}}\rangle$, if $\mathscr {A}\Subset \mathscr {B}$ and for any $\mathscr {C}\in \mathbf{LFil}(M)$, $\mathscr {A}\Subset \mathscr {C}$ implies $\mathscr {B}\Subset \mathscr {C}$.

Theorem 4

Let M be an $R_0$-algebra, L a completely distributive lattice and $\mathscr {A}$ an L-fuzzy subset on M. An L-fuzzy subset $\mathscr {B}$ on M is defined as follows:

$$\begin{aligned} \mathscr {B}(a)=\bigvee \left\{ \mathscr {A}(x_{1})\wedge \cdots \mathscr {A}(x_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } {x_1}\otimes \cdots \otimes {x_n}\leqslant {a}\right\} , \end{aligned}$$

(4)

for all $a\in {M}$. Then $\mathscr {B}=\langle {\mathscr {A}}\rangle$.

Proof

Firstly, we prove that $\mathscr {B}\in \mathbf{LFil}(M)$. For all $a, b\in {M}$, let $a\leqslant {b}$. Then

$$\begin{aligned} \mathscr {A}(a) =&\bigvee \left\{ \mathscr {A}(x_{1})\wedge \cdots \mathscr {A}(x_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } {x_1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a}\right\} \\ \leqslant&\bigvee \left\{ \mathscr {A}(x_{1})\wedge \cdots \mathscr {A}(x_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } {x_1}\otimes \cdots \otimes {x_n}\leqslant {b}\right\} =\mathscr {B}(b). \end{aligned}$$

Thus $\mathscr {B}$ satisfies (LF3). Assume that there are $x_{1}, x_{2}, \cdots , x_{n}\in {M}$ and $y_{1}, \cdots , y_{m}\in {M}$ such that ${x_1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a}$ and ${y_1}\otimes {y_2}\otimes \cdots \otimes {y_m}\leqslant {b}$, we have that ${x_1}\otimes {x_2}\otimes \cdots \otimes {x_n}\otimes {y_1}\otimes {y_2}\otimes \cdots \otimes {y_m}\leqslant {a}\otimes {b}$ by (P11). Thus, we can obtain that

$$\begin{aligned}&\mathscr {B}(a)\wedge \mathscr {B}(b)\\ =&\bigvee \left\{ \mathscr {A}(x_{1})\wedge \cdots \mathscr {A}(x_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } {x_1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a}\right\} \\&\ \ \ \ \ \ \wedge \bigvee \left\{ \mathscr {A}(y_{1})\wedge \cdots \mathscr {A}(y_{m})| y_{1}, y_{2}, \cdots , y_{m}\in {M} \text{ and } {y_1}\otimes {y_2}\otimes \cdots \otimes {y_m}\leqslant {b}\right\} \\ =&\bigvee \{\mathscr {A}(x_{1})\wedge \cdots \wedge \mathscr {A}(x_{n})\wedge \mathscr {A}(y_{1})\wedge \cdots \wedge \mathscr {A}(y_{m})| x_{1}, \cdots , x_{n}, y_{1}, \cdots , y_{m}\in {M}\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ such } \text{ that } {x_1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a} \text{ and } {y_1}\otimes {y_2}\otimes \cdots \otimes {y_m}\leqslant {b}\}\\ \leqslant&\bigvee \{\mathscr {A}(x_{1})\wedge \cdots \wedge \mathscr {A}(x_{n})\wedge \mathscr {A}(y_{1})\wedge \cdots \wedge \mathscr {A}(y_{m})| x_{1}, \cdots , x_{n}, y_{1}, \cdots , y_{m}\in {M}\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ such } \text{ that } {x_1}\otimes {x_2}\otimes \cdots \otimes {x_n}\otimes {y_1}\otimes {y_2}\otimes \cdots \otimes {y_m}\leqslant {a}\otimes {b}\}\\ \leqslant&\bigvee \left\{ \mathscr {A}(z_{1})\wedge \cdots \mathscr {A}(z_{k})| z_{1}, z_{2}, \cdots , z_{k}\in {M} \text{ and } {z_1}\otimes \cdots \otimes {z_k}\leqslant {a}\otimes {b}\right\} \\ =&\mathscr {B}(a\otimes {b}). \end{aligned}$$

Hence $\mathscr {B}$ also satisfies (LF4). It follows from Theorem 1 that $\mathscr {B}\in \mathbf{LFil}(M)$.

Secondly, For any $a\in {M}$, it follows from $a\leqslant {a}$ and the definition of $\mathscr {B}$ that $\mathscr {A}(a)\leqslant \mathscr {B}(a)$. This means that $\mathscr {A}\Subset \mathscr {B}$.

Finally, assume that $\mathscr {C}\in \mathbf{LFil}(M)$ with $\mathscr {A}\Subset \mathscr {C}$. Then for any $a\in {M}$, we have

$$\begin{aligned} \mathscr {B}(a) =&\bigvee \left\{ \mathscr {A}(x_{1})\wedge \cdots \wedge \mathscr {A}(x_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } x_{1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a}\right\} \\ \leqslant&\bigvee \left\{ \mathscr {C}(x_{1})\wedge \cdots \wedge \mathscr {C}(x_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } x_{1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a}\right\} \\ \leqslant&\bigvee \left\{ \mathscr {C}(x_{1}\otimes \cdots \otimes {x}_{n})| x_{1}, x_{2}, \cdots , x_{n}\in {M} \text{ and } x_{1}\otimes {x_2}\otimes \cdots \otimes {x_n}\leqslant {a}\right\} \\ \leqslant&\bigvee \left\{ \mathscr {C}(a)\right\} =\mathscr {C}(a). \end{aligned}$$

Hence $\mathscr {B}\Subset \mathscr {C}$ holds. To sum up, we have that $\mathscr {B}=\langle {\mathscr {A}}\rangle$.

Example 1

Let $M=\{0, a, b, c, d, 1\}$, $\lnot 0=1, \lnot {a}=c, \lnot {b}=d, \lnot {c}=a, \lnot {d}=b, \lnot 1=0$, the Hasse diagram of lattice $(M, \vee , \wedge , \leqslant )$ be defined as Fig. 1, and the binary operator $\rightarrow$ of M be defined as Table 1.

Fig. 1.

The Hasse diagram of M

Table 1. Def. of “$\rightarrow$”

Then $(M, \lnot , \vee , \rightarrow , 1)$ is an $R_0$-algebra. Take $L=([0,1], \max , \min )$ and define an [0,1]-fuzzy subset $\mathscr {A}$ on M by $\mathscr {A}(1)=\mathscr {A}(c)=\alpha , \mathscr {A}(a)=\mathscr {A}(b)=\mathscr {A}(d)=\mathscr {A}(0)=\beta$, $0\leqslant \beta <\alpha \leqslant 1$. Since $c\leqslant {b}$ but $\mathscr {A}(b)=\beta \not \geqslant \alpha =\mathscr {A}(c)$, we know that $\mathscr {A}\not \in \mathbf{LFil}(M)$. It is easy to verify that $\langle {\mathscr {A}}\rangle \in \mathbf{LFil}(M)$ from Theorem 4, where $\langle {\mathscr {A}}\rangle (1)=\langle {\mathscr {A}}\rangle (b)=\langle {\mathscr {A}}\rangle (c)=\alpha , \langle {\mathscr {A}}\rangle (a)=\langle {\mathscr {A}}\rangle (d)=\langle {\mathscr {A}}\rangle (0)=\beta$.

5 The Lattice of L-fuzzy Filters in a Given $R_0$-algebra

In this section, we investigate the lattice structural feature of the set $\mathbf{LFil}(M)$ under the L-fuzzy set-inclusion order $\Subset$.

Theorem 5

Let M be an $R_0$-algebra and L a complete lattice. Then $(\mathbf{LFil}(M), \Subset )$ is a complete lattice.

Proof

For any $\{\mathscr {A}_{\alpha }\}_{\alpha \in \Lambda }\subseteq \mathbf{LFil}(M)$, where $\Lambda$ is an indexed set. It is easy to verify that $\Cap _{\alpha \in \Lambda }\mathscr {A}_{\alpha }\in \mathbf{LFil}(M)$ is infimum of $\{\mathscr {A}_{\alpha }\}_{\alpha \in \Lambda }$, where $\left( \Cap _{\alpha \in \Lambda }\mathscr {A}_{\alpha }\right) (a)=\bigwedge \limits _{\alpha \in \Lambda }\mathscr {A}_{\alpha }(a)$ for all $a\in {M}$. i.e., $\bigwedge \limits _{\alpha \in \Lambda }\mathscr {A}_{\alpha }=\Cap _{\alpha \in \Lambda }\mathscr {A}_{\alpha }$. Define $\Cup _{\alpha \in \Lambda }\mathscr {A}_{\alpha }$ such that $\left( \Cup _{\alpha \in \Lambda }\mathscr {A}_{\alpha }\right) (a)=\bigvee \limits _{\alpha \in \Lambda }\mathscr {A}_{\alpha }(a)$ for all $a\in {M}$. Then $\left\langle \Cup _{\alpha \in \Lambda }\mathscr {A}_{\alpha }\right\rangle$ is supermun of $\{\mathscr {A}_{\alpha }\}_{\alpha \in \Lambda }$, where $\left\langle \Cup _{\alpha \in \Lambda }\mathscr {A}_{\alpha }\right\rangle$ is the L-fuzzy filter generated by $\Cup _{\alpha \in \Lambda }\mathscr {A}_{\alpha }$ of M. i.e., $\bigvee \limits _{\alpha \in \Lambda }\mathscr {A}_{\alpha }=\left\langle \Cup _{\alpha \in \Lambda }\mathscr {A}_{\alpha }\right\rangle$. Therefor $(\mathbf{LFil}(M), \Subset )$ is a complete lattice. The proof is completed.

Remark 1

Let M be an $R_0$-algebra and L a complete lattice. For all $\mathscr {A}, \mathscr {B}\in \mathbf{LFil}(M)$, by Theorem 5 we know that $\mathscr {A}\wedge \mathscr {B}=\mathscr {A}\Cap \mathscr {B}$ and $\mathscr {A}\vee \mathscr {B}=\left\langle {\mathscr {A}\Cup \mathscr {B}}\right\rangle$.

Theorem 6

Let M be an $R_0$-algebra and L a completely distributive lattice. Then for all $\mathscr {A}, \mathscr {B}\in \mathbf{LFil}(M)$, $\mathscr {A}\vee \mathscr {B}=\left\langle {\mathscr {A}\Cup \mathscr {B}}\right\rangle =\mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}$ in the complete lattice $(\mathbf{LFil}(M), \Subset )$.

Proof

For all $\mathscr {A}, \mathscr {B}\in \mathbf{LFil}(M)$, it is obvious that $\mathscr {A}\Subset \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}$ and $\mathscr {B}\Subset \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}$, that is, $\mathscr {A}(a)\leqslant \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) (a)$ and $\mathscr {B}(a)\leqslant \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) (a)$ for all $a\in {M}$. Thus$(\mathscr {A}\Cup \mathscr {B})(a)=\mathscr {A}(a)\vee \mathscr {B}(a) \leqslant \left( \mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}\right) (a)$, that is, $\mathscr {A}\Cup \mathscr {B}\Subset \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}$, and thus $\left\langle {\mathscr {A}\Cup \mathscr {B}}\right\rangle \Subset \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\in \mathbf{LFil}(M)$ by Theorem 3. Let $\mathscr {C}\in \mathbf{LFil}(M)$ such that $\mathscr {A}\Cup \mathscr {B}\Subset \mathscr {C}$. For all $a\in {M}$, we consider the following two cases:

• (i) If $a=1$, then $\left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) (1) =\bigvee \limits _{x\otimes {y}\leqslant 1}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] =\mathscr {A}^{\mathscr {B}}(1)\wedge \mathscr {B}^{\mathscr {A}}(1) =\mathscr {A}(1)\vee \mathscr {B}(1) =\left( \mathscr {A}\Cup \mathscr {B}\right) (1)\leqslant \mathscr {C}(1)$.

• (ii) If $a<1$, then we have

  $$\begin{aligned}&\left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) (a) =\bigvee \limits _{x\otimes {y}\leqslant {a}}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1} \left[ \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] \vee \bigvee \limits _{x\leqslant {a}}\left\{ \mathscr {A}(x)\wedge \left[ \mathscr {A}(1)\vee \mathscr {B}(1)\right] \right\} \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vee \bigvee \limits _{y\leqslant {a}}\left\{ \left[ \mathscr {A}(1)\vee \mathscr {B}(1)\right] \wedge \mathscr {B}(y)\right\} \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] \vee \left[ \bigvee \limits _{x\leqslant {a}}\mathscr {A}(x)\right] \vee \left[ \bigvee \limits _{y\leqslant {a}}\mathscr {B}(y)\right] \\ \leqslant&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1}\left[ \mathscr {C}(x)\wedge \mathscr {C}(y)\right] \vee \left[ \bigvee \limits _{x\leqslant {a}}\mathscr {C}(x)\right] \vee \left[ \bigvee \limits _{y\leqslant {a}}\mathscr {C}(y)\right] \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}}\left[ \mathscr {C}(x)\wedge \mathscr {C}(y)\right] \leqslant \bigvee \limits _{x\otimes {y}\leqslant {a}}\mathscr {C}(x\otimes {y}) \leqslant \mathscr {C}(a), \end{aligned}$$

  thus $\mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\Subset \mathscr {C}$ for above two cases.

By Definition 6 and Theorem 4 we have that $\mathscr {A}\vee \mathscr {B}=\left\langle {\mathscr {A}\Cup \mathscr {B}}\right\rangle =\mathscr {A}^{\mathscr {B}}\uplus {\mathscr {B}}^{\mathscr {A}}$.

Theorem 7

Let M be an $R_0$-algebra and L a completely distributive lattice. Then $(\mathbf{LFil}(M), \Subset )$ is a distributive lattice, where, $\mathscr {A}\wedge \mathscr {B} =\mathscr {A}\Cap \mathscr {B}$ and $\mathscr {A}\vee \mathscr {B}=\left\langle {\mathscr {A}\Cup \mathscr {B}}\right\rangle$, for all $\mathscr {A}, \mathscr {B}\in \mathbf{LFil}(M)$.

Proof

To finish the proof, it suffices to show that $\mathscr {C}\wedge \left( \mathscr {A}\vee \mathscr {B}\right) = \left( \mathscr {C}\wedge \mathscr {A}\right) \vee \left( \mathscr {C}\wedge \mathscr {B}\right)$, for all $\mathscr {A}, \mathscr {B}, \mathscr {C}\in \mathbf{LFil}(M)$. Since the inequality $\left( \mathscr {C}\wedge \mathscr {A}\right) \vee \left( \mathscr {C}\wedge \mathscr {B}\right) \Subset \mathscr {C}\wedge \left( \mathscr {A}\vee \mathscr {B}\right)$ holds automatically in a lattice, we need only to show the inequality $\mathscr {C}\wedge \left( \mathscr {A}\vee \mathscr {B}\right) \Subset \left( \mathscr {C}\wedge \mathscr {A}\right) \vee \left( \mathscr {C}\wedge \mathscr {B}\right)$. i.e., we need only to show that $\left( \mathscr {C}\Cap \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) \right) (a)\leqslant \left( \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}} \uplus \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}\right) (a)$, for all $a\in {M}$. For these, we consider the following two cases:

• (i) If $a=1$, we have

  $$\begin{aligned}&\left( \mathscr {C}\Cap \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) \right) (1) =\mathscr {C}(1)\wedge \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) (1)\\ =&\mathscr {C}(1)\wedge \bigvee \limits _{x\otimes {y}\leqslant 1}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] =\mathscr {C}(1)\wedge \left[ \mathscr {A}^{\mathscr {B}}(1)\wedge \mathscr {B}^{\mathscr {A}}(1)\right] \\ =&\left[ \mathscr {C}(1)\wedge \mathscr {A}(1)\right] \vee \left[ \mathscr {C}(1)\wedge \mathscr {B}(1)\right] =\left( \mathscr {C}\Cap \mathscr {A}\right) (1)\vee \left( \mathscr {C}\Cap \mathscr {B}\right) (1)\\ =&\left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(1)\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(1)\\ =&\bigvee \limits _{x\otimes {y}\leqslant 1}\left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(x)\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(y)\right] \\ =&\left( \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}} \uplus \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}\right) (1). \end{aligned}$$

• (ii) If $a<1$, we have

  $$\begin{aligned}&\left( \mathscr {C}\Cap \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) \right) (a) =\mathscr {C}(a)\wedge \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) (a)\\ =&\mathscr {C}(a)\wedge \bigvee \limits _{x\otimes {y}\leqslant {a}}\left[ \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] =\bigvee \limits _{x\otimes {y}\leqslant {a}}\left[ \mathscr {C}(a)\wedge \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1} \left[ \mathscr {C}(a)\wedge \mathscr {A}^{\mathscr {B}}(x)\wedge \mathscr {B}^{\mathscr {A}}(y)\right] \vee \\&\ \ \ \ \ \ \ \ \ \bigvee \limits _{y\leqslant {a}}\left[ \mathscr {C}(a)\wedge \mathscr {A}^{\mathscr {B}}(1)\wedge \mathscr {B}(y)\right] \vee \bigvee \limits _{x\leqslant {a}}\left[ \mathscr {C}(a)\wedge \mathscr {A}(x)\wedge \mathscr {B}^{\mathscr {A}}(1)\right] \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1} \left\{ \left[ \mathscr {C}(a)\wedge \mathscr {A}(x)\right] \wedge \left[ \mathscr {C}(a)\wedge \mathscr {B}(y)\right] \right\} \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{y\leqslant {a}} \left\{ \left[ \mathscr {C}(a)\wedge \mathscr {A}^{\mathscr {B}}(1)\right] \wedge \left[ \mathscr {C}(a)\wedge \mathscr {B}(y)\right] \right\} \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{x\leqslant {a}} \left\{ \left[ \mathscr {C}(a)\wedge \mathscr {A}(x)\right] \wedge \left[ \mathscr {C}(a)\wedge \mathscr {B}^{\mathscr {A}}(1)\right] \right\} \\ \leqslant&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1} \left\{ \left[ \mathscr {C}(a\vee {x})\wedge \mathscr {A}(a\vee {x})\right] \wedge \left[ \mathscr {C}(a\vee {y})\wedge \mathscr {B}(a\vee {y})\right] \right\} \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{y\leqslant {a}} \left\{ \left[ \mathscr {C}(1)\wedge \left( \mathscr {A}(1)\vee \mathscr {B}(1)\right) \right] \wedge \left[ \mathscr {C}(a\vee {y})\wedge \mathscr {B}(a\vee {y})\right] \right\} \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{x\leqslant {a}} \left\{ \left[ \mathscr {C}(a\vee {x})\wedge \mathscr {A}(a\vee {x})\right] \wedge \left[ \mathscr {C}(1)\wedge \left( \mathscr {B}(1)\vee \mathscr {A}(1)\right) \right] \right\} \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) (a\vee {x})\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) (a\vee {y})\right] \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{y\leqslant {a}} \left\{ \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) (1)\vee \left( \mathscr {C}\Cap \mathscr {B}\right) (1) \right] \wedge \left( \mathscr {C}\Cap \mathscr {B}\right) (a\vee {y})\right\} \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{x\leqslant {a}} \left\{ \left( \mathscr {C}\Cap \mathscr {A}\right) (a\vee {x})\wedge \left[ \left( \mathscr {C}\Cap \mathscr {B}\right) (1)\vee \left( \mathscr {C}\Cap \mathscr {A}\right) (1)\right] \right\} \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}, x\ne 1, y\ne 1} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(a\vee {x})\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(a\vee {y})\right] \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{y\leqslant {a}} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(1)\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(a\vee {y})\right] \vee \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigvee \limits _{x\leqslant {a}} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(a\vee {x})\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(1)\right] \\ =&\bigvee \limits _{x\otimes {y}\leqslant {a}} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(a\vee {x})\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(a\vee {y})\right] . \end{aligned}$$

Let $a\vee {x}=u$ and $a\vee {y}=v$, since $x\otimes {y}\leqslant {a}$, using Lemma 2 we get that

$$\begin{aligned} u\otimes {v}=(a\vee {x})\otimes (a\vee {y}) =&((a\vee {x})\otimes {a})\vee ((a\vee {x})\otimes {y})\\ =&(a\otimes {a})\vee (a\otimes {x})\vee (a\otimes {y})\vee (x\otimes {y})\\ \leqslant&{a}\vee {a}\vee {a}\vee (x\otimes {y})\\ =&a\vee (x\otimes {y}) \leqslant {a}\vee {a}=a. \end{aligned}$$

Hence we can conclude that

$$\begin{aligned} \left( \mathscr {C}\Cap \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) \right) (a) \leqslant&\bigvee \limits _{x\otimes {y}\leqslant {a}} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(a\vee {x})\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(a\vee {y})\right] \\ \leqslant&\bigvee \limits _{u\otimes {v}\leqslant {a}} \left[ \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}}(u)\wedge \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}(v)\right] \\ =&\left( \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}} \uplus \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}\right) (a). \end{aligned}$$

To sum up, we have that

$$\left( \mathscr {C}\Cap \left( \mathscr {A}^{\mathscr {B}}\uplus \mathscr {B}^{\mathscr {A}}\right) \right) (a)\leqslant \left( \left( \mathscr {C}\Cap \mathscr {A}\right) ^{\mathscr {C}\Cap \mathscr {B}} \uplus \left( \mathscr {C}\Cap \mathscr {B}\right) ^{\mathscr {C}\Cap \mathscr {A}}\right) (a),$$

for all $a\in {M}$. The proof is completed.

6 Conclusion

As well known, filters is an important concept for studying the structural features of $R_0$-algebras. In this paper, the L-fuzzy filter theory in $R_0$-algebras is further studied. Some new properties of L-fuzzy filters are given. Representation theorem of L-fuzzy filter which is generated by an L-fuzzy subset is established. It is proved that the set consisting of all L-fuzzy filters in a given $R_0$-algebra, under the L-fuzzy set-inclusion order $\Subset$, forms a complete distributive lattice. Results obtained in this paper not only enrich the content of L-fuzzy filters theory in $R_0$-algebras, but also show interactions of algebraic technique and L-fuzzy sets method in the studying of logic problems. We hope that more links of fuzzy sets and logics emerge by the stipulating of this work.