Prime fuzzy (implicative) MP-filters of R 0 -algebras

Abstract

The filter, also called a “deductive systems”, is the basic tool in the study of logical algebras and the completeness of the corresponding non-classical logics. The aim of this paper is to investigate prime fuzzy MP-filters and prime fuzzy implicative MP-filters of R₀-algebras. In the absence of the condition that $\tilde{A}$ is a fuzzy filter, the equivalent conditions of $\tilde{A}$ being a (prime) fuzzy implicative filter which it necessitates to satisfy are concretely investigated and whose some related properties are specifically enumerated. Meanwhile, we also point out a little mistake occurred in Liu and Li’s paper.

Keywords

Filters fuzzy MP-filters prime fuzzy (implicative) MP-filters

1 Introduction

In non-classical logics, residuated lattice introduced by Ward and Dilworth [11] in 1939, was not only an important and fundamental algebraic structure, but also a popular theory and method. Based on this, some related researches have been prosperously developed. The MTL -algebra presented by Esteva and Godo [3], an algebra induced by using a left continuous t-norm and its corresponding residuum, is undoubtedly one of the most important derivatives of it, by adding the law of prelinearity [denoted by (a → b) ∨ (b → a) =1]. The corresponding of that is, BL -algebras presented by Hájek [4] in 1998, an algebra system induced by using a continuous t-norm and its corresponding residuum. It also can be viewed a formation which was attached the law of divisibility [denoted by a ∧ b = a ⊗ (a → b)] to MTL -algebras. On the other hand, the MV -algebra proposed by Chang [2] in 1958, was an algebraic formation in processing the completeness of axiom systems of Lukasiewicz’s many-valued logics. Of course, it is also one of the most well known classes of BL -algebras. Furthermore, definitions of different algbras systems, such as fuzzy implication algebras by Wu [12], lattice implication algebras by Xu [13] and R₀-algebras by Wang [10] etc, have different forms. Interestingly, Pei [9] proved that R₀-algebras are equivalent to the logic systems NM, which are achieved by attaching the condition [denoted by (a ⊗ b → 0) ∨ (a ∧ b → a ⊗ b) =1] to IMTL -algebras, which are achieved by attaching the law of reflexivity [denoted by (a → 0) →0 = a] to MTL -algebras. Thus, it shall show that an R₀-algebra is, in particular, an MTL -algebra in which its t-norm ⊗₀ is a nilpotent minimum t-norm.

The filter theory of the logical algebras plays a key role in studying these algebras and the completeness of the corresponding non-classical logics. Since a filter is also viewed as a “deductive systems”, which applies some specific inference rules (also called Modus Ponens rules), we take it as an MP -filter, especially, in R₀-algebras. Introduction of the fuzzy MP -filter aims to study the MP -filter by the methodes of fuzzy mathematics, which riches and improves the filter theory. There are many significant results have been achieved successfully. For instance, Zhang et al. [17] investigated IMTL(MV)-filters and fuzzy IMTL(MV)-filters of residuated lattices. Liu and Li [8] proved that fuzzy implicative filters and fuzzy Boolean filters coincide in R₀-algebras. In [16], Zhang and Jun gave the Prime Filter Theorem in R₀-algebras and BL -algebras. Meanwhile, some researchers introduced several special types of filters, such as the regular filter, the fantastic filter, the associative filter and it’s likes. A notion of a t-filter was introduced by Víta [6] on residuated lattices which is a generalization of several special types of filters. Interestingly, Haveshki in [5] proved the notion of associative filter proposed recently by Borzooei et al. [1] is useless. As a consequence, he put forward an open problem: Is there a suitable and useful definition of associative filter in MTL -algebras (residuated lattices)? To solve the problem, the answer can only to be found from the original notion of the filter itself. we understand humbly that the concept of the filter can be depicted by the rule of transitivity. Naturally, the reason of our study to the filter theory focused on R₀-algebras is that, it not only has a good linearity, but also a good transitivity, for example, yet which is not possessed by Lukasiewicz’s implicative operation, even they have the same background of many-valued logics. Moreover, the study of primer and maximality can help us understand the conception of the filter.

The remainder of the paper is organized as follows. Section 2 briefly reviews some basic concepts and results of R₀-algebras. Section 3 investigates prime (fuzzy) MP-filters and their related properties. In Section 4, we introduce the definition of prime (fuzzy) implicative MP-filters and investigate their some related properties. Finally, Section 5 concludes this paper.

2 Preliminaries

Assume that L be a structure of (¬ , ∨ , →) -type algebra, whose ¬, ∨ and → are logic connective. If there exists partial order ≤ such that the partial order set (L, ≤) is a bounded lattice, → is a binary implication operator (also called pseudocomplement, for instance, a maximal element is defined as the pseudocomplement of a relevant to b if the maximal element of set {c∈ L|a ∧ c ≤ b } exists and is denoted a → b). ∨ is a binary supremum operation relevant to order ≤ and ¬ is an unary involution negation operation that ¬a : = a → 0.

In this section, we recall some basic concepts and results, which we shall need in the subsequentsections.

Definition 2.1. [18] An algebraic structure L is a weak R₀-algebra if it satisfies the following axioms: (∀ a, b, c ∈ L)

¬b → ¬ a = a → b;

1 → a = a, a → a = 1, in which 1 is the greatest element of (L, ≤);

a→ b ≤ (c → a) → (c → b) ;

a → (b → c) = b → (a → c);

a → (b ∨ c) = (a → b) ∨ (a → c);

a → (b ∧ c) = (a → b) ∧ (a → c).

Also, if L is a weak R₀-algebra, and it satisfies the following axiom

(a → b) ∨ ((a → b) → ¬ a ∨ b) = 1

for any a, b ∈ L, then L is an R₀-algebra.

Note 2.2. The definition of weak R₀-algebra was put forward by Wang [10] in which it shall also satisfy the properties of bounded distributive lattice, whereas those conditions can be deduced by conditions(R₁) - (R₆).

Proposition 2.3. A weak R₀-algebra is a bounded distributive lattice.

Proof. Since ¬1 =0, it is easily shown that a weak R₀-algebra is bounded. Notice that a ∧ (b ∨ c) ≥ (a ∧ b) ∨ (a ∧ c) is straight, we only verify that a ∧ (b ∨ c) ≤ (a ∧ b) ∨ (a ∧ c). In fact, $\begin{matrix} a \land (b \lor c) \to [(a \land b) \lor (a \land c)] \\ = [a \land (b \lor c) \to (a \land b)] \lor [a \land (b \lor c) \to (a \land c)] \\ = [(a \land (b \lor c) \to a) \land (a \land (b \lor c) \to b)] \lor [(a \\ \land (b \lor c) \to a) \land (a \land (b \lor c) \to c)] \\ = [1 \land (a \land (b \lor c) \to b)] \lor [1 \land (a \land (b \lor c) \to c)] \\ = [a \land (b \lor c) \to b] \lor [a \land (b \lor c) \to c] \\ = a \land (b \lor c) \to (b \lor c) = 1 . \end{matrix}$

This completes the proof of a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c). Similar to the above argument, we have a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) . Hence, a weak R₀-algebra is a bounded distributive lattice. □

Example 2.4. Consider the bounded distributive lattice L = {0, a, b, c, 1} with the partial order≤ defined by 0 < a < b < c < 1, whose ¬ and → are defined, respectively, as follows:

¬			→	0	a	b	c	1
0	1	and	0	1	1	1	1	1
a	c		a	c	1	1	1	1
b	b		b	b	c	1	1	1
c	a		c	a	b	b	1	1
1	0		1	0	a	b	c	1

Then {L, ∧ , ∨ , ¬ , →} is an R₀-algebra. Define a fuzzy subset $\tilde{A}$ in L by $\tilde{A} (1) = \tilde{A} (c) = s$ , $\tilde{A} (0) = \tilde{A} (a) = \tilde{A} (b) = t$ (0 ≤ t < s ≤ 1). It is easily checked that $\tilde{A}$ is an PM -filter. (In the subsequent Section 3, we will discourse the relatedconcepts.)

Proposition 2.5. [18] Let L be an R₀-algebra. Then the following properties hold: (∀ a, b, c ∈ L)

a → b = 1 iff a ≤ b;

a ≤ a ∧ b → b;

a → b ≤ (b → c) → (a → c);

a ≤ b → c iff b ≤ a → c;

a ∨ b → c = (a → c) ∧ (b → c) , a ∧ b → c = (a → c) ∨ (b → c);

If b ≤ c, then a → b ≤ a → c; Ifa ≤ b, then b → c ≤ a → c;

a → b ≥ ¬ a ∨ b;

(a → b) ∨ (b → a) =1;

a ∧ ¬ a ≤ b ∨ ¬ b;

a → (b → a) =1;

((a → b) → b) → b = a → b;

a ∨ b ≤ ((a → b) → b) ∧ ((b → a) → a);

a → b ≤ (a ∨ c) → (b ∨ c), a → b ≤ (a ∧ c) → (b ∧ c);

a → b ≤ (a → c) ∨ (c → b) .

Proposition 2.6. [18] Let L be an R₀-algebra. If a ⊗ b = ¬ (a → ¬ b) (∀ a, b ∈ L), then the following properties hold:

(L, ⊗ , 1) is commutative monoid and in which 1 is the identity element;

a ⊗ b ≤ a ∧ b;

a ⊗ b ≤ c iff a ≤ b → c;

a ⊗ b → c = a → (b → c) , a → (b → (a ⊗ b)) =1;

a ⊗ ¬ a = 0;

a ⊗ (a → b) ≤ b;

b ≤ a → (a ⊗ b) , a ≤ b → (a ⊗ b);

(a → b) ⊗ (b → c) ≤ a → c.

3 Fuzzy MP-filters and prime fuzzy MP-filters

3.1 FMP-filters

Definition 3.1. [8] Let L be an R₀-algebra. A subset F of L is called an MP -filter if for all a, b ∈ L

1∈ F ;

a ∈ F, a → b ∈ F ⇒ b ∈ F .

Definition 3.2. Let L be an R₀-algebra. If $\tilde{A} : L \to [0, 1]_{R_{0}}$ is a mapping of L to R₀- unit interval, then $\tilde{A}$ is called a fuzzy set on R₀-algebra L. Meanwhile, $F (L)$ denotes universal set of R₀-algebras.

Definition 3.3. [8] Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . If the following hold:

$\tilde{A} (1) \geq \tilde{A} (a), \forall a \in L;$

$\tilde{A} (b) \geq \tilde{A} (a) \land \tilde{A} (a \to b), \forall a, b \in L,$

then

\tilde{A}

is called a fuzzy MP -filter (denoted as the FMP -filter for convenience).

Proposition 3.4. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . Then the following statements are equivalent: (∀ a, b ∈ L)

$\tilde{A}$ is an FMP -filter.

$\tilde{A} (a \otimes b) \geq \tilde{A} (a) \land \tilde{A} (b)$ and $\tilde{A}$ is monotonely nondecreasing.

$\tilde{A} (a \otimes b) = \tilde{A} (a) \land \tilde{A} (b)$ and $\tilde{A} (a \land b) = \tilde{A} (a) \land \tilde{A} (b) .$

Proof. (1) ⇒ (2) Assume that $\tilde{A}$ is an FMP -filter. Firstly, we verify that $\tilde{A}$ is monotonely nondecreasing. For all a, b ∈ L, a ≤ b, it follows from (p₁) that a → b = 1 . Then it follows from (FM₁) and (FM₂) that $\tilde{A} (b) \geq \tilde{A} (a) \land \tilde{A} (a \to b) = \tilde{A} (a) \land \tilde{A} (1) = \tilde{A} (a) .$ So, $\tilde{A}$ is monotonely nondecreasing. Then, it follows from (FM₂) that $\tilde{A} (a \otimes b) \geq \tilde{A} (a) \land \tilde{A} (a \to (a \otimes b)),$ and by (p₂₁), we have $\tilde{A} (a \to (a \otimes b)) \geq \tilde{A} (b)$ . Thus, $\tilde{A} (a \otimes b) \geq \tilde{A} (a) \land \tilde{A} (b)$ .

(2) ⇒ (1) Suppose that condition (2) is valid, we consider the $\tilde{A}$ is an FMP -filter. In fact, $\tilde{A}$ is monotonely nondecreasing, it is easily checked that (FM₁) holds. According to condition (2), $\tilde{A} (a \otimes (a \to b)) \geq \tilde{A} (a) \land \tilde{A} (a \to b),$ then by (p₂₀), we have $\tilde{A} (b) \geq \tilde{A} (a \otimes (a \to b)) .$ Thus, $\tilde{A} (b) \geq \tilde{A} (a) \land \tilde{A} (a \to b)$ , that is to say, (FM₂) holds. So, the $\tilde{A}$ is an FMP filter.

(1) ⇒ (3) Assume that $\tilde{A}$ is an FMP -filter. On the one hand, we know that $\tilde{A} (a \otimes b) \geq \tilde{A} (a) \land \tilde{A} (b)$ from condition (2). On the other hand, by (p₁₆), we have a ⊗ b ≤ a, a ⊗ b ≤ b. i.e., $\tilde{A} (a \otimes b) \leq \tilde{A} (a), \tilde{A} (a \otimes b) \leq \tilde{A} (b)$ . Then $\tilde{A} (a \otimes b) \leq \tilde{A} (a) \land \tilde{A} (b)$ . Hence, $\tilde{A} (a \otimes b) = \tilde{A} (a) \land \tilde{A} (b)$ . Meanwhile, since $\tilde{A}$ is monotonely nondecreasing, we get $\tilde{A} (a \land b) \leq \tilde{A} (a) \land \tilde{A} (b)$ . Then using (p₁₆), we note that $\tilde{A} (a \land b) \geq \tilde{A} (a \otimes b) = \tilde{A} (a) \land \tilde{A} (b)$ . Thus, $\tilde{A} (a \land b) = \tilde{A} (a) \land \tilde{A} (b)$ .

(3) ⇒ (2) Suppose that condition (3) is valid, it is clearly noted that $\tilde{A} (a \land b) \geq \tilde{A} (a \otimes b)$ , then using (p₁₆), we can verify that the $\tilde{A}$ is monotonely nondecreasing. □

Definition 3.5. [15] Let L be a lattice. A mapping ϱ : L → L is defined the ∧ -morphism, if it satisfies ϱ (a ∧ b) = ϱ (a) ∧ ϱ (b) (∀ a, b ∈ L); Dually, the ∨ -morphism is defined if it satisfies ϱ (a ∨ b) = ϱ (a) ∨ ϱ (b) . we call ϱ is a lattice homomorphism if it satisfies both the ∧ -morphism and ∨ -morphism.

Corollary 3.6. If $\tilde{A}$ is an FMP -filter on R₀-algebra L, then $\tilde{A}$ is a ∧ -morphism.

Proposition 3.7. Let L be an R₀-algebra, $\tilde{A} \in F (L)$ . Then the following statements are equivalent: (∀ a, b, c ∈ L)

$\tilde{A}$ is an FMP -filter.

a ≤ b → cimply $\tilde{A} (c) \geq \tilde{A} (a) \land \tilde{A} (b) .$

$\tilde{A} (a \to b) \geq \tilde{A} (b)$ and $\tilde{A} (a \to c) \geq \tilde{A} (a \to b) \land \tilde{A} (b \to c) .$

Proof. (1) ⇒ (2) Assume that $\tilde{A}$ is an FMP -filter. If a ≤ b → c, it follows from (p₁₇) and Proposition 3.4(2) that $\tilde{A} (c) \geq \tilde{A} (a \otimes b) \geq \tilde{A} (a) \land \tilde{A} (b)$ . Hence, (2) holds.

(2) ⇒ (1) By (p₂₁) and condition (2), we get $\tilde{A} (a \otimes b) \geq \tilde{A} (a) \land \tilde{A} (b)$ . According to Proposition 3.4(2), we only need to verify that $\tilde{A}$ is monotonely nondecreasing. In fact, using (p₂) and (2), we have $\tilde{A} (b) \geq \tilde{A} (a) \land \tilde{A} (a \land b)$ . Let a ≤ b, it means that $\tilde{A} (a) \leq \tilde{A} (b)$ . Thus $\tilde{A}$ is an FMP -filter.

(1) ⇒ (3) Assume that $\tilde{A}$ is an FMP -filter. On the one hand, according to Definition 3.3(1) and (p₁₀), we have $\tilde{A} (a \to (b \to a)) = \tilde{A} (1) \geq \tilde{A} (b \to a)$ . It shows that $\tilde{A} (a \to b) \geq \tilde{A} (b)$ holds. On the other hand, it follows from (p₃) that $\tilde{A} ((b \to c) \to (a \to c)) \geq \tilde{A} (a \to b)$ . Then using Definition 3.3(2), we have $\tilde{A} (a \to c) \geq \tilde{A} (b \to c) \land \tilde{A} ((b \to c) \to (a \to c)) \geq \tilde{A} (b \to c) \land \tilde{A} (a \to b) .$

(3) ⇒ (1) In condition (3) taking a = 1, b = a and c = b, it is shown that $\tilde{A} (1 \to b) \geq \tilde{A} (1 \to a) \land \tilde{A} (a \to b)$ , i.e., $\tilde{A} (b) \geq \tilde{A} (a) \land \tilde{A} (a \to b)$ holds. Then $\forall a, b \in L, \tilde{A} (a \to (b \to a)) \geq \tilde{A} (b \to a)$ means that $\tilde{A} (1) \geq \tilde{A} (a)$ . □

Proposition 3.8. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . Then the following statements are equivalent:

$\tilde{A}$ is an FMP -filter.

If α- cut set ${\tilde{A}}_{α} = {a \in L | \tilde{A} (a) \geq α} \neq \emptyset$ (∀ α ∈ [0, 1]), then ${\tilde{A}}_{α}$ is an MP -filter.

∀a ∈ L, ${\tilde{A}}_{α_{0}}$ is an MP -filter (where $α_{0} = \tilde{A} (a)$ ).

Proof. (1) ⇒ (2) Assume that $\tilde{A}$ is an FMP -filter, α ∈ [0, 1] and ${\tilde{A}}_{α} \neq \emptyset$ . Since ${\tilde{A}}_{α} \neq \emptyset$ , then ∃a ∈ L, $\tilde{A} (a) \geq α$ . It follows from (FM₁) that $\tilde{A} (1) \geq \tilde{A} (a) \geq α$ . Thus, $1 \in {\tilde{A}}_{α}$ . Let $a \in {\tilde{A}}_{α}, a \to b \in {\tilde{A}}_{α}$ , we have $\tilde{A} (a) \geq α, \tilde{A} (a \to b) \geq α$ . It follows from (FM₂) that $\tilde{A} (b) \geq \tilde{A} (a) \land \tilde{A} (a \to b) \geq α$ , i.e., $b \in {\tilde{A}}_{α}$ . According to Definition 3.1, ${\tilde{A}}_{α}$ is an MP -filter.

(2) ⇒ (1) Suppose that ${\tilde{A}}_{α}$ is an MP -filter, we consider the $\tilde{A}$ is an FMP -filter. In fact, ∀a ∈ L, taking $α_{1} = \tilde{A} (a)$ , then $a \in {\tilde{A}}_{α_{1}},$ i.e., ${\tilde{A}}_{α_{1}} \neq \emptyset$ . Thus ${\tilde{A}}_{α}$ is an MP -filter. Subsequently, by (MP₁), we have $1 \in {\tilde{A}}_{α_{1}}$ . Thus $\tilde{A} (1) \geq α_{1} = \tilde{A} (a)$ . So (FM₁) holds. Meanwhile, ∀a, b ∈ L, taking $α_{2} = \tilde{A} (a) \land \tilde{A} (a \to b)$ , then we have $a, a \to b \in {\tilde{A}}_{α_{2}}$ . Then it follows from (MP₂) that $b \in {\tilde{A}}_{α_{2}}$ . Thus $\tilde{A} (b) \geq α_{2} = \tilde{A} (a) \land \tilde{A} (a \to b)$ , that is to say, (FM₂) holds. Hence, $\tilde{A}$ is an FMP -filter.

(2) ⇒ (3) Since $a \in {\tilde{A}}_{α_{0}}$ , it means that ${\tilde{A}}_{α_{0}} \neq \emptyset$ . Thus, ${\tilde{A}}_{α_{0}}$ is an MP -filter.

(3) ⇒ (1) Suppose that for any a ∈ L, ${\tilde{A}}_{α_{0}}$ is an MP -filter. On the one hand, according to (MP₁) in Definition 3.1, $1 \in {\tilde{A}}_{α_{0}}$ , then $\tilde{A} (1) \geq α_{0} = \tilde{A} (a)$ , that is to say, (FM₁) holds. On the other hand, let $α_{0} = \tilde{A} (a), α_{1} = \tilde{A} (a \to b)$ , taking α₂ = α₀ ∧ α₁, we have $a \in {\tilde{A}}_{α_{2}}, a \to b \in {\tilde{A}}_{α_{2}}$ , then $b \in {\tilde{A}}_{α_{2}}$ , it means that $\tilde{A} (b) \geq α_{2} = \tilde{A} (a) \land \tilde{A} (a \to b)$ , that is to say, (FM₂) holds. Hence, $\tilde{A}$ is an FMP -filter. □

Example 3.9.L is an R₀-algebra, ∅ ≠ F ⊆ L, s, t ∈ [0, 1] and s > t. Taking $\tilde{A} (a) = {\begin{matrix} s, & a \in F, \\ t, & a \in L - F . \end{matrix}$

Then, F is an MP -filter if and only if $\tilde{A}$ is an FMP -filter.

In fact, when using the α- cut set of $\tilde{A}$ , we consider ${\tilde{A}}_{α} = {\begin{matrix} \emptyset, & α > s, \\ F, & t < α \leq s, \\ L, & α \leq t . \end{matrix}$

By Proposition 3.8, it is easily checked that F is an MP -filter if and only if $\tilde{A}$ is an FMP -filter.

Theorem 3.10. Let L be an R₀-algebra, ∅ ≠ F ⊆ L. If χ_{_F is the characteristic function of F, then F is an MP -filter if and only if χ_{_F is an FMP -filter.}}

Proof. In Example 3.9, taking s = 1, t = 0, it can be straightforward to obtain the result. □

Corollary 3.11. [8] Let L be an R₀-algebra, ∅ ≠ F ⊆ L. Then F is an MP -filter if and only if the following hold:(∀ a, b ∈ L)

1∈ F ;

a → b ∈ F, b → c ∈ F ⇒ a → c ∈ F .

Proof. According to Theorem 3.10, we only need to verify that χ_{_F is an FMP -filter if and only if (MP₃) and (MP₄) hold. Suppose that χ_{_F is an FMP -filter. Since F≠ ∅, we know ∃a ∈ F such that χ_{_F (a) =1. Then it follows from (MP₁) that χ_{_F (1) ≥ χ_{_F (a) =1, which means that 1 ∈ F. Thus (MP₃) holds. Subsequently, according to Proposition 3.7(3), it note that χ_{_F (a → c) ≥ χ_{_F (a → b) ∧ χ_{_F (b → c), that is to say, (MP₄) holds.}}}}}}}}

Conversely, If (MP₃) and (MP₄) hold, then it is easily checked that 1 = χ_{_F (1) = χ_{_F (a → (b → a)) ≥ χ_{_F (b → a). Assume that Proposition 3.7(3) doesn’t hold, we have χ_{_F (a → c) < χ_{_F (a → b) ∧ χ_{_F (b → c). This is only a case that χ_{_F (a → c) =0, while χ_{_F (a → b) = χ_{_F (b → c) =1. It means that a → b ∈ F, b → c ∈ F, while a → c ∉ F, which contradicts with the fact of (MP₄). □}}}}}}}}}

Note 3.12. The notion of the filter can be described by transitivity. The implication operator of R₀ has a very good transitivity, that is to say, if $R_{0} (a, b) \geq T > \frac{1}{2}, R_{0} (b, c) \geq U > \frac{1}{2}$ , then R₀ (a, c) ≥ T ∧ U.

Corollary 3.13. Let L be an R₀-algebra, ∅ ≠ F ⊆ L. Then F is an MP -filter if and only if the following hold:

a, b \in F \Rightarrow ↑ (a \otimes b) \subseteq F,

where, ↑ (a ⊗ b) = {c ∈ M ∣ a ⊗ b ≤ c} .

Proof. It can be easily proved by Theorem 3.10, Proposition 3.4(3) and Proposition 3.7(2). □

Example 3.14. Let L = [0, 1] _{R
₀}. ∀a, b ∈ L, ¬a = 1 - a, we define $a \otimes b = {\begin{matrix} 0, & a \leq \neg b, \\ a \land b, & otherwise . \end{matrix}$ $a \to b = {\begin{matrix} 1, & a \leq b, \\ \neg a \lor b, & otherwise . \end{matrix}$

Then (L, ∧ , ∨ , ⊗ , → , 0, 1) is a standard R₀-algebra.

Furthermore, suppose that $\tilde{A}$ is an FMP -filter on [0, 1] _{R
₀}, it follows from Proposition 3.4(3) that $\tilde{A} (a \otimes a) = \tilde{A} (a) \land \tilde{A} (a)$ . Thus when $a \in [0, \frac{1}{2}]$ , we get $\tilde{A} (a) = \tilde{A} (0)$ , that is to say, $\tilde{A}$ is a constant on interval $[0, \frac{1}{2}]$ . And we know that $\tilde{A}$ is monotonely nondecreasing on [0, 1]. Thus, we obtain the general form about $\tilde{A}$ as follows: $\tilde{A} (a) = {\begin{matrix} t + ψ (a), & a \in [\frac{1}{2}, 1], \\ t, & a \in [0, \frac{1}{2}], \end{matrix}$ where t ∈ [0, 1] is a constant, ψ (a) ≥0 and ψ (a) is monotonely nondecreasing.

Note 3.15. In the Example 3.2 of [7], Jun et al. consider the case ψ (a) =0.5a² and the case ψ (a) =0.3a, and them assume that $t \in [0, \frac{1}{2})$ . In fact, $t \in [\frac{1}{2}, 1]$ is also valid under the circumstance of t + ψ (a) ≤1 .

3.2 Prime FMP-filters

Definition 3.16. [15] Let L be an R₀-algebra. An MP -filter F of L is called a primeMP -filter if

a ∨ b ∈ F ⇒ a ∈ F or b ∈ F, ∀ a, b ∈ L .

Definition 3.17. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . If $\tilde{A}$ is an FMP -filter and also satisfies the following hold:

$\tilde{A} (a \lor b) \leq \tilde{A} (a) \lor \tilde{A} (b), \forall a, b \in L,$

then

\tilde{A}

is called a primeFMP -filter.

Proposition 3.18. Let L be an R₀-algebra and $\tilde{A}$ be an FMP -filter. Then the following statements are equivalent:

$\tilde{A}$ is a prime FMP -filter.

$\tilde{A} (a \lor b) = \tilde{A} (a) \lor \tilde{A} (b)$ , ∀ a, b ∈ L.

$\tilde{A} (a \to b) \lor \tilde{A} (b \to a) = \tilde{A} (1),$ ∀ a, b ∈ L.

Proof. (1) ⇒ (2) Assume that $\tilde{A}$ is a prime FMP -filter, then (PF₁) holds. Using Proposition 3.4(2), $\tilde{A}$ is monotonely nondecreasing. Thus $\tilde{A} (a \lor b) \geq \tilde{A} (a) \lor \tilde{A} (b) (\forall a, b \in L)$ .

(2) ⇒ (1) Straightforward.

(2) ⇒ (3) It follows from (p₈) that $\tilde{A} (a \to b) \lor \tilde{A} (b \to a) = \tilde{A} ((a \to b) \lor (b \to a)) = \tilde{A} (1)$ .

(3) ⇒ (1) Suppose that $\tilde{A}$ is an FMP -filter and condition (3) holds. It follows from Proposition 3.4 and (p₁₂) that $\tilde{A} (a \lor b) \leq \tilde{A} ((a \to b) \to b) \land \tilde{A} ((b \to a) \to a)$ . By (FM₂), we have $\tilde{A} (b) \geq \tilde{A} (a \to b) \land \tilde{A} ((a \to b) \to b)$ and $\tilde{A} (a) \geq \tilde{A} (b \to a) \land \tilde{A} ((b \to a) \to a) .$ Thus $\tilde{A} (b) \geq \tilde{A} (a \to b) \land \tilde{A} (a \lor b)$ and $\tilde{A} (a) \geq \tilde{A} (b \to a) \land \tilde{A} (a \lor b)$ . So, $\tilde{A} (a) \lor \tilde{A} (b) \geq \tilde{A} (a \lor b) \land (\tilde{A} (a \to b) \lor \tilde{A} (b \to a))$ . Using (3), we have $\tilde{A} (a) \lor \tilde{A} (b) \geq \tilde{A} (a \lor b) \land \tilde{A} (1) = \tilde{A} (a \lor b)$ , That is to say, (PF₁) holds. □

Corollary 3.19. If $\tilde{A}$ is a prime FMP -filter onR₀-algebra L, then $\tilde{A}$ is a lattice homomorphism.

Proof. An immediate consequence of Definition 3.5, Corollary 3.6 and Proposition 3.18(2). □

Proposition 3.20. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . Then the following statements are equivalent:

$\tilde{A}$ is a prime FMP -filter.

If ${\tilde{A}}_{α} \neq \emptyset (\forall α \in [0, 1])$ , then ${\tilde{A}}_{α}$ is a prime MP -filter.

∀a ∈ L, ${\tilde{A}}_{α_{0}}$ is a prime MP -filter (where $α_{0} = \tilde{A} (a)$ ).

Proof. The proof is similar to that of Proposition 2.14. □

Example 3.21. Let L be an R₀-algebra, ∅ ≠ F ⊆ L, s, t ∈ [0, 1] and s > t. Taking $\tilde{A} (a) = {\begin{matrix} s, & a \in F, \\ t, & a \in L - F . \end{matrix}$

Then, F is a prime MP -filter if and only if $\tilde{A}$ is a prime FMP -filter.

In fact, when using the α- cut set of $\tilde{A}$ , we consider ${\tilde{A}}_{α} = {\begin{matrix} \emptyset, & α > s, \\ F, & t < α \leq s, \\ L, & α \leq t . \end{matrix}$

By Proposition 3.20, it is easily checked that F is a prime MP -filter if and only if $\tilde{A}$ is a prime FMP -filter.

Theorem 3.22. Let L be an R₀-algebra, ∅ ≠ F ⊆ L. If χ_{_F is the characteristic function of F, then F is a prime MP -filter if and only if χ_{_F is a prime FMP -filter.}}

Proof. Similar to Theorem 3.10. □

Theorem 3.23. [15] Let L be an R₀-algebra and F be an MP -filter. Then F is a prime MP -filter if and only if

∀a, b ∈ L, a → b ∈ F or b → a ∈ F .

Proof. Assume that F is a prime MP -filter. 1 ∈ F and χ_{_F (1) =1. By Theorem 3.22, we have χ_{_F a prime FMP -filter. It follows from Proposition 3.18(2) and (p₈) that χ_{_F (a → b) ∨ χ_{_F (b → a) = χ_{_F (1) =1. This means that χ_{_F (a → b) =1 or χ_{_F (b → a) =1, i.e., ∀a, b ∈ L, a → b ∈ F or b → a ∈ F.}}}}}}}

Conversely, If (PM₂) holds. Since F is anMP -filter, 1 ∈ F and χ_{_F (1) =1. a → b ∈ F or b → a ∈ F show that χ_{_F (a → b) =1 or χ_{_F (b → a) =1. Thus χ_{_F (a → b) ∨ χ_{_F (b → a) = χ_{_F (1) =1, that is to say, F is a prime MP -filter from Proposition 2.18(3). □}}}}}}

Note 3.24. It can be directly checked that (PM₂) implies (PM₁) regardless of L is an R₀-algebra. In fact, for the general commutative residual lattices, if ∀a, b ∈ L, a ∨ b ∈ F and a → b ∈ F or b → a ∈ F. Without loss of generality, we suppose that a → b ∈ F. It follows from (p₁₂) that a ∨ b ≤ (a → b) → b, i.e., (a → b) → b ∈ F. And then b ∈ F. Thus (PM₁) holds.

However, under the condition of without L is an R₀-algebra, (PM₁) does not imply (PM₂). This problem is to utilize the condition (p₈) (called prelinearity), which is not contained by general commutative residual lattices. The following example is given byZhang [15].

Example 3.25. Suppose the commutative residual lattice L = {0, a, b, c, d, 1} with the binary operator ⊗ defined by a ⊗ b = a ∧ b, whose partial order relation ≤ is shown in Fig. 1 and implication operator → is defined as follows:

→	0	a	b	c	d	1
0	1	1	1	1	1	1
a	0	1	b	c	d	1
b	0	1	1	c	d	1
c	d	1	1	1	d	1
d	c	1	1	c	1	1
1	0	a	b	c	d	1

Taking F = {1, a}, we can obtain that F is a filter. Moreover, it can be easily checked that a ∨ b ∈ F implies a ∈ F or b ∈ F, while c → d ∉ F and d → c ∉ F, i.e., (PM₁) ≱ (PM₂).

Definition 3.26. [15] Let L be an R₀-algebra, ∅ ≠ F ⊆ L. Given a ∈ L, we define a^-1F = {b ∈ L ∣ a ∨ b ∈ F}.

Theorem 3.27. [15] Let L be an R₀-algebra and F be an MP -filter. Then F is a prime MP -filter if and only if a^-1F = F for all a ∈ L - F.

Theorem 3.28. [16] Let L be an R₀-algebra. F is an MP -filter and S is a ∨ -closed subset of L such that F∩ S = ∅. Then there exists a prime MP -filter P of L such that F ⊆ P and P∩ S = ∅.

4 Fuzzy implicative MP-filters and primefuzzy implicative MP-filters

4.1 FI -filters

Definition 4.1. [8] Let L be an R₀-algebra and F ⊂ L. If the following hold:

1∈ F ;

a → b ∈ F, a → (b → c) ∈ F ⇒ a → c ∈ F, ∀ a, b, c ∈ L,

then F is called an implicative MP -filter (denoted as the I -filter for convenience).

Definition 4.2. [8] Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . If the following hold:

$\tilde{A} (1) \geq \tilde{A} (a), \forall a \in L;$

$\tilde{A} (a \to c) \geq \tilde{A} (a \to b) \land \tilde{A} (a \to (b \to c)), \forall a, b, c \in L,$

then

\tilde{A}

is called a fuzzy implicative MP -filter (denoted as the FI -filter for convenience).

Example 4.3. Assume L = {0, a, b, c, 1} with the partial order ≤ defined by 0 < a < b < c < d < 1, whose ¬ and → are defined, respectively, as follows:

¬			→	0	a	b	c	d	1
0	1	and	0	1	1	1	1	1	1
a	d		a	d	1	1	1	1	1
b	c		b	c	d	1	1	1	1
c	b		c	b	b	b	1	1	1
d	a		d	a	a	b	c	1	1
1	0		1	0	a	b	c	d	1

Then {L, ∧ , ∨ , ¬ , →} is an R₀-algebra. Define a fuzzy subset $\tilde{A}$ in L by $\tilde{A} (1) = \tilde{A} (d) = \tilde{A} (c) = s$ , $\tilde{A} (0) = \tilde{A} (a) = \tilde{A} (b) = t$ (0 ≤ t < s ≤ 1). It is easily checked that $\tilde{A}$ is an FI -filter.

Theorem 4.4. [8] Let L be an R₀-algebra, then every FI -filter is an FMP -filter.

Proposition 4.5. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . Then the following statements are equivalent: (∀ a, b, c ∈ L)

$\tilde{A}$ is an FI -filter.

$\tilde{A} (a \to b) \geq \tilde{A} (c) \land \tilde{A} (c \to (a \to (\neg b \to b)))$ and $\tilde{A}$ is monotonely nondecreasing.

$\tilde{A} (a \to c) \geq \tilde{A} (c)$ and $\tilde{A} (a \to c) \geq \tilde{A} (b \to c) \land \tilde{A} (a \to (\neg c \to b))$ .

$\tilde{A} (1) \geq \tilde{A} (a), \tilde{A} (a \to b) = \tilde{A} (a \to (\neg b \to b))$ and $\tilde{A} (a \to b) \geq \tilde{A} (a \to c) \land \tilde{A} (c \to b)$ .

Proof. (1) ⇒ (3) Assume that $\tilde{A}$ is an FI -filter. By Theorem 4.4, $\tilde{A}$ is an FMP -filter. This shows that $\tilde{A}$ is monotonely nondecreasing, which means that $\tilde{A} (a \to c) \geq \tilde{A} (c)$ . For any a, b, c ∈ L, we have $\tilde{A} (a \to (\neg c \to b)) \land \tilde{A} (b \to c) = \tilde{A} (\neg c \to (\neg b \to \neg a)) \land \tilde{A} (\neg c \to \neg b) \leq \tilde{A} (\neg c \to \neg a) = \tilde{A} (a \to c)$ . Thus condition (3) holds.

(3) ⇒ (1) If condition (3) is valid, then $\forall a, b, c \in L, \tilde{A} (a \to b) \land \tilde{A} (a \to (b \to c)) = \tilde{A} (\neg b \to \neg a) \land \tilde{A} (a \to (\neg c \to \neg b)) = \tilde{A} (\neg b \to \neg a) \land \tilde{A} (\neg c \to (\neg \neg a \to \neg b)) \leq \tilde{A} (\neg c \to \neg a) = \tilde{A} (a \to c)$ , that is to say, (FI₂) holds. And for all a, c ∈ L, let a ≤ c, $\tilde{A} (a \to c) \geq \tilde{A} (c)$ implies that $\tilde{A} (1) \geq \tilde{A} (c)$ . So (FI₁) holds. Thus, $\tilde{A}$ is an FI -filter.

(3) ⇒ (4) Suppose that condition (3) is valid, it is clearly shown that $\tilde{A} (a \to b) \geq \tilde{A} (b \to b) \land \tilde{A} (a \to (\neg b \to b)) = \tilde{A} (a \to (\neg b \to b))$ . On the other hand, we have $\tilde{A} (a \to b) \leq \tilde{A} (\neg b \to (a \to b)) = \tilde{A} (a \to (\neg b \to b))$ . Thus, $\tilde{A} (a \to b) = \tilde{A} (a \to (\neg b \to b))$ . Meanwhile, it follows from condition (3) that $\tilde{A}$ is an FI -filter, then which is an FMP -filter. By Proposition 3.4(2) and (p₂₂), we have $\tilde{A} (a \to c) \land \tilde{A} (c \to b) \leq \tilde{A} ((a \to c) \otimes (c \to b)) \leq \tilde{A} (a \to b)$ .

(4) ⇒ (2) Assume that condition (4) is valid. We firstly verify that $\tilde{A}$ is monotonely nondecreasing. Taking a = 1, we have $\tilde{A} (b) \geq \tilde{A} (c) \land \tilde{A} (c \to b)$ . For all b, c ∈ L, c ≤ b, then $\tilde{A} (b) \geq \tilde{A} (c) \land \tilde{A} (1) = \tilde{A} (c)$ . Moreover, we also have $\tilde{A} (a \to b) = \tilde{A} (a \to (\neg b \to b)) \geq \tilde{A} (c) \land \tilde{A} (c \to (a \to (\neg b \to b)))$ . Thus, condition (2) holds.

(2) ⇒ (3) Suppose that condition (2) is valid. Since $\tilde{A}$ is monotonely nondecreasing, it notes that $\forall a, b, c \in L, \tilde{A} (1) = \tilde{A} (a \to (b \to a)) \geq \tilde{A} (b \to a)$ and also follows from (p₃) that $\tilde{A} ((b \to c) \to (a \to c)) \geq \tilde{A} (a \to b)$ . On the other hand, using ¬b → b ≥ b, we have $\tilde{A} (a \to b) \geq \tilde{A} (c) \land \tilde{A} (c \to (a \to (\neg b \to b))) \geq \tilde{A} (c) \land \tilde{A} (c \to (a \to b))$ . It implies that $\tilde{A} (a \to c) \geq \tilde{A} (b \to c) \land \tilde{A} ((b \to c) \to (a \to c))$ . Thus $\tilde{A} (a \to c) \geq \tilde{A} (b \to c) \land \tilde{A} (a \to b)$ . Then $\tilde{A} (b \to c) \land \tilde{A} (a \to (\neg c \to b)) = \tilde{A} (b \to c) \land \tilde{A} ((a \otimes \neg c) \to b)) \leq \tilde{A} ((a \otimes \neg c) \to c)) = \tilde{A} (a \to (\neg c \to c))$ . Since $\tilde{A} (a \to (\neg c \to c)) = \tilde{A} (1) \land \tilde{A} (1 \to (a \to (\neg c \to c)))$ , by condition (2), we have $\tilde{A} (a \to (\neg c \to c)) \leq \tilde{A} (a \to c)$ . Therefore, $\tilde{A} (b \to c) \land \tilde{A} (a \to (\neg c \to b)) \leq \tilde{A} (a \to c)$ . □

Similarly, we have the following proposition.

Proposition 4.6. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . Then the following statements are equivalent: (∀ a, b, c ∈ L)

$\tilde{A}$ is an FI -filter.

$\tilde{A} (a \to b) \geq \tilde{A} (c) \land \tilde{A} (c \to (a \to (a \to b)))$ and $\tilde{A}$ is monotonely nondecreasing.

$\tilde{A} (1) \geq \tilde{A} (a), \tilde{A} (a \to b) = \tilde{A} (a \to (a \to b))$ and $\tilde{A} (a \to b) \geq \tilde{A} (a \to c) \land \tilde{A} (c \to b)$ .

Corollary 4.7. Let L be an R₀-algebra. If $\tilde{A}$ is an FI -filter, then the following hold:

$\tilde{A} (\neg a \to a) = \tilde{A} (a)$ and $\tilde{A} (a \to \neg a) = \tilde{A} (\neg a)$ ∀ a ∈ L.

Theorem 4.8.Let L be an R₀-algebra and $\tilde{A}$ be an FMP -filter. If it satisfies

$\tilde{A} (a) \geq \tilde{A} (c \to ((a \to b) \to a)) \land \tilde{A} (c), \forall a, b, c \in L,$

then

\tilde{A}

is an FI -filter.

Proof. Since $\tilde{A}$ is an FMP -filter, it is directly checked that (FI₁) holds and follows from Proposition 3.7(3) that $\tilde{A} (a \to c) \geq \tilde{A} (a \to b) \land \tilde{A} (b \to c)$ . Subsequently, we only need to prove (FI₂) holds. In fact, by (p₁₁), we have ((a → c) → c) → (a → c) = a → (((a → c) → c) → c) = a → (a → c). Then it follows from (FI₃) that $\tilde{A} (a \to (b \to c)) \land \tilde{A} (a \to b) = \tilde{A} (b \to (a \to c)) \land \tilde{A} (a \to b) \leq \tilde{A} (a \to (a \to c)) = \tilde{A} ((a \to c) \to c) \to (a \to c) = \tilde{A} (1 \to ((a \to c) \to c) \to (a \to c)) \land \tilde{A} (1) \leq \tilde{A} (a \to c)$ . Thus $\tilde{A}$ is an FI -filter. □

Note 4.9. In paper [8], the proof of Theorem 4.5 given by Liu and li had a little mistake. Under the condition of $\tilde{A}$ is a fuzzy subset in L, $\tilde{A} (b \to (a \to c)) \land \tilde{A} (a \to b) \leq \tilde{A} (a \to (a \to c))$ is not valid.

Definition 4.10. [15] Let L be an R₀-algebra, ∅ ≠ F ⊆ L. We define

[F) = {b ∈ L ∣ ∃ a₁, a₂, …, a_n ∈ F such that a₁ ⊗ a₂ ⊗ ⋯ ⊗ a_n ≤ b},

which is called a filter generated by F. This is easily checked that it is the least filter and contains F.

We denote aⁿ with $\underset{︸}{a \otimes a \otimes \dots \otimes a}$ , and have the following propositions.

Proposition 4.11. [15] Let L be an R₀-algebra and F be an MP -filter and a ∈ L. Then

[F ∪ {a}) = {b ∈ L ∣ aⁿ → b ∈ F for some nonnegative integer n}.

Proposition 4.12. Let L be an R₀-algebra and F be an I -filter. Then F = (a → b) ^-1F ∩ [F ∪ {a → b}) for all a → b ∈ L - F, where (a → b) ^-1F = {c ∈ L ∣ (a → b) ∨ c ∈ F}.

Proof.F ⊆ (a → b) ^-1F ∩ [F ∪ {a → b}) Given c ∈ F, we know that c ≤ (a → b) ∨ c. Since F is an I -filter, it is easily shown from Definition 3.26 that c ∈ (a → b) ^-1F. So, F ⊆ (a → b) ^-1F.

F ⊇ (a → b) ^-1F ∩ [F ∪ {a → b}) Suppose that c ∈ (a → b) ^-1F ∩ [F ∪ {a → b}), we have (a → b) ∨ c ∈ F and c ∈ [F ∪ {a → b}). It follows from Proposition 4.11 that there exists a nonnegative integer n such that (a → b) ⁿ → c ∈ F. If n = 0, then c = 1 - c = (a → b) ⁿ → c ∈ F. If n > 0, then it follows from Proposition 2.6(p₁₈) that (a → b) → [(a → b) ^n-1 → c] = (a → b) ⊗ (a → b) ^n-1 → c = (a → b) ⁿ → c ∈ F . Thus, $\begin{matrix} (a \to b) \to [(a \to b)^{n - 1} \to c] \\ = {(a \to b) \to [(a \to b)^{n - 1} \to c]} \land 1 \\ = {(a \to b) \to [(a \to b)^{n - 1} \to c]} \\ \land {[(a \to b)^{n - 1} \to c] \to [(a \to b)^{n - 1} \to c]} \\ = {(a \to b) \lor [(a \to b)^{n - 1} \to c]} \\ \to [(a \to b)^{n - 1} \to c] . (Proposition 2.5 (p_{5})) \end{matrix}$

On the other hand, since c ≤ (a → b) ^n-1 → c, we have (a → b) ∨ c ≤ (a → b) ∨ [(a → b) ^n-1 → c]. According to above results, it can be easily checked from Definition 3.1(MP₂) that [(a → b) ^n-1 → c] ∈ F. So by principle of induction, c ∈ F. Hence,(a → b) ^-1F ∩ [F ∪ {a → b}) ⊆ F. □

Proposition 4.13. Let L be an R₀-algebra and $\tilde{A}$ be an FMP -filter. Then the following are equivalent:

$\tilde{A}$ is an FI -filter.

$\tilde{A} (a^{n} \to b) = \tilde{A} (a \to b)$ (∀ a, b ∈ L, n ≥ 1).

Proof. (1) ⇒ (2) Assume that $\tilde{A}$ is an FI -filter. By (FI₂), we have $\tilde{A} (a \to b) \geq \tilde{A} (a \to (a \to b)) \land \tilde{A} (1) = \tilde{A} (a \to (a \to b))$ . On the other hand, since b ≤ a → b, we have $\tilde{A} (a \to b) \leq \tilde{A} (a \to (a \to b))$ . Thus, $\tilde{A} (a \to b) = \tilde{A} (a \to (a \to b))$ , that is to say, $\tilde{A} (a \to b) = \tilde{A} ((a \otimes a) \to b) = \tilde{A} (a^{2} \to b)$ . Suppose the proposition holds for n = k. Let n = k + 1, then $\tilde{A} (a^{k} \to b) = \tilde{A} (a^{k} \to (a \to b)) = \tilde{A} (a^{k} \otimes a \to b) = \tilde{A} (a^{k + 1} \to b)$ . Hence by principle of induction, $\tilde{A} (a^{n} \to b) = \tilde{A} (a \to b)$ for all n ≥ 1.

(2) ⇒ (1) If condition (2) holds, it suffices to prove (FI₂). In fact, since $\tilde{A}$ is an FMP -filter, we have $\tilde{A} (a \to b) \land \tilde{A} (a \to (b \to c)) = \tilde{A} (a \to b) \land \tilde{A} (b \to (a \to c)) \leq \tilde{A} (a \to (a \to c)) = \tilde{A} (a \to c)$ . Therefore, $\tilde{A}$ is an FI -filter. □

4.2 Prime FI -filters

Definition 4.14. Let L be an R₀-algebra. An implicative MP -filter F of L is called a primeI -filter if

a → (b ∨ c) ∈ F ⇒ a → b ∈ F or a → c ∈ F, ∀ a, b, c ∈ L .

Definition 4.15. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . An FI -filter $\tilde{A}$ is called a primeFI -filter if it satisfies

$\tilde{A} (a \to (b \lor c)) \leq \tilde{A} (a \to b) \lor \tilde{A} (a \to c), \forall a, b, c \in L .$

Proposition 4.16. Let L be an R₀-algebra and $\tilde{A}$ be an FI -filter. Then the following statements are equivalent:

$\tilde{A}$ is a prime FI -filter.

$\tilde{A} (a \to (b \lor c)) = \tilde{A} (a \to b) \lor \tilde{A} (a \to c), \forall a, b, c \in L,$

$\tilde{A} (a \to b) \lor \tilde{A} (b \to a) = \tilde{A} (1),$ ∀ a, b ∈ L.

Proof. (1) ⇒ (2) Assume that $\tilde{A}$ is a prime FI -filter, then (PFI₁) holds. Using Proposition 3.4(2), $\tilde{A}$ is monotonely nondecreasing. Thus $\tilde{A} (a \to (b \lor c)) = \tilde{A} ((a \to b) \lor (a \to c)) \geq \tilde{A} (a \to b) \lor \tilde{A} (a \to c)$ .

(2) ⇒ (3) If condition (2) holds. It follows from (p₈) that $\tilde{A} (a \to b) \lor \tilde{A} (b \to a) = \tilde{A} ((a \to b) \lor (b \to a)) = \tilde{A} (1)$ .

(3) ⇒ (1) Suppose that $\tilde{A}$ is an FI -filter and condition (3) holds. It follows from (p₁₂) that $\tilde{A} ((a \to b) \lor (a \to c)) \leq \tilde{A} (((a \to b) \to (a \to c)) \to (a \to c)) \land \tilde{A} (((a \to c) \to (a \to b)) \to (a \to b))$ . By (FM₂), we have $\tilde{A} (a \to c) \geq \tilde{A} ((a \to b) \to (a \to c)) \land \tilde{A} (((a \to b) \to (a \to c)) \to (a \to c))$ and $\tilde{A} (a \to b) \geq \tilde{A} ((a \to c) \to (a \to b)) \land \tilde{A} (((a \to c) \to (a \to b)) \to (a \to b)) .$ Thus $\tilde{A} (a \to c) \geq \tilde{A} ((a \to b) \to (a \to c)) \land \tilde{A} ((a \to b) \lor (a \to c))$ and $\tilde{A} (a \to b) \geq \tilde{A} ((a \to c) \to (a \to b)) \land \tilde{A} ((a \to b) \lor (a \to c))$ . So, $\tilde{A} (a \to b) \lor \tilde{A} (a \to c) \geq \tilde{A} ((a \to b) \lor (a \to c)) \land (\tilde{A} ((a \to b) \to (a \to c)) \lor \tilde{A} ((a \to c) \to (a \to b)))$ . Using condition (3), we have $\tilde{A} (a \to b) \lor \tilde{A} (a \to c) \geq \tilde{A} ((a \to b) \lor (a \to c)) \land \tilde{A} (1) = \tilde{A} (a \to (b \lor c))$ , That is to say, (PFI₁) holds. □

Corollary 4.17. If $\tilde{A}$ is a prime FI -filter on R₀-algebra L, then $\tilde{A}$ is a lattice homomorphism.

Proposition 4.18. Let L be an R₀-algebra and $\tilde{A}$ be a prime FI -filter. Then the following holds:

$\tilde{A} (a \to c) \lor \tilde{A} (c \to b) \geq \tilde{A} (a \to b) \geq \tilde{A} (a \to c) \land \tilde{A} (c \to b), \forall a, b, c \in L$ .

Proof. An immediate consequence of Proposition 4.6(3) and (p₁₄). □

Proposition 4.19. Let L be an R₀-algebra and $\tilde{A}$ be a FI -filter. Then $\tilde{A}$ is a prime FI -filter if and only if

$\tilde{A} (a) \lor \tilde{A} (\neg a) = \tilde{A} (1), \forall a \in L$ .

Proof. Since $\tilde{A}$ is an FI -filter, we have $\tilde{A} (a \lor \neg a) = \tilde{A} (1)$ . In fact, for ¬a → ((¬ a → a) → a) =1 and ¬a → (((¬ a → a) → a) → ¬ (¬ a → a)) = ((¬ a → a) → a) → ((¬ a → a) → a) =1, then $\tilde{A} ((\neg a \to a) \to a) = \tilde{A} (\neg a \to \neg (\neg a \to a)) \geq \tilde{A} (\neg a \to ((\neg a \to a) \to a)) \land \tilde{A} (\neg a \to (((\neg a \to a) \to a)) \to \neg (\neg a \to a)) = \tilde{A} (1)$ . On the other hand, it follows from ¬a → a ≥ a that $\tilde{A} ((\neg a \to a) \to a) \leq \tilde{A} (1)$ . Thus $\tilde{A} ((\neg a \to a) \to a) = \tilde{A} (1)$ . Similarly, $\tilde{A} ((a \to \neg a) \to \neg a) = \tilde{A} (1)$ . Therefore it follows from (p₁₂) that $\tilde{A} (a \lor \neg a) = \tilde{A} (((a \to \neg a) \to \neg a) \land ((\neg a \to a) \to a)) = \tilde{A} ((a \to \neg a) \to \neg a) \land \tilde{A} ((\neg a \to a) \to a) = \tilde{A} (1)$ . If $\tilde{A}$ is prime, according to Proposition 4.16(2), we have $\tilde{A} (a) \lor \tilde{A} (\neg a) = \tilde{A} (a \lor \neg a) = \tilde{A} (1)$ , i.e., (PFI₂) holds.

Conversely, if condition (PFI₂) holds, it shows that $\tilde{A} (a) \lor \tilde{A} (\neg a) = \tilde{A} (1) = \tilde{A} (a \lor \neg a)$ , it implies that $\tilde{A} (a \to b) \lor \tilde{A} (a \to c) = \tilde{A} ((a \to b) \lor (a \to c)) = \tilde{A} (a \to (b \lor c))$ . Obviously, (PFI₁) holds. So $\tilde{A}$ is prime. □

Proposition 4.20. Let L be an R₀-algebra and $\tilde{A} \in F (L)$ . Then the following statements are equivalent:

$\tilde{A}$ is a prime FI -filter.

If ${\tilde{A}}_{α} = {a \in L | \tilde{A} (a) \geq α} \neq \emptyset$ (∀ α ∈ [0, 1]), then ${\tilde{A}}_{α}$ is a prime I -filter.

∀a ∈ L, ${\tilde{A}}_{α_{0}}$ is a prime I -filter (there $α_{0} = \tilde{A} (a)$ ).

Proposition 4.21. Let L be an R₀-algebra, ∅ ≠ F ⊆ L. If χ_{_F is the characteristic function of F, then F is a prime I -filter if and only if χ_{_F is a prime FI -filter.}}

Proposition 4.22. Let L be an R₀-algebra and F be a I -filter. Then F is a prime I -filter if and only if (a → b) ^-1F = F for all a → b ∈ L - F.

Proof. Suppose that F is a prime I -filter and a → c ∈ F. Since (a → c) → ((a → b) ∨ (a → c)) =1 ∈ F, we have (a → b) ∨ (a → c) ∈ F, that is to say, a → c ∈ (a → b) ^-1F. Thus F ⊆ (a → b) ^-1F. On the other hand, let a → c ∈ (a → b) ^-1F, then we have (a → b) ∨ (a → c) ∈ F. Since a → b ∉ F, we get (a → c) ∈ F. Thus (a → b) ^-1F ⊆ F. Therefore, (a → b) ^-1F = F.

Conversely, Assume that (a → b) ^-1F = F for all a → b ∈ L - F. If a → (b ∨ c) ∈ F, while a → b ∉ F, it is easily shown that a → c ∈ (a → b) ^-1F = F. So F is prime. □

Proposition 4.23. Let L be an R₀-algebra. F is an I -filter and S is a ∨ -closed subset of L such that F∩ S = ∅, then there exists a prime I -filter P of L such that F ⊆ P and P∩ S = ∅.

Proof. By Zorn’s Lemma, we know that the existence of a filter P being the maximal element of the family Γ of all I -filters containing F has empty intersection with S. we only need to verify that it is prime. In fact, if P is not prime, according to Proposition 4.22, there exists an element a → b ∈ L - P such that (a → b) ^-1P ≠ P. Then P is contained in both (a → b) ^-1P and [P ∪ {a → b}). Thus the maximal element of P implies that (a→ b) ^-1P ∩ S ≠ ∅ and [P∪ {a → b}) ∩ S ≠ ∅ (Otherwise, (a → b) ^-1P ∈ Γ and [P ∪ {a → b}) ∈ Γ. This is a contradiction with that P is maximal). Let a → c ∈ (a → b) ^-1P ∩ S and a → d ∈ [P ∪ {a → b}) ∩ S, then we have a → c ∈ (a → b) ^-1P and a → d ∈ [P ∪ {a → b}). Thus it follows from Proposition 4.12 that (a → c) ∨ (a → d) ∈ (a → b) ^-1P ∩ [P ∪ {a → b}) = P. We also have (a → c) ∨ (a → d) ∈ S since S is ∨ -closed. Hence, (a → c) ∨ (a → d) ∈ P ∩ S and then P∩ S ≠ ∅, a contradiction. Therefore P is prime. □

5 Conclusion

As we can see above, MP-filters, fuzzy (implicative) MP-filters and prime fuzzy (implicative) MP-filters’ inherent relations are shown and their some properties are investigated. Although Masoud Haveshki’s open problem (i.e., giving a suitable and useful definition for the associative filter) could not be solved, starting from the essence of the filter, we make some retrospect and rational discourse to express our strong interest on this problem.

In the paper, it is shown that (fuzzy) MP -filter is ∧ -closed and prime (fuzzy) MP -filter is ∨ -closed. Simultaneously, from Liu and li’s paper, we know that fuzzy implicative filters and fuzzy Boolean filters coincide in R₀-algebras. Beyond that, it is easily checked that (implicative) MP -filter can be extended to become a maximal (implicative) MP -filter, which can be a prime (implicative) MP -filter in R₀-algebras. Furthermore, a super MP -filter is equivalent to maximal implicative MP -filter. Thus, we summarize a framework diagram shown in Fig. 2.

Footnotes

Acknowledgments

This research was supported by grants from the National Nature Science Foundation of China (Grant Nos. 11571010, 61179038).

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¬			→	0	a	b	c	d	1
0	1	and	0	1	1	1	1	1	1
a	d		a	d	1	1	1	1	1
b	c		b	c	d	1	1	1	1
c	b		c	b	b	b	1	1	1
d	a		d	a	a	b	c	1	1
1	0		1	0	a	b	c	d	1

¬			→	0	a	b	c	d	1
0	1	and	0	1	1	1	1	1	1
a	d		a	d	1	1	1	1	1
b	c		b	c	d	1	1	1	1
c	b		c	b	b	b	1	1	1
d	a		d	a	a	b	c	1	1
1	0		1	0	a	b	c	d	1

¬			→	0	a	b	c	d	1
0	1	and	0	1	1	1	1	1	1
a	d		a	d	1	1	1	1	1
b	c		b	c	d	1	1	1	1
c	b		c	b	b	b	1	1	1
d	a		d	a	a	b	c	1	1
1	0		1	0	a	b	c	d	1