Further Complete Solutions to Four Open Problems on Filter of Logical Algebras
- DOI
- 10.2991/ijcis.2019.125905652How to use a DOI?
- Keywords
- Pseudo BCK-algebra; BL-algebra; Artificial intelligence; Filter; Boolean filter; Implicative filter; Normal filter; Fantastic filter
- Abstract
This paper focuses on the investigation of filters of pseudo BCK-algebra and BL-algebra, important and popular generic commutative and non-commutative logical algebras. By characterizing Boolean filter and implicative filter in pseudo BCK-algebra, the essential equivalent relation between these two filters is revealed. An open problem that “In pseudo BCK-algebra or bounded pseudo BCK-algebra, is the notion of implicative pseudo-filter equivalent to the notion of Boolean filter?” is solved. Based on this, this paper explores the essential relations between the implicative (Boolean) filter and implicative pseudo BCK-algebra. A complete solution to an open problem that “Prove or negate that pseudo BCK-algebras is implicative BCK-algebras if and only if every filter of them is implicative filters (or Boolean filter)” is derived. This paper further characterizes the fantastic filter and normal filter in BL-algebra, then gets the equivalent relation between the two filters, and completely solves two open problems regarding the relationship between these two filters: 1. Under what suitable condition a normal filter becomes a fantastic filter? and 2. (Extension property for a normal filter) Under what suitable condition extension property for normal filter holds?
- Copyright
- © 2019 The Authors. Published by Atlantis Press SARL.
- Open Access
- This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).
1. INTRODUCTION
The rapid development of computer science, technology, and mathematical logic put forward many new requirements, thus contributing to the nonclassical logic and the rapid development of modern logic [1]. The study of fuzzy logic has become a hot topic with scientific information and artificial intelligence, which makes fuzzy logic study of algebra and logic inseparable. Fuzzy logic is both the mathematical basis of the artificial intelligence and fuzzy reasoning. Based on the actual background, different forms of fuzzy logic system are proposed.
Logical algebras are the algebraic counterparts of the nonclassical logic and the algebraic foundation of reasoning mechanism in information sciences, computer sciences, theory of control, artificial intelligence, and other important fields. For example, BCK-algebra, BL-algebras, pseudo MTL-algebras, and noncommutative residuated lattice are algebraic counterparts of BCK Logic, Basic Logic, monoidal t-norm-based logic, and monoidal logic, respectively [1–3].
Hájek introduced BL-algebra as algebraic structure for his Basic Logic [3, 5]. Di Nola generalized BL-algebra in a noncommutative form and introduced the notion of pseudo BL-algebra as a common extension of BL-algebra in order to express the noncommutative reasoning [4, 6].
In 1966, BCK-algebra was introduced by Iséki and Imai from BCK/BCI Logic [7–9]. Iorgulescu established the connections between BCK-algebra and BL-algebra in [10]. Afterward, Georgescu and Iorgulescu introduced the notion of pseudo BCK-algebra as an extension of BCK-algebra to express the noncommutative reasoning [11, 12]. Iorgulescu established the connections between pseudo BL-algebra and pseudo BCK-algebra [12]. In [13], Wang and Zhang presented the necessary and sufficient conditions for residuated lattice and bounded pseudo BCK-algebra to be Boolean algebra.
Filter theory plays a vital role not only in studying of algebraic structure, but also in nonclassical logic and computer science [14, 15]. From logical point of view, various filters correspond to various sets of provable formulae [16, 17]. For example, based on filter and prime filter in BL-algebra, Hájek proved the completeness of Basic Logic [3]. In [18], Turunen proposed the notions of implicative filter and Boolean filter and proved that implicative filter is equivalent to Boolean filter in BL-algebra. In [19], some types of filters in BL-algebra were proposed. In [20–22], filters of pseudo MV-algebra, pseudo BL-algebra, pseudo effect algebra, and pseudo hoops were further studied. Literatures [8, 18, 19, 23–28] further studied filters of BL-algebra, lattice implication algebra, pseudo BL-algebra, pseudo BCK-algebra, R0-algebra, residuated lattice, triangle algebra, and the corresponding algebraic structures.
In [29], there is an open problem that “In pseudo BCK-algebra or bounded pseudo BCK algebra, is the notion of implicative pseudo-filter equivalent to the notion of Boolean filter?” Based on this, [25] proposed another open problem that “Prove or negate that pseudo BCK-algebras is implicative BCK-algebras if and only if every filter of them is implicative filters (or Boolean filter).” The two kinds of filters are important in pseudo BCK-algebra, thus the open problems are interesting and meaningful topics for further research and are the motivation of this paper.
We have explored the properties and relations of fuzzy pseudo filter (filter) of pseudo BCK-algebra. After we discussed the equivalent conditions of fuzzy normal filter in pseudo BCK-algebra (pP), we proposed fuzzy implicative pseudo filter and its relation with fuzzy Boolean filter in (bounded) pseudo BCK-algebra (pP). Then the open problem are partly solved [1]. Based on this, having further investigated the Boolean filter and implicative filter in pseudo BCK-algebra, we found the essentially equivalent relation between them, and the relation between them and implicative BCK-algebra, then we completely solved the open problems.
The role of filters are important not only in pseudo BCK-algebra, but also in related domains. In [26], there are two open problems: 1. Ünder what suitable condition a normal filter becomes a fantastic filter?” and 2. “(Extension property for a normal filter) Under what suitable condition extension property for normal filter holds?”
In our previous work, we have characterized the fuzzy fantastic filter and normal filter of BL-algebra, and discussed the relation between them, then partly solved the two open problems [30]. But so far we have not got sufficient and necessary condition for a normal filter to be fantastic. Based on this, we further obtained the relation between them and completely solved the two open problems.
This paper is organized as follows: In Section 2, we present some basic definitions and results in BL-algebra and pseudo BCK-algebra. In Section 3, we focus on the relation between implicative filter and Boolean filter of pseudo BCK-algebra or bounded pseudo BCK-algebra and give a complete solution to an open problem. In Section 4, based on the result we obtained in Section 3, we investigate the relation between implicative filter (Boolean filter) and implicative pseudo BCK-algebra and give a complete solution to another open problem. In Section 5, we recall the concept of filter and the corresponding properties of filter in BL-algebra and we propose complete solutions to two open problems of filter in BL-algebra.
2. PRELIMINARIES
Here we recall some definitions and results which will be needed. Reader can refer to [9, 11, 12, 19, 31–37].
Definition 1.
(Birkhoof [32]) Suppose L is a nonempty set with two binary operations ∧ and ∨. L is called a lattice if for x, y, z ∈ L, the following conditions hold
x ∧ x = x, x ∨ x = x,
x ∧ y = y ∧ x, x ∨ y = y ∨ x,
(x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z),
(x ∧ y) ∨ x = x, (x ∨ y) ∧ x = x.
Definition 2.
(Balbes and Dwinger [31]) A lattice L is called a distributive lattice if for x, y, z ∈ L, the following conditions hold
x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z),
x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z).
In lattices, (1) and (2) are equivalent.
Suppose L be a lattice. A binary relation ≤ is defined as for x, y ∈ L, x ≤ y if x ∧ y = x or x ∨ y = y. Then we can find that binary relation ≤ is a partially ordered relation.
Definition 3.
(Meng and Jun [9]) An algebraic structure (A, →, 1) is called an BCK-algebra if for all x, y, z ∈ A
(z → x) → (y → x) ≥ (y → z),
(y → x) → x ≥ y,
x ≥ x,
x ≥ y and y ≥ x imply x = y,
x → 1 = 1,
Definition 4.
(Georgescu and Iorgulescu [11]) A(reversed left-) pseudo BCK-algebra is a structure (A, ≥, →, ↪, 1), where ≥ is a binary relation on A, → and ↪ are binary operations on A and 1 is an element of A, verifying, for all x, y, z ∈ A, the axioms
(z → x) ↪ (y → x) ≥ y → z, (z ↪ x) → (y ↪ x) ≥ y ↪ z,
(y → x) ↪ x ≥ y, (y ↪ x) → x ≥ y,
x ≥ x,
1 ≥ x,
x ≥ y and y ≥ x imply x = y,
x ≥ y iff y → x = 1 iff y ↪ x = 1.
Example 1.
(Jun, Kim and Neggers [38]) Let X = [0, ∞] and let ≤ be the usual order on X. Define → and ↪ on X as follows:
x → y = 0 (if y ≤ x) or
x ↪ y = 0 (if y ≤ x) or
for all x, y ∈ X. Then (X, ≤, →, ↪, 0) is a pseudo BCK-algebra.
Definition 5.
(Iorgulescu [12]) A pseudo BCK-algebra (A, ≥, →, ↪, 1) is called bounded if there exits unique element 0 such that 0 → x = 1 or 0 ↪ x = 1 for any x ∈ A.
In a pseudo-BCK-algebra A we can define x− = x → 0, x∼ = x ↪ 0 for any x ∈ A.
Proposition 1.
(Iorgulescu [35]) Let (A, ≥, →, ↪, 1) be a pseudo BCK-algebra. Then the following properties hold for any x, y, z ∈ A
x ≤ y ⇒ y → z ≤ x → z and y ↪ z ≤ x ↪ z,
x ≤ y ⇒ z → x ≤ z → y and z ↪ x ≤ z ↪ y,
z ↪ (y → x) = y → (z ↪ x),
x ≤ y iff x → y = 1 iff x ↪ y = 1,
x → y = x → x ∧ y, x ↪ y = x ↪ x ∧ y,
x → y ≤ (z → x) → (z → y), x ↪ y ≤ (z ↪ x) ↪ (z ↪ y),
y ≤ x → y, y ≤ x ↪ y,
1 → x = x = 1 ↪ x,
(x ∨ y) → z = (x → z) ∧ (y → z), (x ∨ y) ↪ z = (x ↪ z) ∧ (y ↪ z),
x ∨ y = ((x → y) ↪ y) ∧ ((y → x) ↪ x), x ∨ y = ((x ↪ y) → y) ∧ ((y ↪ x) → x).
Definition 6.
(Ciungu [34]) A pseudo BCK-algebra with condition (pP) (i.e., with pseudo product) is a pseudo BCK-algebra (A, ≥, →, ↪, 1) satisfying the condition (pP)
(pP) there exists, for all x, y ∈ A,
Theorem 2.
(Ciungu [34]) Let (A, ≥, →, ↪, 1) be a pseudo BCK-algebras with condition (pP), x ⊙ y is defined as
(x ⊙ y) → z = x → (y → z),
(y ⊙ x) ↪ z = x ↪ (y ↪ z),
(x → y) ⊙ x ≤ x, y, x ⊙ (x ↪ y) ≤ x, y,
x ⊙ y ≤ x ∧ y ≤ x, y.
In the sequel, we shall agree that the operations ∨, ∧, ⊙ have priority towards the operations →, ↪.
Definition 7.
(Blount and Tsinakis [33]) A lattice-ordered residuated monoid is an algebra (A, ∨, ∧, ⊙, →, ↪, e) satisfying the following conditions:
(A, ∨, ∧) is a lattice,
(A, ⊙, e) is a monoid,
x ⊙ y ≤ z iff x ≤ y → z iff y ≤ x ↪ z for all x, y, z ∈ A.
A lattice-ordered residuated monoid A is called integral if x ≤ e for all x ∈ A. In an integral lattice-ordered residuated monoid, we use “1” instead of e.
Lemma 3.
(Jipsen and Tsinakis [36]) Pseudo BCK-algebra with condition (pP) is category equivalent to partially ordered residuated integral monoid.
Definition 8.
(Zhang [37]) A BCK-algebra (A, →, 1) is called an implicative BCK-algebra if it satisfies (x → y) → x = x for any x, y ∈ A.
Definition 9.
(Zhang [37]) A pseudo BCK-algebra (A, ≥, →, ↪, 1) is called a 1-type implicative pseudo BCK-algebra if it satisfies (x → y) → x = (x ↪ y) ↪ x = x for any x, y ∈ A.
Definition 10.
(Zhang [37]) A pseudo BCK-algebra (A, ≥, →, ↪, 1) is called a 2-type implicative pseudo BCK-algebra if it satisfies (x ↪ y) → x = (x → y) ↪ x = x for any x, y ∈ A.
Definition 11.
(Haveshki, Saeid, and Eslami [19]) A BL-algebra is an algebra (A, ∨, ∧, ⊙, →, 0, 1) of type (2, 2, 2, 2, 2, 0, 0) such that (A, ∨, ∧, 0, 1) is a bounded lattice, (A, ⊙, 1) is a commutative monoid, and the following conditions hold for all x, y, z ∈ A:
(A1) x ⊙ y ≤ z if and only if x ≤ y → z,
(A2) x ∧ y = x ⊙ (x → y),
(A3) (x → y) ∨ (y → x) = 1.
Example 2.
Let A = [0, 1]. Define ⊙ and → as follows:
x ⊙ y = minx, y and x → y = 1 (if x < y) or x → y = y (if x > y).
Then (A, ∨, ∧, ⊙, →, 0, 1) is a BL-algebra.
An MV-algebra A is a BL-algebra satisfying x−− = x for any x ∈ A. A Gödel-algebra is a BL-algebra satisfying x ⊙ x = x for any x ∈ A.
Proposition 4.
(Zhang [37]) In a BL-algebra A, the following properties hold for all x, y, z ∈ A
y → (x → z) = x → (y → z),
1 → x = x,
x ≤ y iff x → y = 1,
x ∨ y = ((x → y) → y) ∧ ((y → x) → x),
x ≤ y ⇒ y → z ≤ x → z,
x ≤ y ⇒ z → x ≤ z → y,
x → y ≤ (z → x) → (z → y),
x → y ≤ (y → z) → (x → z),
x ≤ (x → y) → y,
x ⊙ (x → y) = x ∧ y.
We shall agree that the operations ∨, ∧, ⊙ have priority toward the operations →.
3. THE RELATION BETWEEN IMPLICATIVE FILTER AND BOOLEAN FILTER OF PSEUDO BCK-ALGEBRA OR BOUNDED PSEUDO BCK-ALGEBRA
In this section, we recall the definitions of filter, positive implicative pseudo filter, Boolean filter, normal filter, and implicative filter of pseudo BCK-algebra.
Definition 12.
(Zhang [29]) A nonempty subset F of pseudo BCK-algebra A is called a (pseudo) filter of A if it satisfies
(F1) x ∈ F, y ∈ A, x ≤ y ⇒ y ∈ F,
(F2) x ∈ F, x → y ∈ F or x ↪ y ∈ F ⇒ y ∈ F.
Theorem 5.
(Wang [1]) A nonempty subset F of a pseudo BCK-algebra A is a filter of A if and only if it satisfies
(F3) 1 ∈ F,
(F4) x ∈ F, x → y ∈ F or x ↪ y ∈ F ⇒ y ∈ F.
In example 1, we can find {0} is a filter.
Theorem 6.
(Wang [1]) A nonempty subset F of a pseudo BCK-algebra A with condition (pP) is a filter of A if and only if it satisfies
(F5) x ∈ F, y ∈ F ⇒ x ⊙ y ∈ F,
(F6) x ∈ F, y ∈ A, x ≤ y ⇒ y ∈ F.
Definition 13.
(Zhang [29]) Let F be a filter of A, for all x, y ∈ A, F is called a(an)
Boolean filter if (x → y) ↪ x ∈ F and (x ↪ y) → x ∈ F, then x ∈ F,
Prime filter if x ∨ y ∈ F implies x ∈ F or y ∈ F,
Maximal filter if x ∈ F or x− ∈ F (and x∼ ∈ F),
Loyal filter if x
Normal filter if x → y ∈ F iff x ↪ y ∈ F,
Implicative filter if (x → y) → x ∈ F and (x ↪ y) ↪ x ∈ F, then x ∈ F.
Definition 14.
(Zhang and Jun [13]) A nonempty subset F of a pseudo BCK-algebra A is called a positive implicative filter of A if it satisfies (F1) and for all x, y ∈ A.
(F7) x ↪ (y → z) ∈ F, x ↪ y ∈ F implies x ↪ z ∈ F,
(F8) x → (y ↪ z) ∈ F, x → y ∈ F implies x → z ∈ F.
Note that any filter of a BCK-algebra is normal.
Theorem 7.
(Zhang [29]) Let (A; ≤, →, ↪, 0, 1) be a bounded pseudo BCK-algebra and F implicative pseudo filter of A. Then
∀x ∈ A, ((x → 0) → x) ↪ x ∈ F, that is, (x− → x) ↪ x ∈ F,
∀x ∈ A, ((x ↪ 0) ↪ x) → x ∈ F, that is, (x∼ ↪ x) → x ∈ F,
∀x, y ∈ A, ((x → y) → x) ↪ x ∈ F,
∀x, y ∈ A, ((x ↪ y) ↪ x) → x ∈ F,
∀x, y ∈ A, if (x → y) → y ∈ F, then (y → x) ↪ x ∈ F,
∀x, y ∈ A, if (x ↪ y) ↪ y ∈ F, then (y ↪ x) → x ∈ F,
∀x, y ∈ A, if x ↪ y ∈ F, then ((y ↪ x) → x) ↪ y ∈ F,
∀x, y ∈ A, if x → y ∈ F, then ((y → x) ↪ x) → y ∈ F.
In [29], there is an open problem: “In pseudo BCK-algebra or bounded pseudo BCK algebra, is the notion of implicative pseudo filter equivalent to the notion of Boolean filter?”
To solve the open problem, we recall the results of the relation between the two filters and then get a new solution to the open problem in pseudo BCK-algebra.
Theorem 8.
(Zhang [29]) Let (A, ≥, →, ↪, 1) be a pseudo BCK-algebra and F a normal pseudo filter of A. Then F is implicative if and only if F is Boolean.
With the help of the equivalent conditions of fuzzy normal filter of pseudo BCK-algebra (pP), [1, 39] get the following results and partly solve the open problem.
Theorem 9.
(Wang [1]) In bounded pseudo BCK-algebra, every implicative pseudo filter is a Boolean filter. In pseudo BCK-algebras(pP), every Boolean filter is an implicative pseudo filter.
We further investigate the properties of Boolean filter and implicative filter which make the relation between the two filters much clear, and get the solution for the open problem.
Theorem 10.
Implicative pseudo filter is Boolean filter in pseudo BCK-algebra.
Proof.
Let F be an implicative pseudo filter of A. Then ∀x ∈ A, suppose (x → y) ↪ x ∈ F,
from x ≤ ((x → y) ↪ x) → x,
so (((x → y) ↪ x) → x) → y ≤ x → y,
and ((((x → y) ↪ x) → x) → y) → (x → y) = 1.
On the other hand, x → y ≤ ((x → y) ↪ x) → x,
so we get ((((x → y) ↪ x) → x) → y) → (x → y) ≤ ((((x → y) ↪ x) → x) → y) → (((x → y) ↪ x) → x).
Then ((((x → y) ↪ x) → x) → y) → (((x → y) ↪ x) → x) = 1 ∈ F,
and ((x → y) ↪ x) → x ∈ F, since F is an implicative filter.
Combine that (x → y) ↪ x ∈ F, according to the definition of filter, we get x ∈ F.
Similarly, suppose (x ↪ y) → x ∈ F,
from x ≤ ((x ↪ y) → x) ↪ x,
so (((x ↪ y) → x) ↪ x) ↪ y ≤ x ↪ y,
and (((x ↪ y) → x) ↪ x) ↪ y) ↪ (x ↪ y) = 1.
By x ↪ y ≤ ((x ↪ y) → x) ↪ x,
so we get ((((x ↪ y) → x) ↪ x) ↪ y) ↪ (x ↪ y) ≤ ((((x ↪ y) → x) ↪ x) ↪ y) ↪ (((x ↪ y) → x) ↪ x).
Then ((((x ↪ y) → x) ↪ x) ↪ y) ↪ (((x ↪ y) → x) ↪ x) = 1 ∈ F,
and ((x ↪ y) → x) ↪ x ∈ F, since F is an implicative filter.
Combine that (x ↪ y) → x ∈ F, according to the definition of filter, then we get x ∈ F.
According to the definition, then F is a Boolean filter of A.
Similarly, we can get
Theorem 11.
In pseudo BCK-algebra, every Boolean filter is an implicative pseudo filter.
Proof.
Let F be an Boolean filter of A. Then ∀x ∈ A, suppose (x → y) → x ∈ F,
from x ≤ ((x → y) → x) ↪ x,
so (((x → y) → x) ↪ x) → y ≤ x → y,
and (((x → y) → x) ↪ x) → y) ↪ (x → y) = 1.
On the other hand, x → y ≤ ((x → y) → x) ↪ x,
so we get ((((x → y) → x) ↪ x) → y) ↪ (x → y) ≤ ((((x → y) → x) ↪ x) → y) ↪ ((x → y) → x) ↪ x).
Then ((((x → y) → x) ↪ x) → y) ↪ (((x → y) → x) ↪ x) = 1 ∈ F and ((x → y) → x) ↪ x ∈ F, since F is an implicative filter.
Combine that (x → y) → x ∈ F, according to the definition of filter, then we get x ∈ F.
Similarly, suppose (x ↪ y) ↪ x ∈ F,
from x ≤ ((x ↪ y) ↪ x) → x,
so (((x ↪ y) ↪ x) → x) ↪ y ≤ x ↪ y,
and (((x ↪ y) ↪ x) → x) ↪ y) → (x ↪ y) = 1.
On the other hand, x ↪ y ≤ ((x ↪ y) ↪ x) → x,
so we get ((((x ↪ y) ↪ x) → x) ↪ y) → (x ↪ y) ≤ ((((x ↪ y) ↪ x) → x) ↪ y) → ((x ↪ y) ↪ x) → x).
Then ((((x ↪ y) ↪ x) → x) ↪ y) → (((x ↪ y) ↪ x) → x) = 1 ∈ F and (x ↪ y) ↪ x) → x ∈ F, since F is a Boolean filter.
Combine that (x ↪ y) ↪ x ∈ F, according to the definition of filter, we get x ∈ F.
Thus F is an implicative pseudo filter of A.
From the above results, we can get the following results as a solution for the open problem:
Theorem 12.
In pseudo BCK-algebra or bounded pseudo BCK-algebra, the notion of implicative pseudo filter is equivalent to the notion of Boolean filter.
Remark 1.
The equivalent relation between the implicative filter and Boolean filter is of importance in the study of logical algebras. For example, when studying of the pseudo BCK-algebra, implicative filter and Boolean filter can reflect the algebraic structure of the pseudo BCK-algebra. When we get the equivalent relation between them, and based on some other results we obtained [1, 16, 30], we can completely solve some other problems like this.
4. THE RELATION BETWEEN IMPLICATIVE FILTER (BOOLEAN FILTER) AND IMPLICATIVE PSEUDO BCK-ALGEBRA
The filters play a vital role in representing the algebras, such as in a pseudo BL-algebra A,
A is pseudo MV-algebra if and only if every filter of A is a pseudo MV filter,
A is a Godel-algebra if and only if every filter of A is a pseudo G filter,
A is a Boolean algebra if and only if every filter of A is a Boolean filter.
The similar relation between implicative or Boolean filter and implicative pseudo BCK-algebra is not obtained, yet. For this reason, [25] set it as an open problem.
Prove or negate that pseudo BCK-algebras is implicative BCK-algebras if and only if every filter of them is implicative filters (or Boolean filter).
[1] partly solve the open problem. Based on this, we can further get the following results as a solution for the open problem:
Proposition 13.
Let A be a pseudo BCK algebra. Then the following statements are equivalent:
A is an implicative BCK-algebras,
Every filter of A is an implicative filters (or Boolean filter),
{1} is an implicative filters (or Boolean filter).
Proof.
(1) ⇒ (2) Based on the results of [9] and the previous result, pseudo BCK-algebra A is implicative BCK-algebra if and only if A is a 1-type (or 2-type) implicative pseudo BCK-algebra. Then for every filter of them, if (x → y) → x, (x ↪ y) ↪ x ∈ F ⇒ x ⇒ F, (x ↪ y) → x, (x → y) ↪ x ∈ F ⇒ x ∈ F. Then every pseudo filters of them is implicative pseudo filters (Boolean filter), so necessity is obvious.
(2) ⇒ (3) obvious.
(3) ⇒ (1) Now suppose every pseudo filter of a pseudo BCK-algebra A is an implicative pseudo filters (Boolean filter), then pseudo filter {1} is an implicative pseudo filters (Boolean filter).
For any x, y ∈ A, from x ≤ ((x ↪ y) → x) ↪ x (by Theorem (7)),
we get (((x ↪ y) → x) ↪ x) ↪ y ≤ x ↪ y (by Theorem (1)),
then ((((x ↪ y) → x) ↪ x) ↪ y) → (x ↪ y) = 1 (by Definition (6)).
From x ↪ y ≤ ((x ↪ y) → x) ↪ x (by Definition (2)),
we get ((((x ↪ y) → x) ↪ x) ↪ y) → (x ↪ y) ≤ ((((x ↪ y) → x) ↪ x) ↪ y) → (((x ↪ y) → x) ↪ x) (by Theorem (2)),
then ((((x ↪ y) → x) ↪ x) ↪ y) → (((x ↪ y) → x) ↪ x) = 1 ∈ {1}, (by Definition (4)),
we get ((x ↪ y) → x) ↪ x = 1 ∈ {1}, since {1} is an implicative pseudo filters, that is, (x ↪ y) → x ≤ x (by Definition (6)).
On the other hand, x ≤ (x ↪ y) → x, then (x ↪ y) → x = x.
For the same reason, from x ≤ ((x → y) ↪ x) → x, so (((x → y) ↪ x) → x) → y ≤ x → y and (((x → y) ↪ x) → x) → y) ↪ (x → y) = 1. And x → y ≤ ((x → y) ↪ x) → x, so we get ((((x → y) ↪ x) → x) → y) ↪ (x → y) ≤ ((((x → y) ↪ x) → x) → y) ↪ ((x → y) ↪ x) → x). Then ((((x → y) ↪ x) → x) → y) ↪ (((x → y) ↪ x) → x) = 1 ∈ {1}, then (x → y) ↪ x) → x = 1 ∈ {1}, that is, (x → y) ↪ x ≤ x. On the other hand, x ≤ (x → y) ↪ x, then (x → y) ↪ x = x.
From above results, we find that A is a 1-type implicative pseudo BCK-algebra, then A is implicative BCK-algebra.
5. THE RELATION BETWEEN FANTASTIC FILTER AND NORMAL FILTER IN BL-ALGEBRA
Here we recall some kinds of filters in BL-algebra. Similar with the pseudo BCK-algebra, here we recall some definitions and results which will be needed. Reader can refer to [3, 16, 30, 37, 40–42].
Definition 15.
A filter of a BL-algebra A is a nonempty subset F of A such that for all x, y ∈ A,
(F1) if x, y ∈ F, then x ⊙ y ∈ F,
(F2) if x ∈ F and x ≤ y, then y ∈ F.
Proposition 14.
Let F be a nonempty subset of a BL-algebra A. Then F is a filter of A if and only if the following conditions hold
1 ∈ F,
x, x → y ∈ F implies y ∈ F.
A filter F of a BL-algebra A is proper if F ≠ A, that is,
In example 2, we can find
Definition 16.
A proper filter F is prime if for any x, y, z ∈ A, x ∨ y ∈ F implies x ∈ F or y ∈ F.
Theorem 15.
A proper filter F is prime if for any x, y ∈ A, x → y ∈ F or y → x ∈ F.
Definition 17.
A filter F of A is called normal if for any x, y, z ∈ A, z → ((y → x) → x) ∈ F and z ∈ F imply (x → y) → y ∈ F.
Definition 18.
Let F be a nonempty subset of a BL-algebra A. Then F is called a fantastic filter of A if for all x, y, z ∈ A, the following conditions hold:
1 ∈ F,
z → (y → x) ∈ F, z ∈ F implies ((x → y) → y) → x ∈ F.
Definition 19.
Let F be a nonempty subset of a BL-algebra A. Then F is called an implicative filter of A if for all x, y, z ∈ A, the following conditions hold:
1 ∈ F,
x → (y → z) ∈ F, x → y ∈ F imply x → z ∈ F.
Definition 20.
Let F be a nonempty subset of a BL-algebra A. Then F is called a positive implicative filter of A if for all x, y, z ∈ A, the following conditions hold:
1 ∈ F,
x → ((y → z) → y) ∈ F, x ∈ F imply y ∈ F.
Definition 21.
A filter F of A is called Boolean if x ∨ x− ∈ F for any x ∈ A.
Definition 22.
Let F be a filter of A. F is called an ultra filter of A if it satisfies x ∈ F or x− ∈ F for all x ∈ A.
Definition 23.
Let F be a filter of A. F is called an obstinate filter of A if it satisfies x ∉ F and y ∉ F implies x → y ∈ F for all x, y ∈ A.
Definition 24.
A proper filter of a BL-algebra A is called maximal if it is not properly contained in any other proper filter of A.
Proposition 16.
A proper filter F of a BL-algebra A is maximal if and only if
In order to investigate the essential relations among the filters, and based on the past work [1, 16, 30], we characterize the following filters.
Theorem 17.
Let F be a filter of a BL-algebra A. F is a normal filter if and only if one of the followings holds for all x, y ∈ A
(y → x) → x ∈ F implies (x → y) → y ∈ F,
x−− ∈ F implies x ∈ F.
Theorem 18.
Let F be a filter of A. Then the followings are equivalent for all x, y, z ∈ A
F is a Boolean filter of A,
(x → y) → x ∈ F implies x ∈ F,
x− → x ∈ F implies x ∈ F.
Theorem 19.
Let F be a Boolean filter of A, then for all x, y ∈ A
x → y ∈ F implies ((y → x) → x) → y ∈ F,
(x → x−) → x− ∈ F,
x−− ∈ F implies x ∈ F,
x → (x → y) ∈ F implies x → y ∈ F.
Theorem 20.
Let F be a filter of a BL-algebra A. Then the followings are equivalent for all x, y, z ∈ A:
F is a fantastic filter,
y → x ∈ F ⇒ ((x → y) → y) → x ∈ F,
x−− → x ∈ F.
Theorem 21.
Let F be a filter of a BL-algebra A. Then for any x, y, z ∈ A the followings are equivalent:
F is an implicative filter of A,
y → (y → x) ∈ F implies y → x ∈ F,
x → x ⊙ x ∈ F.
By the above results, we can get the following results:
Corollary 22.
In an MV-algebra, every filter is a fantastic filter.
Corollary 23.
In a Gödel-algebra, every filter is an implicative filter.
Based on this, we get some relations among the filters in BL-algebras.
Theorem 24.
Each Boolean filter is equivalent to a positive implicative filter in BL-algebras.
Theorem 25.
Each ultra filter is equivalent to an obstinate filter in BL-algebras.
Theorem 26.
Each ultra filter f a BL-algebra A is a fantastic filter.
Proof.
Suppose F is an ultra filter. If x ∈ F, we have x ≤ x−− → x ∈ F. If x− ∈ F, we get (x → 0) → (x−− → x) = x−− → ((x → 0) → x) ≥ 0 → x = 1, we get x−− → x ∈ F, then F is a fantastic filter.
Theorem 27.
Let F be a filter of A. Then F is a Boolean filter if and only if it is an implicative and normal filter.
Proof.
If F is a Boolean filter, then by Theorem 17, 19, and 21, we know that F is an implicative and normal filter.
Suppose F is an implicative and normal filter. Since x− → x ≤ x− → x−− = x− → (x− → 0), we have x− → 0 ∈ F since F is an implicative filter. Then since F is also a normal filter, we have x ∈ F, thus we get then F is a Boolean filter.
Theorem 28.
Let F be a filter of A. Then F is a Boolean filter if and only if it is an implicative and fantastic filter.
Based on the previous work and the above results, we get the essential relation.
Theorem 29.
Let F be a filter of A. Then F is an implicative and normal filter if and only if it is an implicative and fantastic filter.
In [26], there are two open problems in BL-algebras:
Under what suitable condition a normal filter becomes a fantastic filter?
(Extension property for a normal filter) Under what suitable condition extension property for normal filter holds?
[30] proposed solutions for the two open problems by the fuzzy filters, respectively as follows:
Suitable condition should be
BL-algebra is an MV-algebra,
The filter is an implicative filter,
The filter is an obstinate filter,
The filter is an ultra filter.
Under the condition
If A is an MV-algebra,
F is a normal and implicative filter of A,
F is a normal and obstinate filter of A,
F is a normal and ultra filter of A.
Extension property for normal filter holds.
According to the above theorem and corollary, we can get the equivalent relation between the two filters and give answers to the open problems.
The suitable condition should be
A normal filter is equivalent to a fantastic filter.
Extension property for a normal filter holds.
We further characterized the filters in BL-algebra. Compared with the solutions in [30], the condition that the filter is an obstinate filter or the filter is an ultra filter is redundant.
6. CONCLUSION
We discuss the properties of implicative filters and Boolean filters in pseudo BCK-algebra. Based on the results and previous work, we completely solve an open problem which is important to deep study of the algebraic structure of pseudo BCK-algebra. Based on this, we prove that pseudo BCK-algebra is implicative BCK-algebra if and only if every filter of them is implicative filter (or Boolean filter).
We further characterize the filters in BL-algebra. Compared with the solutions in [30], the condition that the filter is an obstinate filter or the filter is an ultra filter is redundant.
In the future work, we will extend the corresponding filter theory to different algebraic structures, and study the congruence relations induced by the filters.
ACKNOWLEDGMENTS
This Research work is supported by the National Natural Science Foundation of P. R. China (Grant No. 11571281, 61673320); Xi’an Shiyou University College Students Innovation and Entrepreneurship Training Program Funding Project (Grant No. 201819062); the Fundamental Research Funds for the Central Universities (Grant No. 2682017ZT12).
REFERENCES
Cite this article
TY - JOUR AU - Wei Wang AU - Pengxi Yang AU - Yang Xu PY - 2019 DA - 2019/02/04 TI - Further Complete Solutions to Four Open Problems on Filter of Logical Algebras JO - International Journal of Computational Intelligence Systems SP - 359 EP - 366 VL - 12 IS - 1 SN - 1875-6883 UR - https://doi.org/10.2991/ijcis.2019.125905652 DO - 10.2991/ijcis.2019.125905652 ID - Wang2019 ER -