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Further Complete Solutions to Four Open Problems on Filterof Logical Algebras

Wei Wang1,2,3,*, Pengxi Yang1, Yang Xu2,3

1College of Sciences, Xi’an Shiyou University, Xi’an, Shaanxi, 710065, China2National-Local Joint Engineering Laboratory of System Credibility Automatic Verification, Southwest Jiaotong University, Chengdu, 610031, China3System Credibility Automatic Verification Engineering Lab of Sichuan Province, Southwest Jiaotong University, Chengdu, 610031, China

ART I C L E I N FOArticle History

Received 03 Jan 2019Revised 11 Jan 2019Accepted 18 Jan 2019

Keywords

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Pseudo BCK-algebraBL-algebraArtificial intelligenceFilterBoolean filterImplicative filterNormal filterFantastic filter

ABSTRACT

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This paper focuses on the investigation of filters of pseudo BCK-algebra and BL-algebra, important and popular generic com-mutative 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 boundedpseudo BCK-algebra, is the notion of implicative pseudo-filter equivalent to the notion of Boolean filter?” is solved. Based onthis, this paper explores the essential relations between the implicative (Boolean) filter and implicative pseudo BCK-algebra. Acomplete solution to an open problem that “Prove or negate that pseudo BCK-algebras is implicative BCK-algebras if and onlyif every filter of them is implicative filters (or Boolean filter)” is derived. This paper further characterizes the fantastic filter andnormal filter in BL-algebra, then gets the equivalent relation between the two filters, and completely solves two open problemsregarding 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?

© 2019 The Authors. Published by Atlantis Press SARL.This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).

1.

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INTRODUCTION

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The rapid development of computer science, technology, andmath-ematical logic put forwardmany new requirements, thus contribut-ing to the nonclassical logic and the rapid development of modernlogic [1]. The study of fuzzy logic has become a hot topic with sci-entific information and artificial intelligence, which makes fuzzylogic study of algebra and logic inseparable. Fuzzy logic is both themathematical basis of the artificial intelligence and fuzzy reason-ing. Based on the actual background, different forms of fuzzy logicsystem are proposed.

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Logical algebras are the algebraic counterparts of the nonclassi-cal logic and the algebraic foundation of reasoning mechanismin information sciences, computer sciences, theory of control,artificial intelligence, and other important fields. For example,BCK-algebra, BL-algebras, pseudoMTL-algebras, and noncommu-tative residuated lattice are algebraic counterparts of BCK Logic,Basic Logic, monoidal t-norm-based logic, and monoidal logic,respectively [1–3].

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Hájek introduced BL-algebra as algebraic structure for his BasicLogic [3, 5]. Di Nola generalized BL-algebra in a noncommutativeform and introduced the notion of pseudo BL-algebra as a common

*Corresponding author. Email: wwmath@xsyu.edu.cn

extension of BL-algebra in order to express the noncommutativereasoning [4, 6].

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In 1966, BCK-algebra was introduced by Iséki and Imai fromBCK/BCI Logic [7–9]. Iorgulescu established the connec-tions 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 noncommu-tative reasoning [11, 12]. Iorgulescu established the connectionsbetween pseudo BL-algebra and pseudo BCK-algebra [12]. In [13],Wang and Zhang presented the necessary and sufficient condi-tions for residuated lattice and bounded pseudo BCK-algebra to beBoolean algebra.

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Filter theory plays a vital role not only in studying of algebraic struc-ture, but also in nonclassical logic and computer science [14, 15].From logical point of view, various filters correspond to varioussets of provable formulae [16, 17]. For example, based on filter andprime filter in BL-algebra, Hájek proved the completeness of BasicLogic [3]. In [18], Turunen proposed the notions of implicative fil-ter and Boolean filter and proved that implicative filter is equivalentto 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 werefurther studied. Literatures [8, 18, 19, 23–28] further studied fil-ters of BL-algebra, lattice implication algebra, pseudo BL-algebra,

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International Journal of Computational Intelligence SystemsVol. 12(1); 2019, pp. 359–366

DOI: https://doi.org/10.2991/ijcis.2019.125905652; ISSN: 1875-6891; eISSN: 1875-6883 https://www.atlantis-press.com/journals/ijcis/

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pseudo BCK-algebra, R0-algebra, residuated lattice, triangle alge-bra, and the corresponding algebraic structures.

In [29], there is an open problem that “In pseudo BCK-algebraor bounded pseudo BCK algebra, is the notion of implicativepseudo-filter equivalent to the notion of Boolean filter?” Based onthis, [25] proposed another open problem that “Prove or negatethat pseudo BCK-algebras is implicative BCK-algebras if and only ifevery filter of them is implicative filters (or Boolean filter).” The twokinds of filters are important in pseudo BCK-algebra, thus the openproblems are interesting and meaningful topics for further researchand are the motivation of this paper.

We have explored the properties and relations of fuzzy pseudo fil-ter (filter) of pseudo BCK-algebra. After we discussed the equiva-lent conditions of fuzzy normal filter in pseudo BCK-algebra (pP),we proposed fuzzy implicative pseudo filter and its relation withfuzzy Boolean filter in (bounded) pseudo BCK-algebra (pP). Thenthe open problem are partly solved [1]. Based on this, having fur-ther investigated the Boolean filter and implicative filter in pseudoBCK-algebra, we found the essentially equivalent relation betweenthem, 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 fantasticfilter?” and 2. “(Extension property for a normal filter) Under whatsuitable condition extension property for normal filter holds?”

In our previouswork, we have characterized the fuzzy fantastic filterand normal filter of BL-algebra, and discussed the relation betweenthem, then partly solved the two open problems [30]. But so far wehave not got sufficient and necessary condition for a normal filter tobe fantastic. Based on this, we further obtained the relation betweenthem and completely solved the two open problems.

This paper is organized as follows: In Section 2, we present somebasic definitions and results inBL-algebra andpseudoBCK-algebra.In Section 3, we focus on the relation between implicative fil-ter and Boolean filter of pseudo BCK-algebra or bounded pseudoBCK-algebra and give a complete solution to an open problem. InSection 4, based on the result we obtained in Section 3, we inves-tigate the relation between implicative filter (Boolean filter) andimplicative pseudo BCK-algebra and give a complete solution toanother open problem. In Section 5, we recall the concept of fil-ter and the corresponding properties of filter in BL-algebra and wepropose 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 twobinary operations ∧ and ∨. L is called a lattice if for x, y, z ∈ L, thefollowing conditions hold

(1) x ∧ x = x, x ∨ x = x,

(2) x ∧ y = y ∧ x, x ∨ y = y ∨ x,

(3) (x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z),(4) (x ∧ y) ∨ x = x, (x ∨ y) ∧ x = x.

Definition 2. (Balbes and Dwinger [31]) A lattice L is called a dis-tributive lattice if for x, y, z ∈ L, the following conditions hold

(1) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z),(2) 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

(1) (z → x) → (y → x) ≥ (y → z),(2) (y → x) → x ≥ y,

(3) x ≥ x,

(4) x ≥ y and y ≥ x imply x = y,

(5) x → 1 = 1,where x ≤ y means x → y = 1.Definition 4. (Georgescu and Iorgulescu [11]) A(reversed left-)pseudo BCK-algebra is a structure (A, ≥,→,↪, 1), where ≥ is abinary relation on A,→ and↪ are binary operations on A and 1 isan element of A, verifying, for all x, y, z ∈ A, the axioms

(1) (z → x) ↪ (y → x) ≥ y → z, (z ↪ x) → (y ↪ x) ≥ y ↪ z,

(2) (y → x) ↪ x ≥ y, (y ↪ x) → x ≥ y,

(3) x ≥ x,

(4) 1 ≥ x,

(5) x ≥ y and y ≥ x imply x = y,

(6) 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 = 2y𝜋 arctan (ln (

yx)) (if x < y) and

x ↪ y = 0 (if y ≤ x) or x ↪ y = ye–tan(

𝜋x2y

)(if x < y),

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 0such that 0 → x = 1 or 0 ↪ x = 1 for any x ∈ A.

In a pseudo-BCK-algebraAwe can define x– = x → 0, x∼ = x ↪ 0for any x ∈ A.

Proposition 1. (Iorgulescu [35]) Let (A, ≥,→,↪, 1) be a pseudoBCK-algebra. Then the following properties hold for any x, y, z ∈ A

(1) x ≤ y ⇒ y → z ≤ x → z and y ↪ z ≤ x ↪ z,

(2) x ≤ y ⇒ z → x ≤ z → y and z ↪ x ≤ z ↪ y,Pdf_Folio:2

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(3) z ↪ (y → x) = y → (z ↪ x),

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(4) x ≤ y iff x → y = 1 iff x ↪ y = 1,

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(5) x → y = x → x ∧ y, x ↪ y = x ↪ x ∧ y,

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(6) x → y ≤ (z → x) → (z → y), x ↪ y ≤ (z ↪ x) ↪ (z ↪ y),

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(7) y ≤ x → y, y ≤ x ↪ y,

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(8) 1 → x = x = 1 ↪ x,

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(9) (x ∨ y) → z = (x → z) ∧ (y → z), (x ∨ y) ↪ z = (x ↪ z) ∧(y ↪ z),

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(10) x ∨ y = ((x → y) ↪ y) ∧ ((y → x) ↪ x), x ∨ y =((x ↪ y) → y) ∧ ((y ↪ x) → x).Definition 6.

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(Ciungu [34]) A pseudo BCK-algebra with con-dition (pP) (i.e., with pseudo product) is a pseudo BCK-algebra(A, ≥,→,↪, 1) satisfying the condition (pP)

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(pP) there exists, for all x, y ∈ A, x ⊙ y = min{z|x ≤ y → z} =min{z|y ≤ x ↪ z}.Theorem 2.

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(Ciungu [34]) Let (A, ≥,→,↪, 1) be a pseudo BCK-algebras with condition (pP), x⊙ y is defined as min{z|x ≤ y → z}or min{z|y ≤ x ↪ z}, then the followings hold in A

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(1) (x⊙ y) → z = x → (y → z),

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(2) (y⊙ x) ↪ z = x ↪ (y ↪ z),

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(3) (x → y) ⊙ x ≤ x, y, x⊙ (x ↪ y) ≤ x, y,

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(4) x⊙ y ≤ x ∧ y ≤ x, y.

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In the sequel, we shall agree that the operations∨, ∧,⊙have prioritytowards the operations→,↪.

Definition 7.

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(Blount and Tsinakis [33]) A lattice-ordered resid-uated monoid is an algebra (A, ∨, ∧,⊙,→,↪, e) satisfying the fol-lowing conditions:

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(1) (A, ∨, ∧) is a lattice,

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(2) (A,⊙, e) is a monoid,

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(3) x⊙ y ≤ z iff x ≤ y → z iff y ≤ x ↪ z for all x, y, z ∈ A.

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A lattice-ordered residuated monoid A is called integral if x ≤ e forall x ∈ A. In an integral lattice-ordered residuated monoid, we use“1” instead of e.

Lemma 3.

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(Jipsen and Tsinakis [36]) Pseudo BCK-algebra with con-dition (pP) is category equivalent to partially ordered residuated inte-gral monoid.

Definition 8.

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(Zhang [37]) A BCK-algebra (A,→, 1) is called animplicative BCK-algebra if it satisfies (x → y) → x = x for anyx, y ∈ A.

Definition 9.

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(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.

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(Zhang [37]) A pseudo BCK-algebra(A, ≥,→,↪, 1) is called a 2-type implicative pseudo BCK-algebraif it satisfies (x ↪ y) → x = (x → y) ↪ x = x for any x, y ∈ A.

Definition 11.

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(Haveshki, Saeid, and Eslami [19]) A BL-algebrais 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 commutativemonoid, and the following conditions hold for all x, y, z ∈ A:

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(A1) x⊙ y ≤ z if and only if x ≤ y → z,

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(A2) x ∧ y = x⊙ (x → y),

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(A3) (x → y) ∨ (y → x) = 1.

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Example 2. Let A = [0, 1]. Define⊙ and→ as follows:

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x⊙ y = minx, y and x → y = 1 (if x < y) or x → y = y (if x > y).

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Then (A, ∨, ∧,⊙,→, 0, 1) is a BL-algebra.

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An MV-algebra A is a BL-algebra satisfying x–– = x for any x ∈ A.A G ̈odel-algebra is a BL-algebra satisfying x⊙ x = x for any x ∈ A.

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Proposition 4. (Zhang [37]) In a BL-algebra A, the following prop-erties hold for all x, y, z ∈ A

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(1) y → (x → z) = x → (y → z),

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(2) 1 → x = x,

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(3) x ≤ y iff x → y = 1,

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(4) x ∨ y = ((x → y) → y) ∧ ((y → x) → x),

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(5) x ≤ y ⇒ y → z ≤ x → z,

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(6) x ≤ y ⇒ z → x ≤ z → y,

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(7) x → y ≤ (z → x) → (z → y),

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(8) x → y ≤ (y → z) → (x → z),

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(9) x ≤ (x → y) → y,

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(10) x⊙ (x → y) = x ∧ y.

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We shall agree that the operations ∨, ∧,⊙ have priority toward theoperations→.

3.

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THE RELATION BETWEEN IMPLICATIVEFILTER AND BOOLEAN FILTER OFPSEUDO BCK-ALGEBRA OR BOUNDEDPSEUDO BCK-ALGEBRA

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In this section, we recall the definitions of filter, positive implicativepseudo filter, Boolean filter, normal filter, and implicative filter ofpseudo BCK-algebra.

Definition 12.

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(Zhang [29]) A nonempty subset F of pseudo BCK-algebra A is called a (pseudo) filter of A if it satisfies

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(F1) x ∈ F, y ∈ A, x ≤ y ⇒ y ∈ F,

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(F2) x ∈ F, x → y ∈ F or x ↪ y ∈ F ⇒ y ∈ F.

Theorem 5.

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(Wang [1]) A nonempty subset F of a pseudo BCK-algebra A is a filter of A if and only if it satisfies

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(F3) 1 ∈ F,

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(F4) x ∈ F, x → y ∈ F or x ↪ y ∈ F ⇒ y ∈ F.

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In example 1, we can find {0} is a filter.

Theorem 6.

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(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

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(F5) x ∈ F, y ∈ F ⇒ x⊙ y ∈ F,

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(F6) x ∈ F, y ∈ A, x ≤ y ⇒ y ∈ F.

Definition 13.

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(Zhang [29]) Let F be a filter of A, for all x, y ∈ A,F is called a(an)

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(1) Boolean filter if (x → y) ↪ x ∈ F and (x ↪ y) → x ∈ F, thenx ∈ F,

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(2) Prime filter if x ∨ y ∈ F implies x ∈ F or y ∈ F,

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(3) Maximal filter if x ∈ F or x– ∈ F (and x∼ ∈ F),

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(4) Loyal filter if x∈ F and y∈ F implies x → y ∈ F and x ↪ y ∈ F,

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(5) Normal filter if x → y ∈ F iff x ↪ y ∈ F,

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(6) Implicative filter if (x → y) → x ∈ F and (x ↪ y) ↪ x ∈ F,then x ∈ F.

Definition 14.

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(Zhang and Jun [13]) A nonempty subset F of apseudo BCK-algebra A is called a positive implicative filter of A if itsatisfies (F1) and for all x, y ∈ A.

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(F7) x ↪ (y → z) ∈ F, x ↪ y ∈ F implies x ↪ z ∈ F,

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(F8) x → (y ↪ z) ∈ F, x → y ∈ F implies x → z ∈ F.

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Note that any filter of a BCK-algebra is normal.

Theorem 7.

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(Zhang [29]) Let (A; ≤,→,↪, 0, 1) be a boundedpseudo BCK-algebra and F implicative pseudo filter of A. Then

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(1) ∀x ∈ A, ((x → 0) → x) ↪ x ∈ F, that is, (x– → x) ↪ x ∈ F,

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(2) ∀x ∈ A, ((x ↪ 0) ↪ x) → x ∈ F, that is, (x∼ ↪ x) → x ∈ F,

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(3) ∀x, y ∈ A, ((x → y) → x) ↪ x ∈ F,

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(4) ∀x, y ∈ A, ((x ↪ y) ↪ x) → x ∈ F,

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(5) ∀x, y ∈ A, if (x → y) → y ∈ F, then (y → x) ↪ x ∈ F,

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(6) ∀x, y ∈ A, if (x ↪ y) ↪ y ∈ F, then (y ↪ x) → x ∈ F,

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(7) ∀x, y ∈ A, if x ↪ y ∈ F, then ((y ↪ x) → x) ↪ y ∈ F,

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(8) ∀x, y ∈ A, if x → y ∈ F, then ((y → x) ↪ x) → y ∈ F.

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In [29], there is an open problem: “In pseudo BCK-algebra orbounded pseudo BCK algebra, is the notion of implicative pseudofilter equivalent to the notion of Boolean filter?”

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To solve the open problem, we recall the results of the relationbetween the two filters and then get a new solution to the openproblem in pseudo BCK-algebra.

Theorem 8.

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(Zhang [29]) Let (A, ≥,→,↪, 1) be a pseudo BCK-algebra and F a normal pseudo filter of A. Then F is implicative if andonly if F is Boolean.

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With the help of the equivalent conditions of fuzzy normal filterof pseudo BCK-algebra (pP), [1, 39] get the following results andpartly solve the open problem.

Theorem 9.

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(Wang [1]) In bounded pseudo BCK-algebra, everyimplicative pseudo filter is a Boolean filter. In pseudo BCK-algebras(pP), every Boolean filter is an implicative pseudo filter.

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We further investigate the properties of Boolean filter and implica-tive filter which make the relation between the two filters muchclear, and get the solution for the open problem.

Theorem 10.

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Implicative pseudo filter is Boolean filter in pseudoBCK-algebra.

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Proof. Let F be an implicative pseudo filter of A. Then ∀x ∈ A,suppose (x → y) ↪ x ∈ F,

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from x ≤ ((x → y) ↪ x) → x,

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so (((x → y) ↪ x) → x) → y ≤ x → y,

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and ((((x → y) ↪ x) → x) → y) → (x → y) = 1.

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On the other hand, x → y ≤ ((x → y) ↪ x) → x,

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so we get ((((x → y) ↪ x) → x) → y) → (x → y) ≤((((x → y) ↪ x) → x) → y) → (((x → y) ↪ x) → x).

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Then ((((x → y) ↪ x) → x) → y) → (((x → y) ↪ x) → x) =1 ∈ F,

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and ((x → y) ↪ x) → x ∈ F, since F is an implicative filter.

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Combine that (x → y) ↪ x ∈ F, according to the definition offilter, we get x ∈ F.

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Similarly, suppose (x ↪ y) → x ∈ F,

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from x ≤ ((x ↪ y) → x) ↪ x,

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so (((x ↪ y) → x) ↪ x) ↪ y ≤ x ↪ y,

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and (((x ↪ y) → x) ↪ x) ↪ y) ↪ (x ↪ y) = 1.

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By x ↪ y ≤ ((x ↪ y) → x) ↪ x,

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so we get ((((x ↪ y) → x) ↪ x) ↪ y) ↪ (x ↪ y) ≤((((x ↪ y) → x) ↪ x) ↪ y) ↪ (((x ↪ y) → x) ↪ x).

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Then ((((x ↪ y) → x) ↪ x) ↪ y) ↪ (((x ↪ y) → x) ↪ x) =1 ∈ F,

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and ((x ↪ y) → x) ↪ x ∈ F, since F is an implicative filter.

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Combine that (x ↪ y) → x ∈ F, according to the definition offilter, then we get x ∈ F.

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According to the definition, then F is a Boolean filter of A.

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Similarly, we can get

Theorem 11.

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In pseudo BCK-algebra, every Boolean filter is animplicative pseudo filter.

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Proof. Let F be an Boolean filter of A. Then ∀x ∈ A, suppose(x → y) → x ∈ F,

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from x ≤ ((x → y) → x) ↪ x,

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so (((x → y) → x) ↪ x) → y ≤ x → y,

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and (((x → y) → x) ↪ x) → y) ↪ (x → y) = 1.

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On the other hand, x → y ≤ ((x → y) → x) ↪ x,

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so we get ((((x → y) → x) ↪ x) → y) ↪ (x → y) ≤((((x → y) → x) ↪ x) → y) ↪ ((x → y) → x) ↪ x).

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Then ((((x → y) → x) ↪ x) → y) ↪ (((x → y) → x) ↪ x) =1 ∈ F and ((x → y) → x) ↪ x ∈ F, since F is an implicative filter.

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Combine that (x → y) → x ∈ F, according to the definition offilter, then we get x ∈ F.

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Similarly, suppose (x ↪ y) ↪ x ∈ F,Pdf_Folio:4

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from x ≤ ((x ↪ y) ↪ x) → x,

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so (((x ↪ y) ↪ x) → x) ↪ y ≤ x ↪ y,

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and (((x ↪ y) ↪ x) → x) ↪ y) → (x ↪ y) = 1.

ID:p0965

On the other hand, x ↪ y ≤ ((x ↪ y) ↪ x) → x,

ID:p0970

so we get ((((x ↪ y) ↪ x) → x) ↪ y) → (x ↪ y) ≤((((x ↪ y) ↪ x) → x) ↪ y) → ((x ↪ y) ↪ x) → x).

ID:p0975

Then ((((x ↪ y) ↪ x) → x) ↪ y) → (((x ↪ y) ↪ x) → x) =1 ∈ F and (x ↪ y) ↪ x) → x ∈ F, since F is a Boolean filter.

ID:p0980

Combine that (x ↪ y) ↪ x ∈ F, according to the definition offilter, we get x ∈ F.

ID:p0985

Thus F is an implicative pseudo filter of A.

ID:p0990

From the above results, we can get the following results as a solutionfor the open problem.

Theorem 12.

ID:p0995

In pseudo BCK-algebra or bounded pseudo BCK-algebra, the notion of implicative pseudo filter is equivalent to thenotion of Boolean filter.

ID:p1000

Remark 1. The equivalent relation between the implicative filterand Boolean filter is of importance in the study of logical algebras.For example, when studying of the pseudo BCK-algebra, implica-tive filter and Boolean filter can reflect the algebraic structure of thepseudo BCK-algebra. When we get the equivalent relation betweenthem, and based on some other results we obtained [1, 16, 30], wecan completely solve some other problems like this.

4.

ID:TI0035

THE RELATION BETWEEN IMPLICATIVEFILTER (BOOLEAN FILTER) ANDIMPLICATIVE PSEUDO BCK-ALGEBRA

ID:p1005

The filters play a vital role in representing the algebras, such as in apseudo BL-algebra A,

ID:p1010

A is pseudo MV-algebra if and only if every filter of A is a pseudoMV filter,

ID:p1015

A is aGodel-algebra if and only if every filter ofA is a pseudoG filter,

ID:p1020

A is a Boolean algebra if and only if every filter of A is a Booleanfilter.

ID:p1025

The similar relation between implicative or Boolean filter andimplicative pseudo BCK-algebra is not obtained, yet. For this rea-son, [25] set it as an open problem.

ID:p1030

Prove or negate that pseudo BCK-algebras is implicative BCK-algebras if and only if every filter of them is implicative filters (orBoolean filter).

ID:p1035

[1] partly solve the open problem. Based on this, we can further getthe following results as a solution for the open problem.

Proposition 13.

ID:p1040

Let A be a pseudo BCK algebra. Then the followingstatements are equivalent:

ID:p1045

(1) A is an implicative BCK-algebras,

ID:p1050

(2) Every filter of A is an implicative filters (or Boolean filter),

ID:p1055

(3) {1} is an implicative filters (or Boolean filter).

ID:p1065

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 Ais a 1-type (or 2-type) implicative pseudo BCK-algebra. Then forevery 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 fil-ters of them is implicative pseudo filters (Boolean filter), so neces-sity is obvious.

ID:p1070

(2) ⇒ (3)obvious.

ID:p1075

(3) ⇒ (1)Now suppose every pseudo filter of a pseudoBCK-algebraA is an implicative pseudo filters (Boolean filter), then pseudo filter{1} is an implicative pseudo filters (Boolean filter).

ID:p1080

For any x, y ∈ A, from x ≤ ((x ↪ y) → x) ↪ x (by Theorem (7)),

ID:p1085

we get (((x ↪ y) → x) ↪ x) ↪ y ≤ x ↪ y (by Theorem (1)),

ID:p1090

then ((((x ↪ y) → x) ↪ x) ↪ y) → (x ↪ y) = 1 (by Definition(6)).

ID:p1095

From x ↪ y ≤ ((x ↪ y) → x) ↪ x (by Definition (2)),

ID:p1100

we get ((((x ↪ y) → x) ↪ x) ↪ y) → (x ↪ y) ≤((((x ↪ y) → x) ↪ x) ↪ y) → (((x ↪ y) → x) ↪ x) (byTheorem (2)),

ID:p1105

then ((((x ↪ y) → x) ↪ x) ↪ y) → (((x ↪ y) → x) ↪ x) =1 ∈ {1}, (by Definition (4)),

ID:p1110

we get ((x ↪ y) → x) ↪ x = 1 ∈ {1}, since {1} is an implicativepseudo filters, that is, (x ↪ y) → x ≤ x (by Definition (6)).

ID:p1115

On the other hand, x ≤ (x ↪ y) → x, then (x ↪ y) → x = x.

ID:p1120

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. Onthe other hand, x ≤ (x → y) ↪ x, then (x → y) ↪ x = x.

ID:p1125

From above results, we find that A is a 1-type implicative pseudoBCK-algebra, then A is implicative BCK-algebra.

5.

ID:TI0040

THE RELATION BETWEEN FANTASTICFILTER AND NORMAL FILTER INBL-ALGEBRA

ID:p1130

Here we recall some kinds of filters in BL-algebra. Similar with thepseudo BCK-algebra, here we recall some definitions and resultswhich will be needed. Reader can refer to [3, 16, 30, 37, 40–42].

Definition 15.

ID:p0200

A filter of a BL-algebra A is a nonempty subset F ofA such that for all x, y ∈ A,

ID:p1145

(F1) if x, y ∈ F, then x⊙ y ∈ F,

ID:p1150

(F2) if x ∈ F and x ≤ y, then y ∈ F.

Proposition 14.

ID:p1155

Let F be a nonempty subset of a BL-algebra A. ThenF is a filter of A if and only if the following conditions hold

ID:p1160

(1) 1 ∈ F,

ID:p1165

(2) x, x → y ∈ F implies y ∈ F.Pdf_Folio:5

W.Wang et al. / International Journal of Computational Intelligence Systems 12(1) 359–366 363

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ID:p1170

A filter F of a BL-algebra A is proper if F ≠ A, that is, 0∈A.

ID:p1175

In example 2, we can find ( 12 , 1) is a filter.

Definition 16.

ID:p1185

A proper filter F is prime if for any x, y, z ∈ A,x ∨ y ∈ F implies x ∈ F or y ∈ F.

Theorem15.

ID:p1190

A proper filter F is prime if for any x, y ∈ A, x → y ∈ For y → x ∈ F.

Definition 17.

ID:p1200

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.

ID:p1210

Let F be a nonempty subset of a BL-algebraA. ThenF is called a fantastic filter of A if for all x, y, z ∈ A, the followingconditions hold:

ID:p1215

(1) 1 ∈ F,

ID:p1220

(2) z → (y → x) ∈ F, z ∈ F implies ((x → y) → y) → x ∈ F.

Definition 19.

ID:p1230

Let F be a nonempty subset of a BL-algebraA. ThenF is called an implicative filter ofA if for all x, y, z ∈ A, the followingconditions hold:

ID:p1235

(1) 1 ∈ F,

ID:p1240

(2) x → (y → z) ∈ F, x → y ∈ F imply x → z ∈ F.

Definition 20.

ID:p1250

Let F be a nonempty subset of a BL-algebraA. ThenF is called a positive implicative filter of A if for all x, y, z ∈ A, thefollowing conditions hold:

ID:p1255

(1) 1 ∈ F,

ID:p1260

(2) x → ((y → z) → y) ∈ F, x ∈ F imply y ∈ F.

Definition 21.

ID:p1270

A filter F of A is called Boolean if x∨x– ∈ F for anyx ∈ A.

Definition 22.

ID:p1280

Let F be a filter of A. F is called an ultra filter of Aif it satisfies x ∈ F or x– ∈ F for all x ∈ A.

Definition 23.

ID:p1290

Let F be a filter ofA. F is called an obstinate filter ofA if it satisfies x ∉ F and y ∉ F implies x → y ∈ F for all x, y ∈ A.

Definition 24.

ID:p1300

A proper filter of a BL-algebra A is called maximalif it is not properly contained in any other proper filter of A.

Proposition 16.

ID:p1305

A proper filter F of a BL-algebra A is maximal ifand only if ∀x∈F, ∃n ∈ N, such that xn ∈ F.

ID:p1310

In order to investigate the essential relations among the filters, andbased on the past work [1, 16, 30], we characterize the followingfilters.

Theorem 17.

ID:p1315

Let F be a filter of a BL-algebra A. F is a normal filterif and only if one of the followings holds for all x, y ∈ A

ID:p1320

(1) (y → x) → x ∈ F implies (x → y) → y ∈ F,

ID:p1325

(2) x–– ∈ F implies x ∈ F.

Theorem18.

ID:p1330

Let F be a filter of A. Then the followings are equivalentfor all x, y, z ∈ A

ID:p1335

(1) F is a Boolean filter of A,

ID:p1340

(2) (x → y) → x ∈ F implies x ∈ F,

ID:p1345

(3) x– → x ∈ F implies x ∈ F.

Theorem 19.

ID:p1350

Let F be a Boolean filter of A, then for all x, y ∈ A

ID:p1355

(1) x → y ∈ F implies ((y → x) → x) → y ∈ F,

ID:p1360

(2) (x → x–) → x– ∈ F,

ID:p1365

(3) x–– ∈ F implies x ∈ F,

ID:p1370

(4) x → (x → y) ∈ F implies x → y ∈ F.

Theorem 20.

ID:p1375

Let F be a filter of a BL-algebra A. Then the followingsare equivalent for all x, y, z ∈ A:

ID:p1380

(1) F is a fantastic filter,

ID:p1385

(2) y → x ∈ F ⇒ ((x → y) → y) → x ∈ F,

ID:p1390

(3) x–– → x ∈ F.

Theorem 21.

ID:p1395

Let F be a filter of a BL-algebra A. Then for anyx, y, z ∈ A the followings are equivalent:

ID:p1400

(1) F is an implicative filter of A,

ID:p1405

(2) y → (y → x) ∈ F implies y → x ∈ F,

ID:p1410

(3) x → x⊙ x ∈ F.

ID:p1415

By the above results, we can get the following results.

Corollary 22.

ID:p1420

In an MV-algebra, every filter is a fantastic filter.

Corollary 23.

ID:p1425

In a G ̈odel-algebra, every filter is an implicative filter.

ID:p1430

Based on this, we get some relations among the filters in BL-algebras.

Theorem 24.

ID:p1435

Each Boolean filter is equivalent to a positive implica-tive filter in BL-algebras.

Theorem 25.

ID:p1440

Each ultra filter is equivalent to an obstinate filter inBL-algebras.

Theorem 26.

ID:p1445

Each ultra filter f a BL-algebra A is a fantastic filter.

ID:p1450

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 afantastic filter.

Theorem 27.

ID:p1455

Let F be a filter of A. Then F is a Boolean filter if andonly if it is an implicative and normal filter.

ID:p1460

Proof. If F is a Boolean filter, then by Theorem 17, 19, and 21, weknow that F is an implicative and normal filter.

ID:p1465

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 implicativefilter. Then since F is also a normal filter, wehave x ∈ F, thus we getthen F is a Boolean filter.

Theorem 28.

ID:p1470

Let F be a filter of A. Then F is a Boolean filter if andonly if it is an implicative and fantastic filter.

ID:p1475

Based on the previous work and the above results, we get the essen-tial relation.

Theorem 29.

ID:p1480

Let F be a filter of A. Then F is an implicative andnormal filter if and only if it is an implicative and fantastic filter.

ID:p1485

In [26], there are two open problems in BL-algebras:

Pdf_Folio:6

364 W.Wang et al. / International Journal of Computational Intelligence Systems 12(1) 359–366ID:p1170ID:p1175ID:p1185ID:p1190ID:p1200ID:p1210ID:p1215ID:p1220ID:p1230ID:p1235ID:p1240ID:p1250ID:p1255ID:p1260ID:p1270ID:p1280ID:p1290ID:p1300ID:p1305ID:p1310ID:p1315ID:p1320ID:p1325ID:p1330ID:p1335ID:p1340ID:p1345ID:p1350ID:p1355ID:p1360ID:p1365ID:p1370ID:p1375ID:p1380ID:p1385ID:p1390ID:p1395ID:p1400ID:p1405ID:p1410ID:p1415ID:p1420ID:p1425ID:p1430ID:p1435ID:p1440ID:p1445ID:p1450ID:p1455ID:p1460ID:p1465ID:p1470ID:p1475ID:p1480ID:p1485

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1.

ID:p1490

Under what suitable condition a normal filter becomes a fan-tastic filter?

2.

ID:p1495

(Extension property for a normal filter) Under what suitablecondition extension property for normal filter holds?

ID:p1500

[30] proposed solutions for the two open problems by the fuzzy fil-ters, respectively as follows:

1.

ID:p1505

Suitable condition should be

ID:p1510

(1) BL-algebra is an MV-algebra,

ID:p1515

(2) The filter is an implicative filter,

ID:p1520

(3) The filter is an obstinate filter,

ID:p1525

(4) The filter is an ultra filter.

2.

ID:p1530

Under the condition

ID:p1535

(1) If A is an MV-algebra,

ID:p1540

(2) F is a normal and implicative filter of A,

ID:p1545

(3) F is a normal and obstinate filter of A,

ID:p1550

(4) F is a normal and ultra filter of A.

ID:p1555

Extension property for normal filter holds.

ID:p1560

According to the above theorem and corollary, we can get the equiv-alent relation between the two filters and give answers to the openproblems.

ID:p1565

The suitable condition should be

ID:p1570

(1) A normal filter is equivalent to a fantastic filter.

ID:p1575

(2) Extension property for a normal filter holds.

ID:p1580

We further characterized the filters in BL-algebra. Compared withthe solutions in [30], the condition that the filter is an obstinate filteror the filter is an ultra filter is redundant.

6.

ID:TI0045

CONCLUSION

ID:p1585

We discuss the properties of implicative filters and Boolean filtersin pseudo BCK-algebra. Based on the results and previous work, wecompletely solve an open problemwhich is important to deep studyof the algebraic structure of pseudo BCK-algebra. Based on this, weprove that pseudo BCK-algebra is implicative BCK-algebra if andonly if every filter of them is implicative filter (or Boolean filter).

ID:p1590

We further characterize the filters in BL-algebra. Compared withthe solutions in [30], the condition that the filter is an obstinate filteror the filter is an ultra filter is redundant.

ID:p1595

In the future work, we will extend the corresponding filter theoryto different algebraic structures, and study the congruence relationsinduced by the filters.

ID:TI0050

ACKNOWLEDGMENTS

ID:p1600

This Research work is supported by the National Natural Science Founda-tion of P. R. China (Grant No. 11571281, 61673320); Xi’an Shiyou Univer-sity College Students Innovation and Entrepreneurship Training ProgramFunding Project (Grant No. 201819062); the Fundamental Research Fundsfor the Central Universities (Grant No. 2682017ZT12).

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ID:TI0055

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