EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 6190
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Pseudo-Dual B-Algebra
Jessa Mae Sale Leuveras1,∗, Katrina Belleza Fuentes1
1 Department of Computer, Information Sciences and Mathematics, School of Arts and
Sciences, University of San Carlos, 6000 Cebu City, Philippines

Abstract. This paper introduces the structure of a pseudo-dual B-algebra and some of its subsets,
specifically pseudo-dual B-subalgebra and pseudo-dual B-filter. Furthermore, it presents some of
their properties and gives a characterization of a pseudo-dual B-subalgebra. This study also pro-
vides the relationship between a pseudo-dual B-subalgebra and a pseudo-dual B-filter. Moreover,
this paper presents a Python program that was used to perform the calculations needed to ver-
ify the example of a pseudo-dual B-algebra and to verify the independence of the axioms in a
pseudo-dual B-algebra.
2020 Mathematics Subject Classifications: 03G25, 08A05, 08A30
Key Words and Phrases: Dual B-algebra, pseudo-dual B-algebra, pseudo-dual B-subalgebra,
pseudo-dual B-filter

1. Introduction

In 1966, Y. Imai and K. Iseki [1] introduced two classes of algebra: BCK-algebra
and BCI -algebra. In 2001, G. Georgescu and A. Iorgulescu [2] extended the concept of
BCK-algebra to pseudo-BCK-algebra. In 2008, W. Dudek and Y. Jun [3] introduced the
notion of pseudo-BCI algebra as an extension of BCI -algebra and investigated some of
its properties. In 2002, J.Neggers and H.S. Kim [4] introduced and investigated a new
class of algebra which is related to several classes of algebra such as BCH/BCI/BCK-
algebra called B-algebra. In 2007, A. Walendziak [5] introduced a generalization of B-
algebra called BF-algebra and investigated some properties of ideals and normal-ideals
in BF-algebra and gave some characterization of them. Simultaneously, H.S. Kim and
Y.H. Kim [6] introduced BE-algebra as a generalization of BCK-algebra and studied the
filters of BE-algebra. In 2020, H. Al-Malki and D. Al-Kadi [7] introduced the structure
of pseudo-BF/BF ∗-algebra which is a generalization of BF-algebra together with pseudo-
subalgebra, pseudo-ideal, pseudo-normal-ideal, and pseudo-atoms as its subsets. On the

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6190
Email addresses: 23104906@usc.edu.ph (J. M. S. Leuveras),
kebelleza@usc.edu.ph (K. B. Fuentes)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 2 of 12

other hand, in 2013, R. Borzooei and et al., [8] introduced the notion of pseudo BE-
algebra which is a generalization of BE-algebra. They also studied the concepts of pseudo-
subalgebra, pseudo-filter, and pseudo-upper-set and proved that, under some conditions,
pseudo-subalgebra can be a pseudo-filter. Furthermore, in 2019, K. Belleza and J. Vilela
[9] introduced and characterized the notion of a dual B-algebra. In 2022, K. Belleza and
J. Albaracin [10] introduced the concepts of dual B-subalgebra and dual B-filter.

Motivated by the aforementioned studies, this paper aims to introduce the structure
of pseudo-dual B-algebra and some of its subsets, particularly pseudo-dual B-subalgebra
and pseudo-dual B-filter. Some of their properties will also be investigated. Furthermore,
the relationship between pseudo-dual B-subalgebra and pseudo-dual B-filter will be given.
Moreover, this study can contribute to the understanding of how the term “pseudo” is
used in algebraic structures, and the results of this study could be used to develop further
studies.

2. Preliminaries

Definition 1. [9] A dual B-algebra (or dB-algebra) X is a triple (X, •, 1) where X is a
nonempty set with a binary operation “•” and a constant 1 satisfying the following axioms
for all x, y, z in X:
(DB1) x • x = 1;
(DB2) 1 • x = x; and
(DB3) x • (y • z) = ((y • 1) • x) • z.

Definition 2. [10] Let X be a dB-algebra and S a nonempty subset of X. Then S is
called a dual B-subalgebra (or dB-subalgebra) of X if S itself is a dB-algebra with binary
operation of X on S.

Definition 3. [10] Let X be a dB-algebra. A subset F of X is called a dual B-filter (or
dB-filter) if it satisfies the following axioms for all x, y in X:
(i) 1 ∈ F ; and
(ii) x • y ∈ F and x ∈ F imply y ∈ F .

3. Pseudo-Dual B-Algebra

Definition 4. A pseudo-dual B-algebra (or pseudo-dB-algebra)X is a quadruple (X, •, ∗, 1)
where X is a nonempty set with two binary operations “ • ” and “ ∗ ” and a constant 1
satisfying the following axioms for all x, y, z in X:
(PDB1) x • x = 1 and x ∗ x = 1;
(PDB2) 1 • x = x and 1 ∗ x = x; and
(PDB3) x • (y ∗ z) = ((y ∗ 1) • x) • z and x ∗ (y • z) = ((y • 1) ∗ x) ∗ z.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 3 of 12

In a pseudo-dB-algebra X, define a binary relation “ ≤ ” by x ≤ y ⇐⇒ x • y =
1 ⇐⇒ x ∗ y = 1, for any x, y ∈ X.

Note: A Python program (see Appendix) was used to perform the calculations needed to
verify Example 1, Example 2, Example 3, and Example 4.

Example 1. Consider (X, •, ∗, 1), where X = {1,−1}, “ • ” is the usual multiplication,
and “ ∗ ” is the usual division. Then X is a pseudo-dB-algebra.

In Definition 4, the three axioms (PDB1), (PDB2), and (PDB3) are independent,
which ensures that one axiom cannot be derived from the other to avoid redundancy.
This is illustrated in Example 2, Example 3, and Example 4, where the axioms (PDB1)
and (PDB2) hold but axiom (PDB3) fails, the axioms (PDB1) and (PDB3) hold but
axiom (PDB2) fails, and the axioms (PDB2) and (PDB3) hold but axiom (PDB1) fails,
respectively. Moreover, Example 2, Example 3, and Example 4 are counterexamples of a
pseudo-dB-algebra.

Example 2. Let X = {1, a, b}. Define the binary operations “ • ” and “ ∗ ” on X by the
following Cayley tables:

• 1 a b
1 1 a b
a a 1 b
b b b 1

∗ 1 a b
1 1 a b
a a 1 a
b b a 1

Then the axioms (PDB1) and (PDB2) hold. However, (PDB3) fails since a • (b ∗ b) =
a • 1 = a ̸= 1 = b • b = (b • a) • b = ((b ∗ 1) • a) • b.

Example 3. Let X = {1, a, b}. Define the binary operations “ • ” and “ ∗ ” on X by the
following Cayley tables:

• 1 a b
1 1 1 1
a 1 1 1
b 1 1 1

∗ 1 a b
1 1 1 1
a 1 1 a
b 1 a 1

Then the axioms (PDB1) and (PDB3) hold. However, axiom (PDB2) fails since 1 • a =
1 ̸= a.

Example 4. Let X = {1, a}. Define the binary operations “•” and “∗” on X by x•y = y
and x ∗ y = y for all x, y ∈ X. Then the axioms (PDB2) and (PDB3) hold. However,
axiom (PDB1) fails since a • a = a ̸= 1.

Remark 1. In any pseudo-dB-algebra X, if x • y = x ∗ y for all x, y in X, then X is a
dB-algebra.

Remark 2. Any two dB-algebras do not necessarily construct a pseudo-dB-algebra. This
is illustrated in the next example.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 4 of 12

Example 5. Let X = {1, a, b, c}. Define the binary operations “ • ” and “ ∗ ” on X by
the following Cayley tables:

• 1 a b c
1 1 a b c
a a 1 c b
b b c 1 a
c c b a 1

∗ 1 a b c
1 1 a b c
a c 1 a b
b b c 1 a
c a b c 1

Then (X, •, 1) and (X, ∗, 1) are dB-algebras by [9] and [11], respectively. Since a• (a∗c) =
a • b = c ̸= a = b • c = (c • a) • c = ((a ∗ 1) • a) • c, (X•, ∗, 1) is not a pseudo-dB-algebra.

The following lemma provides some properties of a pseudo-dB-algebra.

Lemma 1. In a pseudo-dB-algebra X, the following properties hold for any x, y, z in X:

(i) If 1 ≤ x, then x = 1;

(ii) x • 1 = 1 if and only if x ∗ 1 = 1;

(iii) (x • 1) ∗ (x • y) = y and (x ∗ 1) • (x ∗ y) = y;

(iv) (x • y) ∗ 1 = y ∗ x and (x ∗ y) • 1 = y • x;

(v) If z • x = z • y (or z ∗ x = z ∗ y), then x = y;

(vi) If x • y = 1 (or x ∗ y = 1), then x = y;

(vii) If x • y = 1, then (x • z) ∗ (y • z) = 1;

(viii) If x ∗ y = 1, then (x ∗ z) • (y ∗ z) = 1;

(ix) x = (x • 1) ∗ 1 and x = (x ∗ 1) • 1; and

(x) If x • 1 = y • 1 (or x ∗ 1 = y ∗ 1), then x = y.

Proof. Let X be a pseudo-dB-algebra and x, y, z ∈ X.

(i) Suppose 1 ≤ x. Then 1 • x = 1 ∗ x = 1. By (PDB2), x = 1 • x = 1 ∗ x. Hence, x = 1.

(ii) Suppose x•1 = 1. By (PDB1), (PDB3), and (PDB2), x∗1 = x∗(x•x) = [(x • 1) ∗ x]∗
x = (1∗x)∗x = x∗x = 1. For the converse, suppose x∗1 = 1. By (PDB1), (PDB3),
and (PDB2), x • 1 = x • (x ∗ x) = [(x ∗ 1) • x] • x = (1 • x) • x = x • x = 1. Hence,
x • 1 = 1 if and only if x ∗ 1 = 1.

(iii) By (PDB3), (PDB1), and (PDB2), (x • 1) ∗ (x • y) = [(x • 1) ∗ (x • 1)] ∗ y = 1 ∗ y = y
and (x ∗ 1) • (x ∗ y) = [(x ∗ 1) • (x ∗ 1)] • y = 1 • y = y.

(iv) By (PDB1) and (PDB3), and (iii), (x•y)∗1 = (x•y)∗(x•x) = [(x • 1) ∗ (x • y)]∗x =
y ∗ x and (x ∗ y) • 1 = (x ∗ y) • (x ∗ x) = [(x ∗ 1) • (x ∗ y)] • x = y • x.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 5 of 12

(v) Suppose z • x = z • y. Then (z • 1) ∗ (z • x) = (z • 1) ∗ (z • y) implies x = y by (iii).
Suppose also that z ∗ x = z ∗ y. Then (z ∗ 1) • (z ∗ x) = (z ∗ 1) • (z ∗ y) implies x = y
by (iii).

(vi) Suppose x • y = 1. By (PDB1), x • y = 1 = x • x. Hence, x • y = x • x implies y = x
by (v). By a similar way, if x ∗ y = 1, then x ∗ y = x ∗ x, which implies y = x.

(vii) Suppose x • y = 1 Then x = y by (vi). Hence, (x • z) ∗ (y • z) = (x • z) ∗ (x • z) = 1
by (PDB1).

(viii) Suppose x ∗ y = 1. Then x = y by (vi). Hence, (x ∗ z) • (y ∗ z) = (x ∗ z) • (x ∗ z) = 1
by (PDB1).

(ix) By (PDB1), (PDB2), and (PDB3), 1 = x • x = 1 ∗ (x • x) = [(x • 1) ∗ 1] ∗ x. Hence,
x = (x • 1) ∗ 1 by (vi). By a similar way, 1 = x ∗ x = 1 • (x ∗ x) = [(x ∗ 1) • 1] • x,
which implies x = (x ∗ 1) • 1.

(x) Suppose x • 1 = y • 1. By (PDB1), (PDB2),(PDB3), and (ix) we have 1 = x • x =
1∗(x•x) = [(x • 1) ∗ 1]∗x = [(y • 1) ∗ 1]∗x = y∗x. Hence, y = x by (vi). By a similar
way, if x∗1 = y∗1, then 1 = x∗x = 1•(x∗x) = [(x ∗ 1) • 1]•x = [(y ∗ 1) • 1]•x = y•x,
which implies y = x.

The following proposition shows that the relation “ ≤ ” on a pseudo-dB-algebra X is
an equivalence relation.
Proposition 1. Let X be a pseudo-dB-algebra. Then the relation “ ≤ ” on a set X is an
equivalence relation.

Proof. Let X be a pseudo-dB-algebra and x, y, z ∈ X. Note that x • x = x ∗ x = 1
by (PDB1). Hence, x ≤ x and so “ ≤ ” is reflexive. Suppose x ≤ y. Then x • y = 1 and
x ∗ y = 1. By Lemma 1(vi), x = y. Thus, y ≤ x and so “ ≤ ” is symmetric. Moreover,
suppose x ≤ y and y ≤ z. Since we have shown that x ≤ y implies x = y, x ≤ z. Hence,
“ ≤ ” is transitive. Therefore, “ ≤ ” is an equivalence relation.

Theorem 1. Let X = (X, •, ∗, 1) be any algebra of type (2, 2, 0). If X is a pseudo-dB-
algebra, then for any x, y, z in X,
(i) x ≤ x;

(ii) x = (x • 1) ∗ 1 and x = (x ∗ 1) • 1; and

(iii) (x ∗ y) • (x ∗ z) = y • z and (x • y) ∗ (x • z) = y ∗ z.
Proof. Suppose X is a pseudo-dB-algebra and and x, y, z ∈ X. Then x • x = 1 and

x ∗ x = 1 by (PDB1). Hence, x ≤ x. Moreover, by Lemma 1(ix), x = (x • 1) ∗ 1 and
x = (x∗1)•1. By (PDB3) and Lemma 1(iii), (x∗ y)• (x∗ z) = [(x ∗ 1) • (x ∗ y)]• z = y • z
and (x • y) ∗ (x • z) = [(x • 1) ∗ (x • y)] ∗ z = y ∗ z. Therefore, X satisfies (i), (ii), and
(iii).

The next Proposition gives a condition for the converse of Theorem 1 to hold.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 6 of 12

Proposition 2. Let X = (X, •, ∗, 1) be any algebra of type (2, 2, 0). For any x, y, z in X,
if X satisfies x • (y ∗ z) = ((y ∗ 1) • x) • z, x ∗ (y • z) = ((y • 1) ∗ x) ∗ z, and (i), (ii), (iii)
of Theorem 1, then X is a pseudo-dB-algebra.

Proof. It remains to show (PDB2). By (iii), (i), and (ii), 1 • x = (x ∗ 1) • (x ∗ x) =
(x ∗ 1) • 1 = x and 1 ∗ x = (x • 1) ∗ (x • x) = (x • 1) ∗ 1 = x. Thus, X satisfies (PDB2).
Therefore, X is a pseudo-dB-algebra.

4. Pseudo-Dual B-Subalgebra and Pseudo-Dual B-Filter

In this section, X is a pseudo-dB-algebra, unless otherwise is stated.

Definition 5. Let S be a nonempty subset of X. Then S is called a pseudo-dual B-
subalgebra (or pseudo-dB-subalgebra) of X if S itself is a pseudo-dB-algebra with the
binary operations of X on S.

Remark 3. If S is a pseudo-dB-subalgebra of X, then 1 ∈ S.

The following theorem is a characterization of a pseudo-dB-subalgebra.

Theorem 2. Let S be a nonempty subset of X. Then S is a pseudo-dB-subalgebra of X
if and only if for all x, y in S, x • y ∈ S and x ∗ y ∈ S.

Proof. Let S be a nonempty subset of X with x • y ∈ S and x ∗ y ∈ S for all x, y in S.
Note that S satisfies (PDB1), (PDB2), and (PDB3) with 1 = x • x = x ∗ x ∈ S. Thus, S
is itself a pseudo-dB-algebra. The converse follows immediately by definition of a binary
operator.

Proposition 3. Let S be a pseudo-dB-subalgebra of X. Then for all x, y in S:

(i) If x ∗ y ∈ S, then y • x ∈ S; and

(ii) If x • y ∈ S, then y ∗ x ∈ S.

Proof. For x, y ∈ S, let x ∗ y ∈ S and x • y ∈ S. By Lemma 1(iv),(x ∗ y) • 1 = y • x
and (x • y) ∗ 1 = y ∗ x. Since x ∗ y, x • y, 1 ∈ S, (x ∗ y) • 1 ∈ S and (x • y) ∗ 1 ∈ S. Hence,
y • x ∈ S and y ∗ x ∈ S.

Proposition 4. Let S be a pseudo-dB-subalgebra of X. If T is a pseudo-dB-subalgebra of
S, then T is a pseudo-dB-subalgebra of X as well.

Proof. Suppose S is a pseudo-dB-subalgebra of X. Suppose further that T is a pseudo-
dB-subalgebra of S. Then 1 ∈ T by Remark 3. Hence, T ̸= ∅. Let x, y ∈ T . Since T
is a pseudo-dB-subalgebra of S, x • y, x ∗ y ∈ T for all x, y ∈ T . Therefore, T is a
pseudo-dB-subalgebra of X.

The next theorem shows that the intersection of a nonempty collection of pseudo-dB-
subalgebra of X is a pseudo-dB-subalgebra.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 7 of 12

Theorem 3. Let {Sα : α ∈ I} be a nonempty collection of pseudo-dB-subalgebra of X.
Then

∩
α∈I

Sα is also a pseudo-dB-subalgebra of X.

Proof. Since Sα is a pseudo-dB-subalgebra for each α, 1 ∈ Sα for all α ∈ I. Hence,
1 ∈

∩
α∈I

Sα and
∩
α∈I

Sα ̸= ∅. Let x, y ∈
∩
α∈I

Sα. Then x, y ∈ Sα for all α ∈ I. Since Sα is a

pseudo-dB-subalgebra for each α, x•y, x∗y ∈ Sα for all α ∈ I. Thus, x•y, x∗y ∈
∩
α∈I

Sα.

Therefore,
∩
α∈I

Sα is a pseudo-dB-subalgebra of X.

Definition 6. Let F be a nonempty subset of X. Then F is called a pseudo-dual B-filter
(or pseudo-dB-filter) of X if it satisfies the following axioms for all x, y in X:
(PDBF1) 1 ∈ F ; and
(PDBF2) x • y ∈ F , x ∗ y ∈ F , and x ∈ F imply y ∈ F .

Lemma 2. If F is a pseudo-dB-filter of X, then for all x, y, z in X,

(i) If x ≤ y and x ∈ F , then y ∈ F ; and

(ii) If x ≤ (y • z), x ≤ (y ∗ z), and x, y ∈ F , then z ∈ F .

Proof. Let F be a pseudo-dB-filter of X and x, y, z ∈ X.

(i) Suppose x ≤ y and x ∈ F . Since x ≤ y, x • y = 1 and x ∗ y = 1. By (PDBF1),
1 ∈ F . Hence, x • y ∈ F and x ∗ y ∈ F . Thus, y ∈ F by (PDBF2).

(ii) Suppose x ≤ (y • z), x ≤ (y ∗ z), and x, y ∈ F . By (i), y • z, y ∗ z ∈ F . Since
y • z, y ∗ z, y ∈ F , z ∈ F by (PDBF2).

Theorem 4. Let F be a subset of X containing 1. Then F is a pseudo-dB-filter of X if
and only if for all x, y, z in X, if x ≤ (y • z), x ≤ (y ∗ z), and x, y ∈ F , then z ∈ F .

Proof. Let F be a subset of X containing 1. Suppose F is a pseudo-dB-filter of X.
Then for all x, y, z in X, if x ≤ (y • z), x ≤ (y ∗ z), and x, y ∈ F , then z ∈ F by Lemma
2 (ii). Conversely, 1 ∈ F by assumption. Hence, (PDBF1) holds. Let x, x • y, x ∗ y ∈ F
and y ∈ X. By (PDB1), (x • y) • (x • y) = 1, (x • y) ∗ (x • y) = 1, (x ∗ y) • (x ∗ y) = 1, and
(x ∗ y) ∗ (x ∗ y) = 1. Hence, (x • y) ≤ (x • y) and (x ∗ y) ≤ (x ∗ y). Since x, x • y, x ∗ y ∈ F ,
y ∈ F . Thus, (PDBF2) holds. Therefore, F is a pseudo-dB-filter of X.

Proposition 5. Let F be a pseudo-dB-filter of X. If G is a pseudo-dB-filter of F , then
G is a pseudo-dB-filter of X as well.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 8 of 12

Proof. Suppose G is a pseudo-dB-filter of F . Then 1 ∈ G by (PDBF1). Let x, x •
y, x ∗ y ∈ G for any y ∈ X. Note that x, x • y, x ∗ y ∈ G ⊆ F . Hence, x, x • y, x ∗ y ∈ F .
Since F is a pseudo-dB-filter of X, y ∈ F by (PDBF2). Now, y ∈ F implies y ∈ G since
G is a pseudo-dB-filter of F . Therefore, G is a pseudo-dB-filter of X as well.

The next theorem shows that the intersection of a nonempty collection of pseudo-dB-
filter of X is a pseudo-dB-filter.

Theorem 5. Let {Fα : α ∈ I} be a nonempty collection of pseudo-dB-filter of X. Then∩
α∈I

Fα is also a pseudo-dB-filter of X.

Proof. Let {Fα : α ∈ I} be a nonempty collection of pseudo-dB-filter of X. Since Fα

is a pseudo-dB-filter for each α, 1 ∈ Fα for all α ∈ I. Hence, 1 ∈
∩
α∈I

Fα and
∩
α∈I

Fα ̸= ∅.

Suppose x, x • y, x ∗ y ∈
∩
α∈I

Fα and y ∈ X. Then x, x • y, x ∗ y ∈ Fα for all α ∈ I. Since

Fα is a pseudo-dB-filter for each α, y ∈ Fα for all α ∈ I. Thus, y ∈
∩
α∈I

Fα. Therefore,∩
α∈I

Fα is a pseudo-dB-filter of X.

The following theorem presents the relationship between a pseudo-dB-filter and a
pseudo-dB-subalgebra.

Theorem 6. Any pseudo-dB-filter of X is a pseudo-dB-subalgebra.

Proof. Suppose F is a pseudo-dB-filter of X and let x, y ∈ F . By (PDBF1), 1 ∈ F .
Since 1 ∈ F and F is a pseudo-dB-filter, 1 • (x • y) ∈ F and 1 ∗ (x ∗ y) ∈ F imply
x • y, x ∗ y ∈ F .

The following theorem provides a condition for a pseudo-dB-subalgebra to be a pseudo-
dB-filter.

Theorem 7. Let F be a pseudo-dB-subalgebra of X. Then F is a pseudo-dB-filter of X
if and only if for all x, y ∈ X if x ∈ F and y /∈ F then x • y, x ∗ y /∈ F .

Proof. Let F be a pseudo-dB-subalgebra of X. Suppose x, y ∈ X and F is a pseudo-
dB-filter of X where x ∈ F and y /∈ F . If x • y, x ∗ y ∈ F , then y ∈ F by (PDBF2),
which is a contradiction. Hence, x • y, x ∗ y /∈ F . Conversely, note that 1 ∈ F by Remark
3. Hence, (PDBF1) holds. It remains to show (PDBF2). Suppose that for any x, y ∈ X,
x•y, x∗y, x ∈ F . If y /∈ F , then x•y, x∗y /∈ F by the hypothesis, which is a contradiction.
Thus, for any x, y ∈ X, x•y, x∗y, x ∈ F imply y ∈ F . Hence, (PDBF2) holds. Therefore,
F is a pseudo-dB-filter.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 9 of 12

5. Conclusion

In this paper, the structure of pseudo-dB-algebra and some of its subsets, particu-
larly pseudo-dB-subalgebra and pseudo-dB-filter, are introduced. It was illustrated that
any two dB-algebras do not necessarily construct a pseudo-dB-algebra. This study also
presented some properties of pseudo-dB-algebra and its subsets under consideration, and
established a characterization of a pseudo-dB-subalgebra. Furthermore, it was shown that
any pseudo-dB-filter is a pseudo-dB-subalgebra, and under some condition, a pseudo-
dB-subalgebra is a pseudo-dB-filter. Moreover, interested researchers may explore the
relationship of the pseudo-dB-algebra with the other pseudo-algebras.

Acknowledgements

The authors would like to thank the Department of Science and Technology - Acceler-
ated Science and Technology Human Resource Development Program (DOST-ASTHRDP)
and the University of San Carlos for funding this research.

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[10] Katrina Belleza and Jimboy Albaracin. On dual b-filters and dual b-subalgebras in a
topological dual b-algebra. Journal of Mathematics and Computer Science, 28:1–10,
04 2022.

[11] Jethro Elijah Bolima. The isomorphism theorems for the dual b-algebra. Master’s
thesis, University of San Carlos, 2023.



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 10 of 12

Appendix

The following Python program was used to verify Example 1. The same script was
used to perform the calculations needed in Example 2, Example 3, and Example 4.
# Define the e lements in the s e t X
X = [1 , ’−1 ’ ]
# Define the b u l l e t opera t ion
def bu l l e t (x , y ) :

bu l l e t_tab l e = {
(1 , 1) : 1 , (1 , ’−1 ’ ) : ’−1 ’ ,
( ’−1 ’ , 1) : ’−1 ’ , ( ’−1 ’ , ’−1 ’ ) : 1 ,

}
return bu l l e t_tab l e [ ( x , y ) ]

# Define the a s t opera t ion
def as t (x , y ) :

as t_table = {
(1 , 1) : 1 , (1 , ’−1 ’ ) : ’−1 ’ ,
( ’−1 ’ , 1) : ’−1 ’ , ( ’−1 ’ , ’−1 ’ ) : 1 ,

}
return ast_table [ ( x , y ) ]

# Define the cons tant 1
constant = 1
# Ver i fy Axiom PDB1: x b u l l e t x = 1 and x a s t x = 1
def check_pdb1 ( ) :

print ( ”Checking␣PDB1: ␣x␣ bu l l e t ␣x␣=␣1␣and␣x␣ as t ␣x␣=␣1” )
for x in X:

bu l l e t_ r e s u l t = bu l l e t (x , x )
a s t_re su l t = as t (x , x )
print ( f ”x␣=␣{x } : ␣{x}␣ bu l l e t ␣{x}␣=␣{ bu l l e t_ r e s u l t } ,␣{x}␣ as t ␣{x}␣

=␣{ a s t_re su l t }” )
i f bu l l e t_ r e s u l t != constant or a s t_re su l t != constant :

print ( ”PDB1␣ f a i l s . ” )
return False

print ( ”PDB1␣holds . ” )
return True

# Ver i fy Axiom PDB2: 1 b u l l e t x = x and 1 as t x = x
def check_pdb2 ( ) :

print ( ”\nChecking␣PDB2: ␣1␣ bu l l e t ␣x␣=␣x␣and␣1␣ as t ␣x␣=␣x” )
for x in X:

bu l l e t_ r e s u l t = bu l l e t ( constant , x )
a s t_re su l t = as t ( constant , x )
print ( f ”x␣=␣{x } : ␣1␣ bu l l e t ␣{x}␣=␣{ bu l l e t_ r e s u l t } ,␣1␣ as t ␣{x}␣=␣{

a s t_re su l t }” )
i f bu l l e t_ r e s u l t != x or a s t_re su l t != x :

print ( ”PDB2␣ f a i l s . ” )
return False

print ( ”PDB2␣holds . ” )
return True



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 11 of 12

# Ver i fy Axiom PDB3
# x b u l l e t ( y a s t z ) = (( y a s t 1) b u l l e t x ) b u l l e t z
# x as t ( y b u l l e t z ) = (( y b u l l e t 1) a s t x ) a s t z
def check_pdb3 ( ) :

print ( ”\nChecking␣PDB3: ” )
for x in X:

for y in X:
for z in X:

# Check the f i r s t par t o f PDB3
l e f t 1 = bu l l e t (x , a s t (y , z ) )
r i gh t 1 = bu l l e t ( b u l l e t ( a s t (y , constant ) , x ) , z )
print ( f ”x␣=␣{x} ,␣y␣=␣{y} ,␣z␣=␣{z } : ␣x␣ bu l l e t ␣ ( y␣ as t ␣z )␣=

␣{ l e f t 1 } ,␣ ( ( y␣ as t ␣1)␣ bu l l e t ␣x )␣ bu l l e t ␣z␣=␣{ r i gh t1 }”
)

i f l e f t 1 != r i gh t1 :
print ( ”PDB3␣ f a i l s ␣on␣ f i r s t ␣part . ” )
return False

# Check the second par t o f PDB3
l e f t 2 = ast (x , b u l l e t (y , z ) )
r i gh t 2 = ast ( a s t ( b u l l e t (y , constant ) , x ) , z )
print ( f ”x␣=␣{x} ,␣y␣=␣{y} ,␣z␣=␣{z } : ␣x␣ as t ␣ ( y␣ bu l l e t ␣z )␣=

␣{ l e f t 2 } ,␣ ( ( y␣ bu l l e t ␣1)␣ as t ␣x )␣ as t ␣z␣=␣{ r i gh t2 }” )
i f l e f t 2 != r i gh t2 :

print ( ”PDB3␣ f a i l s ␣on␣ second␣part . ” )
return False

print ( ”PDB3␣holds . ” )
return True

# Main func t i on to check a l l axioms
def verify_pseudo_dual_b_algebra ( ) :

print ( ” Ver i f y ing ␣pseudo−dual␣B−algebra ␣axioms . . . \ n” )
pdb1 = check_pdb1 ( )
pdb2 = check_pdb2 ( )
pdb3 = check_pdb3 ( )
i f pdb1 and pdb2 and pdb3 :

print ( ”\nX␣ i s ␣a␣pseudo−dB−algebra . ” )
else :

print ( ”\nX␣ i s ␣not␣a␣pseudo−dB−algebra . ” )
print ( f ”PDB1: ␣{pdb1} ,␣PDB2: ␣{pdb2} ,␣PDB3: ␣{pdb3}” )

# Run the v e r i f i c a t i o n
verify_pseudo_dual_b_algebra ( )
Output:
Checking PDB1: x bu l l e t x = 1 and x as t x = 1
x = 1 : 1 bu l l e t 1 = 1 , 1 as t 1 = 1
x = −1: −1 bu l l e t −1 = 1 , −1 as t −1 = 1
PDB1 holds .
Checking PDB2: 1 bu l l e t x = x and 1 as t x = x
x = 1 : 1 bu l l e t 1 = 1 , 1 as t 1 = 1
x = −1: 1 bu l l e t −1 = −1, 1 as t −1 = −1



J. M. S. Leuveras, K. B. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6190 12 of 12

PDB2 holds .
Checking PDB3:
x = 1 , y = 1 , z = 1 : x bu l l e t ( y as t z ) = 1 , ( ( y as t 1) b u l l e t x )

b u l l e t z = 1
x = 1 , y = 1 , z = 1 : x as t ( y bu l l e t z ) = 1 , ( ( y bu l l e t 1) a s t x ) a s t z

= 1
x = 1 , y = 1 , z = −1: x bu l l e t ( y as t z ) = −1, ( ( y as t 1) b u l l e t x )

b u l l e t z = −1
x = 1 , y = 1 , z = −1: x as t ( y bu l l e t z ) = −1, ( ( y bu l l e t 1) a s t x ) a s t

z = −1
x = 1 , y = −1, z = 1 : x bu l l e t ( y as t z ) = −1, ( ( y as t 1) b u l l e t x )

b u l l e t z = −1
x = 1 , y = −1, z = 1 : x as t ( y bu l l e t z ) = −1, ( ( y bu l l e t 1) a s t x ) a s t

z = −1
x = 1 , y = −1, z = −1: x bu l l e t ( y as t z ) = 1 , ( ( y as t 1) b u l l e t x )

b u l l e t z = 1
x = 1 , y = −1, z = −1: x as t ( y bu l l e t z ) = 1 , ( ( y bu l l e t 1) a s t x ) a s t

z = 1
x = −1, y = 1 , z = 1 : x bu l l e t ( y as t z ) = −1, ( ( y as t 1) b u l l e t x )

b u l l e t z = −1
x = −1, y = 1 , z = 1 : x as t ( y bu l l e t z ) = −1, ( ( y bu l l e t 1) a s t x ) a s t

z = −1
x = −1, y = 1 , z = −1: x bu l l e t ( y as t z ) = 1 , ( ( y as t 1) b u l l e t x )

b u l l e t z = 1
x = −1, y = 1 , z = −1: x as t ( y bu l l e t z ) = 1 , ( ( y bu l l e t 1) a s t x ) a s t

z = 1
x = −1, y = −1, z = 1 : x bu l l e t ( y as t z ) = 1 , ( ( y as t 1) b u l l e t x )

b u l l e t z = 1
x = −1, y = −1, z = 1 : x as t ( y bu l l e t z ) = 1 , ( ( y bu l l e t 1) a s t x ) a s t

z = 1
x = −1, y = −1, z = −1: x bu l l e t ( y as t z ) = −1, ( ( y as t 1) b u l l e t x )

b u l l e t z = −1
x = −1, y = −1, z = −1: x as t ( y bu l l e t z ) = −1, ( ( y bu l l e t 1) a s t x )

a s t z = −1
PDB3 holds .
X i s a pseudo−dB−algebra .