EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6974 ISSN 1307-5543 – ejpam.com Published by New York Business Global Group-Derived and Non-Group-Derived Dual BG-Algebra Clive Martin G. Chan1,∗, Katrina B. Fuentes1 1 Department of Computer, Information Sciences and Mathematics, School of Arts and Sciences, University of San Carlos, Cebu City, Cebu, Philippines Abstract. This study introduces the notion of the dual BG-algebra. The axioms are presented and are shown to be independent. Fundamental properties of the dual BG-algebra are also pro- vided. The concept of a group-derived dual BG-algebra and its characterization were established. Moreover, a non-group-derived dual BG-algebra can be constructed from a set containing at least 3 elements. Lastly, this paper also presented a Python script used to verify whether a given Cayley table is a dual BG-algebra. This was utilized throughout the process of this study. 2020 Mathematics Subject Classifications: 06F35, 47L45, 08C05, 20A05, 20F14 Key Words and Phrases: BG-algebra, dual BG-algebra, dual algebra, group-derived algebra 1. Introduction Since the 1960s, numerous classes of algebras have been introduced, beginning with BCK/BCI-algebras [1], later extended to BCH-algebras [2, 3], and further generalized to BH-algebras [4]. Neggers and Kim subsequently developed d-algebras [5] and B-algebras [6], while Kim and Kim introduced BG-algebras as a generalization of B-algebras [7]. In parallel, dual algebras were also studied. Kim and Yon investigated dual BCK- algebras and their relation to MV -algebras [8], Kim and Kim proposed BE-algebras [9], Walendziak showed commutative BE-algebras coincide with dual BCK-algebras [10], Meng defined dual BCI-algebras and CI-algebras [11], with Saeid proved the equiva- lence of CI-algebras and dual Q-algebras [12], and Belleza and Vilela introduced the dual B-algebra and established its relationship to BCK-algebra, CI-algebra, and the dual BCI-algebra [13]. While many algebras and dual algebras have been established and interconnected, no work has addressed the dual BG-algebra. This study introduces its definition, examines its properties, and establishes key characterizations, with particular emphasis on group- derived and non-group-derived dual BG-algebras. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6974 Email addresses: clivemartinchan@gmail.com (C.M. Chan), kebelleza@usc.edu.ph (K. Fuentes) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 2 of 16 2. Preliminaries Definition 1. [14] A binary operation on a nonempty set B (or simply an operation on B) is a function f : B × B → B. Commonly, the symbol ∗ is used instead of f to denote the operation and write a ∗ b instead of f(a, b). Definition 2. [14] A group is a nonempty set G equipped with a binary operation “∗” that satisfies the following axioms: (i) associativity: a ∗ (b ∗ c) = (a ∗ b) ∗ c for all a, b, c ∈ G; (ii) identity element: there is an element e ∈ G such that a ∗ e = a = e ∗ a for every a ∈ G; and (iii) inverse element: for each a ∈ G, there is an element d ∈ G such that a ∗ d = e and d ∗ a = e. The number of elements in G is called the order of G and is denoted by |G|. Theorem 1. [14] Let G be a group and let a, b, c ∈ G. Then (i) G has a unique identity element e; (ii) Each element a ∈ G has a unique inverse denoted by a−1; (iii) cancellation law: for all a, b, c ∈ G, if either a ∗ b = a ∗ c or b ∗ a = c ∗ a, then b = c; (iv) (a ∗ b)−1 = b−1 ∗ a−1; and (v) ( a−1 )−1 = a for all a ∈ G. Definition 3. [13] A dual B-algebra X is a triple (X, ◦, 1) where X is a nonempty set with a binary operation ◦ and a constant 1 satisying the following axioms for all x, y, z ∈ X: (DB1) x ◦ x = 1 (DB2) 1 ◦ x = x (DB3) x ◦ (y ◦ z) = ((y ◦ 1) ◦ x) ◦ z Example 1. [13] Let X = {e, a, b, c} be a set with the following Cayley table: Table 1: Cayley table of the dual B-algebra (X, ◦, e) ◦ e a b c e e a b c a a e c b b b c e a c c b a e Then (X, ◦, e) is a dual B-algebra. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 3 of 16 Lemma 1. [13] Let X be a dual B-algebra. For any x, y ∈ X, (y ◦ 1) ◦ (y ◦ x) = x. Definition 4. [7] A BG-algebra is a nonempty set X with a constant 0 and a binary operation ∗ satisfying the following axioms for all x, y ∈ X: (BG1) x ∗ x = 0 (BG2) x ∗ 0 = x (BG3) (x ∗ y) ∗ (0 ∗ y) = x Example 2. [7] Let X = {0, 1, 2} be a set with the following Cayley table: Table 2: Cayley table of the BG-algebra (X, ∗, 0) ∗ 0 1 2 0 0 1 2 1 1 0 1 2 2 2 0 Then (X, ∗, 0) is a BG-algebra. 3. Dual BG-Algebra In this section, all finite dual BG-algebra examples were verified using a Python script developed by the author found in the Appendix section. Definition 5. A dual BG-algebra X is a triple (X, ◦, 1) where X is a nonempty set with a binary operation ◦ and a constant 1 satisying the following axioms for all x, y ∈ X: (DBG1) x ◦ x = 1 (DBG2) 1 ◦ x = x (DBG3) (y ◦ 1) ◦ (y ◦ x) = x Example 3. Let X = {1, a, b, c, d, e} be a set with the following Cayley table: Table 3: Cayley table of the dual BG-algebra (X, ◦, 1) ◦ 1 a b c d e 1 1 a b c d e a b 1 a d e c b a b 1 e c d c c d e 1 a b d d e c b 1 a e e c d a b 1 Then (X, ◦, 1) is a dual BG-algebra. Example 4. Let X = R\{0}. Define the binary operation ◦ as x ◦ y = y x for all x, y ∈ X. Note that X satisfies (DBG1): x ◦ x = x x = 1, (DBG2): 1 ◦ x = x 1 = x, and (DBG3): (y ◦ 1) ◦ (y ◦ x) = y ◦ x y ◦ 1 = x y 1 y = x 1 = x. Therefore, (X, ◦, 1) is a dual BG-algebra. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 4 of 16 Example 5 shows that the axioms are independent. Example 5. Let X1 = (X, ◦1, 1), X2 = (X, ◦2, 1), and X3 = (Y, ◦3, 1) where X = {1, a, b} and Y = {1, a, b, c}. The Cayley tables of the binary operations ◦1, ◦2, and ◦3 are shown in Table 4. Table 4: Cayley tables of X1, X2, and X3 ◦1 1 a b 1 1 a b a a 1 b b 1 a b ◦2 1 a b 1 1 b a a b 1 a b a b 1 ◦3 1 a b c 1 1 a b c a b 1 a c b c a 1 b c a b c 1 The axioms (DBG2) and (DBG3) hold for X1. However, (DBG1) does not hold since b ◦1 b = b ̸= 1. For X2, (DBG1) and (DBG3) are satisfied but not (DBG2) since 1 ◦2 a = b ̸= a. (DBG1) and (DBG2) are satisfied in X3 but fails on (DBG3) since (b ◦3 1) ◦3 (b ◦3 a) = c ◦3 a = b ̸= a. Example 6. Consider the dual BG-algebra (X, ◦, 1) in Example 3 where X = {1, a, b, c, d, e}. Define the binary operation “∗” as x ∗ y = y ◦ x for all x, y ∈ X. The Cayley table of (X, ∗, 1) is shown below. Table 5: Cayley table of the BG-algebra (X, ∗, 1) ∗ 1 a b c d e 1 1 b a c d e a a 1 b d e c b b a 1 e c d c c d e 1 b a d d e c a 1 b e e c d b a 1 Note that (X, ∗, 1) satisfies (BG1): x∗x = x◦x = 1 by (DBG1), (BG2): x∗1 = 1◦x = x by (DBG2), and (BG3): (x∗y)∗ (1∗y) = (1∗y)◦ (x∗y) = (y ◦1)◦ (y ◦x) = x by (DBG3). Thus, (X, ∗, 1) is a BG-algebra where 1 acts as the constant element. Example 7. Consider the BG-algebra (X, ∗, 1) in Example 6 where X = {1, a, b, c, d, e}. Let “◦” be a binary operation where x ◦ y = y ∗ x for all x, y ∈ X. Then the Cayley table of (X, ◦, 1) is the same as the Cayley table in 3 and so (X, ◦, 1) is a dual BG-algebra. Every dual BG-algebra corresponds to a BG-algebra by commuting the operation. This is formalized in the next proposition. Proposition 1. Let (X, ◦, 1) be a dual BG-algebra. Then (X, ∗, 1) is a BG-algebra where x ∗ y = y ◦ x for all x, y ∈ X and 1 corresponds to the constant element. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 5 of 16 Proof. Suppose (X, ◦, 1) is a dual BG-algebra and define “∗” as follows: x ∗ y = y ◦ x for all x, y in X. Then (X, ◦, 1) satisfies (BG1): x ∗ x = x ◦ x = 1 by (DBG1), (BG2): x∗1 = 1◦x = x by (DBG2), and (BG3): (x∗y)∗(1∗y) = (1∗y)◦(x∗y) = (y◦1)◦(y◦x) = x by (DBG3). Thus, (X, ∗, 1) is a BG-algebra. Example 8. The dual BG-algebra (R\{0}, ◦, 1) where x ◦ y = y x for all x, y in R\{0} from Example 4 is not a BG-algebra since x ◦ 1 = 1 x ̸= x, failing to satisfy (BG2). By Example 8, there exists a dual BG-algebra that is not a BG-algebra, which leads to the next remark. Remark 1. Not every dual BG-algebra is a BG-algebra. Example 9. Let X = {1, a, b, c}. Consider the following Cayley table for the binary operation “◦”. Then (X, ◦, 1) is a dual BG-algebra. Note that for any x, y ∈ X, x◦y = y◦x. Table 6: Cayley table of the dual BG-algebra (X, ◦, 1) ◦ 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 Now, let x ∗ y = y ◦ x for a binary operation “∗” where x, y in X. Then (X, ∗, 1) is a BG-algebra by Proposition 1. Because (X, ◦, 1) satisfies x ◦ y = y ◦ x for all x, y ∈ X, it follows that it is also a BG-algebra. Thus, there exists a dual BG-algebra that is a BG-algebra at the same time. This is formalized in the next theorem. Theorem 2. Let (X, ◦, 1) be a dual BG-algebra satisfying x ◦ y = y ◦ x for all x, y in X. Then (X, ◦, 1) is also a BG-algebra. Proof. Suppose (X, ◦, 1) is a dual BG-algebra where x ◦ y = y ◦ x for all x, y in X. Then (X, ◦, 1) satisfies (BG1): x ◦x = 1 by (DBG1), (BG2): x ◦ 1 = 1 ◦x = x by (DBG2), and (BG3): (x ◦ y) ◦ (1 ◦ y) = (1 ◦ y) ◦ (x ◦ y) = (y ◦ 1) ◦ (y ◦ x) = x by (DBG3). Thus, (X, ◦, 1) is also a BG-algebra. Example 10. Consider the BG-algebra (X, ∗, 0) in Example 2. This is not a dual BG- algebra because (2 ∗ 0) ∗ (2 ∗ 1) = 2 ∗ 2 = 0 ̸= 1, failing to satisfy (DBG3). By Example 10, there exists a BG-algebra that is not a dual BG-algebra, which leads to the next remark. Remark 2. Not every BG-algebra is a dual BG-algebra. Lemma 2 shows some properties of the dual BG-algebra. Lemma 2. Let (X, ◦, 1) be a dual BG-algebra. Then for any x, y, z in X, C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 6 of 16 (i) x = (x ◦ 1) ◦ 1; (ii) x = y ◦ [(y ◦ 1) ◦ x]; (iii) x ◦ y = x ◦ z implies y = z; (iv) if x ◦ y = 1, then x = y; (v) if x ◦ 1 = y ◦ 1, then x = y; and (vi) if x ◦ y = 1, then (x ◦ z) ◦ (y ◦ z) = 1. Proof. Let (X, ◦, 1) be a dual BG-algebra and x, y, z in X. By replacing y with x in (DBG3) and applying (DBG1), (i) is proved as shown: x = (y ◦ 1) ◦ (y ◦ x) = (x ◦ 1) ◦ (x ◦x) = (x ◦ 1) ◦ 1. For (ii), replace y with y ◦ 1 in (DBG3) and apply (i), that is, x = [(y ◦ 1) ◦ 1] ◦ [(y ◦ 1) ◦ x] = y ◦ [(y ◦ 1) ◦ x]. To show (iii), replace y with x in (ii) and so x = x ◦ [(x ◦ 1) ◦ x]. Using (i), (iv) immediately follows. By hypothesis and (DBG3), y = (x◦1)◦(x◦y) = (x◦1)◦(x◦z) = z, proving (v). For (vi), if x◦y = 1, then x◦y = x◦x by (DBG1). This implies x = y using (v). By (DBG3), y = (y ◦ 1) ◦ (y ◦ y). This implies y = (x ◦ 1) ◦ 1 after applying the hypothesis and (DBG1). Using (i), x = y as needed in (vii). Finally, if x ◦ y = 1, then x = y by (vi). So, (x ◦ z) ◦ (y ◦ z) = (x ◦ z) ◦ (x ◦ z) = 1 by (DBG1), which proves (viii). For any dual BG-algebra, replacing y with x on the right-hand side in Lemma 2(ii) yields x = x ◦ [(x ◦ 1) ◦ x]. Also, using Lemma 2(i), x ◦ y = [(x ◦ 1) ◦ 1] ◦ y follows. This is formalized in the next remark. Remark 3. Let (X, ◦, 1) be a dual BG-algebra. Then for any x, y, z in X, (i) x = x ◦ [(x ◦ 1) ◦ x] ; (ii) x ◦ y = [(x ◦ 1) ◦ 1] ◦ y. The next theorem characterizes the dual BG-algebra given any algebra with a binary operation and a constant element, which will be referred to as an algebra of type (2, 0). Theorem 3. Let (X, ◦, 1) be an algebra of type (2, 0). Then (X, ◦, 1) is a dual BG-algebra if and only if for any x, y ∈ X, (i) 1 ◦ x = x; (ii) (y ◦ 1) ◦ (y ◦ x) = x; (iii) x ◦ y = 1 if and only if x = y. Proof. Let (X, ◦, 1) be a dual BG-algebra. Then (i) and (ii) immediately follow from (DBG2) and (DBG3). If x ◦ y = 1, then x = y by Lemma 2(iv). If x = y, then x◦y = y◦y = 1 by (DBG1). This proves (iii). So, X satisfies (i), (ii), and (iii). Conversely, (DBG1) is implied from (iii), (DBG2) is equivalent to (i), and (DBG3) is equivalent to (ii). Hence, (X, ◦, 1) is a dual BG-algebra. Every group can generate a dual BG-algebra given a condition. This is formalized in the next proposition. Proposition 2. Let (X, ∗, 1) be a group where 1 is the identity element. Then (X, ◦, 1) is a dual BG-algebra assuming x ◦ y = x−1 ∗ y for any x, y ∈ X. The dual BG-algebra (X, ◦, 1) is said to be group-derived. Proof. Let (X, ∗, 1) be a group where x, y in X. Then (X, ◦, 1) satisfies (DBG1): x ◦ x = x−1 ∗ x = 1, (DBG2): 1 ◦ x = 1−1 ∗ x = 1 ∗ x = x, and (DBG3): (y ◦ 1) ◦ (y ◦ x) = C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 7 of 16 (y ◦ 1)−1 ∗ (y ◦ x) = ( y−1 ∗ 1 )−1 ∗ (y−1 ∗ x) = y ∗ ( y−1 ∗ x ) = ( y ∗ y−1 ) ∗ x = 1 ∗ x = x. Thus, (X, ◦, 1) is a dual BG-algebra. Example 11 shows a dual BG-algebra that is non-group-derived. Example 11. Let (X, ◦, 1) with the following Cayley table: Table 7: Cayley table of the non-group-derived dual BG-algebra (X, ◦, 1) ◦ 1 a b 1 1 a b a a 1 b b b a 1 Then (X, ◦, 1) is a dual BG-algebra. To show that it is non-group-derived, assume first that (X, ◦, 1) is group-derived. So, x ◦ y = x−1 ∗ y for any x, y from a group (X, ∗, 1) where 1 is the identity element. Since X has only 3 elements, the inverse of a has to be b, that is, a−1 = b. Hence, b = a ◦ b = a−1 ∗ b = b ∗ b = a, which is a contradiction. By Example 11, there exists a dual BG-algebra that is non-group-derived, which leads to the next remark. Remark 4. Not all dual BG-algebras are group-derived. The next theorem shows that a dual BG-algebra is group-derived when it satisfies a certain identity. To establish this, define a binary operation, show that it is a group, and then use Proposition 2. Theorem 4. Let (X, ◦, 1) be a dual BG-algebra with the identity x ◦ (y ◦ z) = ((x ◦ (y ◦ 1)) ◦ 1) ◦ z for all x, y, z ∈ X. Then (X, ◦, 1) is group-derived. Proof. Define a binary operation “∗” on X as x ∗ y = (x ◦ 1) ◦ y. Note that x ∗ 1 = (x ◦ 1) ◦ 1 = x by Lemma 2(i) and 1 ∗ x = (1 ◦ 1) ◦ x = 1 ◦ x = x by (DBG1) and (DBG2). So 1 is the identity element with respect to the binary operation “∗”. Also x ∗ (x ◦ 1) = (x ◦ 1) ◦ (x ◦ 1) = 1 by (DBG1) and (x ◦ 1) ∗x = ((x ◦ 1) ◦ 1) ◦x = x ◦x = 1 by Lemma 2(i) and (DBG1). Thus, x◦1 is the inverse for x. Now, x∗(y∗z) = (x◦1)◦(y∗z) = (x◦1)◦ [(y ◦ 1) ◦ z] = [[(x ◦ 1) ◦ [(y ◦ 1) ◦ 1]] ◦ 1]◦z by replacing x with x◦1 and y with y◦1 in the given identity. By Lemma 2(i), x ∗ (y ∗ z) = [[(x ◦ 1) ◦ y] ◦ 1] ◦ z. The continuation is as follows: x ∗ (y ∗ z) = [(x ∗ y) ◦ 1] ◦ z = (x ∗ y) ∗ z. Hence, (X, ∗, 1) is a group by Definition 2. Observe that x−1 ∗ y = ( x−1 ◦ 1 ) ◦ y = [(x ◦ 1) ◦ 1] ◦ y = x ◦ y. Therefore, (X, ◦, 1) is a group-derived dual BG-algebra by Proposition 2. Consider Example 11. Note that b ◦ (a ◦ b) = b ◦ b = 1 while ((b ◦ (a ◦ 1)) ◦ 1) ◦ b = b. So, the condition x ◦ (y ◦ z) = ((x ◦ (y ◦ 1)) ◦ 1) ◦ z in Theorem 4 is not necessarily true in general. Example 12. Consider the dual BG-algebra (X, ◦, 1) from Example 9. It satisfies the identity x ◦ (y ◦ z) = ((x ◦ (y ◦ 1)) ◦ 1) ◦ z. Hence, (X, ◦, 1) is a group-derived dual BG- algebra by Theorem 4. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 8 of 16 Theorem 5 characterizes a group-derived dual BG-algebra through the dual B-algebra. Theorem 5. Let (X, ◦, 1) be an algebra of type (2, 0). Then (X, ◦, 1) is a dual B-algebra if and only if it is a group-derived dual BG-algebra. Proof. Let (X, ◦, 1) be a dual B-algebra. (DBG1), (DBG2), and (DBG3) immediately follow from (DB1), (DB2), and Lemma 1, respectively. Thus, (X, ◦, 1) is a dual BG- algebra. Define a binary operation “∗” on X as x∗y = (x◦1)◦y. Note that x∗1 = (x◦1)◦1 = x by Lemma 2(i) and 1 ∗ x = (1 ◦ 1) ◦ x = 1 ◦ x = x by (DBG1) and (DBG2). So 1 is the identity element with respect to the binary operation “∗”. Also x ∗ (x ◦ 1) = (x ◦ 1) ◦ (x ◦ 1) = 1 by (DBG1) and (x ◦ 1) ∗ x = ((x ◦ 1) ◦ 1) ◦ x = x ◦ x = 1 by Lemma 2(i) and (DBG1). Thus, x ◦ 1 is the inverse for x. Now, x ∗ (y ∗ z) = (x ◦ 1) ◦ (y ∗ z) = (x ◦ 1) ◦ [(y ◦ 1) ◦ z] = [[(y ◦ 1) ◦ 1] ◦ (x ◦ 1)] ◦ z by replacing x with x ◦ 1 and y with y ◦ 1 in (DB3). By Lemma 2(i), x ∗ (y ∗ z) = [y ◦ (x ◦ 1)] ◦ z. Using (DB3), y ◦ (x ◦ 1) = [(x ◦ 1) ◦ y] ◦ 1 and so x ∗ (y ∗ z) = [[(x ◦ 1) ◦ y] ◦ 1] ◦ z. The continuation is as follows: x ∗ (y ∗ z) = [(x ∗ y) ◦ 1] ◦ z = (x ∗ y) ∗ z. Thus, (X, ∗, 1) is a group by Definition 2. Note that x−1 ∗ y = ( x−1 ◦ 1 ) ◦ y = [(x ◦ 1) ◦ 1] ◦ y = x ◦ y. Therefore, (X, ◦, 1) is a group-derived dual BG-algebra by Proposition 2. Conversely, let X = (X, ◦, 1) be a group-derived dual BG-algebra. Then (DB1) and (DB2) follow from (DBG1) and (DBG2), respectively. Now, since X is group-derived, then it was generated from a group (X, ∗, 1) where x−1 ∗ y = x ◦ y where x, y ∈ X. Now, x ◦ (y ◦ z) = x−1 ∗ (y ◦ z) = x−1 ∗ ( y−1 ∗ z ) = ( x−1 ∗ y−1 ) ∗ z. Associative property can be applied since (X, ∗, 1) is a group. Using Theorem 1, the continuation is as follows: x◦(y◦z) = (y ∗ x)−1∗z = [( y−1 )−1 ∗ x ]−1 ∗z = [( y−1 ∗ 1 )−1 ∗ x ]−1 ∗z = [ (y ◦ 1)−1 ∗ x ]−1 ∗ z = [(y ◦ 1) ◦ x]−1 ∗ z = ((y ◦ 1) ◦ x)◦ z. Hence, (DB3) is satisfied and so (X, ◦, 1) is a dual B-algebra. This proves the theorem. Example 13. Consider the dual BG-algebra (X, ◦, 1) from Example 9. It was shown that it is a group-derived dual BG-algebra in Example 12. Thus, (X, ◦, 1) is also a dual B-algebra by Theorem 5. Indeed, treating 1 as e, (X, ◦, 1) is a dual B-algebra by Example 1. It is evident from the proof of Theorem 5 that every dual B-algebra is a dual BG- algebra. Consequently, non-group-derived dual BG-algebras are not dual B-algebras. Theorem 6 can be used to construct infinitely many dual BG-algebras. These dual BG-algebras are also shown to be non-group-derived. Theorem 6. Define a binary operation “◦” on a set X where 1 ∈ X by x ◦ y =  x if y = 1 1 if x = y y otherwise , for any x, y ∈ X, then (X, ◦, 1) is a dual BG-algebra. Moreover, if X has at least 3 elements, then (X, ◦, 1) is non-group-derived. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 9 of 16 Proof. Let x, y ∈ X. Note that (X, ◦, 1) satisfies (DBG1): x ◦ x = 1 and (DBG2): 1 ◦ x = x. Now, if y = 1, then (y ◦ 1) ◦ (y ◦ x) = (1 ◦ 1) ◦ (1 ◦ x) = 1 ◦ x = x. Assume y ̸= 1. If x = y, then (y ◦ 1) ◦ (y ◦ x) = (x ◦ 1) ◦ (x ◦ x) = x ◦ 1 = x. If x ̸= y, then (y ◦ 1) ◦ (y ◦ x) = y ◦ x = x. Hence, (X, ◦, 1) satisfies (DBG3). Therefore, (X, ◦, 1) is a dual BG-algebra. Now, assume X has at least 3 elements. Let x, y, z ∈ X be unique elements and z ̸= 1. Then x ◦ z = y ◦ z = z. Assume (X, ◦, 1) is a group-derived dual BG-algebra obtained from the group (X, ∗), then z = x◦z = x−1 ∗z and z = y ◦z = y−1 ∗z. This implies x = y, which is a contradiction since they are unique by hypothesis. So, the dual BG-algebra (X, ◦, 1) has to be non-group-derived. The condition for X in Theorem 6 to have at least 3 elements is necessary. To see this, if X has only one element, then it has to be 1 and so X = {1}. If X has two elements, say X = {1, a}, then 1◦1 = 1, 1◦a = a, a◦1 = a, and a◦a = 1. In any case, (X, ◦, 1) satisfies x ◦ (y ◦ z) = ((x ◦ (y ◦ 1)) ◦ 1) ◦ z. Therefore, (X, ◦, 1) is a group-derived dual BG-algebra by Theorem 4 when X has only 1 or 2 elements. Example 14. Let X1 = (X, ◦1, 1), Y1 = (Y, ◦2, 1), and Z1 = (Z, ◦3, 1) where X = {1, a, b, c}, Y = {1, a, b, c, d}, and Z = {1, a, b, c, d, e}. The Cayley tables of the binary operations ◦1, ◦2, and ◦3 are shown in Table 8. Table 8: Cayley tables of X1, Y1, and Z1 ◦1 1 a b c 1 1 a b c a a 1 b c b b a 1 c c c a b 1 ◦2 1 a b c d 1 1 a b c d a a 1 b c d b b a 1 c d c c a b 1 d d d a b c 1 ◦3 1 a b c d e 1 1 a b c d e a a 1 b c d e b b a 1 c d e c c a b 1 d e d d a b c 1 e e e a b c d 1 Then X1, Y1, and Z1 are all non-group-derived dual BG-algebras by Theorem 6. Example 14 shows some of the dual BG-algebras that can be generated using Theorem 6. 4. Conclusion It was shown that the axioms of the dual BG-algebra are independent. Not every dual BG-algebra is a BG-algebra and not every BG-algebra is dual BG-algebra. But it is possible that a dual BG-algebra is also a BG-algebra. A characterization for the dual BG-algebra was established. The notion of a group-derived and non-group-derived dual BG-algebras were also introduced. It was shown that a group-derived dual BG- algebra is characterized by a dual B-algebra and infinitely many non-group-derived dual BG-algebras can be constructed. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 10 of 16 5. Recommendations In one of the theorems, it was shown that every dual B-algebra is a dual BG-algebra. The authors recommend exploring the relationship of the dual BG-algebra to other alge- bras and dual algebras, and not just the dual B-algebra. Other structural properties may also be considered such as ideals, filters, homomorphisms, among others. References [1] Y Imai and K Iséki. On axiom systems of propositional calculi. Proceedings of the Japan Academy, 42(1):19–22, 1966. [2] Q P Hu and X Li. On BCH-algebras. Mathematics Seminar Notes (Kobe University), 11(2):313–320, 1983. [3] Q P Hu and X Li. On proper BCH-algebras. Mathematica Japonica, 30(4):659–661, 1985. [4] Y B Jun, E H Roh, and H S Kim. On BH-algebras. Scientiae Mathematicae, 1(3):347–354, 1998. [5] J Neggers and H S Kim. On d-algebras. Mathematica Slovaca, 49(1):19–26, 1999. [6] J Neggers and H S Kim. On B-algebras. Matematički Vesnik, 54(1-2):21–29, 2002. [7] C B Kim and H S Kim. On BG-algebras. Demonstratio Mathematica, 41(3):497–506, 2008. [8] K H Kim and Y H Yon. Dual BCK-algebra and MV -algebra. Scientiae Mathematicae Japonicae, 66(2):393–399, 2007. [9] H S Kim and Y H Kim. On BE-algebras. Scientiae Mathematicae Japonicae, 66(1):113–116, 2007. [10] A Walendziak. On commutative BE-algebras. Scientiae Mathematicae Japonicae, 69(2):281–284, 2009. [11] B L Meng. CI-algebras. Scientiae Mathematicae Japonicae, 71(1):11–17, 2010. [12] A B Saeid. CI-algebra is equivalent to dual Q-algebra. Journal of the Egyptian Mathematical Society, 21(1):1–2, 2013. [13] K Belleza and J Vilela. The dual B-algebra. European Journal of Pure and Applied Mathematics, 12(4):1497–1507, 2019. [14] T Hungerford. Abstract Algebra: An Introduction. Cengage Learning, 3rd edition, 2012. C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 11 of 16 Appendix The following Python script developed by the author was used to verify if a given Cayley table is a dual BG-algebra. The verification result of Example 3 using the script is shown. def dbg1(X, tbl): # x o x = 1 for all x in X shp = len(tbl) flag = True print(f"X = {X}") print("\nCayley table") print(np.array(tbl)) print("") constant = X[0] ctr = 0 for i in range(shp): ctr += 1 left = tbl[i][i] right = constant if (left != right): print(f"{ctr}.\tx = {X[i]}:\t{X[i]} o {X[i]} != {constant}\t-> \t {left} != {right}") flag = False break print(f"{ctr}.\tx = {X[i]}:\t{X[i]} o {X[i]} = {constant}\t-> \t {left} = {right}") if (flag == True): print(f"\nx o x = {constant} for all x in X") else: print("\nX does not satisfy axiom dbg1.") return flag def dbg2(X, tbl): # 1 o x = x for all x in X C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 12 of 16 shp = len(tbl) flag = True print(f"X = {X}") print("\nCayley table") print(np.array(tbl)) print("") constant = X[0] ctr = 0 for i in range(shp): ctr += 1 left = tbl[0][i] right = X[i] if (left != right): print(f"{ctr}.\tx = {X[i]}:\t{constant} o {X[i]} != {X[i]}\t-> \t {left} != {right}") flag = False break print(f"{ctr}.\tx = {X[i]}:\t{constant} o {X[i]} = {X[i]}\t-> \t {left} = {right}") if (flag == True): print(f"\n{constant} o x = x for all x in X") else: print("\nX does not satisfy axiom dbg2.") return flag def dbg3(X, tbl): # (y o 1) o (y o x) = x for all x, y in X shp = len(tbl) flag = True print(f"X = {X}") print("\nCayley table") print(np.array(tbl)) print("") C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 13 of 16 constant = X[0] ctr = 0 for i in range(shp): for j in range(shp): ctr += 1 y_0 = tbl[j][0] y_x = tbl[j][i] left = tbl[X.index(y_0)][X.index(y_x)] right = X[i] if (left != right): print(f"{ctr}.\tx = {X[i]}, y = {X[j]}: \t({X[j]} o {constant}) o ({X[j]} o {X[i]}) != {X[i]}\t-> \t{y_0} o {y_x} != {X[i]}\t->\t {left} != {right}") flag = False break if (left == right): print(f"{ctr}.\tx = {X[i]}, y = {X[j]}: \t({X[j]} o {constant}) o ({X[j]} o {X[i]}) = {X[i]}\t-> \t{y_0} o {y_x} = {X[i]}\t->\t {left} = {right}") if (flag == False): break if (flag == True): print(f"\n(y o {constant}) o (y o x) = x for all x, y in X") else: print("\nX does not satisfy axiom dbg3.") return flag def dbg(X, tbl): dbg_flag = False constant = X[0] print(f"checking dbg1: x o x = {constant} for all x in X ...") dbg1_flag = dbg1(X, tbl) C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 14 of 16 print("-----------------------------------------------------------\n") print(f"checking dbg2: {constant} o x = x for all x in X ...") dbg2_flag = dbg2(X, tbl) print("-----------------------------------------------------------\n") print("checking dbg3: \n") print(f"(y o {constant}) o (y o x) = x for all x, y in X ...") dbg3_flag = dbg3(X, tbl) print("-----------------------------------------------------------\n") if (dbg1_flag, dbg2_flag, dbg3_flag) == (True, True, True): dbg_flag = True if (dbg_flag == True): print("X satisfies dbg1, dbg2, and dbg3.") print("\nTherefore, X is a Dual BG-algebra.") else: print("\nX is NOT a Dual BG-algebra.") return dbg_flag C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 15 of 16 DBG1 Verification DBG2 Verification C.M. Chan, K. Fuentes / Eur. J. Pure Appl. Math, 18 (4) (2025), 6974 16 of 16 DBG3 Verification Example 3 is a Dual BG-Algebra