EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 4, 2019, 1497-1507 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global The Dual B-Algebra Katrina E. Belleza1,∗, Jocelyn P. Vilela2 1 Department of Mathematics, School of Arts and Sciences, University of San Carlos, 6000 Cebu City, Philippines 2 Department of Mathematics and Statistics, College of Science and Mathematics, Center of Graph Theory, Algebra and Analysis, Premier Research Institute of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. This paper introduces and characterizes the notion of a dual B-algebra. Moreover, this study investigates the relationship between a dual B-algebra and a BCK-algebra. Commutativity of a dual B-algebra is also discussed and its relation to some algebras such as CI-algebra and dual BCI-algebra is examined. 2010 Mathematics Subject Classifications: 06F35, 47L45, 08C05 Key Words and Phrases: B-algebra, dual B-algebra, dual algebra 1. Introduction In 2002, J.Neggers and H.S. Kim [9] introduced and investigated B-algebras which is related to several classes of algebras such as BCH/BCI/BCK-algebras and established that B-algebras are related to groups. In the same year, M.Kondo and Y.B. Jun [4] showed that every B-algebra is group-derived. In 2010, N.O. Al-Shehrie [1] introduced the left-right (resp. right-left) derivation on a B-algebra and some related properties were investigated. In 1996, Y.Imai and K.Iseki [2] introduced two classes of algebras: BCK- algebras and BCI-algebras. It is known that a BCI-algebra is a generalization of a BCK- algebra. In 2007, dual BCK-algebra was introduced by K.H. Kim and Y.H. Yon [3] and some properties were also studied. Moreover, K.H. Kim and Y.H. Yon [3] investigated the relationship between a dual BCK-algebra and an MV -algebra. On the other hand, A. Walendziak [12] defined commutative BE-algebras in 2008 and proved that these are equivalent to the commutative dual BCK-algebras. In 2009, the notions of dual BCI- algebra and CI-algebra were introduced by B.L. Meng [5] together with some of their properties. It is shown that CI-algebra is a generalization of dual BCK/BCI/BCH- algebras. In 2013, A.B. Saeid [11] established the relationship between CI-algebra and dual Q-algebra. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i4.3526 Email addresses: kebelleza@usc.edu.ph (K. Belleza), jocelyn.vilela@g.msuiit.edu.ph (J. Vilela) http://www.ejpam.com 1497 c© 2019 EJPAM All rights reserved. K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1498 This paper aims to characterize a dual B-algebra and to investigate the relationship between a dual B-algebra and BCK-algebra. Moreover, commutativity of a dual B- algebra will also be considered. Relationships between commutative dual B-algebra and other algebras such as CI-algebra and dual BCI-algebra will be investigated in this paper. 2. Preliminaries An algebra of type (2,0) is an algebra with a binary operation and a constant element. Definition 1. [9] A B-algebra is a non-empty set X with a constant 0 and a binary operation “ ∗ ” satisfying the following axioms for all x, y, z in X: (B1) x ∗ x = 0 (B2) x ∗ 0 = x (B3) (x ∗ y) ∗ z = x ∗ [z ∗ (0 ∗ y)] Example 1. [8] Let X := {0, 1, 2, 3, 4, 5} be a set with the following Cayley table: ∗ 0 1 2 3 4 5 0 0 2 1 3 4 5 1 1 0 2 4 5 3 2 2 1 0 5 3 4 3 3 4 5 0 2 1 4 4 5 3 1 0 2 5 5 3 4 2 1 0 Then (X; ∗, 0) is a B-algebra. Definition 2. [6] An algebra (X, ∗, 0) of type (2, 0) is called a BCK-algebra if for all x, y, z in X, the following hold: (BCK1) [(x ∗ y) ∗ (x ∗ z)] ∗ (z ∗ y) = 0 (BCK4) x ∗ y = 0 and y ∗ x = 0 imply x = y (BCK2) [x ∗ (x ∗ y)] ∗ y = 0 (BCK5) 0 ∗ x = 0 (BCK3) x ∗ x = 0 Lemma 1. [2] In any BCK-algebra (X, ∗, 0), the following hold for all x, y, z in X: (i) x ∗ 0 = x (ii) (x ∗ y) ∗ z = (x ∗ z) ∗ y Definition 3. [7] A Q-algebra is a nonempty set X with a constant 0 and a binary operation ∗ satisfying the following axioms: for all x, y, z in X, (Q1) x ∗ x = 0 (Q2) x ∗ 0 = x (Q3) (x ∗ y) ∗ z = (x ∗ z) ∗ y Definition 4. [11] Let (X, ∗, 0) be a Q-algebra and a binary operation ◦ on X is defined as: x◦y = y ∗x. Then (X, ◦, 1) is called a dual Q-algebra. In fact, its axioms are as follows for all x, y, z in X: (DQ1) x ◦ x = 1 (DQ2) 1 ◦ x = x (DQ3) x ◦ (y ◦ z) = y ◦ (x ◦ z) K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1499 Definition 5. [5] A CI-algebra is an algebra (X, ∗, 1) of type (2, 0) satisfying the following axioms: for all x, y, z in X, (CI1) x ∗ x = 1 (CI2) 1 ∗ x = x (CI3) x ∗ (y ∗ z) = y ∗ (x ∗ z) Theorem 1. [11] Any CI-algebra is equivalent to a dual Q-algebra. Definition 6. [5] A dual BCI-algebra is an algebra (X, ∗, 1) of type (2,0) satisfying the following axioms: for all x, y, z in X, (DBCI1) x ∗ x = 1 (DBCI3) (x ∗ y) ∗ [(y ∗ z) ∗ (x ∗ z)] = 1 (DBCI2) x ∗ y = y ∗ x = 1 implies x = y (DBCI4) x ∗ [(x ∗ y) ∗ y] = 1 Proposition 1. [5] Let (X, ∗, 1) be a dual BCI-algebra. Then for all x, y, z in X, the following hold: (i) x ∗ y = 1 implies (y ∗ z) ∗ (x ∗ z) = 1 (iii) y ∗ (z ∗ x) = z ∗ (y ∗ x) (ii) x ∗ y = 1 and y ∗ z = 1 imply x ∗ z = 1 (iv) 1 ∗ x = x 3. Dual B-Algebra Definition 7. A dual B-algebra XD is a triple (X, ◦, 1) where X is a non-empty set with a binary operation “ ◦ ” and a constant 1 satisfying the following axioms for all x, y, z in XD: (DB1) x ◦ x = 1 (DB2) 1 ◦ x = x (DB3) x ◦ (y ◦ z) = ((y ◦ 1) ◦ x) ◦ z Remark 1. If (X, ∗, 0) is a B-algebra, define “ ◦ ” as follows: x ◦ y = y ∗ x for all x, y in X. Then (X, ◦, 0) is a dual B-algebra, called the derived dual B-algebra. Example 2. Consider the B-algebra X = {0, 1, 2, 3, 4, 5} in Example 1. The dual B- algebra of X is XD = (X, ◦, 0) with the following table: ◦ 0 1 2 3 4 5 0 0 1 2 3 4 5 1 2 0 1 4 5 3 2 1 2 0 5 3 4 3 3 4 5 0 1 2 4 4 5 3 2 0 1 5 5 3 4 1 2 0 Define “ · ” as follows: x · y = y ◦x. Then XDD = (X, ·, 0) is the B-algebra X with Cayley table in Example 1. Proposition 2. Let XD = (X, ◦, 0) be a dual B-algebra. Then XDD = (X, ·, 0) is a B-algebra where x · y = y ◦ x for all x, y in XD. Proof: Suppose XD is a dual B-algebra and define “ · ” as follows: x · y = y ◦ x for all x, y in XD. Then the axioms of XDD = (X, ·, 0) coincide with that of a B-algebra. Hence, XDD is a B-algebra. K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1500 Example 3. Let X = R and ◦ be defined as x ◦ y = y x for all x, y in X with x 6= 0. Note that X satisfies (DB1): x ◦ x = x x = 1, (DB2): 1 ◦ x = x 1 = x, and (DB3): x ◦ (y ◦ z) = y ◦ z x = z xy = z x y ◦ 1 = z (y ◦ 1) ◦ x = ((y ◦ 1) ◦ x) ◦ z. Hence, (R, ◦, 1) is a dual B-algebra. Observe that (R, ◦, 1) is not a B-algebra since 4 ◦ 1 = 1 4 6= 4. This leads to the next remark. Remark 2. Not every dual B-algebra is a B-algebra. Example 4. Let X = {e, a, b, c} be the Klein-4 B-algebra with the following table: ◦ 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 the dual XD of X is itself. Hence, the Klein-4 B-algebra is a dual B-algebra. Observe that the Klein-4 B-algebra has a symmetric Cayley table and is a dual B-algebra itself. Hence, there exists a B-algebra that is also a dual B-algebra. This is generalized in the next theorem. Let (X, ∗, 0) be any algebra of type (2, 0) satisfying x∗y = y ∗x for all x, y in X. Then we say that (X, ∗, 0) satisfies a symmetric condition. Theorem 2. Let X be a B-algebra satisfying a symmetric condition. Then X itself is a dual B-algebra, that is, X = XD. Proof: Suppose X is a B-algebra satisfying a symmetric condition. Then the dual B-algebra axioms hold, namely (DB1): x ∗ x = 0 by (B1), (DB2): 0 ∗ x = x ∗ 0 = x by (B2), and (DB3): x ∗ (y ∗ z) = (z ∗ y) ∗x = z ∗ [x ∗ (0 ∗ y)] = [(y ∗ 0) ∗x] ∗ z by (B3). Hence, X is a dual B-algebra. Example 5. Let X = {0, 1, 2} be a set with the following table: ∗ 0 1 2 0 0 2 1 1 1 0 2 2 2 1 0 Then (X, ∗, 0) is a B-algebra [9]. Observe that in this example, 1 ∗ (2 ∗ 0) = 1 ∗ 2 = 2 6= 1 = 1 ∗ 0 = (2 ∗ 1) ∗ 0 = [(2 ∗ 0) ∗ 1] ∗ 0. This implies that X is not a dual B-algebra. Remark 3. Not every B-algebra is a dual B-algebra. K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1501 Lemma 2. Let XD be a dual B-algebra. Then for any x, y, z in XD, we have (i) x ◦ y = [(x ◦ 1) ◦ 1] ◦ y (vi) x ◦ 1 = y ◦ 1 implies x = y (ii) (x ◦ 1) ◦ (x ◦ y) = y (vii) x = (x ◦ 1) ◦ 1 (iii) (y ◦ z) ◦ x = z ◦ [(y ◦ 1) ◦ x] (viii) (y ◦ x) ◦ (y ◦ 1) = x ◦ 1 (iv) z ◦ x = z ◦ y implies x = y (ix) x ◦ [(x ◦ 1) ◦ x] = x (v) x ◦ y = 1 implies x = y (x) x ◦ y = 1 implies (x ◦ z) ◦ (y ◦ z) = 1. Proof: Let XD be a dual B-algebra and x, y, z ∈ XD. (i) By (DB2) and (DB3), x ◦ y = 1 ◦ (x ◦ y) = [(x ◦ 1) ◦ 1] ◦ y. (ii) By (DB3), (DB1), and (DB2), (x ◦ 1) ◦ (x ◦ y) = [(x ◦ 1) ◦ (x ◦ 1)] ◦ y = 1 ◦ y = y. (iii) By (i) and (DB3), (y ◦ z) ◦ x = [ ((y ◦ 1) ◦ 1) ◦ z ] ◦ x = z ◦ [(y ◦ 1) ◦ x]. (iv) Suppose z ◦ x = z ◦ y. Then (z ◦ 1) ◦ (z ◦ x) = (z ◦ 1) ◦ (z ◦ y) implies x = y by (ii). (v) Suppose x ◦ y = 1. By (DB1) and (iv), we get x ◦ y = x ◦ x implying x = y. (vi) Suppose x◦1 = y◦1.By (DB1), (DB2),(DB3), and (i) we have 1 = x◦x = 1◦(x◦x) = [(x ◦ 1) ◦ 1] ◦ x = [(y ◦ 1) ◦ 1] ◦ x = y ◦ x. Hence, y = x by (v). (vii) By (DB2), (DB3), and (vi), x◦1 = 1◦(x◦1) = [(x◦1)◦1]◦1 implies that x = (x◦1)◦1. (viii) By (iii) and (DB1), (y ◦ x) ◦ (y ◦ 1) = x ◦ [(y ◦ 1) ◦ (y ◦ 1)] = x ◦ 1. (ix) Take y = z = x in (iii). Then apply (DB1) and (DB2). (x) By (v), x◦y = 1 implies x = y. Hence by (DB1), (x◦z)◦(y◦z) = (x◦z)◦(x◦z) = 1. The following theorem is a characterization of a dual B-algebra given any algebra with a binary operation and a constant element. Theorem 3. Let X = (X, ◦, 1) be any algebra of type (2, 0). Then X is a dual B-algebra if and only if for any x, y, z in X, (i) x ◦ x = 1; (ii) x = (x ◦ 1) ◦ 1; (iii) (x ◦ y) ◦ (x ◦ z) = y ◦ z. Proof: Suppose X = (X, ◦, 1) is a dual B-algebra. Then X satisfies (DB1) and Lemma 2(vii). By (DB3), (DB1), and (DB2), (x◦y)◦(x◦z) = [(x◦1)◦(x◦y)]◦z = [( (x◦1)◦(x◦1) ) ◦ y ] ◦z = (1◦y)◦z = y◦z. It follows that X satisfies (i), (ii), and (iii). Conversely by (iii), (i), and (ii), 1◦x = (x◦1)◦(x◦x) = (x◦1)◦1 = x. Hence, X satisfies (DB2). For X to satisfy (DB3), we have x◦(y◦z) = [(y◦1)◦x]◦ [(y◦1)◦(y◦z)] = [(y◦1)◦x]◦(1◦z) = [(y◦1)◦x]◦z by (iii) and (DB2). Therefore, X is a dual B-algebra. Comparing the axioms of a dual B-algebra and a BCK-algebra, we have the following remark. Remark 4. (DB1) is equivalent to (BCK3) and Lemma 2(v) is equivalent to (BCK4) where the constant 1 corresponds to the constant 0 in a dual B-algebra and BCK-algebra, respectively. Example 6. Consider the dual B-algebra X = {0, 1, 2, 3, 4, 5} in Example 2. Note that (X, ◦, 0) is not a BCK-algebra since (BCK2) is not satisfied, that is, [1 ◦ (1 ◦ 5)] ◦ 5 = (1 ◦ 3) ◦ 5 = 4 ◦ 5 = 1 6= 0. Also, 2 ◦ 1 = 2 6= 1 = 1 ◦ 2. K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1502 Example 7. Consider the Klein-4 dual B-algebra XD in Example 4. Observe that this example satisfies the symmetric condition but is not a BCK-algebra since e ◦x 6= e for all x ∈ X. Lemma 3. Let XD = (X, ◦, 1) be a dual B-algebra satisfying a symmetric condition. Then for all x, y, z in X, (x ◦ y) ◦ (z ◦ y) = x ◦ z. Proof: By (DB3), hypothesis, Lemma 2(iii) and (i), (DB1), and (DB2), we have (x ◦ y) ◦ (z ◦ y) = [(z ◦ 1) ◦ (x ◦ y)] ◦ y = [z ◦ (x ◦ y)] ◦ y = [(x ◦ y) ◦ z] ◦ y = ( [(x ◦ y) ◦ 1] ◦ z ) ◦ y = z ◦ [(x◦y)◦y] = z ◦ ( y ◦ [(x◦1)◦y] ) = z ◦ [y ◦(x◦y)] = z ◦ [y ◦(y ◦x)] = z ◦ ( y ◦ [(y ◦1)◦x] ) = z ◦ [( [(y ◦ 1) ◦ 1] ◦ y ) ◦ x ] = z ◦ [(y ◦ y) ◦ x] = z ◦ (1 ◦ x) = z ◦ x = x ◦ z. Proposition 3. Let XD = (X, ◦, 1) be a dual B-algebra satisfying a symmetric condition. Then XD satisfies (BCK1), (BCK2), (BCK3), and (BCK4) of a BCK-algebra. Proof: Suppose XD is a dual B-algebra satisfying a symmetric condition. Then by (DB2) and the hypothesis, x = 1 ◦ x = x ◦ 1 for all x in XD. By Remark 4, it remains to show that XD satisfies (BCK1) and (BCK2). Let x, y, z ∈ XD. By (DB3), hypothesis, (DB1) and (DB2), [x◦(x◦y)]◦y = ( [(x◦1)◦x]◦y ) ◦y = [(x◦x)◦y]◦y = (1◦y)◦y = y◦y = 1. Thus, XD satisfies (BCK2). By Lemma 2 (iii) and hypothesis, [(x ◦ y) ◦ (x ◦ z)] ◦ (z ◦ y) = (x ◦ z) ◦ ( [(x ◦ y) ◦ 1] ◦ (z ◦ y) ) = (x ◦ z) ◦ [(x ◦ y) ◦ (z ◦ y)]. By the hypothesis, Lemma 3 and (DB1), [(x ◦ y) ◦ (x ◦ z)] ◦ (z ◦ y) = [(y ◦ x) ◦ (z ◦ x)] ◦ (y ◦ z) = (y ◦ z) ◦ (y ◦ z) = 1. So, XD satisfies (BCK1). Example 8. Let X = {0, a, b, c, d} be a BCK-algebra [10] with the following Cayley table: ∗ 0 a b c d 0 0 0 0 0 0 a a 0 a 0 0 b b b 0 b 0 c c c c 0 c d d d d d 0 Note that b ∗ a = b 6= a = a ∗ b. In fact, 0 ∗ b = 0 6= b. So, X does not satisfy (DB2) and hence, is not a dual B-algebra. The following theorem shows that if the symmetric condition holds in a BCK-algebra X, then X is a dual B-algebra. Theorem 4. If (X, ◦, 1) is a BCK-algebra satisfying a symmetric condition, then X is a dual B-algebra. Proof: Suppose X is a BCK-algebra satisfying x ◦ y = y ◦ x for all x, y in X. By Remark 4, it remains to show that X satisfies (DB3) and (DB2). By Lemma 1(i) and (ii) of a BCK-algebra, [(y ◦1)◦x]◦ z = (y ◦x)◦z = (y ◦ z)◦x. Since x◦y = y ◦x for all x, y in X, (y ◦ z) ◦ x = x ◦ (y ◦ z). Hence, X satisfies (DB3). By Lemma 1(i) and the hypothesis, x = x ◦ 1 = 1 ◦ x. This implies that X satisfies (DB2). K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1503 4. Commutativity in a Dual B-algebra Definition 8. Let XD be a dual B-algebra. Define a binary operation “ + ” on X as follows: x + y = (x ◦ 1) ◦ y for all x, y in XD. A dual B-algebra is said to be commutative if x + y = y + x, that is, (x ◦ 1) ◦ y = (y ◦ 1) ◦ x for all x, y in XD. Example 9. The dual B-algebra X = R in Example 3 is commutative since for all x, y in R, (x ◦ 1) ◦ y = y x ◦ 1 = y 1 x = xy = x 1 y = x y ◦ 1 = (y ◦ 1) ◦x. However, the dual B-algebra in Example 2 is not commutative since (1 ◦ 0) ◦ 4 = 2 ◦ 4 = 3 6= 5 = 4 ◦ 1 = (4 ◦ 0) ◦ 1. Observe that (1 ◦ 0) ◦ (3 ◦ 0) = 2 ◦ 3 = 5 6= 4 = 3 ◦ 1 and (2 ◦ 5) ◦ 5 = 4 ◦ 5 = 1 6= 2. However, for a commutative dual B-algebra, the following proposition holds. Proposition 4. Suppose XD is a commutative B-algebra. Then the following hold for all x, y in XD: (i) (x ◦ 1) ◦ (y ◦ 1) = y ◦ x (ii) (y ◦ x) ◦ x = y. Proof: Let XD be a commutative B-algebra. (i)By Definition 8 and Lemma 2(i), (x ◦ 1) ◦ (y ◦ 1) = [(y ◦ 1) ◦ 1] ◦ x = y ◦ x. (ii)Applying Lemma 2(iii), Definition 8, (DB3), Lemma 2(i), (DB1), and (DB2), (y ◦ x) ◦ x = x ◦ [(y ◦ 1) ◦ x] = x ◦ [(x ◦ 1) ◦ y] =( [(x ◦ 1) ◦ 1] ◦ x ) ◦ y = (x ◦ x) ◦ y = 1 ◦ y = y. Lemma 4. If XD is a commutative dual B-algebra, then the right cancellation law holds, that is, x ◦ z = y ◦ z implies x = y for all x, y, z in XD. Proof: Suppose XD is commutative and x ◦ z = y ◦ z for any x, y, z in XD. Then by Proposition 4(ii), we can write x = (x ◦ z) ◦ z = (y ◦ z) ◦ z = y. Proposition 5. If XD is a commutative dual B-algebra, then the following hold for all x, y, z in XD: (i) x ◦ (y ◦ z) = y ◦ (x ◦ z) (iii) x ◦ (y ◦ x) = (x ◦ y) ◦ (x ◦ 1) (ii) (x ◦ y) ◦ z = (z ◦ y) ◦ x (iv) y ◦ [(y ◦ x) ◦ x] = 1. Proof: Suppose XD is commutative and x, y, z ∈ XD. (i) By (DB3) and Definition 8, x ◦ (y ◦ z) = [(y ◦ 1) ◦ x] ◦ z = [(x ◦ 1) ◦ y] ◦ z = y ◦ (x ◦ z). (ii) Applying Lemma 2(iii) and since XD is commutative, (x ◦ y) ◦ z = y ◦ [(x ◦ 1) ◦ z] = y ◦ [(z ◦ 1) ◦ x] = (z ◦ y) ◦ x. (iii) Write x ◦ (y ◦ x) = y ◦ (x ◦ x) by (i). Then y ◦ (x ◦ x) = y ◦ 1 = (x ◦ y) ◦ (x ◦ 1) by (DB1) and Lemma 2(viii). (iv) Follows directly from Proposition 4(ii) and (DB1). Corollary 1. If XD is a dual B-algebra satisfying a symmetric condition, then XD is commutative. Proof: Let XD be a dual B-algebra satisfying a symmetric condition. Then (x◦1)◦y = (1◦x)◦y = x◦y = y◦x = (1◦y)◦x = (y◦1)◦x. This implies that XD is commutative. The following corollary follows from Theorem 4 and Corollary 1. K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1504 Corollary 2. Suppose X is a BCK-algebra satisfying a symmetric condition. Then X is a commutative dual B-algebra. The following results present the relationship between a commutative dual B-algebra and some algebras, namely, CI-algebra and dual BCI-algebra. Comparing the axioms and properties of commutative dual B-algebra, CI-algebra and dual BCI-algebra, we have the following remarks. Remark 5. (i) The class of commutative dual B-algebras is a subclass of CI-algebras since (DB1) is equivalent to (CI1), (DB2) is equivalent to (CI2), and Proposition 5(i) is equivalent to (CI3). (ii) (DB1) is equivalent to (DBCI1), Lemma 2(v) is equivalent to (DBCI2), Proposition 5(iv) is equivalent to (DBCI4), (DB2) is equivalent to Proposition 1(iv) Example 10. Consider the non-commutative dual B-algebra X = {0, 1, 2, 3, 4, 5} in Ex- ample 2. Now 2 ◦ (4 ◦ 5) = 2 ◦ 1 = 2 6= 0 = 4 ◦ 4 = 4 ◦ (2 ◦ 5). Hence, X does not satisfy (CI3). The following corollaries follow from Remark 5 and Theorem 1. Corollary 3. If XD is a commutative dual B-algebra, then XD is a CI-algebra. Corollary 4. Every commutative dual B-algebra is a dual Q-algebra. The converse of Corollary 3 is not always true as shown in the following example. Example 11. Let X = {1, a, b, c, d} be a set with the following Cayley table: ∗ 1 a b c d 1 1 a b c d a 1 1 b b d b 1 a 1 a d c 1 1 1 1 d d d d d d 1 Then (X, ∗, 1) is a CI-algebra [5] but is not a dual B-algebra since it does not satisfy (DB3). Indeed, a ◦ (b ◦ c) = a ◦ a = 1 6= b = a ◦ c = (1 ◦ a) ◦ c = [(b ◦ 1) ◦ a] ◦ c. Theorem 5. If X is a CI-algebra satisfying a symmetric condition, then X is a commu- tative dual B-algebra. Proof: Suppose X is a CI-algebra satisfying a symmetric condition. By Remark 5, it remains to show that X satisfies (DB3) and that X is commutative. Applying (CI3) and the hypothesis, x ◦ (y ◦ z) = y ◦ (x ◦ z) = (y ◦ 1) ◦ (z ◦ x) = z ◦ [(y ◦ 1) ◦ x] = [(y ◦ 1) ◦ x] ◦ z. Hence, X satisfies (DB3). By Corollary 1, it follows that X is commutative. K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 12 (4) (2019), 1497-1507 1505 Example 12. Consider the non-commutative dual B-algebra X = {0, 1, 2, 3, 4, 5} in Ex- ample 2. Observe that (1 ◦ 2) ◦ [(2 ◦ 4) ◦ (1 ◦ 4)] = 1 ◦ (3 ◦ 5) = 1 ◦ 2 = 1 6= 0. Hence, XD does not satisfy (DBCI3) and so XD is not a dual BCI-algebra. However, if commutativity holds for a dual B-algebra, then it is also a dual BCI- algebra as shown in the next theorem. Theorem 6. Every commutative dual B-algebra is a dual BCI-algebra. Proof: Let XD be a commutative dual B-algebra. By Remark 5, it remains to show that XD satisfies (DBCI3). By Proposition 5(ii), Proposition 4(ii), and (DB1), (x ◦ y) ◦ [(y ◦ z) ◦ (x ◦ z)] = (x ◦ y) ◦ ( [(x ◦ z) ◦ z] ◦ y ) = (x ◦ y) ◦ (x ◦ y) = 1. Hence, X satisfies (DBCI3). Therefore, X is a dual BCI-algebra. Note that the converse of Theorem 6 is not always true as shown in the following example. Example 13. Let X = {0, 1, a, b, c} with binary operation “ ∗ ” on X defined by the following table on the left: ∗ 0 1 a b c 0 0 0 a a a 1 1 0 a a a a a a 0 0 0 b b a 1 0 1 c c a 1 1 0 ◦ 0 1 a b c 0 0 1 a b c 1 0 0 a a a a a a 0 1 1 b a a 0 0 1 c a a 0 1 0 Then X = (X, ∗, 0) is a BCI-algebra [13]. Note that (X, ◦, 0) is a dual BCI-algebra. Now, 1 ◦ (b ◦ c) = 1 ◦ 1 = 0 6= 1 = a ◦ c = (a ◦ 1) ◦ c = [(b ◦ 0) ◦ 1] ◦ c. Thus, X does not satisfy (DB3). Hence, X is not a dual B-algebra. However if a dual BCI-algebra X satisfies the symmetric condition, then X is also a dual B-algebra as shown in the next theorem. Theorem 7. If X is a dual BCI-algebra satisfying a symmetric condition, then X is a commutative dual B-algebra. Proof: Suppose X is a dual BCI-algebra satisfying a symmetric condition. Then Proposition 1(iv) becomes x = 1 ◦ x = x ◦ 1. By Remark 5, it remains to show that X satisfies (DB3) and is commutative. Applying the hypothesis, Propositon 1(iii) and (iv), x ◦ (y ◦ z) = x ◦ (z ◦ y) = z ◦ (x ◦ y) = z ◦ [x ◦ (1 ◦ y)] = [x ◦ (1 ◦ y)] ◦ z = [(1 ◦ y) ◦ x] ◦ z = [(y ◦ 1) ◦ x] ◦ z. Hence, X satisfies (DB3). Also by the hypothesis and Proposition 1(iii), (x ◦ 1) ◦ y = y ◦ (x ◦ 1) = x ◦ (y ◦ 1) = (y ◦ 1) ◦ x. Therefore, X is commutative. REFERENCES 1506 5. Conclusion In this paper, the notion of a dual B-algebra is presented together with some of its properties and characterizations. Not every B-algebra is a dual B-algebra and not every dual B-algebra is a B-algebra. However, there exists an algebra that is both a B-algebra and a dual B-algebra. 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