ON BO-ALGEBRAS 1. Introduction

DOI: 10.2478/s12175-012-0050-9 Math. Slovaca 62 (2012), No. 5, 855–864

ON BO-ALGEBRAS Chang Bum Kim* — Hee Sik Kim** (Communicated by Anatolij Dvureˇ censkij ) ABSTRACT. In this paper, we introduce the notion of a BO-algebra, and we prove that every BO-algebra is 0-commutative, and we show that BO-algebras, 0-commutative B-algebras, BM-algebras, p-semisimple BCI-algebras and abelian groups are logically equivalent. c 2012 Mathematical Institute Slovak Academy of Sciences

1. Introduction The notion of B-algebras was introduced by J. Neggers and H. S. Kim [6]. They defined a B-algebra as an algebra (X, ∗, 0) of type (2, 0) (i.e., a non-empty set with a binary operation ∗ and a constant 0) satisfying the following axioms: (B1) x ∗ x = 0, (B2) x ∗ 0 = x, (B) (x ∗ y) ∗ z = x ∗ [z ∗ (0 ∗ y)] for any x, y, z ∈ X. Recently, the present authors [2] defined a BG-algebra, which is a generalization of B-algebra. An algebra (X, ∗, 0) of type (2, 0) is called a BG-algebra if it satisfies (B1), (B2), and (BG) x = (x ∗ y) ∗ (0 ∗ y) for any x, y ∈ X. Y. B. Jun, E. H. Roh and H. S. Kim [1] introduced the notion of a BH-algebra which is a generalization of BCK/BCI/BCH-algebras. An algebra (X, ∗, 0) of type (2, 0) is called a BH-algebra if it satisfies (B1), (B2), and (BH) x ∗ y = y ∗ x = 0 implies x = y for any x, y ∈ X. Moreover, A. Walendziak [7] introduced the notion of BF/BF1 /BF2 -algebras. An algebra (X, ∗, 0) of type (2, 0) is called a BF -algebra if it satisfies (B1), (B2) and 2010 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 06F35. K e y w o r d s: BO-algebra, BM-algebra, B-algebra, 0-commutative. Research of the first author was supported by Kookmin University Research Fund, 2010.

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CHANG BUM KIM — HEE SIK KIM

(BF) 0 ∗ (x ∗ y) = y ∗ x for any x, y ∈ X. A BF -algebra is called a BF1 -algebra (resp., a BF2 -algebra) if it satisfies (BG) (resp., (BH)).

2. BO-algebras In this section, we introduce the notion of a BO-algebra, and we prove that every BO-algebra is 0-commutative, and we discuss basic properties of BO-algebras. With this properties we show that every BO-algebra is both a BG-algebra and a BH-algebra.

2.1 A BO-algebra is an algebra (X, ∗, 0) of type (2, 0) satisfying (B1), (B2) and (BO), where (BO) x ∗ (y ∗ z) = (x ∗ y) ∗ (0 ∗ z) for any x, y, z ∈ X. Example 2.2. Let X := {0, 1, 2, 3, 4} be a set. If we define a binary operation ∗ on X as follows: ∗ 0 1 2 3 4 0 1 2 3 4

0 1 2 3 4

2 0 4 1 3

1 3 0 4 2

4 2 3 0 1

3 4 1 2 0

then (X, ∗, 0) is a BO-algebra.

2.3

If (X, ∗, 0) is a BO-algebra, then

(i) 0 ∗ (0 ∗ x) = x, (ii) 0 ∗ (x ∗ y) = (0 ∗ x) ∗ (0 ∗ y), (iii) x ∗ (x ∗ y) = y, (iv) x = (x ∗ y) ∗ (0 ∗ y), (v) x ∗ y = z ∗ y =⇒ x = z, (vi) x ∗ y = x ∗ z =⇒ y = z, (vii) 0 ∗ x = 0 ∗ y =⇒ x = y, (viii) the linear equation a ∗ x = b has a unique solution, for any x, y, z ∈ X. P r o o f. (i) If we let x := z and y := z in (BO), then z ∗ (z ∗ z) = (z ∗ z) ∗ (0 ∗ z). By applying (B1) and (B2), we obtain 0 ∗ (0 ∗ z) = z. (ii) Let x := 0 in (BO). Then we obtain 0 ∗ (x ∗ y) = (0 ∗ x) ∗ (0 ∗ y). 856


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(iii) Let y := x in (BO). By applying (i), (B1) and (B2), we obtain x ∗ (x ∗ z) = (x ∗ x) ∗ (0 ∗ z) = (0 ∗ 0) ∗ (0 ∗ z) = z. (iv) Let z := y in (BO). By applying (B1) and (B2) we obtain x = (x ∗ y) ∗ (0 ∗ y). (v) Let x ∗ y = z ∗ y. By (iv), x = (x ∗ y) ∗ (0 ∗ y) = (z ∗ y) ∗ (0 ∗ y) = z. (vi) Let x ∗ y = x ∗ z. Then x ∗ (x ∗ y) = x ∗ (x ∗ z). By (iii), we obtain y = z. (vii) It is trivial by (i). (viii) a∗b is a solution of a∗x = b by (iii). The uniqueness follows immediately from (v).

2.4

x = y.

Let (X, ∗, 0) be a BO-algebra. If x ∗ y = 0, x, y ∈ X, then

P r o o f. It can be proved by Theorem 2.3(v).

2.5

If (X, ∗, 0) is a BO-algebra, then it is a BG-algebra.

P r o o f. By Theorem 2.3(iv).

Remark The converse of Theorem 2.5 does not hold in general. See the following example. Example 2.6. Let X := {0, 1, 2} be a set with the following table: ∗

0 1

2

0 1 2

0 1 1 0 2 2

2 1 0

Then it is a BG-algebra, but not a BO-algebra, since 1 ∗ (1 ∗ 2) = 0 = 2 = (1 ∗ 1) ∗ (0 ∗ 2).

2.7

If (X, ∗, 0) is a BO-algebra, then (X, ∗, 0) is a BH-algebra.

P r o o f. Let x ∗ y = 0 and y ∗ x = 0, then x = y by Theorem 2.3(v).

3. BO-algebras with 0 ∗ x = x In this section, we show that every BO-algebra (X, ∗, 0) with condition 0 ∗ x = x for all x ∈ X is both an abelian group and a 0-commutative B-algebra.

3.1

Let (X, ∗, 0) be a BO-algebra. If 0 ∗ x = x for all x ∈ X, then (X, ∗, 0) is an abelian group. 857



P r o o f. Since (X, ∗, 0) is a BO-algebra, x∗0 = 0∗x = x for any x ∈ X. Hence 0 is the identity element for (X, ∗, 0). Since x ∗ x = 0, the inverse of x is itself, i.e., x−1 = x. The associative law holds, since x ∗ (y ∗ z) = (x ∗ y) ∗ (0 ∗ z) = (x ∗ y) ∗ z. Hence (X, ∗, 0) is a group. On the other hand, x∗y = (x∗y)−1 = y −1 ∗x−1 = y∗x, proving that (X, ∗, 0) is an abelian group. Any groupoid (algebra, binary system) (X, ∗, 0) of type (2, 0) is said to be 0-commutative if x ∗ (0 ∗ y) = y ∗ (0 ∗ x) for all x, y ∈ X. Even though it was proved that the notion of 0-commutative B-algebras is logically equivalent to the notion of abelian groups ([4]), we prove directly that every BO-algebra (X, ∗, 0) with 0 ∗ x = x for all x ∈ X is a 0-commutative B-algebra.

3.2 Let (X, ∗, 0) be a BO-algebra. If 0 ∗ x = x for all x ∈ X, then

(X, ∗, 0) is a 0-commutative B-algebra.

P r o o f. By Theorem 3.1, we must show that the condition (B) and 0-commutativity hold. For any x, y, z ∈ X, x ∗ [z ∗ (0 ∗ y)] = (x ∗ z) ∗ (0 ∗ (0 ∗ y))

[(BO)]

= (x ∗ z) ∗ y

[Theorem 2.3(i)]

= x ∗ (z ∗ y)

[Theorem 3.1]

= x ∗ (y ∗ z)

[Theorem 3.1]

= (x ∗ y) ∗ z.

[Theorem 3.1]

Hence (X, ∗, 0) is a B-algebra. On the other hand, given x, y ∈ X, x ∗ (0 ∗ y)

=

x∗y

=

y∗x

=

y ∗ (0 ∗ x).

Hence (X, ∗, 0) is a 0-commutative B-algebra.

Example 3.3. Let X := {0, 1, 2, 3} be a set with the following table: ∗

0

1

2

3

0 1 2 3

0 1 2 3

1 0 3 2

2 3 0 1

3 2 1 0

Then it is easy to check that (X, ∗, 0) is a BO-algebra with 0 ∗ x = x for any x ∈ X. By Corollary 3.2, it is also a 0-commutative B-algebra. In fact, (X, ∗, 0) is the Klein’s four group. 858


ON BO-ALGEBRAS

4. Logically equivalences In this section, we show that BO-algebras are logically equivalent to several well-known algebras.

4.1

Every 0-commutative B-algebra (X, ∗, 0) is a BO-algebra.

P r o o f. It is enough to show the condition (BO). Given x, y, z ∈ X we have (x ∗ y) ∗ (0 ∗ z) = x ∗ [(0 ∗ z) ∗ (0 ∗ y)]

[(B)]

= x ∗ (y ∗ (0 ∗ (0 ∗ z))

[0-commutative]

= x ∗ (y ∗ z).

[Theorem 2.3(i)]

The condition, 0-commutativity, is very necessary for B-algebras to be BO-algebras. Example 4.2. Let X := {0, 1, 2, 3, 4, 5} be a set with the following table: ∗ 0 1 2 3 4 5 0 0 2 1 3 4 5 1 1 0 2 5 3 4 2 2 1 0 4 5 3 3 3 5 4 0 1 2 4 4 3 5 2 0 1 5 5 4 3 1 2 0 Then (X, ∗, 0) is a non-0-commutative B-algebra, since 1 ∗ (0 ∗ 3) = 1 ∗ 3 = 3 = 3 ∗ (0 ∗ 1) = 3 ∗ 2 = 4. Moreover, (X, ∗, 0) is not a BO-algebra, since 1 ∗ (2 ∗ 5) = 1 ∗ 3 = 5 = (1 ∗ 2) ∗ (0 ∗ 5) = 2 ∗ 5 = 3. Every abelian group can determine a BO-algebra as follow:

4.3

Let (X, ◦, 0) be an abelian group. If we define x ∗ y := x ◦ y −1 , then (X, ∗, 0) is a BO-algebra. P r o o f. We see that x ∗ x = x ◦ x−1 = 0 and x ∗ 0 = x ◦ 0−1 = x ◦ 0 = x. For any x, y, z ∈ X, (x ∗ y) ∗ (0 ∗ z)

=

(x ◦ y −1 ) ◦ (0 ◦ z −1 )−1

=

(x ◦ y −1 ) ◦ (z −1 )−1

=

(x ◦ y −1 ) ◦ z

=

x ◦ (y −1 ◦ z)

=

x ◦ (z ◦ y −1 )

=

x ◦ (y ◦ z −1 )−1

=

x ∗ (y ∗ z).

Hence (X, ∗, 0) is a BO-algebra.

859



4.4 Let (X, ∗, 0) be a group with x ∗ x = 0 for any x ∈ X. Then (X, ∗, 0) is a BO-algebra with 0 ∗ x = x for any x ∈ X. P r o o f. For any x ∈ X, x ∗ x = 0, x ∗ 0 = x = 0 ∗ x, since (X, ∗, 0) is a group. For any x, y, z ∈ X, (x ∗ y) ∗ (0 ∗ z) = =

(x ∗ y) ∗ z x ∗ (y ∗ z).

Hence (X, ∗, 0) is a BO-algebra with 0 ∗ x = x for any x ∈ X.

4.5

Every BO-algebra is 0-commutative.

P r o o f. Let (X, ∗, 0) be a BO-algebra. Given x, y ∈ X, we have 3 cases as follow: (i) 0 ∗ x = x and 0 ∗ y = y, (ii) 0 ∗ x = x and 0 ∗ y = y, (iii) 0 ∗ x = x and 0 ∗ y = y. Case (i): If we let a := x ∗ y, b := y ∗ x, then a = b. In fact, if we assume a = b, then y = y ∗ 0 = y ∗ (x ∗ x) = (y ∗ x) ∗ (0 ∗ x) = b ∗ (0 ∗ x) = b ∗ x, i.e., y = b ∗ x. This means that b ∗ a = b ∗ (x ∗ y) = (b ∗ x) ∗ (0 ∗ y) = y ∗ (0 ∗ y) = y ∗ y = 0. By applying Corollary 2.4, we obtain a = b, a contradiction. Hence x ∗ (0 ∗ y) = x ∗ y = a = b = y ∗ x = y ∗ (0 ∗ x). Case (ii): If z := 0 ∗ y, then z = y. By applying Corollary 2.4 and Theorem 2.3(vi) we obtain that z ∈ / {0, x}. Similarly, by applying Corollary 2.4 and Theorem 2.3(v) we have y ∗ x ∈ / {0, x}. Consider a case: y ∗ x = y. Since y ∗ 0 = y, by applying Theorem 2.3(v), we obtain x = 0. Hence x ∗ (0 ∗ y) = 0 ∗ (0 ∗ y) = y = y ∗ 0 = y ∗ (0 ∗ 0) = y ∗ (0 ∗ x). We claim that x ∗ z = 0. In fact, if x ∗ z = 0, then x = z = 0 ∗ y. Since 0 ∗ x = x, by Theorem 2.3(vii), x = y and hence y = 0 ∗ y = 0 ∗ x = x, a contradiction. Similarly, we obtain x ∗ z = x. Consider a case: x ∗ z = y. Since z = 0 ∗ y, we have 0 ∗ z = 0 ∗ (0 ∗ y) = y. By Theorem 2.3(vi), we obtain x = 0, which proves x ∗ (0 ∗ y) = y ∗ (0 ∗ x). We have observed several cases. It remains only to consider the case: x ∗ z ∈ / {0, x, y}, y∗x∈ / {0, x, y}. In this case, we have 4 subcases: (1) x ∗ z = z, y ∗ x = z; (2) x ∗ z = z, y ∗ x = b; (3) x ∗ z = a, y ∗ x = z; (4) x ∗ z = a, y ∗ x = b where a, b ∈ / {0, x, y, z}. 860


ON BO-ALGEBRAS

If we consider (1) x∗z = z, y∗x = z, then x∗(0∗y) = x∗z = z = y∗x = y∗(0∗x). The subcase (2) cannot happen. In fact, y = y ∗ 0 = y ∗ (x ∗ x) = (y ∗ x) ∗ (0 ∗ x) = b ∗ x. It follows that b ∗ z = b ∗ (x ∗ z) = (b ∗ x) ∗ (0 ∗ z) = y ∗ (0 ∗ z) = y ∗ y = 0, i.e., b ∗ z = 0. By applying Corollary 2.4, we obtain b = z ∈ / {0, x, y, z}, a contradiction. Similarly, the subcase (3) cannot happen, and we omit the proof. Consider the subcase (4). We claim that a = b. In fact, a∗b

=

(x ∗ z) ∗ (y ∗ x)

=

((x ∗ z) ∗ y) ∗ (0 ∗ x)

=

((x ∗ z) ∗ (0 ∗ z)) ∗ (0 ∗ x)

= =

(x ∗ (z ∗ z)) ∗ (0 ∗ x) (x ∗ 0) ∗ x

=

x ∗ x = 0.

By applying Corollary 2.4, we obtain a = b. Hence x ∗ (0 ∗ y) = x ∗ z = a = b = y ∗ x = y ∗ (0 ∗ x). Case (iii): Let z := 0 ∗ x, t := 0 ∗ y. Then z = x, t = y. We can see that 0-commutativity holds for x∗t = 0. In fact, if x∗t = 0, then by Corollary 2.4 we have x = t = 0∗y and hence 0 ∗ x = 0 ∗ (0 ∗ y) = y. Thus x ∗ (0 ∗ y) = x ∗ x = 0 = y ∗ y = y ∗ (0 ∗ x). By applying Corollary 2.4 and Theorem 2.3(vi), x ∗ t = x and x ∗ t = y cannot happen. Similarly, we can say that 0-commutativity holds for x ∗ t = 0, and y ∗ z = y and y ∗ z = x cannot happen. Hence, it remains to consider the following case: x ∗ t ∈ / {0, x, y}, y ∗ z ∈ / {0, x, y}. If we consider a case: x ∗ t = a, y ∗ z = b where a, b ∈ / {0, x, y, z, t}, then we obtain a = b. In fact, a∗b

= =

(x ∗ t) ∗ (y ∗ z) ((x ∗ t) ∗ y) ∗ (0 ∗ z)

=

((x ∗ t) ∗ y) ∗ (0 ∗ (0 ∗ x))

=

((x ∗ t) ∗ y) ∗ x

= ((x ∗ (0 ∗ y)) ∗ (0 ∗ t)) ∗ x = [x ∗ ((0 ∗ y) ∗ t)] ∗ x =

(x ∗ 0) ∗ x

=

x ∗ x = 0.

By applying Corollary 2.4, we obtain a = b. Hence x ∗ (0 ∗ y) = x ∗ t = a = b = y ∗ z = y ∗ (0 ∗ x), i.e., 0-commutativity holds. It makes us to classify the following subcases: 861



(1) x ∗ t = z, y ∗ z = t; (2) x ∗ t = z, y ∗ z = a; (3) x ∗ t = t, y ∗ z = z; (4) x ∗ t = t, y ∗ z = a; (5) x ∗ t = a, y ∗ z = z; (6) x ∗ t = a, y ∗ z = t, where a ∈ / {0, x, y, z, t}. If we consider (1) x ∗ t = z, y ∗ z = t, then t = x ∗ (x ∗ t) = x ∗ z. Since t = y ∗ z, by Theorem 2.3(v), we obtain x = y, i.e., 0-commutativity holds. Similarly the subcase (3) holds. Consider the subcase (2) x ∗ t = z, y ∗ z = a. Since y = y ∗ 0 = y ∗ (z ∗ z) = (y ∗ z) ∗ (0 ∗ z) = a ∗ x, we have a∗z

= =

a ∗ (x ∗ t) (a ∗ x) ∗ (0 ∗ t)

=

y ∗ y = 0.

By Corollary 2.4 we have a = z, a contradiction. Thus this subcase cannot happen. Similarly, the subcases (4), (5), (6) cannot happen. This proves that 0-commutativity holds in BO-algebra.

4.6 If (X, ∗, 0) is a BO-algebra, then (X, ∗, 0) is a BF -algebra. P r o o f. For x, y ∈ X, 0 ∗ (x ∗ y) = (0 ∗ x) ∗ (0 ∗ y)

[Theorem 2.3(ii)]

= y ∗ (0 ∗ (0 ∗ x))

[Theorem 4.5]

= y ∗ x,

[Theorem 2.3(i)]

proving the corollary.

4.7 If (X, ∗, 0) is a BO-algebra, then (X, ∗, 0) is a B-algebra. P r o o f. Since (X, ∗, 0) is a BO-algebra, it is enough to show that (x ∗ y) ∗ z = x ∗ [z ∗ (0 ∗ y)] for any x, y, z ∈ X. x ∗ [z ∗ (0 ∗ y)] = x ∗ [y ∗ (0 ∗ z)]

[Theorem 4.5]

= (x ∗ y) ∗ [0 ∗ (0 ∗ z)]

[(BO)]

= (x ∗ y) ∗ z.

[Theorem 2.3(i)]

Thus (X, ∗, 0) is a B-algebra.

From Theorem 4.1, Theorem 4.5 and Corollary 4.7, we have the following corollary. 862


ON BO-ALGEBRAS

4.8 (X, ∗, 0) is a BO-algebra if and only if (X, ∗, 0) is a 0-commutative B-algebra. C. B. Kim and H. S. Kim introduced the notion of BM -algebras. A BM -algebra ([3]) is a non-empty set X with a constant 0 and a binary operation ∗ satisfying the following axioms: (B2) x ∗ 0 = x, (BM) (z ∗ x) ∗ (z ∗ y) = y ∗ x for any x, y, z ∈ X. They proved that an algebra (X, ∗, 0) is a 0-commutative B-algebra if and only if (X, ∗, 0) is a BM -algebra. From the above fact and Corollary 2.16, we have the following corollary.

4.9

(X, ∗, 0) is a BO-algebra if and only if (X, ∗, 0) is a BM -al-

gebra.

If we consider Theorem 3.1, any BO-algebra (X, ∗, 0) with 0 ∗ x = x for any x ∈ X becomes an abelian group (X, ∗, 0). In this case, the condition 0 ∗ x = x is necessary, but the condition is superfluous in the discussion of logically equivalences as follows:

4.10

The following are logically equivalent.

(1) (X, ∗, 0) is a BO-algebra, (2) (X, ∗, 0) is a 0-commutative B-algebra, (3) (X, ∗, 0) is a BM -algebra, (4) (X, ◦, 0) is an abelian group, (5) (X, ∗, 0) is a p-semisimple BCI-algebra. P r o o f. (1) ⇐⇒ (2). By Corollary 4.8. (4) =⇒ (1). By Theorem 4.1. (1) =⇒ (4). Let (X, ∗, 0) be a BO-algebra. If we define x ◦ y := x ∗ (0 ∗ y) for all x, y ∈ X and x−1 := 0 ∗ x for all x ∈ X, then (X, ◦, 0) is an abelian group. In fact, it is easy to show that x ◦ 0 = 0 ◦ x = x and x ◦ x−1 = x−1 ◦ x = 0 for any x ∈ X. Given y, z ∈ X, we have 0 ∗ (y ∗ z −1 )

=

0 ∗ (y ∗ (0 ∗ z))

=

(0 ∗ y) ∗ (0 ∗ (0 ∗ z))

=

(0 ∗ y) ∗ z

=

y −1 ∗ z.

It follows that 863



(x ◦ y) ◦ z

= (x ∗ (0 ∗ y)) ∗ (0 ∗ z) = (x ∗ y −1 ) ∗ (0 ∗ z) = x ∗ (y −1 ∗ z) = x ∗ (0 ∗ (y ∗ z −1 )) = x ∗ (0 ∗ (y ∗ (0 ∗ z))) = x ∗ (0 ∗ (y ◦ z)) = x ◦ (y ◦ z),

proving that (X, ◦, 0) is a group. By Theorem 4.5, we obtain x ◦ y = x ∗ (0 ∗ y) = y ∗ (0 ∗ x) = y ◦ x. (4) ⇐⇒ (5). See [5]. REFERENCES JUN, Y. B.—ROH, E. W.—KIM, H. S.: On BH-algebra, Sci. Math. Jpn. 1 (1998), 347–354. KIM, C. B.—KIM, H. S.: On BG-algebras, Demonstratio Math. 41 (2008), 259–262. KIM, C. B.—KIM, H. S.: On BM-algebras, Sci. Math. Jpn. 63 (2006), 421–427. KIM, H. S.—PARK, H. G.: On 0-commutative B-algebras, Sci. Math. Jpn. Online, e-2005 (2005), 31–36. [5] LEI, T.—XI, C. C.: The p-semisimple BCI-algebras, J. Shaanxi Teachers Univ. 2 (1984), 25–29. [6] NEGGERS, J.—KIM, H. S.: On B-algebras, Mat. Vesnik 54 (2002), 21–29. [7] WALENDZIAK, A.: On BF -algebras, Math. Slovaca 57 (2007), 119–128. [1] [2] [3] [4]

Received 7. 1. 2010 Accepted 23. 2. 2010

* Department of Mathematics Kookmin University Seoul, 136-702 KOREA E-mail : [email protected] ** Corresponding author: Department of Mathematics Research Institute for Natural Sciences Hanyang University Seoul, 133-791 KOREA E-mail : [email protected] Tel.: +82 2 2290 0897 Fax: +82 2 2281 0019

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