The notion of a BE-semigroup is introduced, and related properties are investigated. The concept of left (resp., right) deductive systems of a BE-semigroup is also introduced.

1. Introduction

Hu and Li, Iséki and Tanaka, respectively, introduced two classes of abstract algebras: BCK-algebras and BCI-algebras [1–3]. It is known that the class of BCK-algebras is a proper subclass of the class of BCI-algebras. In [1, 4] Hu and Li introduced a wide class of abstract algebras: BCH-algebras. They have shown that the class of BCI-algebras is a proper subclass of the class of BCH-algebras. We refer to [5] for general information on BCK-algebras. Neggers and Kim [6] introduced the notion of a d-algebra which is a generalization of BCK-algebras, and also they introduced the notion of a B-algebra [7, 8], that is, (I) x*x = 0, (II) x*0 = x, (III) (x*y)*z = x*(z*(0*y)), for any x, y, z ∈ X, which is equivalent to the idea of groups. Moreover, Jun et al. [9] introduced a new notion, called an BH- algebra, which is another generalization of BCH/BCI/BCK-algebras, that is, (I), (II), and (IV) x*y = 0 and y*x = 0 imply that x = y for any x, y ∈ X. Walendziak obtained other equivalent set of axioms for a B-algebra [10]. Kim et al. [11] introduced the notion of a (pre-) Coxeter algebra and showed that a Coxeter algebra is equivalent to an abelian group all of whose elements have order 2, that is, a Boolean group. C. B. Kim and H. S. Kim [12] introduced the notion of a BM-algebra which is a specialization of B-algebras. They proved that the class of BM-algebras is a proper subclass of B-algebras and also showed that a BM-algebra is equivalent to a 0-commutative B-algebra. In [13], H. S. Kim and Y. H. Kim introduced the notion of a BE-algebra as a generalization of a BCK-algebra. Using the notion of upper sets, they gave an equivalent condition of the filter in BE-algebras. In [14, 15], Ahn and So introduced the notion of ideals in BE-algebras and proved several characterizations of such ideals.

In this paper, by combining BE-algebras and semigroups, we introduce the notion of BE-semigroups. We define left (resp., right) deductive systems (LDS (resp., RDS) for short) of a BE-semigroup, and then we describe LDS generated by a nonempty subset in a BE-semigroup as a simple form.

2. Preliminaries

We recall some definitions and results discussed in [13].

Definition 2.1 (see [13].)An algebra (X; *, 1) of type (2, 0) is called a BE-algebra if

(BE1)
x*x = 1 for all x ∈ X,
(BE2)
x*1 = 1 for all x ∈ X,
(BE3)
1*x = x for all x ∈ X,
(BE4)
x*(y*z) = y*(x*z) for all x, y, z ∈ X (exchange).

We introduce a relation “≤" on X by x ≤ y if and only if x*y = 1.

Proposition 2.2 (see [13].)If (X; *, 1) is a BE-algebra, then x*(y*x) = 1 for any x, y ∈ X.

Example 2.3 (see [13].)Let X : = {1, a, b, c, d, 0} be a set with the following table:

(2.1)

Then (X; *, 1) is a BE-algebra.

Definition 2.4 (see [13].)A BE-algebra (X; *, 1) is said to be self-distributive if x*(y*z) = (x*y)*(x*z) for all x, y, z ∈ X.

Example 2.5 (see [13].)Let X : = {1, a, b, c, d} be a set with the following table:

(2.2)

Then it is easy to see that X is a self-distributive BE-algebra.

Note that the BE-algebra in Example 2.3 is not self-distributive, since d*(a*0) = d*d = 1, while (d*a)*(d*0) = 1*a = a.

Proposition 2.6. Let X be a self-distributive BE-algebra. If x ≤ y, then z*x ≤ z*y and y*z ≤ x*z for any x, y, z ∈ X.

Proof. The proof is straightforward.

3. BE-Semigroups

Definition 3.1. An algebraic system (X; ⊙, *, 1) is called a BE-semigroup if it satisfies the following:

(i)
(X; ⊙) is a semigroup,
(ii)
(X; *, 1) is a BE-algebra,
(iii)
the operation “⊙” is distributive (on both sides) over the operation “*”.

Example 3.2. (1) Define two operations “⊙” and “*” on a set X : = {1, a, b, c} as follows:

(3.1)

It is easy to see that (X; ⊙, *, 1) is a BE-semigroup.

(2) Define two binary operations “⊙” and “*” on a set A : = {1, a, b, c} as follows:

(3.2)

It is easy to show that (A; ⊙, *, 1) is a BE-semigroup.

Proposition 3.3. Let (X; ⊙, *, 1) be a BE-semigroup. Then

(i)
(∀x ∈ X) (1⊙x = x⊙1 = 1),
(ii)
(∀x, y, z ∈ X) (x ≤ y⇒x⊙z ≤ y⊙z, z⊙x ≤ z⊙y).

Proof. (i) For all x ∈ X, we have that 1⊙x = (1*1)⊙x = (1⊙x)*(1⊙x) = 1 and x⊙1 = x⊙(1*1) = (x⊙1)*(x⊙1) = 1.

(ii) Let x, y, z ∈ X be such that x ≤ y. Then

(3.3)

Hence x⊙z ≤ y⊙z and z⊙x ≤ z⊙y.

Definition 3.4. An element a(≠1) in a BE-semigroup (X; ⊙, *, 1) is said to be a left (resp., right) unit divisor if

(3.4)

A unit divisor is an element of X which is both a left and a right unit divisors.

Theorem 3.5. Let (X; ⊙, *, 1) be a BE-semigroup. If it satisfies the left (resp., right ) cancellation law for the operation ⊙, that is,

(3.5)

then X contains no left (resp., right) unit divisors.

Proof. Let (X; ⊙, *, 1) satisfy the left cancellation law for the operation ⊙ and assume that x⊙y = 1 where x ≠ 1. Then x⊙y = 1 = x⊙1 by Proposition 3.3(i), which implies y = 1. Similarly it holds for the right case. Hence there is no left (resp., right) unit divisors in X.

Now we consider the converse of Theorem 3.5.

Theorem 3.6. Let (X; ⊙, *, 1) be a BE-semigroup in which there are no left (resp., right ) unit divisors. Then it satisfies the left (resp., right) cancellation law for the operation ⊙.

Proof. Let x, y, z ∈ X be such that x⊙y = x⊙z and x ≠ 1. Then

(3.6)

Since X has no left unit divisor, it follows that y*z = 1 = z*y so that y = z. The argument is the same for the right case.

Definition 3.7. Let (X; ⊙, *, 1) be a BE-semigroup. A nonempty subset D of X is called a left (resp., right) deductive system (LDS (resp., RDS), for short) if it satisfies

(ds1)
X⊙D⊆D (resp., (D⊙X⊆D)),
(ds2)
(∀a ∈ D) ((∀x ∈ X) (a*x ∈ D⇒x ∈ D).

Example 3.8. Let X : = {x, y, z, 1} be a set with the following Cayley tables:

(3.7)

It is easy to show that (X; ⊙, *, 1) is a BE-semigroup. We know that D : = {1, x} is an LDS of X, but E : = {1, y} is not an LDS of X, since z⊙y = z ∉ E and/or y*x = 1 ∈ E, y ∈ E but x ∉ E.

Let (X; *, 1) be a BE-algebra, and let a, b ∈ X. Then the set

(3.8)

is nonempty, since 1, a, b ∈ A(a, b).

Proposition 3.9. If D is an LDS of a BE-semigroup (X; ⊙, *, 1), then

(3.9)

Proof. Let x ∈ A(a, b) where a, b ∈ D. Then a*(b*x) = 1 ∈ D and so x ∈ D by (ds2). Therefore A(a, b)⊆D.

Theorem 3.10. Let {D_i} be an arbitrary collection of LDSs of a BE-semigroup (X; ⊙, *, 1), where i ranges over some index set I. Then ∩_i∈ID_i is also an LDS of A.

Proof. The proof is straightforward.

Let (X; ⊙, *, 1) be a BE-semigroup. For any subset D of X, the intersection of all LDSs (resp., RDSs) of X containing D is called the LDSs (resp., RDSs) generated by D, and is denoted by 〈D〉 _l (resp., 〈D〉 _r). It is clear that if D and E are subsets of a BE-semigroup (X; ⊙, *, 1) satisfying D⊆E, then 〈D〉 _l⊆〈E〉 _l (resp., 〈D〉 _r⊆〈E〉 _r), and if D is an LDS (resp., RDS) of X, then 〈D〉 _l = D (resp., 〈D〉 _r = D).

A BE-semigroup (X; ⊙, *, 1) is said to be self-distributive if (X; *, 1) is a self-distributive BE-algebra.

Theorem 3.11. Let (X; ⊙, *, 1) be a self-distributive BE-semigroup and let D be a nonempty subset of X such that A⊙D⊆D. Then 〈D〉 _l : = {a ∈ X∣y_n*(⋯*(y₁*a)⋯) = 1 for some y₁, …, y_n ∈ D}.

Proof. Denote

(3.10)

Let a ∈ X and b ∈ B. Then there exist y₁, …, y_n ∈ D such that y_n*(⋯*(y₁*b)⋯) = 1. It follows that

(3.11)

Since x⊙y_i ∈ D for i = 1, …, n, we have that x⊙b ∈ B. Let x, a ∈ X be such that a*x ∈ B and a ∈ B. Then there exist y₁, …, y_n, z₁, …, z_m ∈ D such that

(3.12)

(3.13)

Using (BE4), it follows from (3.12) that a*(y_n*(⋯*(y₁*x)⋯)) = 1, that is, a ≤ y_n*(⋯*(y₁*x)⋯), and so from (3.13) and Proposition 2.6 it follows that

(3.14)

Thus z_m*(⋯*(z₁*(y_n*(⋯*(y₁*x)⋯)))⋯) = 1, which implies x ∈ B. Therefore B is an LDS of X. Obviously D⊆B. Let G be an LDS containing D. To show B⊆G, let a be any element of B. Then there exist y₁, …, y_n ∈ D such that y_n*(⋯*(y₁*a)⋯) = 1. It follows from (ds2) that a ∈ G so that B⊆G. Consequently, we have that 〈D〉 _l = B.

In the following example, we know that the union of any LDSs (resp., RDSs) D and E may not be an LDS (resp., RDS) of a self-distributive BE-semigroup (X; ·, *, 1).

Example 3.12. Let X : = {1, a, b, c, d} be a set with the following Cayley tables:

(3.15)

It is easy to check that (X; ⊙, *, 1) is a self-distributive BE-semigroup. We know that D : = {1, a} and E : = {1, b} are LDSs of X, but D ∪ E = {1, a, b} is not an LDS of X, since b*c = a ∈ D ∪ E, c ∉ D ∪ E.

Theorem 3.13. Let D and E be LDSs of a self-distributive BE-semigroup (X; ·, *, 1). Then

(3.16)

Proof. Denote

(3.17)

Obviously, K⊆〈D ∪ E〉 _l. Let b ∈ 〈D ∪ E〉 _l. Then there exist y₁, …, y_n ∈ D ∪ E such that y_n*(⋯*(y₁*b)⋯) = 1 by Theorem 3.11. If y_i ∈ D (resp., E) for all i = 1, …, n, then b ∈ D (resp., E). Hence b ∈ K since b*(1*b) = 1 (resp., 1*(b*b) = 1). If some of y₁, …, y_n belong to D and others belong to E, then we may assume that y₁, …, y_k ∈ D and y_k+1, …, y_n ∈ E for 1 ≤ k < n, without loss of generality. Let p = y_k*(⋯*(y₁*b)⋯). Then

(3.18)

and so p ∈ E. Now let q = p*b = (y_k*(⋯*(y₁*b)⋯))*b. Then

(3.19)

which implies that q ∈ D. Since p*(q*b) = q*(p*b) = q*q = 1, it follows that b ∈ K so that 〈D ∪ E〉 _l⊆K. This completes the proof.

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