#### Abstract

We introduce the notions of the commutative and bounded BE-Algebras. We give some related properties of them.

#### 1. Introduction

Imai and Iséki introduced two classes of abstract algebras called BCK-algebras and BCI-algebras [1, 2]. It is known that the class of BCK-algebras is a proper subclass of BCI-algebras. In [3, 4], Hu and Li introduced a wide class of abstract algebras called BCH-algebras. They have shown that the class of BCI-algebras is a proper subclass of BCH-algebras. Neggers and Kim [5] introduced the notion of d-algebras which is another generalization of BCK-algebras, and also they introduced the notion of B-algebras [6, 7]. Jun et al. [8] introduced a new notion called BH-algebra which is another generalization of BCH/BCI/BCK-algebras. Walendziak obtained some equivalent axioms for B-algebras [9]. C. B. Kim and H. S. Kim [10] introduced the notion of 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 [11], H. S. Kim and Y. H. Kim introduced the notion of 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 [12, 13], Ahn and So introduced the notion of ideals in BE-algebras and proved several characterizations of such ideals. Also they generalized the notion of upper sets in BE-algebras and discussed some properties of the characterizations of generalized upper sets related to the structure of ideals in transitive and self-distributive BE-algebras. In [14], Ahn et al. introduced the notion of terminal section of BE-algebras and provided the characterization of the commutative BE-algebras.

In this paper we introduce the notion of bounded BE-algebras and investigate some properties of them.

#### 2. Preliminaries

*Definition 1 (see [11]). *An algebra of type (2, 0) is called a BE-algebra if, for all , , and in ,(BE1),(BE2),(BE3),(BE4).

In , a binary relation “” is defined by if and only if .

*Example 2 (see [11]). *Let be a set with the following table:
Then is a BE-algebra.

*Definition 3. *A BE-algebra is said to be self-distributive if for all , , and .

*Example 4 (see [11]). *Let be a set with the following table:
Then is a self-distributive BE-algebra.

Proposition 5 (see [14]). *Let be a self-distributive BE-algebra. If , then, for all , , and in , the following inequalities hold:*(i),(ii).

*Definition 6 (see [15]). *A dual BCK-algebra is an algebra of type (2,0) satisfying (BE1) and (BE2) and the following axioms: (dBCK1) implies , (dBCK2) , (dBCK3) .

Proposition 7 (see [16]). *Any dual BCK-algebra is a BE-algebra.*

The converse of Proposition 7 does not hold in general [16].

*Definition 8 (see [16]). *Let be a BE-algebra or dual BCK-algebra. is said to be commutative if the following identity holds:
for all , .

Theorem 9 (see [16]). *If is a commutative BE-algebra, then is a dual BCK-algebra.*

Corollary 10 ([16]). * is a commutative BE-algebra if and only if it is a commutative dual BCK-algebra.*

If is a commutative BE-algebra, then the relation “” is a partial order on [16].

In the following, we abbreviate as .

#### 3. Bounded BE-Algebras

The following definition introduces the notion of boundedness for BE-algebras.

*Definition 11. *Let be a BE-algebra. If there exists an element satisfying for all , then the element “” is called unit of . A BE-algebra with unit is called a bounded BE-algebra.

In a bounded BE-algebra, denoted by .

*Example 12. *The BE-algebra in Example 2 is a bounded BE-algebra and its unit element is .

*Example 13. *The BE-algebra in Example 4 is a bounded BE-algebra and its unit element is .

*Example 14. *Let be a set with the following table:
It is clear that is a BE-algebra, but it is not a bounded BE-algebra.

Theorem 15. *In a bounded BE-algebra with unit , the following properties hold for all , :*(i)*, ,*(ii)*,*(iii)*,*(iv)*, .*

*Proof. *(i) Using (BE3) and (BE1), we have and .

(ii) Since
by (BE4) and (BE1), we get .

(iii) Using (BE4), we have

(iv) By routine operations, we have and .

Theorem 16. *In a bounded and self-distributive BE-algebra with unit , the following properties hold for all , :*(i)*,*(ii)* implies .*

*Proof. *(i) Since
we have .

(ii) It is trivial by Proposition 5.

Proposition 17. *Let be a BE-algebra. Then .*

*Proof. *Since
we have .

Proposition 18. *Let be a self-distributive BE-algebra. Then, the next properties are valid for all , :*(i),(ii).

*Proof. *(i) Since
we have . By Proposition 5(i), we have (.

(ii) Using Definition 3, (BE1) and (BE3), we have

Corollary 19. *If is a self-distributive and commutative BE-algebra, then .*

*Proof. *It is clear by Propositions 17 and 18 and the property (dBCK1).

Corollary 20. *If is a self-distributive, commutative, and bounded BE-algebra with unit , then .*

*Proof. *In Corollary 19, taking , we have . Then we get ; that is, .

*Definition 21. *In a bounded BE-algebra, the element such that is called an involution.

Let where is a bounded BE-algebra. is the set of all involutions in . Moreover, since and , we have and so .

*Example 22. *For the BE-algebra in Example 2, it is clear that .

*Example 23. *For the BE-algebra in Example 4, it is clear that .

Proposition 24. *Let be a bounded BE-algebra with unit and let be the set of all involutions in . Then, for all , , the following conditions hold:*(i),(ii).

*Proof. *(i) The proof is obvious by the definition of .

(ii) By Theorem 15(iii), we have

In a bounded BE-algebra , we denote where for all , .

Theorem 25. *In a bounded and commutative BE-algebra the following identities hold:*(i)*,*(ii)*,*(iii)*,*(iv)*.*

*Proof. *(i) Using (BE3), it is obtained that

(ii) By the definition of “” and (i), we have that .

(iii) By the definition of “” and (i), we have that .

(iv) We have

Corollary 26. *If is a bounded commutative BE-algebra, then .*

*Definition 27. *Each of the elements and in a bounded BE-algebra is called the complement of the other if and .

Theorem 28. *Let be a bounded and commutative BE-algebra. If there exists a complement of any element of , then it is unique.*

*Proof. *Let and , be two complements of . Then we know that and . Also since and , we have and . So we get . Similarly . Hence
With similar operations, we have . Hence we obtain which gives that the complement of is unique.

Note that for a bounded and commutative BE-algebra, an element does not need to have any complement.

*Example 29. *In Example 2, the complement of is , but has no complement in .

Theorem 30. *Let be a commutative and bounded BE-algebra. Then the following conditions are equivalent, for all , :*(i)*,*(ii)*,*(iii)*,*(iv)*,*(v)*.*

*Proof. *(i)*⇒*(ii) Let . Then it follows that

(ii)*⇒*(iii) Let . Then, since and , we get by (dBCK1).

(iii)*⇒*(iv) Let . Substituting for and using Theorem 25 (i), we obtain the result.

(iv)*⇒*(v) Let . Then we get . Hence we have . Using Theorem 25(iv), we obtain .

(v)*⇒*(ii) Let . Then we have

(ii)*⇒*(i) Let . Then we obtain

Note that, if is a self-distributive BE-algebra, then by Proposition 18. In this case, Theorem 30(v) is true. If is also commutative and bounded, then is the complement of by Theorem 30(i) and (ii).

Now we obtain a bounded BE-algebra from a nonbounded BE-algebra as the following theorem.

Theorem 31. *Let be a BE-algebra and . Define the operation on as follows:
**
Then ( is a bounded BE-algebra with unit .*

*Proof. *It is clear that BE1, BE2, and BE3 are satisfied. It suffices to verify BE4. Note that if and , , then and ; if and , , then and ; if and , , then and .

For the remain situations, it is clearly seen that BE4 is satisfied with similar arguments.

We call the BE-algebra ( in the previous theorem the extension of . Note that for the BE-algebra (, if for , we have since . Hence .

Theorem 32. *The extension of a self-distributive BE-algebra is also self-distributive.*

*Proof. *Let and denote =. Let be defined as in Theorem 31. We want to see that for all , , and . For all , , we know , , , and . So we obtain the following, for all , , and :
The above results show that ( is self-distributive.

Theorem 33. *If is a commutative, self-distributive, and bounded BE-algebra, then it is a lattice with respect to and where and for any .*

*Proof. *From Theorem 3.6 in [14], we know that is a semilattice with respect to . Then we need to show that the set for all , has a greatest lower bound. We know that and . By Proposition 5, we have and . Since and by Theorem 25(i), we have and . So is a lower bound of and . Next we must show that if and then . Let and . We have and by Proposition 5. Since is an upper semilattice with respect to , then we obtain and hence . So we have . Then is the greatest lower bound of and . So, we can say that is a lower semilattice. Consequently () is a lattice with respect to and .