Commutative and Bounded BE-algebras
We introduce the notions of the commutative and bounded BE-Algebras. We give some related properties of them.
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  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.  introduced a new notion called BH-algebra which is another generalization of BCH/BCI/BCK-algebras. Walendziak obtained some equivalent axioms for B-algebras . C. B. Kim and H. S. Kim  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 , 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 , 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.
Definition 1 (see ). 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 ). 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 ). Let be a set with the following table: Then is a self-distributive BE-algebra.
Proposition 5 (see ). Let be a self-distributive BE-algebra. If , then, for all , , and in , the following inequalities hold:(i),(ii).
Definition 6 (see ). 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 ). Any dual BCK-algebra is a BE-algebra.
Definition 8 (see ). Let be a BE-algebra or dual BCK-algebra. is said to be commutative if the following identity holds: for all , .
Theorem 9 (see ). If is a commutative BE-algebra, then is a dual BCK-algebra.
Corollary 10 (). 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 .
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).
Corollary 19. If is a self-distributive and commutative BE-algebra, then .
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
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 , 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 .