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Volume 2013 (2013), Article ID 473714, 5 pages
Commutative and Bounded BE-algebras
Department of Mathematics, Faculty of Arts and Sciences, Süleyman Demirel University, 32260 Isparta, Turkey
Received 2 August 2013; Accepted 28 October 2013
Academic Editor: Yao-Zhong Zhang
Copyright © 2013 Zekiye Çiloğlu and Yılmaz Çeven. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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 .
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