Strong GE-Filters and GE-Ideals of Bordered GE-Algebras

Abstract

The notion of strong GE-filters and GE-ideals (generated) is introduced, and the related properties are investigated. The intersection of strong GE-filters (resp., GE-ideals) is proved to be a strong GE-filter (resp., GE-ideal), and the union of strong GE-filters (resp., GE-ideals) is generally not a strong GE-filters (resp., GE-ideal) by example. Conditions for a subset of a bordered GE-algebra to be a strong GE-filter are provided, and a characterization of a strong GE-filter is considered. In order to do so, irreducible GE-filter is defined first and its properties are examined. Conditions for a GE-filter to be irreducible are discussed. Given a GE-filter, and a subset in a bordered GE-algebra, the existence of an irreducible GE-filter, which contains the given GE-filter and is disjoint to the given subset, is considered. Conditions under which any subset of a bordered GE-algebra can be a GE-ideal are provided, and GE-ideal that is generated from a subset in a bordered GE-algebra is discussed. Also, what element it is formed into is stated. Finally, the smallest GE-ideal which contains a given GE-ideal and an element in a bordered GE-algebra is established.

1. Introduction

In 1966, Imai and Iséki introduced BCK-algebras (see [1]) as the algebraic semantics for a non classical logic possessing only implication. Since then, the generalized concepts of BCK-algebras have been studied by various scholars. Kim and Kim introduced the notion of a BE-algebra as a generalization of a dual BCK-algebra (see [2]). Hilbert algebras were introduced by Henkin and Skolem in the fifties for investigations in intuitionistic and other non classical logics. Diego proved that Hilbert algebras form a variety which is locally finite (see [3]). Rezaei et al. discussed relations between Hilbert algebras and BE-algebras (see [4]). The generalization process in the study of algebraic structures is also an important area of study. As a generalization of Hilbert algebras, Bandaru et al. introduced the notion of GE-algebras and investigated several properties (see [5, 6]).

In this article, we introduce the notions of strong GE-filters and GE-ideals (generated) and investigate the related properties in bordered GE-algebras. In Section 3, we introduce the concept of strong GE-filters and investigate its properties. We show that the intersection of strong GE-filters is a strong GE-filters, but the union of strong GE-filters is not a strong GE-filter by example. We provide conditions for a subset of a bordered GE-algebra to be a strong GE-filter. We consider a characterization of a strong GE-filter. In order to do so, we define irreducible GE-filter first and examine its properties. We provide conditions for a GE-filter to be irreducible. Given a GE-filter and a subset in a bordered GE-algebra, we consider the existence of an irreducible GE-filter which contains the given GE-filter and is disjoint to the given subset. In Section 4, we introduce the notion of GE-ideals in bordered GE-algebras and study its properties. We show that the intersection of GE-ideals is a GE-ideal, but the union of GE-ideals is not a GE-ideal by example. We suggest conditions under which any subset of a bordered GE-algebra can be a GE-ideal. We discuss GE-ideal that is generated from a subset in a bordered GE-algebra and investigate what element it is formed into. We construct the smallest GE-ideal which contains a given GE-ideal and an element in a bordered GE-algebra.

2. Preliminaries

Definition 1. (see [5]). A GE-algebra is a nonempty set with a constant 1 and a binary operation , satisfying the following axioms:(GE1)(GE2)(GE3) for all

In a GE-algebra , a binary relation “” is defined by

Definition 2. (see [5]). A GE-algebra is said to be(1)Transitive if it satisfies(2)Antisymmetric if the binary relation “” is antisymmetric.

Definition 3. (see [6]). If a GE-algebra has a special element, say , which satisfies for all , we call the bordered GE-algebra.

Definition 4. (see [6]). If a bordered GE-algebra satisfies condition (2), we say that is a transitive bordered GE-algebra.

Definition 5. (see [6]). A bordered GE-algebra is said to be antisymmetric if the binary operation “” is antisymmetric.

Proposition 1 (see [5]). Every GE-algebrasatisfies the following items:If is transitive, then

Lemma 1 (see [5]). In a GE-algebra, the following facts are equivalent to each other:

Definition 6. (see [5]). A subset of a GE-algebra is called a GE-filter of if it satisfies

Lemma 2 (see [5]). In a GE-algebra, every GE-filterofsatisfies

Proposition 2 (see [6]). In a bordered GE-algebra, the following assertions are valid:If is a transitive bordered GE-algebra, thenIf is an antisymmetric bordered GE-algebra, thenIf is a transitive and antisymmetric bordered GE-algebra, then

Definition 7. (see [6]). By a duplex bordered element in a bordered GE-algebra , we mean an element of which satisfies .

The set of all duplex bordered elements of a bordered GE-algebra is denoted by and is called the duplex bordered set of . It is clear that .

Definition 8. (see [6]). A bordered GE-algebra is said to be duplex if every element of is a duplex bordered element, that is, .

3. Strong GE-Filters

Given a subset of a bordered GE-algebra , we define the set

Definition 9. A GE-filter of a bordered GE-algebra is said to be strong if it satisfiesi.e., whenever .

It is clear that every GE-filter is strong in a duplex bordered GE-algebra.

Example 1. Let be a set with a binary operation given in the following table:Then, is a bordered GE-algebra, and it is easy to check that is a strong GE-filter of .

Example 2. Let be a set with a binary operation given in the following table:Then, is a bordered GE-algebra, and , , , , and are all strong GE-filters of .

Theorem 1. Ifandare strong GE-filters of a bordered GE-algebra, then so is.

Proof. Let and be strong GE-filters of a bordered GE-algebra . Then, is a GE-filter of (see [5]). Let . Then, , so and , that is, and . Thus, , which shows that . Therefore, is a strong GE-filter of .
The following example shows that the union of strong GE-filters may not be a GE-filter.

Example 3. Consider a bordered GE-algebra with a binary operation given in the following table:It is routine to verify that and are strong GE-filters of . However, is not a GE-filter of since and but , so it is not a strong GE-filter of .

Lemma 3. In a transitive and antisymmetric bordered GE-algebra, we have the next properties:where .

Proof. Let be a transitive and antisymmetric bordered GE-algebra and . Equation (30) is easily obtained. Since , it follows from (11) that and . Hence, by (4) and (10), so . Using (GE3) and (3), we obtainHence, . Similarly, we haveby (GE3), (3), and (30). Hence, . We know that which implies that by (11). Therefore, by (11) and (25). Hence, . Now, we have by (12) and (24), which implies from (22) and (25) that . Therefore, , which proves (33). We have (34) by the following calculation:Using the above facts, we obtainThis completes the proof.
For every subset of a bordered GE-algebra , consider the following:We find subsets of a bordered GE-algebra satisfying condition (40).

Example 4. Using the bordered GE-algebra in Example 2, we find subsets that satisfy (40) as follows: , , , , , , , , , , , , , , , , , for , and for .

We provide conditions for a subset of a bordered GE-algebra to be a strong GE-filter.

Theorem 2. Letbe a transitive and antisymmetric bordered GE-algebra. If a nonempty subsetofsatisfies (15), (26), and (40), thenis a strong GE-filter of.

Proof. Let be a nonempty subset of satisfying (15), (26), and (40). Since , there exists . As , we have by (15). Let be such that and . Then, and by (9) and (15), which inducesby (25) and (40). It follows from (24) and (25) that . Since by (2), we have by (22), and hence, by (15). Thus, by (26), and thus, is a GE-filter of . Therefore, is a strong GE-filter of .

Definition 10. Let be a GE-algebra. A GE-filter of is said to be irreducible if, for every GEfilters and of ,

Example 5. (1)Let be a set with a binary operation given as follows: Then, is a GE-algebra. Clearly, , , , , , , and are all GE-filters of , and is an irreducible GE-filter of .(2)Let be a set with a binary operation given as follows:Then, is a GE-algebra. We know that , , and are GE-filters of , and is an irreducible GE-filter of .

We provide conditions for a GE-filter to be irreducible.

Lemma 4. Letbe a GE-filter of a GE-algebra. For every, if there existssuch thatand, thenis irreducible.

Proof. Let and be GE-filters of such that . Assume that and . Then, there exist and which imply from the hypothesis that there exists such that and . Hence, and . It follows that , which is a contradiction. Thus, or , and therefore, is an irreducible GE-filter of .
Let be a GE-filter of a transitive GE-algebra , and let be a subset of such thatIf and are disjoint, then there exists an irreducible GE-filter of such that and . In fact, let denote the set of GE-filter such that and , i.e.,where is the set of all GE-filters of . Then, , so is nonempty. It is obvious that the union of a chain of elements of is also contained in , so has a maximal element, say , by Zorn’s lemma. For any , consider the GE-filter and . Then, and . Hence, , so there exists such that and . It follows from (45) that and for some . Hence, and by (11), so and . Therefore, is an irreducible GE-filter of by Lemma 4. Consequently, we have the following theorem.

Theorem 3. Letbe a GE-filter of a transitive GE-algebraand letbe a subset ofsatisfying (44) and (45). If, then there exists an irreducible GE-filterofsuch thatand.

Corollary 1. Letbe an irreducible GE-filter of a transitive GE-algebra. For every,if and only if there exists an irreducible GE-filtersuch that,, and.

Proposition 3. In a transitive bordered GE-algebra, every GE-filtersatisfies (40).

Proof. Suppose that there exists such that . Then, there exists an irreducible GE-filter such that by Theorem 3. It follows from Corollary 1 that there exists an irreducible GE-filter such that and . Since , we have , so . This is a contradiction, and therefore, , that is, satisfies (40).
The above mentioned induces the characterization of a strong GE-filter as follows.

Theorem 4. Letbe a transitive and antisymmetric bordered GE-algebra. A nonempty subsetofis a strong GE-filter ofif and only if it satisfies (15), (26), and (40).

4. GE-Ideals

Definition 11. Let be a bordered GE-algebra. A subset of is called a GE-ideal of if it satisfies

It is clear that is a GE-ideal of a bordered GE-algebra .

Example 6. Let be a set with a binary operation given as follows:Then, is a bordered GE-algebra, and it is a routine to verify that is a GE-ideal of .

Theorem 5. Ifandare GE-ideals of a bordered GE-algebra, then so is.

Proof. It is clear that . Let be such that and . Then, , , , and . It follows from (48) that and . Hence, . This completes the proof.

The following example shows that the union of GE-ideals may not be a GE-ideal.

Example 7. In Example 6, we can observe that and are GE-ideals of . However, is not a GE-ideal of since and , but .

Proposition 4. Letbe a bordered GE-algebra. Every GE-idealofsatisfies

Proof. Let and be such that . Then, , so . It follows from (48) that .
The following corollary is induced directly by (22) and Proposition 4.

Corollary 2. Every GE-idealof a transitive bordered GE-algebrasatisfies

Proposition 5. Every GE-idealof a transitive bordered GE-algebrasatisfies

Proof. Let be a GE-ideal of a transitive bordered GE-algebra and assume that and for all . Using (22) and (23), we have for all . Hence, by Corollary 2. Since , it follows from (48) that .
We suggest the conditions under which any subset of a bordered GE-algebra can be a GE-ideal of .

Theorem 6. In an antisymmetric and duplex bordered GE-algebra, every subsetofsatisfying (47) and (52) is a GE-ideal of.

Proof. Let and be such that . Since is antisymmetric and duplex, it follows from (24) that . Hence, by (52), and therefore, is a GE-ideal of .

Lemma 5 (see [6]). Given an antisymmetric bordered GE-algebra, the following are equivalent:(i) is duplex.(ii) satisfies(iii) satisfies(iv) satisfies

The combination of Theorem 6 and Lemma 5 induces the next corollary.

Corollary 3. If an antisymmetric bordered GE-algebrasatisfies either condition (53), (54), or (55), then every subsetofsatisfying (47) and (52) is a GE-ideal of.

Theorem 7. In a transitive and antisymmetric bordered GE-algebra, every subsetofsatisfying (47) and (52) is a GE-ideal of.

Proof. Let and be such that . Using (24), (25), and (35) inducesIt follows from (52) that . Therefore, is a GE-ideal of .

Definition 12. Let be a subset of a bordered GE-algebra . The GE-ideal generated by is defined to be the intersection of all GE-ideals containing , and it is denoted by .

Example 8. Let be a set with a binary operation given as follows:Then, is a bordered GE-algebra. If we take a subset of , then the GE-ideal of generated by is .

Let be a GE-algebra. For any natural number and , is recursively defined as follows:

Let be a transitive GE-algebra. By (6) and induction, it is easy to prove that

If we put for all in (59), thenProof. It is straightforward by (9).

Proposition 6. Every GE-algebrasatisfiesfor all .

Theorem 8. Letbe a subset of a bordered GE-algebra