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

Authors: Mehmet Ali Öztürk; Jeong-Gon Lee; Ravikumar Bandaru; Young Bae Jun
Journal: Journal of Mathematics
 (2021)

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## 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.


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## Body

## 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 Section3, 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 X with a constant 1 and a binary operation ∗, satisfying the following axioms:(GE1)
u∗u=1(GE2)
1∗u=u(GE3)
u∗v∗w=u∗v∗u∗w for all u,v,w∈XIn a GE-algebraX, a binary relation “≤” is defined by(1)∀x,y∈X,x≤y⇔x∗y=1.Definition 2.
(see [5]). A GE-algebra X is said to be(1
)Transitive if it satisfies(2)∀x,y,z∈X,x∗y≤z∗x∗z∗y.(2)
Antisymmetric if the binary relation “≤” is antisymmetric.Definition 3.
(see [6]). If a GE-algebra X has a special element, say 0, which satisfies 0≤x for all x∈X, we call X the bordered GE-algebra.Definition 4.
(see [6]). If a bordered GE-algebra X satisfies condition (2), we say that X is a transitive bordered GE-algebra.Definition 5.
(see [6]). A bordered GE-algebra X is said to be antisymmetric if the binary operation “≤” is antisymmetric.Proposition 1 (see [5]).
Every GE-algebraXsatisfies the following items:(3)∀u∈X,u∗1=1,(4)∀u,v∈X,u∗u∗v=u∗v,(5)∀u,v∈X,u≤v∗u,(6)∀u,v,w∈X,u∗v∗w≤v∗u∗w,(7)∀u∈X,1≤u⇒u=1,(8)∀u,v∈X,u≤v∗u∗u,(9)∀u,v∈X,u≤u∗v∗v,(10)∀u,v,w∈X,u≤v∗w⇔v≤u∗w.
IfX is transitive, then(11)∀u,v,w∈X,u≤v⇒w∗u≤w∗v,v∗w≤u∗w,(12)∀u,v,w∈X,u∗v≤v∗w∗u∗w.Lemma 1 (see [5]).
In a GE-algebraX, the following facts are equivalent to each other:(13)∀x,y,z∈X,x∗y≤z∗x∗z∗y,∀x,y,z∈X,x∗y≤y∗z∗x∗z.Definition 6.
(see [5]). A subset F of a GE-algebra X is called a GE-filter of X if it satisfies(14)1∈F,∀x,y∈X,x∗y∈F,x∈F⇒y∈F.Lemma 2 (see [5]).
In a GE-algebraX, every GE-filterFofXsatisfies(15)∀x,y∈X,x≤y,x∈F⇒y∈F.Proposition 2 (see [6]).
In a bordered GE-algebraX, the following assertions are valid:(16)10=0,00=1,(17)∀x∈X,x≤x00,0≤x00,(18)∀x∈X,0≤x∗x0=x0,(19)∀x,y∈X,x∗y0≤y∗x0,(20)∀x,y∈X,x≤y0⇔y≤x0,(21)∀x,y∈X,x∗y0=x∗y∗x0.
IfX is a transitive bordered GE-algebra, then(22)∀x,y∈X,x≤y⇒y0≤x0,(23)∀x,y∈X,x∗y≤y0∗x0.
IfX is an antisymmetric bordered GE-algebra, then(24)∀x,y∈X,x∗y0=y∗x0.
IfX is a transitive and antisymmetric bordered GE-algebra, then(25)∀x∈X,x000=x0.Definition 7.
(see [6]). By a duplex bordered element in a bordered GE-algebra X, we mean an element x of X which satisfies x00=x.The set of all duplex bordered elements of a bordered GE-algebraX is denoted by 02X and is called the duplex bordered set of X. It is clear that 0,1∈02X.Definition 8.
(see [6]). A bordered GE-algebra X is said to be duplex if every element of X is a duplex bordered element, that is, X=02X.


## 3. Strong GE-Filters

Given a subsetF of a bordered GE-algebra X, we define the set 02X,F:=x∈X|x00∈F.Definition 9.
A GE-filterF of a bordered GE-algebra X is said to be strong if it satisfies(26)02X,F⊆F,i.e., x∈F whenever x00∈F.It is clear that every GE-filter is strong in a duplex bordered GE-algebra.Example 1.
LetX=0,1,a,b,c be a set with a binary operation ∗ given in the following table:(27)
Then,X is a bordered GE-algebra, and it is easy to check that F:=1,a,b is a strong GE-filter of X.Example 2.
LetX=0,1,a,b,c,d,e,f,g,h,x,y,u,v be a set with a binary operation ∗ given in the following table:(28)
Then,X is a bordered GE-algebra, and 1, 1,a,b,c,g,y,u, 1,c,d,e,g,y,u, 1,c,g,y,u, and X are all strong GE-filters of X.Theorem 1.
IfFandGare strong GE-filters of a bordered GE-algebraX, then so isF∩G.Proof.
LetF and G be strong GE-filters of a bordered GE-algebra X. Then, F∩G is a GE-filter of X (see [5]). Let x∈02X,F∩G. Then, x00∈F∩G, so x00∈F and x00∈G, that is, x∈02X,F⊆F and x∈02X,G⊆G. Thus, x∈F∩G, which shows that 02X,F∩G⊆F∩G. Therefore, F∩G is a strong GE-filter of X.
The following example shows that the union of strong GE-filters may not be a GE-filter.Example 3.
Consider a bordered GE-algebraX=0,1,a,b,c,d,e with a binary operation ∗ given in the following table:(29)
It is routine to verify thatF:=1,a,d,e and G:=1,b,c,d,e are strong GE-filters of X. However, F∪G=1,a,b,c,d,e is not a GE-filter of X since a∈F∪G and a∗0=b∈F∪G but 0∉F∪G, so it is not a strong GE-filter of X.Lemma 3.
In a transitive and antisymmetric bordered GE-algebraX, we have the next properties:(30)∀x,y,z∈X,y∗x∗z=x∗y∗z=x∗y∗x∗z,(31)∀x∈X,x0∗x=x00,(32)∀x,y∈X,x≤x0∗y,y≤x0∗y,(33)∀x,y∈X,x+^y=x00+^y00,(34)∀x,y∈X,x∗x∗y00=x∗y00,(35)∀x,y∈X,x∗y00=x00∗y00,where x+^y:=x∗y00.Proof.
LetX be a transitive and antisymmetric bordered GE-algebra and x,y∈X. Equation (30) is easily obtained. Since 0≤x, it follows from (11) that x∗0∗0≤x∗0∗x and x∗0≤x∗0∗x∗x∗0∗0. Hence, x∗0∗x≤x∗0∗x∗0∗0=x∗0∗0 by (4) and (10), so x0∗x=x∗0∗x=x∗0∗0=x00. Using (GE3) and (3), we obtain(36)y∗x0∗y=y∗x∗0∗y=y∗x∗0∗y∗y=y∗x∗0∗1=y∗1=1.
Hence,y≤x0∗y. Similarly, we have(37)x∗x0∗y=x∗x∗0∗y=x∗x∗0∗x∗y,=x∗0∗x∗y=x∗1=1,by (GE3), (3), and (30). Hence, x≤x0∗y. We know that x≤x00 which implies that x00∗y0≤x∗y0 by (11). Therefore, x∗y00≤x00∗y0000 by (11) and (25). Hence, x+^y≤x00+^y00. Now, we have x∗y0=y∗x0≤x00∗y0 by (12) and (24), which implies from (22) and (25) that x00∗y0000=x00∗y00≤x∗y00=x+^y. Therefore, x+^y=x00+^y00, which proves (33). We have (34) by the following calculation:(38)x∗x∗y00=x∗x∗y0∗x∗y=x∗x∗y0∗x∗x∗y=x∗x∗y0∗x∗y=x∗x∗y0∗y=x∗y0∗x∗y=x∗y00.
Using the above facts, we obtain(39)x∗y00=x∗x∗y00=x∗x∗y0∗x0=x∗y∗x0∗x0=x∗y0∗x0=x∗y00=y0∗x0=y0∗x000=x00∗y00.
This completes the proof.
For every subsetF of a bordered GE-algebra X, consider the following:(40)∀x,y∈X,x,y∈F⇒x+^y∈F.
We find subsets of a bordered GE-algebraX satisfying condition (40).Example 4.
Using the bordered GE-algebraX in Example 2, we find subsets that satisfy (40) as follows: 0, 1, a, b, d, e, x, u, 1,a,b, 1,d,u, 1,e,d, 1,e,d,u, 1,a,b,u, 1,c,g,y,u, 1,a,b,c,g,y,u, 1,c,d,e,g,y,u, 1,c,g,y,u, Ft:=0,t for t∈X∖0, and Fq:=1,q for q∈a,b,d,e,x,u.We provide conditions for a subset of a bordered GE-algebra to be a strong GE-filter.Theorem 2.
LetXbe a transitive and antisymmetric bordered GE-algebra. If a nonempty subsetFofXsatisfies (15), (26), and (40), thenFis a strong GE-filter ofX.Proof. LetF be a nonempty subset of X satisfying (15), (26), and (40). Since F≠∅, there exists x∈F. As x≤1, we have 1∈F by (15). Let x,y∈X be such that x∈F and x∗y∈F. Then, x00∈F and x∗y00∈F by (9) and (15), which induces(41)x00∗x∗y00=x00∗x∗y0000=x00+^x∗y00∈F,by (25) and (40). It follows from (24) and (25) that x∗y∗x00∈F. Since y0=y∗0≤x∗y∗x∗0 by (2), we have x∗y∗x∗00≤y00 by (22), and hence, y00∈F by (15). Thus, y∈F by (26), and thus, F is a GE-filter of X. Therefore, F is a strong GE-filter of X.Definition 10.
LetX be a GE-algebra. A GE-filter F of X is said to be irreducible if, for every GEfilters F1 and F2 of X, F=F1∩F2⇒F=F1orF=F2.Example 5.
(1)
LetX=1,a,b,c,d,e be a set with a binary operation ∗ given as follows:(42)Then,X is a GE-algebra. Clearly, 1, 1,b, 1,e, 1,a,b, 1,b,e, 1,a,b,e, and X are all GE-filters of X, and 1,a,b,e is an irreducible GE-filter of X.(2)
LetX=1,a,b,c,d,e be a set with a binary operation ∗ given as follows:(43)
Then,X is a GE-algebra. We know that 1, 1,b,c, and X are GE-filters of X, and 1,b,c is an irreducible GE-filter of X.We provide conditions for a GE-filter to be irreducible.Lemma 4.
LetFbe a GE-filter of a GE-algebraX. For everya,b∉F, if there existsc∉Fsuch thata∗c∈Fandb∗c∈F, thenFis irreducible.Proof.
LetF1 and F2 be GE-filters of X such that F=F1∩F2. Assume that F≠F1 and F≠F2. Then, there exist a∈F1∖F and b∈F2∖F which imply from the hypothesis that there exists c∉F such that a∗c∈F and b∗c∈F. Hence, a,a∗c∈F1 and b,b∗c∈F2. It follows that c∈F1∩F2=F, which is a contradiction. Thus, F=F1 or F=F2, and therefore, F is an irreducible GE-filter of X.
LetF be a GE-filter of a transitive GE-algebra X, and let A be a subset of X such that(44)∀a,b∈X,a≤b,b∈A⇒a∈A,(45)∀a,b∈X,a,b∈A⇒a≤candb≤cfor somec∈A.
IfF and A are disjoint, then there exists an irreducible GE-filter G of X such that F⊆G and G∩A=∅. In fact, let K denote the set of GE-filter H such that F⊆H and H∩A=∅, i.e.,(46)K≔H∈ℱX|F⊆H,H∩A=∅,where ℱX is the set of all GE-filters of X. Then, F∈K, so K is nonempty. It is obvious that the union of a chain of elements of K is also contained in K, so K has a maximal element, say G, by Zorn’s lemma. For any a,b∉G, consider the GE-filter Ga:=G∪a and Gb:=G∪b. Then, G⊆Ga∩Gb and Ga,Gb∉K. Hence, Ga∩A≠∅≠Gb∩A, so there exists x,y∈A such that a∗x∈G and b∗y∈G. It follows from (45) that x≤c and y≤c for some c∈A. Hence, a∗x≤a∗c and b∗y≤b∗c by (11), so a∗c∈G and b∗c∈G. Therefore, G is an irreducible GE-filter of X by Lemma 4. Consequently, we have the following theorem.Theorem 3.
LetFbe a GE-filter of a transitive GE-algebraXand letAbe a subset ofXsatisfying (44) and (45). IfF∩A=∅, then there exists an irreducible GE-filterGofXsuch thatF⊆GandG∩A=∅.Corollary 1.
LetFbe an irreducible GE-filter of a transitive GE-algebraX. For everya,b∈X,a∗b∉Fif and only if there exists an irreducible GE-filterGsuch thatF⊆G,a∈G, andb∉G.Proposition 3.
In a transitive bordered GE-algebraX, every GE-filterFsatisfies (40).Proof.
Suppose that there existsa,b∈F such that a+^b=a∗b00∉F. Then, there exists an irreducible GE-filter G such that a+^b=a∗b00∉G by Theorem 3. It follows from Corollary 1 that there exists an irreducible GE-filter H such that G⊆H and a∗b0∉H. Since a∈F⊆G⊆H, we have b0=b∗0∈H, so 0∈H. This is a contradiction, and therefore, a+^b=a∗b00∈F, that is, F satisfies (40).
The above mentioned induces the characterization of a strong GE-filter as follows.Theorem 4.
LetXbe a transitive and antisymmetric bordered GE-algebra. A nonempty subsetFofXis a strong GE-filter ofXif and only if it satisfies (15), (26), and (40).


## 4. GE-Ideals

Definition 11.
LetX be a bordered GE-algebra. A subset A of X is called a GE-ideal of X if it satisfies(47)0∈A,(48)∀x,y∈X,y0∗x00∈A,y∈A⇒x∈A.It is clear that0 is a GE-ideal of a bordered GE-algebra X.Example 6.
LetX=0,1,a,b,c,d be a set with a binary operation ∗ given as follows:(49)
Then,X is a bordered GE-algebra, and it is a routine to verify that A:=0,a,b is a GE-ideal of X.Theorem 5.
IfAandBare GE-ideals of a bordered GE-algebraX, then so isA∩B.Proof. It is clear that0∈A∩B. Let x,y∈X be such that y∈A∩B and y0∗x00∈A∩B. Then, y∈A, y0∗x00∈A, y∈B, and y0∗x00∈B. It follows from (48) that x∈A and x∈B. Hence, x∈A∩B. This completes the proof.The following example shows that the union of GE-ideals may not be a GE-ideal.Example 7.
In Example6, we can observe that A=0,a,b and B=0,c,d are GE-ideals of X. However, A∪B=0,a,b,c,d is not a GE-ideal of X since b0∗100=c00=c∈A∪B and b∈A∪B, but 1∉A∪B.Proposition 4.
LetXbe a bordered GE-algebra. Every GE-idealAofXsatisfies(50)∀x,y∈X,y∈A,y0≤x0⇒x∈A.Proof.
Letx∈X and y∈A be such that y0≤x0. Then, y0∗x0=1, so y0∗x00=10=0∈A. It follows from (48) that x∈A.
The following corollary is induced directly by (22) and Proposition 4.Corollary 2.
Every GE-idealAof a transitive bordered GE-algebraXsatisfies(51)∀x,y∈X,y∈A,x≤y⇒x∈A.Proposition 5.
Every GE-idealAof a transitive bordered GE-algebraXsatisfies(52)∀x,y∈X,x∗y0∈A,y∈A⇒x∈A.Proof.
LetA be a GE-ideal of a transitive bordered GE-algebra X and assume that x∗y0∈A and y∈A for all x,y∈X. Using (22) and (23), we have y0∗x00≤x∗y0 for all x,y∈X. Hence, y0∗x00∈A by Corollary 2. Since y∈A, it follows from (48) that x∈A.
We suggest the conditions under which any subsetA of a bordered GE-algebra X can be a GE-ideal of X.Theorem 6.
In an antisymmetric and duplex bordered GE-algebraX, every subsetAofXsatisfying (47) and (52) is a GE-ideal ofX.Proof.
Letx∈X and y∈A be such that y0∗x00∈A. Since X is antisymmetric and duplex, it follows from (24) that x∗y0=x∗y000=y0∗x00∈A. Hence, x∈A by (52), and therefore, A is a GE-ideal of X.Lemma 5 (see [6]).
Given an antisymmetric bordered GE-algebraX, the following are equivalent:(i)
X is duplex.(ii)
X satisfies(53)∀x,y∈X,x∗y=y0∗x0.(iii)
X satisfies(54)∀x,y∈X,x0∗y=y0∗x.(iv)
X satisfies(55)∀x,y∈X,x0≤y⇒y0≤x.The combination of Theorem6 and Lemma 5 induces the next corollary.Corollary 3.
If an antisymmetric bordered GE-algebraXsatisfies either condition (53), (54), or (55), then every subsetAofXsatisfying (47) and (52) is a GE-ideal ofX.Theorem 7.
In a transitive and antisymmetric bordered GE-algebraX, every subsetAofXsatisfying (47) and (52) is a GE-ideal ofX.Proof.
Letx∈X and y∈A be such that y0∗x00∈A. Using (24), (25), and (35) induces(56)x∗y0=x∗y000=x00∗y000=y0∗x0000=y0∗x00∈A.
It follows from (52) that x∈A. Therefore, A is a GE-ideal of X.Definition 12.
LetA be a subset of a bordered GE-algebra X. The GE-ideal generated byA is defined to be the intersection of all GE-ideals containing A, and it is denoted by A.Example 8.
LetX=0,1,a,b,c,d be a set with a binary operation ∗ given as follows:(57)
Then,X is a bordered GE-algebra. If we take a subset G:=0,a of X, then the GE-ideal of X generated by G is G=0,a,b.LetX be a GE-algebra. For any natural number k and x,y∈X, x∗ky is recursively defined as follows:(58)x∗1y=x∗y,x∗2y=x∗x∗yandx∗k+1y=x∗x∗ky.LetX be a transitive GE-algebra. By (6) and induction, it is easy to prove that(59)yn∗⋯∗y2∗y1∗z∗x⋯≤z∗yn∗⋯∗y2∗y1∗x⋯.If we putyi=y for all i=1,2,…,n in (59), then(60)y∗nz∗x≤z∗y∗nx.Proof. It is straightforward by (9).Proposition 6.
Every GE-algebraXsatisfies(61)yn∗⋯∗y1∗x∗a⋯≤yn∗⋯∗y1∗x∗a⋯∗a∗a,for all a,x,y1,…,yn∈X.Theorem 8.
LetAbe a subset of a bordered GE-algebraX. Then,(i)
0=0 and ∅=0(ii)
X=X and 1=X(iii)
For every subsetsA and B of X, if A⊆B, then A⊆B(iv)
∀x,y∈Xx≤y⇒y⊆x(v)
IfA is a GE-ideal of X, then A=AProof.
Straightforward.
In the next theorem, we discuss the description of the elements ofA.Theorem 9.
LetAbe a nonempty subset of a transitive and antisymmetric bordered GE-algebraX. Then,(62)A=x∈X|an0∗⋯∗a20∗a10∗x0⋯=1for somea1,a2,…,an∈A.Proof.
LetB:=x∈X|an0∗⋯∗a20∗a10∗x0⋯=1for somea1,a2,…,an∈A. Since A is nonempty, there exists a∈A and a0∗00=a0∗1=1. Hence, 0∈B. Let x,y∈X be such that x0∗y00∈B and x∈B. Then, there are a1,a2,…,an,b1,b2,…,bm∈A such that(63)an0∗⋯∗a20∗a10∗x0∗y000⋯=1,(64)bm0∗⋯∗b20∗b10∗x0⋯=1.
Using (25) and (35) induces x0∗y000=x000∗y000=x0∗y0. If follows from (11), (59), and (63) that(65)1=an0∗⋯∗a20∗a10∗x0∗y0⋯,≤x0∗an0∗⋯∗a20∗a10∗y0⋯.
Hence,x0∗an0∗⋯∗a20∗a10∗y0⋯=1 by (7), so(66)x0≤an0∗⋯∗a20∗a10∗y0⋯.
If we multiply both sides of (66) by b10 from the left, then(67)b10∗x0≤b10∗an0∗⋯∗a20∗a10∗y0⋯.
If we repeat this processm times, then(68)1=bm0∗⋯∗b20∗b10∗x0⋯,≤bm0∗⋯∗b10∗an0∗⋯∗a20∗a10∗y0⋯⋯,so bm0∗⋯∗b10∗an0∗⋯∗a20∗a10∗y0⋯⋯=1. Hence, y∈B, and therefore, B is a GE-ideal of X. It is clear that A⊆B. Let C be a GE-ideal of X containing A. If x∈B, then an0∗⋯∗a20∗a10∗x0⋯=1 for some a1,a2,…,an∈A⊆C. On the contrary, we have(69)1=yn∗⋯∗y1∗x∗a⋯∗yn∗⋯∗y1∗x∗a⋯≤yn∗⋯∗y1∗x∗yn∗⋯∗y1∗x∗a⋯∗a⋯≤yn∗⋯∗y1∗x∗yn∗⋯∗y1∗x∗a⋯∗a∗a∗a⋯≤yn∗⋯∗y1∗x∗a⋯∗a∗a∗yn∗⋯∗y1∗x∗a⋯,by using (6), (9), and (11) repeatedly, which implies from (7) that(70)yn∗⋯∗y1∗x∗a⋯∗a∗a∗yn∗⋯∗y1∗x∗a⋯=1,that is, yn∗⋯∗y1∗x∗a⋯∗a∗a≤yn∗⋯∗y1∗x∗a⋯. The combination of this and (61) induces(71)yn∗⋯∗y1∗x∗a⋯∗a∗a=yn∗⋯∗y1∗x∗a⋯.
If we takea=0 and yk=ak0 for k=1,2,…,n in (71), then(72)an0∗⋯∗a10∗x0⋯000=an0∗⋯∗a10∗x0⋯0=10=0∈C.
Sincean∈A⊆C and C is a GE-ideal of X, we obtain(73)an−10∗⋯∗a10∗x0⋯000=an−10∗⋯∗a10∗x0⋯0∈C.
Repeating this process, we obtainx00∈C. Since x≤x00 by (17), it follows from Corollary 2 that x∈X. Hence, B⊆C, and therefore, B=A. This completes the proof.
If we takeak=a for all k=1,2,…,n in Theorem 9 and use (4), then we have the next corollary.Corollary 4.
For every elementain a transitive and antisymmetric bordered GE-algebraX, we have(74)a=x∈X|a0∗x0=1,that is, a=x∈X|a0≤x0.Given a GE-idealA and an element b in a bordered GE-algebra, we construct the smallest GE-ideal containing A and b.Theorem 10.
LetAbe a GE-ideal of a transitive and antisymmetric bordered GE-algebraXand letbbe any element ofX. Then,(75)A∪b=x∈X|b0∗nx00∈Afor somen∈ℕ.Proof.
LetAb:=x∈X|b0∗nx00∈Afor somen∈ℕ. Since b0∗nb00=10=0∈A, Ab contains the element b. For any x∈A, we have x0≤b0∗x0≤b0∗x000 by (5) and (17). It follows from Proposition 4 that b0∗x00∈A. Thus, x∈Ab, which shows that A⊆Ab. Hence, A∪b⊆Ab. Since b0∗n000=b0∗n10=10=0∈A, we get 0∈Ab. Let x,y∈X be such that x∈Ab and x0∗y00∈Ab. Then,(76)b0∗nx0∗y0000∈A,(77)b0∗mx00∈A,for some natural numbers n and m. Using (4), (25), (30), and (35), we have(78)b0∗nx0∗y0000=b0∗x0∗y0000=b0∗x000∗y0000=b0∗x0∗y00=b0∗x0∗b0∗y00=b0∗x000∗b0∗y0000,so b0∗x000∗b0∗y0000∈A by (76). The combination of (4) and (77) induces b0∗x00∈A. Since A is a GE-ideal of X, it follows that b0∗ny00=b0∗y00∈A. Hence, y∈Ab, and thus, Ab is a GE-ideal of X. Now, we prove that Ab is the smallest GE-ideal containing A∪b. Let B be a GE-ideal of X containing A∪b and x∈Ab. Then, b∈B and b0∗x00∈A⊆B. Hence, x∈B since B is GE-ideal of X. Therefore, Ab⊆B. Hence, Ab=A∪b.


## 5. Conclusions

We introduced the notion of strong GE-filters and GE-ideals (generated) and investigated the related properties. We have shown that the intersection of strong GE-filters (resp., GE-ideals) is a strong GE-filter (resp., GE-ideal), but the union of strong GE-filters (resp., GE-ideals) is not a strong GE-filter (resp., GE-ideal) by example. We provided conditions for a subset of a bordered GE-algebra to be a strong GE-filter. We considered a characterization of a strong GE-filter. In order to do so, we defined irreducible GE-filter first and examined its properties. We provided conditions for a GE-filter to be irreducible. Given a GE-filter and a subset in a bordered GE-algebra, we considered the existence of an irreducible GE-filter which contains the given GE-filter and is disjoint to the given subset. We suggested conditions under which any subset of a bordered GE-algebra can be a GE-ideal, discussed GE-ideal that is generated from a subset in a bordered GE-algebra, and investigated what element it is formed into. We constructed the smallest GE-ideal which contains a given GE-ideal and an element in a bordered GE-algebra. It can be seen that the study of cubic structures on BCK/BCI-algebras, etc., has been actively conducted recently (see [7, 8]). In the future, we will study the cubic structure based on the ideas and results of this paper.



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*Source: 5520023-2021-05-19.xml*