# Γα,β-BE-Algebras

Authors: Ravikumar Bandaru; Jeong Gi Kang; Amal S. Alali; Young Bae Jun
Journal: Journal of Mathematics
 (2023)

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## Abstract

The concept ofΓα,β-BE-algebra is introduced as a generalization of a Γα-BE-algebra and a Γβ-BE-algebra and its properties are studied. We introduced the concepts of α,β-subalgebra and α,β-filter of a Γα,β-BE-algebra and discussed the relation between these two concepts. We provided conditions for an α,β-subalgebra to be an α,β-filter. We provided equivalent conditions for the formation of an α,β-filter from a nonempty subset of a Γα,β-BE-algebra. We introduced the concept of α,β-atomic Γα,β-BE-algebra and studied its properties. We introduced the concept of atomic α,β-filter a Γα,β-BE-algebra and gave the necessary and sufficient condition for an α,β-filter to be atomic.


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## Body

## 1. Introduction

The concept of BE-algebras was introduced by Kim and Kim as a generalization of BCK-algebras (see [1]). Later, several authors introduced and studied several substructures of BE-algebras (see [2–4]). In 1964, Nobusawa studied Γ-rings (see [5]). Later, in the year 1966, Barnes weakened the defining conditions of Γ-ring and studied it (see [6]). The structure of Γ-rings was also studied by Ravisankar and Shukla (see [7]). Sen and Saha studied the structure of Γ-semigroups and obtained various generalizations and analogues of corresponding results in the semigroup theory (see [8]). Later, Rao introduced the concept of Γ-semiring as a generalization of Γ-ring and studied various types of Γ-semirings (see [9, 10]). Most algebraic structures are studied in environments where only one binary operation is given, and so is BE-algebra. However, it can be seen that Γ-semigroup, Γ-semiring, and Γ-ring are being studied as algebraic structures with two or more binary operations and these are regarded as generalizations of semigroup, semiring, and ring, respectively. With this motivation, Jun et al. introduced the concept of Γ-BE-algebra as a generalization of a BE-algebra and investigated its properties (see [11]).In this paper, we introduced the concept ofΓα,β-BE-algebra as a generalization of a Γα-BE-algebra and a Γβ-BE-algebra and studied its properties. We introduced the concepts of α,β-subalgebra and α,β-filter of a Γα,β-BE-algebra and discussed the relation between these two. We provided conditions for an α,β-subalgebra to be an α,β-filter. We provided equivalent conditions for the formation of an α,β-filter from a nonempty subset of a Γα,β-BE-algebra. We defined and explored the properties of α,β-atomic Γα,β-BE-algebra. We presented the idea of an atomic α,β-filter and provided a necessary and sufficient condition for an atomic α,β-filter in the context of the Γα,β-BE-algebra.


## 2. Preliminaries

A BE-algebra, denoted byX,1∗ (see [1]), is defined to be a set X together with a binary operation “∗” and a special element, called the unit, “1” satisfying the following conditions:(BE1)∀a∈Xa∗a=1,(BE2)∀a∈Xa∗1=1,(BE3)∀a∈X1∗a=a,(BE4)∀a,b,c∈Xa∗b∗c=b∗a∗c,Every BE-algebraX,1∗ meets the following requirements (see [1]):(1)∀a,b∈Xa∗b∗a=1,(2)∀a,b∈Xa∗a∗b∗b=1.A subsetA of a BE-algebra X,1∗ is called as follows:(1)
• a BE-subalgebra of X,1∗ if it satisfies the following equation:(3)∀a,b∈Aa∗b∈A.(2)
• a BE-filter of X,1∗ (see [1]) if it satisfies the following equations:(4)1∈A,(5)∀a,b∈Xa∗b∈A,a∈A⟹b∈A.LetX be a nonempty set and let Γ be a set of binary operations on X, that is, every element α∈Γ is given as follows:(6)α:X×X⟶X,x,y↦αx,y.In what follows,αx,y is denoted by xαy, and let X,1Γ be a structure related to a special element ″1″ and Γ.Definition 1 (See [11]).
For a fixedα∈Γ, a structure is called a Γα-BE-algebra if it satisfies the following equations:(7)∀x∈Xxαx=1,(8)∀x∈Xxα1=1,(9)∀x∈X1αx=x,(10)∀x,y,z∈X∀β∈Γxαyβz=yβxαz.
IfX,1Γ is a Γα-BE-algebra for all α∈Γ, we say that X,1Γ is a Γ-BE-algebra.


## 3.Γα,β-BE-Algebras

Definition 2.
Givenα,β∈Γ, a structure X,1Γ is called a Γα,β-BE-algebra if it satisfies the following equations:(11)∀x∈Xxαx=1=xβx,(12)∀x∈Xxα1=1=xβ1,(13)∀x∈X1αx=x=1βx,(14)∀x∈X∀γ∈Γxαyγz=yγxαz,xβyγz=yγxβz.
It is clear thatΓα,β-BE-algebra and Γβ,α-BE-algebra are congruent concepts.
For a fixedα∈Γ, we define a binary relation ″≤α″ on X as follows:(15)∀x,y∈Xx≤αy⇔xαy=1.
Ifx≤αy is valid for all α∈Γ, we present it as x≤Γy.Example 1.
LetX=0,1,2,3 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 1. It is a routine to verify that X,1Γ is a Γα,β-BE-algebra. But, it is neither a Γα,γ-BE-algebra nor Γβ,γ-BE-algebra since 3α3=1≠3=3γ3 and 3β3=1≠3=3γ3. So, X,1Γ is not even a Γ-BE-algebra.
It is clear that everyΓα,β-BE-algebra is both a Γα-BE-algebra and a Γβ-BE-algebra. The following example shows that any Γα-BE-algebra or Γβ-BE-algebra may not be a Γα,β-BE-algebra.Table 1
Cayley tables for the binary operations “α,β,andγ.”
α012301103101232111131111β012301101101232111130121γ012300121101232110330123Example 2.
(i)
LetX=0,1,2 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 2. Then, X,1Γ is a Γα-BE-algebra (see [11]). But it is neither a Γα,β-BE-algebra nor a Γα,γ-BE-algebra since 2α2=1≠0=2β2 and 0α0=1≠2=0γ0.(ii)
LetX=0,1,2 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 3. Then, X,1Γ is a Γβ-BE-algebra. But it is neither a Γα,β-BE-algebra nor a Γα,γ-BE-algebra since 0β0=1≠0=0α0 and 0β0=1≠0=0γ0.Proposition 1.
EveryΓα,β-BE-algebraX,1Γ satisfies the following equations:(16)∀x,y∈Xx≤αyβx,x≤βyαx.(17)∀x,y∈Xx≤αyαx,x≤βyβx.(18)∀x,y∈Xx≤αxαyβy,x≤βxβyαy.(19)∀x,y,z∈Xx≤αyβz⇔y≤βxαz.(20)∀x∈X1≤αxor1≤βx⟹x=1.(21)∀x,y,z∈Xx≤αy⟹x≤αzβy.(22)∀x,y,z∈Xxαy=z⟹y≤βz.(23)∀x,y,z∈Xxβy=z⟹y≤αz.(24)∀x,y∈Xx≠1,xαy=x⟹xβy≠y.(25)∀x,y∈Xx≠1,xαy=y⟹xβy≠x.(26)∀x,y,a,b∈Xxαy=a,xβy=b⟹xαb=xβa.(27)∀x,y,z∈Xxαy=x,xβy=z⟹x≤αz,x≤αyαz,x≤βyαz.(28)∀x,y,z∈Xxαy=y,xβy=z⟹xαz=z,x≤αyαz,x≤αyβz.Table 2
Cayley tables for the binary operations “α,β,andγ.”
α012011110122111β012001211122010γ012021010122212Table 3
Cayley tables for the binary operations “α,β,andγ.”
α012001010102011β012011110122111γ012001010102012Proof.
Letx,y∈X and α,β∈Γ. The combination of equations (11), (12), and (14) induces xαyβx=yβxαx=yβ1=1 and xβyαx=yαxβx=yα1=1. Hence, equation (16) is valid. It is clear that equation (17) is true. Using equations (11) and (14), we have xαxαyβy=xαyβxαy=1 and xβxβyαy=xβyαxβy=1. Thus, x≤αxαyβy and x≤βxβyαy. Equations (19) and (20) are induced from equations (14) and (13), respectively. If x≤αy, then xαy=1 and so xαzβy=zβxαy=zβ1=1 by equations (12) and (14). Hence, x≤αzβy. The conditions (22) and (23) are induced by the combination of equations (11), (12), and (14). Let x,y∈X be such that x≠1 and xαy=x. If xβy=y, then x=xαy=xαxβy=xβxαy=xβx=1 by equations (11) and (14), which is a contradiction. Hence, xβy≠y. Let x,y∈X be such that x≠1 and xαy=y. If xβy=x, then 1=xαx=xαxβy=xβxαy=xβy=x by equations (11) and (14). This is a contradiction, and so xβy≠x. Let x,y,a,b∈X be such that xαy=a and xβy=b. Then, xαb=xαxβy=xβxαy=xβa, which proves equation (26). Let x,y,z∈X be such that xαy=x and xβy=z. Then, xαz=xαxβy=xβxαy=xβx=1. Hence, x≤αz. Now, xαyαz=yαxαz=yα1=1. Therefore, x≤αyαz. Similarly, xβyαz=xβyαxβy=xβxβyαy=xβxβ1=xβ1=1, that is, x≤βyαz. Let x,y,z∈X be such that xαy=y and xβy=z. Then, xαz=xαxβy=xβxαy=xβy=z. Now, xαyαz=yαxαz=yαz=yαxβy=xβyαy=xβ1=1. Therefore, x≤αyαz. Similarly, xαyβz=xαyβxβy=xαxβyβy=xαxβ1=xα1=1. Therefore, x≤αyβz.
In general, aΓα,β-BE-algebraX,1Γ does not satisfy the following assertions:(29)∀x,y∈Xx≤αy⇔x≤βy,(30)∀x,y∈Xx≤αxβyαy,x≤βxαyβy,as seen in the example below.Example 3.
TheΓα,β-BE-algebraX,1Γ in Example 1 does not satisfy equation (29) since 3≤α2 but 3≰β2, and 0≤β3 but 0≰α3. Let X=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 4.
It is a routine to verify thatX,1Γ is a Γα,β-BE-algebra. But it does not satisfy equation (30) since(31)0α0β2α2=0α3α2=0α4=0≠1.and(32)0β0α2β2=0β0β2=0β3=3≠1.Table 4
Cayley tables for the binary operations “α,β,andγ.”
α01234011010101234211111301414411111β01234011331101234211111301414411331γ01234001234101234201234301234411010Definition 3.
AΓα,β-BE-algebraX,1Γ is said to be strong if it fulfils condition (29).Example 4.
LetX=0,1,2,3 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 5. It is a routine to verify that X,1Γ is a strong Γα,β-BE-algebra.Table 5
Cayley tables for the binary operations “α,β,andγ.”
α012301111101232111130121β012301111101232111132121γ012300121111112212132121Proposition 2.
Every strongΓα,β-BE-algebraX,1Γ satisfies condition (30).Proof.
It is induced by the combination of equation (18) and (29).Proposition 3.
Every strongΓα,β-BE-algebraX,1Γ satisfies the following equation:(33)∀x,y,z∈Xxαy=x,xβy=z⟹x≤βz.Proof.
Letx,y,z∈X be such that xαy=x and xβy=z. Then, x≤αz in equation (27), and so x≤βz by equation (29).
The example below shows that ifX,1Γ is not strong, then the condition (33) does not hold.Example 5.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 6.
It is routine to verify thatX,1Γ is a Γα,β-BE-algebra which is not strong. We can observe that 0α2=0 and 0β2=3, but 0β3=3≠1.Table 6
Cayley tables for the binary operations “α,β,andγ.”
α01234011014101234211111311011401231β01234011334101234211114311014411331γ01234001231101234201231301234411334


## 4.Γ-Subalgebras and Γ-Filters

In this section, unless otherwise stated, assume thatX,1Γ is a Γα,β-BE-algebra.Definition 4.
A subsetA of X is called an α,β-subalgebra of X,1Γ if it satisfies the following equation:(34)∀x,y∈Xx,y∈A⟹xαy,xβy∈A.
It is clear that1∈A for every α,β-subalgebra A of X,1Γ.Example 6.
In Example1, we can observe that A=0,1 is a α,β-subalgebra of X,1Γ.
The following example shows that anyα,β-subalgebra A of X,1Γ does not satisfy the following assertion:(35)∀x,y∈X∀γ∈Γ\α,βx,y∈A⇒xγy∈A.Example 7.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 7. It is a routine to verify that X,1Γ is a Γα,β-BE-algebra. Then, B≔1,2 is an α,β-subalgebra of X,1Γ, but it does not satisfy equation (35) since 2∈B but 2γ2=0∉B.Table 7
Cayley tables for the binary operations “α,β,andγ.”
α01234011334101234211111311011401231β01234011014101234211114311014411111γ01234001231101234211011311011401231Definition 5.
If anα,β-subalgebra A of X,1Γ satisfies condition (35), we say A is a Γ-subalgebra of X,1Γ.Example 8.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 8.
It is a routine to verify thatX,1Γ is a Γα,β-BE-algebra, and A≔0,1 is a Γ-subalgebra of X,1Γ.
It is easy to check that the intersection ofα,β-subalgebras is an α,β-subalgebra. But the union of α,β-subalgebras is not an α,β-subalgebra in general as seen in the following example.Table 8
Cayley tables for the binary operations “α,β,andγ.”
α01234011011101234211111301211401211β01234011011101234211111301211411011γ01234001211101213201213301211401211Example 9.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 9. It is routine to verify that X,1Γ is a Γα,β-BE-algebra. Let A=1,2 and B=1,3,4. Then, A and B are α,β-subalgebras of X,1Γ. But A∪B=1,2,3,4 is not an α,β-subalgebra of X,1Γ since 4α2=0∉A∪B.Table 9
Cayley tables for the binary operations “α,β,andγ.”
α01234011011101234211113301213411011β01234011011101234211111301211401211γ01234001211101211201211301211411013Definition 6.
A subsetA of X is called an α,β-filter of X,1Γ if it satisfies equation (4) and the following equations:(36)∀x,y∈Xx∈A,xαy∈A⟹y∈A,(37)∀x,y∈Xx∈A,xβy∈A⟹y∈A.Example 10.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 10. It is a routine to verify that X,1Γ is a Γα,β-BE-algebra. Then, A≔0,1 is an α,β-filter of X,1Γ.
The following example shows that anyα,β-filter A of X,1Γ does not satisfy the following assertion:(38)∀x,y∈X∀γ∈Γ\α,βx∈A,xγy∈A⇒y∈A.Table 10
Cayley tables for the binary operations “α,β,andγ.”
α01234011224101234201111301114401221β01234011224101234201114301114401111γ01234001114101114201111301114401114Example 11.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 11. It is a routine to verify that X,1Γ is a Γα,β-BE-algebra. Then, A≔0,1 is an α,β-filter of X,1Γ. But A does not satisfy equation (38) since 0∈A and 0γ2=0∈A, but 2∉A.Table 11
Cayley tables for the binary operations “α,β,andγ.”
α01234011334101234211111301014401011β01234011334101234211111301011411331γ01234001011101014201011301011411111Definition 7.
If anα,β-subalgebra A of X,1Γ satisfies condition (38), we say A is a Γ-filter of X,1Γ.Example 12.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 12.
It is a routine to verify thatX,1Γ is a Γα,β-BE-algebra and A≔0,1 is a Γ-filter of X,1Γ.Table 12
Cayley tables for the binary operations “α,β,andγ.”
α01234011224101234201111301111401111β01234011224101234201111301114411221γ01234001224101224201111301114401114Theorem 1.
IfX,1Γ is strong, then two conditions (36) and (37) are equivalent to each other for every subset A of X which contains 1.Proof.
LetA be a subset of X which contains 1. Assume that equation (36) is valid, and let x,y∈X be such that x∈A and xβy∈A. Using Proposition 2 leads to xαxβyαy=1∈A, and so xβyαy∈A by equation (36). Using equation (36) again leads to y∈A. Similarly, if A satisfies equation (37), then equation (36) is valid.
The next example shows that two conditions (36) and (37) are independent of each other if X,1Γ is not strong. That is, if X,1Γ is a Γα,β-BE-algebra which is not strong, then there is a subset A of X such that equation (36) is established, but equation (37) is not established, or equation (37) is established, but equation (36) is not established.Example 13.
(i)
LetX,1Γ be a Γα,β-BE-algebra in Example 7. Then, it is not strong and A≔0,1 satisfies equation (36). But it does not satisfy equation (37) since 0β2=0∈A and 0∈A, but 2∉A.(ii)
LetX,1Γ be a Γα,β-BE-algebra in Example 5. Then, it is not strong and A≔0,1 satisfies equation (37). But A does not satisfy equation (37) since 0α2=0∈A and 0∈A, but 2∉A.Proposition 4.
Everyα,β-filter A of X,1Γ satisfies the following equations:(39)∀x∈A∀y∈Xx≤αy⟹y∈Ax≤βy⟹y∈A,(40)∀x∈A∀y∈Xxαxβy∈A⟹y∈Axβxαy∈A⟹y∈A.Proof.
Straightforward.Proposition 5.
Everyα,β-filter A of X,1Γ satisfies the following equations:(41)∀x,y∈A∀z∈Xx≤αyβz⟹z∈Ax≤βyαz⟹z∈A,(42)∀x,y∈Xy∈A⟹yαxβx∈A,yβxαx∈A.Proof.
LetA be an α,β-filter of X,1Γ. Then, 1∈A. Hence, if you use equations (36) and (37), it connects directly to equation (41). Let x∈X and y∈A. Then, yαyαxβx=yαxβyαx=1∈A and yβyβxαx=yβxαyβx=1∈A. It follows from equations (36) and (37) that yαxβx∈A and yβxαx∈A.
We discuss the relationship between anα,β-subalgebra and an α,β-filter.Theorem 2.
Everyα,β-filter is an α,β-subalgebra.Proof.
LetA be an α,β-filter of X,1Γ and let x,y∈A. Since y≤αxβy and y≤βxαy, it follows from Proposition 5 that xαy,xβy∈A. Therefore, A is an α,β-subalgebra of X,1Γ.
The converse of Theorem2 may not be true as seen in the example below.Example 14.
In Example1, we can observe that A=0,1 is an α,β-subalgebra of X,1Γ. But A is not an α,β-filter of X,1Γ since 0∈A and 0α2=0∈A, but 2∉A.
We provide a condition for anα,β-subalgebra to be an α,β-filter.Theorem 3.
If anα,β-subalgebra A of X,1Γ satisfies the following equation:(43)∀x,y∈Xx∈A,y∉A⟹xαy∉A,xβy∉A,then A is an α,β-filter of X,1Γ.Proof.
Straightforward.
The combination of Theorems2 and 3 induces the following corollary.Corollary 1.
Anα,β-subalgebra A of X,1Γ is an α,β-filter of X,1Γ if and only if it satisfies condition (43).Theorem 4.
If everyα,β-subalgebra A of X,1Γ satisfies condition (40), then it is an α,β-filter of X,1Γ.Proof.
LetA be an α,β-subalgebra of X,1Γ that satisfies condition (40). Clearly, 1∈A. Let x,y∈X be such that x∈A, xαy∈A, and xβy∈A. Then, xαxβy∈A and xβxαy∈A, which implies from equation (40) that y∈A. In light of this, A is an α,β-filter of X,1Γ.
We explore the conditions under which we can make anα,β-filter using a nonempty set.Theorem 5.
Given a nonempty subsetA of X, the following are equivalent.(i)
A is an α,β-filter of X,1Γ.(ii)
A satisfies equation (39) and the following equation:(44)∀x∈X∀a,b∈Aaαbβxαx∈Aaβbαxβx∈A.Proof.
LetA be an α,β-filter of X,1Γ. Then, A satisfies equation (39) (see Proposition 4). Let x∈X and a,b∈A. Then,(45)aαaαbβxαbβx=aαbβxαaαbβx=1∈A, (46)aβaβbαxβbαx=aβbαxβaβbαx=1∈A.
It follows from equations (14), (36), and (37) that(47)bβaαbβxαx=aαbβxαbβx∈A, (48)bαaβbαxβx=aβbαxβbαx∈A.
Hence,aαbβxαx∈A and aβbαxβx∈A by equations (36) and (37).
Conversely, suppose thatA satisfies (39) and (44). Since x≤α1 and x≤β1 for all x∈A, we have 1∈A by equation (39). Let x∈A and y∈X be such that xαy∈A and xβy∈A. Using equations (13) and (44) leads toy=1αy=xβyαxβyαy∈A and y=1βy=xαyβxαyβy∈A.
Hence,A is an α,β-filter of X,1Γ.Definition 8.
A nonunit elementb in X,1Γ is called an α,β atom if the following assertion is valid.(49)∀x∈Xb≤αx⟹b=xor x=1b≤βx⟹b=xor x=1.Example 15.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 13.
It is a routine to verify thatX,1Γ is a Γα,β-BE-algebra. The element 0 is the only α,β-atom.Table 13
Cayley tables for the binary operations “α,β,andγ.”
α01234011224101234201111301111401231β01234011224101234201111301111411111γ01234001114101111201114301111411224Theorem 6.
For every nonunit elementb in X,1Γ, if b,1 is an α,β-filter of X,1Γ, then b is an α,β-atom in X,1Γ.Proof.
Assume thatb,1 is an α,β-filter of X,1Γ. Let x∈X be such that b≤αx and b≤βx. Then, bαx=1∈b,1 and bβx=1∈b,1. It follows from equations (36) and (37) that x∈b,1. Hence, b=x or x=1. Thus, b is an α,β-atom in X,1Γ.
The following example shows that the converse of Theorem6 may not be true, that is, there exists an α,β-atom b in X,1Γ for which b,1 is not an α,β-filter of X,1Γ.Example 16.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 14.
It is a routine to verify thatX,1Γ is a Γα,β-BE-algebra. We can observe that 2 is an α,β atom. But 1,2 is not an α,β-filter of X,1Γ since 2∈1,2, 2α0=2∈1,2 and 2β0=2∈1,2, but 0∉1,2.Table 14
Cayley tables for the binary operations “α,β,andγ.”
α01234011111101234221104311111411121β01234011114101234221104311114411111γ01234001231101231201231301231411124Definition 9.
AΓα,β-BE-algebraX,1Γ is said to be α,β-atomic if every nonunit element of X is an α,β-atom in X,1Γ.Theorem 7.
IfX,1Γ is a Γα,β-BE-algebra, then the following assertions are equivalent.(i)
X,1Γ is α,β-atomic.(ii)
X,1Γ satisfies the following equation:(50)∀x,y∈Xx≠y⟹xαy=y=xβy.(iii)
X,1Γ satisfies the following equation:(51)∀x,y∈Xx≠y⟹xαyβy=1=xβyαy.Proof.
(i)
⟹ (ii). Assume that X,1Γ is α,β-atomic. Let x,y∈X be such that x≠y. Note that y≤αxβy and y≤βxαy. Since y is an α,β-atom, we have y=xβy or xβy=1 and y=xαy or xαy=1. If xβy=1, then x≤βy and so x=y or y=1. This is a contradiction. Similarly, xαy=1 induces a contradiction. Hence, xαy=y=xβy.(ii)
⟹ (iii). Straightforward.(iii)
⟹ (i). Suppose that X,1Γ satisfies (4.14). Let x be an arbitrary nonunit element of X such that x≤αy and x≤βy for all nonunit element y∈X. Then, xαy=1 and xβy=1. If x≠y, then 1=xαyβy=1βy=y and 1=xβyαy=1αy=y by equations (13) and (51). This is a contradiction, and so x=y, which shows that x is an α,β-atom in X,1Γ. Consequently, X,1Γ is α,β-atomic.Theorem 8.
In anα,β-atomic Γα,β-BE-algebra, every α,β-subalgebra is an α,β-filter.Proof.
LetX,1Γ be an α,β-atomic Γα,β-BE-algebra. Let A be an α,β-subalgebra of X,1Γ. Clearly, 1∈A. Let x,y∈X be such that x∈A, xαy∈A, and xβy∈A. It is clear that if x=1 or y=1, then y∈A. So we may assume that x and y are nonunit. Note that y≤αxβy and y≤βxαy. Since y is an α,β-atom, we have y=xβy∈A or xβy=1 and y=xαy∈A or xαy=1. If xβy=1 and xαy=1, then x≤βy and x≤αy. Since x is an α,β-atom, it follows that y=x∈A or y=1∈A. Therefore, A is an α,β-filter of X,1Γ.
Given a subsetA of X, we consider a set as follows:(52)TA,1Γ≔b∈A|bis anα,β−atom.
Note thatTX,1Γ is the set of all α,β-atoms in X,1Γ. It is clear that 1∉TA,1Γ and 1∉TX,1Γ.Theorem 9.
IfA is an α,β-subalgebra of X,1Γ, then the set TA,1Γ∪1 is also an α,β-subalgebra of X,1Γ.Proof.
Assume thatA is an α,β-subalgebra of X,1Γ. Let x,y∈TA,1Γ∪1. It is clear that if x=1, y=1, or x=y, then xαy, xβy∈TA,1Γ∪1. Assume that 1≠x≠y≠1. Since y≤αxβy and y≤βxαy, we have y=xβy∈A or xβy=1, and y=xαy∈A or xαy=1. If xβy=1 or xαy=1, then x≤βy or x≤αy. Since x∈TA,1Γ, it follows that x=y, a contradiction. Hence, xβy≠1 and xαy≠1, and so xβy=y∈TA,1Γ∪1 and xαy=y∈TA,1Γ∪1. Therefore, TA,1Γ∪1 is an α,β-subalgebra of X,1Γ.Corollary 2.
The setTX,1Γ∪1 is an α,β-subalgebra of X,1Γ.Question 1.
IfA is an α,β-filter of X,1Γ, then the set TA,1Γ∪1 is an α,β-filter of X,1Γ?
The following example provides a negative answer to Question1.Example 17.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 15. It is a routine to verify that X,1Γ is a Γα,β-BE-algebra. We can observe that A=0,1,2 is an α,β-filter of X,1Γ. But, TA,1Γ∪1=0,1, and it is not an α,β-filter of X,1Γ since 0∈TA,1Γ∪1 and 0α2=0β2=0∈TA,1Γ∪1, but 2∉TA,1Γ∪1.Table 15
Cayley tables for the binary operations “α,β,andγ.”
α01234011034101234211134301211401231β01234011034101234211134301214411031γ01234001214101211201214301211401211Definition 10.
Anα,β-filter A of X,1Γ is said to be atomic if A=TA,1Γ∪1.Example 18.
LetX=0,1,2,3,4 be a set and Γ=α,β,γ be a set of binary operations on X given in Table 16. It is routine to verify that X,1Γ is a Γα,β-BE-algebra. Let A=0,1,2. Then, A is an α,β-filter of X,1Γ and A=TA,1Γ∪1. Hence, A is an atomic α,β-filter of X,1Γ.Table 16
Cayley tables for the binary operations “α,β,andγ.”
α01234011234101234201134301114401131β01234011234101234201134301111401131γ01234001111101114201111301111401234Theorem 10.
In anα,β-atomic Γα,β-BE-algebra, every α,β-filter is atomic.Proof.
Assume thatA is an α,β-filter of an α,β-atomic Γα,β-BE-algebraX,1Γ. It is clear that TA,1Γ∪1⊆A. Let x≠1∈A. Then, x is an α,β-atom since X,1Γ is α,β-atomic. Hence, x∈TA,1Γ, and so A⊆TA,1Γ∪1. Therefore, A is atomic.Theorem 11.
LetA be an α,β-filter of X,1Γ. Then, A is atomic if and only if it satisfies the following equation:(53)∀x,y∈Ax≠y⟹xαy=y=xβy.Proof.
Assume thatA is an atomic α,β-filter of X,1Γ. Let x,y∈A be such that x≠y. If x=1, then y≠1 and so 1αy=y=1βy by equation (13). If y=1, then x≠1 and so xα1=1=xβ1 by equation (12). Suppose that 1≠x≠y≠1. Since y≤αxβy and y≤βxαy, we have y=xβy or xβy=1 and y=xαy or xαy=1. If xβy=1 or xαy=1, then x≤βy or x≤αy. Since x∈TA,1Γ and y≠1, it follows that x=y, a contradiction. Hence, xβy≠1 and xαy≠1, and thus xαy=y=xβy.
Conversely, suppose thatA satisfies condition (53). It is clear that TA,1Γ∪1⊆A. Let x be a nonunit element of A such that x≤αy and x≤βz for all nonunit elements y,z∈X. Then, y,z∈A. If x≠y or x≠z, then 1=xαy=y=xβy=1 or 1=xαz=z=xβz=1 by condition (53). This is a contradiction, and so x=y and x=z. Thus, x∈TA,1Γ, and hence, A⊆TA,1Γ. Therefore, A is an atomic α,β-filter of X,1Γ.


## 5. Conclusion

We have introduced the concept ofΓα,β-BE-algebra as a generalization of a Γα-BE-algebra and a Γβ-BE-algebra and studied its properties. We have introduced the concepts of α,β-subalgebra and α,β-filter of a Γα,β-BE-algebra and discuss the relation between these two concepts. We have provided conditions for an α,β-subalgebra to be an α,β-filter. We have provided equivalent conditions for the formation of an α,β-filter from a nonempty subset of a Γα,β-BE-algebra. We have introduced the concept of α,β-atomic Γα,β-BE-algebra and studied its properties. We have introduced the concept of atomic α,β-filter a Γα,β-BE-algebra and given a necessary and sufficient condition for an α,β-filter to be atomic.



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