# Commutative and Bounded BE-algebras

Authors: Zekiye Çiloğlu; Yılmaz Çeven
Journal: Algebra
 (2013)

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## Abstract

We introduce the notions of the commutative and bounded BE-Algebras. We give some related properties of them.


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## Body

## 1. Introduction

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 [5] 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. [8] introduced a new notion called BH-algebra which is another generalization of BCH/BCI/BCK-algebras. Walendziak obtained some equivalent axioms for B-algebras [9]. C. B. Kim and H. S. Kim [10] 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 [11], 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 [14], 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.


## 2. Preliminaries

Definition 1 (see [11]).
An algebra(X;*,1) of type (2, 0) is called a BE-algebra if, for all x, y, and z in X,(BE1)
x*x=1,(BE2)
x*1=1,(BE3)
1*x=x,(BE4)
x*(y*z)=y*(x*z).InX, a binary relation “≤” is defined by x≤y  if and only if x*y=1.Example 2 (see [11]).
LetX={1,a,b,c,d,0} be a set with the following table:
(1)*1abcd011abcd0a11accdb111cccc1ab1abd11a11a0111111
Then (X;*,1) is a BE-algebra.Definition 3.
A BE-algebra(X;*,1) is said to be self-distributive if x*(y*z)=(x*y)*(x*z) for all x, y, and z∈X.Example 4 (see [11]).
LetX={1,a,b,c,d} be a set with the following table:
(2)*1abcd11abcda11bcdb1a1ccc11b1bd11111
Then (X;*,1) is a self-distributive BE-algebra.Proposition 5 (see [14]).
LetX be a self-distributive BE-algebra. If x≤y, then, for all x, y, and z in X, the following inequalities hold:(i)
z*x≤z*y,(ii)
y*z≤x*z.Definition 6 (see [15]).
A dual BCK-algebra is an algebra(X;*,1) of type (2,0) satisfying (BE1) and (BE2) and the following axioms:(dBCK1)x*y=y*x=1 implies x=y,(dBCK2)(x*y)*((y*z)*(x*z))=1,(dBCK3)x*((x*y)*y)=1.Proposition 7 (see [16]).
Any dual BCK-algebra is a BE-algebra.The converse of Proposition7 does not hold in general [16].Definition 8 (see [16]).
LetX be a BE-algebra or dual BCK-algebra. X is said to be commutative if the following identity holds:
(3)(C)x∨By=y∨Bx,wherex∨By=(y*x)*x
for all x, y∈X.Theorem 9 (see [16]).
IfX is a commutative BE-algebra, then (X;*,1) is a dual BCK-algebra.Corollary 10 ([16]).
X is a commutative BE-algebra if and only if it is a commutative dual BCK-algebra.IfX is a commutative BE-algebra, then the relation “≤” is a partial order on X [16].In the following, we abbreviate∨B as ∨.


## 3. Bounded BE-Algebras

The following definition introduces the notion of boundedness for BE-algebras.Definition 11.
LetX be a BE-algebra. If there exists an element 0 satisfying 0≤x(or0*x=1) for all x∈X, then the element “0” is called unit of X. A BE-algebra with unit is called a bounded BE-algebra.In a bounded BE-algebra,x*0 denoted by xN.Example 12.
The BE-algebra in Example2 is a bounded BE-algebra and its unit element is 0.Example 13.
The BE-algebra in Example4 is a bounded BE-algebra and its unit element is d.Example 14.
LetX={1,a,b,c,d} be a set with the following table:
(4)*1abcd11abcda11bcdb111cdc1111dd11bc1
It is clear that X is a BE-algebra, but it is not a bounded BE-algebra.Theorem 15.
In a bounded BE-algebra with unit0, the following properties hold for all x, y∈X:(i)
1N=0, 0N=1,(ii)
x≤xNN,(iii)
x*yN=y*xN,(iv)
0∨x=xNN, x∨0=x.Proof.
(i) Using (BE3) and (BE1), we have1N=1*0=0 and 0N=0*0=1.
(ii) Since(5)x*xNN=x*((x*0)*0)=(x*0)*(x*0)=1,
by (BE4) and (BE1), we get x≤xNN.
(iii) Using (BE4), we have(6)x*yN=x*(y*0)=y*(x*0)=y*xN.
(iv) By routine operations, we have0∨x=(x*0)*0=xNN and x∨0=(0*x)*x=1*x=x.Theorem 16.
In a bounded and self-distributive BE-algebra with unit0, the following properties hold for all x, y∈X:(i)
x*y≤yN*xN,(ii)
x≤y implies yN≤xN.Proof.
(i) Since(7)(x*y)*(yN*xN)=(x*y)*((y*0)*(x*0))=(y*0)*((x*y)*(x*0))(byBE4)=(y*0)*(x*(y*0))(bydistributivity)=x*((y*0)*(y*0))(byBE4)=x*1(byBE1)=1(byBE2),
we have x*y≤yN*xN.
(ii) It is trivial by Proposition5.Proposition 17.
LetX be a BE-algebra. Then x*y≤(y∨x)*y.Proof.
Since(8)(x*y)*((y∨x)*y)=(y∨x)*((x*y)*y)=(y∨x)*(y∨x)=1,
we have x*y≤(y∨x)*y.Proposition 18.
LetX be a self-distributive BE-algebra. Then, the next properties are valid for all x, y∈X:(i)
(y∨x)*y≤x*y,(ii)
x*(x*y)=x*y.Proof.
(i) Since(9)x*(y∨x)=x*((x*y)*y)=(x*y)*(x*y)=1,
we have x≤y∨x. By Proposition 5(i), we have (y∨x)*y≤x*y.
(ii) Using Definition3, (BE1) and (BE3), we have
(10)x*(x*y)=(x*x)*(x*y)=1*(x*y)=x*y.Corollary 19.
IfX is a self-distributive and commutative BE-algebra, then (y∨x)*y=x*y.Proof.
It is clear by Propositions17 and 18 and the property (dBCK1).Corollary 20.
IfX is a self-distributive, commutative, and bounded BE-algebra with unit 0, then xNNN=xN.Proof.
In Corollary19, taking y=0, we have (0∨x)*0=x*0. Then we get ((x*0)*0)*0=x*0; that is, xNNN=xN.Definition 21.
In a bounded BE-algebra, the elementx such that xNN=x is called an involution.LetS(X)={x∈X:xNN=x} where X is a bounded BE-algebra. S(X) is the set of all involutions in X. Moreover, since 1NN=(1*0)*0=0*0=1 and 0NN=(0*0)*0=1*0=0, we have 0,1∈S(X) and so S(X)≠⌀.Example 22.
For the BE-algebra in Example2, it is clear that S(X)=X.Example 23.
For the BE-algebra in Example4, it is clear that S(X)={1,b,c,d}.Proposition 24.
LetX be a bounded BE-algebra with unit 0 and let S(X) be the set of all involutions in X. Then, for all x, y∈S(X), the following conditions hold:(i)
xNNN=xN,(ii)
xN*y=yN*x.Proof.
(i) The proof is obvious by the definition ofS(X).
(ii) By Theorem15(iii), we have
(11)xN*y=xN*yNN=yN*xNN=yN*x.In a bounded BE-algebraX, we denote x∧y=(xN∨yN)N where x∨y=(y*x)*x for all x, y∈X.Theorem 25.
In a bounded and commutative BE-algebraX the following identities hold:(i)
xNN=x,(ii)
xN∧yN=(x∨y)N,(iii)
xN∨yN=(x∧y)N,(iv)
xN*yN=y*x.Proof.
(i) Using (BE3), it is obtained that(12)xNN=(x*0)*0=(0*x)*x,bycommutativity=1*x=x.
(ii) By the definition of “∧” and (i), we have that xN∧yN=(xNN∨yNN)N=(x∨y)N.
(iii) By the definition of “∧” and (i), we have that (x∧y)N=(xN∨yN)NN=xN∨yN.
(iv) We have(13)xN*yN=(x*0)*(y*0)=y*((x*0)*0)=y*(xNN)=y*x.Corollary 26.
IfX is a bounded commutative BE-algebra, then S(X)=X.Definition 27.
Each of the elementsa and b in a bounded BE-algebra is called the complement of the other if a∨b=1 and a∧b=0.Theorem 28.
Let(X;*,1) be a bounded and commutative BE-algebra. If there exists a complement of any element of X, then it is unique.Proof.
Letx∈X and a, b be two complements of x. Then we know that x∧a=x∧b=0 and x∨a=x∨b=1. Also since x∨a=(x*a)*a=1 and a*(x*a)=x*(a*a)=x*1=1, we have x*a≤a and a≤x*a. So we get x*a=a. Similarly x*b=b. Hence
(14)a*b=(x*a)*(x*b)=(aN*xN)*(bn*xN),byTheorem25(iv)=bN*((aN*xN)*xN),byBE-4=bN*(xN∨aN)=bN*(x∧a)N,byTheorem25(iii)=(x∧a)*b,byTheorem25(iii)=0*b=1.
With similar operations, we have b*a=1. Hence we obtain a=b which gives that the complement of x is unique.Note that for a bounded and commutative BE-algebra, an element does not need to have any complement.Example 29.
In Example2, the complement of b is c, but a has no complement in X.Theorem 30.
Let(X;*,1) be a commutative and bounded BE-algebra. Then the following conditions are equivalent, for all x, y∈X:(i)
x∧xN=0,(ii)
xN∨x=1,(iii)
xN*x=x,(iv)
x*xN=xN,(v)
x*(x*y)=x*y.Proof.
(i)⇒(ii) Let x∧xN=0. Then it follows that
(15)xN∨x=(xN∨x)NN,byTheorem25(i)=(xNN∧xN)N,byTheorem25(ii)=(x∧xN)N,byTheorem25(i)=0N=1.
(ii)⇒(iii) Let xN∨x=1. Then,  since (xN*x)*x=x∨xN=1 and x*(xN*x) = xN*(x*x) = xN*1=1, we get xN*x=x by (dBCK1).
(iii)⇒(iv) Let xN*x=x. Substituting xN for x and using Theorem 25 (i), we obtain the result.
(iv)⇒(v) Let x*xN=xN. Then we get yN*(x*xN)=yN*xN. Hence we have x*(yN*xN)=yN*xN. Using Theorem 25(iv), we obtain x*(x*y)=x*y.
(v)⇒(ii) Let x*(x*y)=x*y. Then we have
(16)xN∨x=(x*(xN))*xN=(x*(x*0))*xN=(x*0)*(x*0)=1.
(ii)⇒(i) Let xN∨x=1. Then we obtain
(17)xN∧x=xN∧xNN=(x∨xN)N,byTheorem25(ii)=1N=0.Note that, ifX is a self-distributive BE-algebra, then x*(x*y)=x*y by Proposition 18. In this case, Theorem 30(v) is true. If X is also commutative and bounded, then xN is the complement of x by Theorem 30(i) and (ii).Now we obtain a bounded BE-algebra from a nonbounded BE-algebra as the following theorem.Theorem 31.
Let(X;*,1) be a BE-algebra and 0∉X. Define the operation ⊛ on X¯=X∪{0} as follows:
(18)x⊛y={x*yifx,y∈X,1ifx=0,y∈X,0ifx∈X,y=0,1ifx=y=0.
Then (X¯;⊛,1) is a bounded BE-algebra with unit 0.Proof.
It is clear that BE1, BE2, and BE3 are satisfied. It suffices to verify BE4. Note thatifx=0 and y, z∈X, then 0⊛(y*z)=1 and y⊛(0⊛z)=y⊛1=1;ify=0 and x, z∈X, then x⊛(0⊛z)=x⊛1=1 and 0⊛(x⊛z)=1;ifz=0 and x, y∈X, then x⊛(y⊛0)=x⊛0=0 and y⊛(x⊛0)=y⊛0=0.
For the remain situations, it is clearly seen that BE4 is satisfied with similar arguments.We call the BE-algebra (X¯;⊛,1) in the previous theorem the extension of (X;1,*). Note that for the BE-algebra (X¯;⊛,1), if  for 1≠x∈X, we have x∉S(X) since xNN=(x⊛0)⊛0=0*0=1≠x. Hence S(X)={0,1}.Theorem 32.
The extension of a self-distributive BE-algebra(X;*,1) is also self-distributive.Proof.
Let0∉X and denote X¯ =X∪{0}. Let ⊛ be defined as in Theorem 31. We want to see that x⊛(y⊛z)=(x⊛y)⊛(x⊛z) for all x, y, and z∈X¯. For all x, y∈X, we know x⊛y=x*y, 0⊛y=1, x⊛0=0, and 0⊛0=1. So we obtain the following, for all x, y, and z∈X:
(19)0⊛(0⊛0)=(0⊛0)⊛(0⊛0),0⊛(y⊛0)=(0⊛y)⊛(0⊛0),x⊛(0⊛0)=(x⊛0)⊛(x⊛0),0⊛(0⊛z)=(0⊛0)⊛(0⊛z),0⊛(y⊛z)=(0⊛y)⊛(0⊛z),x⊛(0⊛z)=(x⊛0)⊛(x⊛z),x⊛(y⊛0)=(x⊛y)⊛(x⊛0),x⊛(y⊛z)=(x⊛y)⊛(x⊛z).
The above results show that (X¯;⊛,1) is self-distributive.Theorem 33.
If(X;*,1) is a commutative, self-distributive, and bounded BE-algebra, then it is a lattice with respect to ∨ and ∧ where x∨y=y*(y*x) and x∧y=(xN∨yN)N for any x,y∈X.Proof.
From Theorem 3.6 in [14], we know that (X;*,1) is a semilattice with respect to ∨. Then we need to show that the set {x,y} for all x, y∈X  has a greatest lower bound. We know that xN≤xN∨yN and yN≤xN∨yN. By Proposition 5, we have (xN∨yN)N≤xNN and (xN∨yN)N≤yNN. Since xNN=x and yNN=y  by Theorem 25(i), we have x∧y≤x and  x∧y≤y. So x∧y is a lower bound of x and y. Next we must show that if z≤x and z≤y then z≤x∧y. Let z≤x and z≤y. We have xN≤zN and yN≤zN by Proposition 5. Since X is an upper semilattice with respect to ∨, then we obtain xN∨yN≤zN and hence zNN≤(xN∨yN)N. So we have z≤x∧y. Then x∧y is the greatest lower bound of x and y. So, we can say that X is a lower semilattice. Consequently (X,≤) is a lattice with respect to ∨ and ∧.



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