# A Construction of MirrorQ-Algebras

Authors: Keum Sook So
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We investigate how to construct mirrorQ-algebras of a
Q-algebra, and we obtain the necessary  conditions for
M(X) to be a Q-algebra.


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## Body

## 1. Introduction

Imai and Iséki introduced two classes of abstract algebras:BCK-algebras and BCI-algebras [1, 2]. It is known that the class of BCK-algebras is a proper subclass of the class of BCI-algebras. We refer the reader for useful textbooks for BCK/BCI-algebra to [3–5]. Neggers et al. [6] introduced the notion of Q-algebras which is a generalization of BCK/BCI/BCH-algebras, obtained several properties, and discussed quadratic Q-algebras. Ahn and Kim [7] introduced the notion of QS-algebras, and Ahn et al. [8] studied positive implicativity in Q-algebras and discussed some relations between R-(L-) maps and positive implicativity. Neggers and Kim introduced the notion of d-algebras which is another useful generalization of BCK-algebras and then investigated several relations between d-algebras and BCK-algebras as well as several other relations between d-algebras and oriented digraphs [9]. After that some further aspects were studied [10–13]. Allen et al. [14] introduced the notion of mirror image of given algebras and obtained some interesting properties: a mirror algebra of a d-algebra is also a d-algebra, and a mirror algebra of an implicative BCK-algebra is a left L-up algebra.In this paper we introduce the notion of mirror algebras toQ-algebras, and we investigate how to construct mirror Q-algebras from a Q-algebra; and we also obtain the necessary conditions for M(X) to be a Q-algebra.


## 2.Q-Algebras and Related Algebras

AQ-algebra [6] is a nonempty set X with a constant 0 and a binary operation “*” satisfying the following axioms:  (I)
x*x=0,(II)
0*x=0,(III)
(x*y)*z=(x*z)*y for all x,y,z∈X.For brevity we also callX a  Q-algebra. In X we can define a binary relation “≤” by x≤y if and only if x*y=0.Example 2.1 (see [6]).
LetX:{0,1,2,3} be a set with the following table:
(2.1)*012300000100322200033330
Then (X,*,0) is a Q-algebra, which is not a BCK/BCI/BCH-algebra.
Ahn and Kim [7] introduced the notion of QS-algebras. They showed that the G-part of an associative QS-algebra is a group in which every element is an involution. A Q-algebra X is said to be a QS-algebra if it satisfies the following condition:  (IV)
(x*y)*(x*z)=z*y, for all x,y,z∈X.Proposition 2.2 (see [6]).
If(X,*,0) is a  Q-algebra, then   (V)
(x*(x*y))*y=0, for all x,y∈X.It was proved that everyBCH-algebra is a Q-algebra and every Q-algebra satisfying some additional conditions is a BCI-algebra.Neggers and Kim [15] introduced the notion of B-algebras which is related to several classes of algebras of interest such as BCH/BCI/BCK-algebras and which seems to have rather nice properties without being excessively complicated otherwise. And they demonstrated some interesting connections between B-algebras and groups.Example 2.3.
LetX:={0,1,2,…,ω} be a set. Define a binary operation “*” on X by
(2.2)x*y:={0,x≤y,ω,y<x,x≠0,x,y<x,y=0.
Then (X,*,0) is a Q-algebra, but not a B-algebra, since (3*ω)*0=0,   3*(0*(0*ω))=3.Example 2.4.
LetX:={0,1,…,5} be a set with the following table:
(2.3)*012345002134511024532210534334502144531025534210
Then (X,*,0) is a B-algebra, but not a Q-algebra, since (5*3)*1=1,   (5*1)*3=0.Example 2.5.
LetX be the set of all real numbers except for a negative integer -n. Define a binary operation * on X by
(2.4)x*y:=n(x-y)n+y
for any x,y∈X. Then (X,*,0) is both a Q-algebra and B-algebra.If we consider several families of abstract algebras including the well-knownBCK-algebras and several larger classes including the class of d-algebras which is a generalization of BCK-algebras, then it is usually difficult and often impossible to obtain a complementation operation and the associated “de Morgan’s laws.” In the sense of this point of view it is natural to construct a “mirror image” of a given algebra which when adjoined to the original algebra permits a natural complementation to take place. The class of BCK-algebras is not closed under this operation but the class of d-algebras is, thus explaining why it may be better to work with this class rather than the class of BCK-algebras. Allen et al. [14] introduced the notion of mirror algebras of a given algebra.Let(X,*,0) be an algebra. Let M(X):=X×{0,1}, and define a binary operation “*” on M(X) as follows:(2.5)(x,0)*(y,0):=(x*y,0),(x,1)*(y,1):=(y*x,0),(x,0)*(y,1):=(x*(x*y),0),(x,1)*(y,0):={(y,1)whenx*y=0,(x,1)whenx*y≠0.
Then we say that M(X):=(M(X),*,(0,0)) is aleft mirror algebra of the algebra (X,*,0). Similarly, if we define (2.6)(x,*)*(y,1):=(y*(y*x),0),
then M(X):=(M(X),*,(0,0)) is a right mirror algebra of the algebra (X,*,0).It was shown [14] that the mirror algebra of a d (resp., d-BH)-algebra is also a d (resp., d-BH)-algebra, but the mirror algebra of a BCK-algebra need not be a BCK-algebra.


## 3. A Construction of MirrorQ-Algebras

In [14] Allen et al. defined (left, right) mirror algebras of an algebra, but it is not known how to construct mirror algebras of any given algebra. In this paper, we investigate a construction of a mirror algebra in Q-algebras.Let(X,*,0) be a Q-algebra, and let M(X):=X×{0,1}. Define a binary operation “⊕” on M(X) by(M1)
(x,0)⊕(y,0)=(x*y,0),(M2)
(x,1)⊕(y,1)=(y*x,0),(M3)
(x,0)⊕(y,1)=(α(x,y),0),(M4)
(x,1)⊕(y,0)=(β(x,y),1),whereα,β:X×X→X are mappings.Consider condition (I). If we letx=y in (1) and (2), then (I) holds trivially. Consider condition (II). For any (x,0)∈M(X), we have (x,0)⊕(0,0)=(x*0,0)=(x,0). For any (x,1)∈M(X), we have (x,1)=(x,1)⊕(0,0)=(β(x,0),1), which shows that the required condition is β(x,0)=x. Consider condition (III). There are 8 cases to check that condition (III) holds.Case 1 ((x,0),(y,0),(z,0)).
It holds trivially.Case 2 ((x,0),(y,1),(z,0)).
Since((x,0)⊕(y,1))⊕(z,0)=(α(x,y),0)⊕(z,0)=(α(x,y)*z,0) and ((x,0)⊕(z,0))⊕(y,1)=(x*z,0)⊕(y,1)=(α(x*z,y),0), we obtain the requirement that α(x,y)*z=α(x*z,y).Case 3 ((x,0),(y,0),(z,1)).
It is the same as Case2.Case 4 ((x,0),(y,1),(z,1)).
Since((x,0)⊕(y,1))⊕(z,1)=(α(x,y),0)⊕(z,1)=(α(α(x,y),z),0) and ((x,0)⊕(z,1))⊕(y,1)=(α(x,z),0)⊕(y,1)=(α(α(x,z),y),0), we obtain the requirement that α(α(x,y),z)=α(α(x,z),y).Case 5 ((x,1),(y,0),(z,0)).
Since((x,1)⊕(y,0))⊕(z,0)=(β(x,y),1)⊕(z,0)=(β(β(x,y),z),0) and ((x,1)⊕(z,0))⊕(y,0)=(β(x,z),1)⊕(y,0)=(β(β(x,z),y),0), we obtain the requirement that β(β(x,y),z)=β(β(x,z),y).Case 6 ((x,1),(y,0),(z,1)).
Since((x,1)⊕(y,0))⊕(z,1)=(β(x,y),1)⊕(z,1)=(z*β(x,y),0) and ((x,1)⊕(z,1))⊕(y,0)=(z*x,0)⊕(y,0)=((z*x)*y,0), we obtain the requirement that z*β(x,y)=(z*x)*y.Case 7 ((x,1),(y,1),(z,0)).
It is the same as Case6.Case 8 ((x,1),(y,1),(z,1)).
Since((x,1)⊕(y,1))⊕(z,1)=(β(x,y),0)⊕(z,1)=(α(β(x,y),z),0) and ((x,1)⊕(z,1))⊕(y,1)=(α(β(x,z),y),0) by exchanging y with z, we obtain the requirement that α(β(x,y),z)=α(β(x,z),y). If we summarize this discussion, we obtain the following theorem.Theorem 3.1.
Let(X,*,0) be a Q-algebra, and let M(X):=X×{0,1} be a set with a binary operation “⊕” on M(X) with (M1)~(M4). Then the necessary conditions for (M(X),⊕,(0,0)) to be a Q-algebra are the following:  (i)
β(x,0)=x,(ii)
α(α(x,y),z)=α(α(x,z),y),(iii)
α(x,y)*z=α(x*z,y),(iv)
β(β(x,y),z)=β(β(x,z),y),(v)
z*β(x,y)=(z*x)*y,(vi)
α(β(x,y),z)=α(β(x,z),y)for any x,y,z∈X.Remark 3.2.
By condition(M1), if we identify (x,0)≡x for any x∈X, then X is a subalgebra of M(X). By applying Theorem 3.1, we obtain many (mirror) Q-algebras: X⊆M(X)⊆M(M(X))=M2(X)⊆M3(X)⊆M4(X)⊆⋯.Example 3.3.
LetZ be the set of all integers. Then (Z,-,0) is a Q-algebra where “−” is the usual subtraction in Z. If we define mappings α,β:Z×Z→Z by α(x,y)=β(x,y)=x+y for any x,y∈Z, then the mirror algebra (M(Z),⊕,(0,0)) is also a Q-algebra, that is, (x,0)⊕(y,0)=(x-y,0), (x,1)⊕(y,1)=(y-x,0), (x,0)⊕(y,1)=(x+y,0), and (x,1)⊕(y,0)=(x+y,1).Example 3.4.
LetX:={0,1} be a set with the following table:
(3.1)*01000110
Then (X,*,0) is a Q-algebra. Using the same method we obtain its mirror algebra as follows: M(X)={0,α,β,γ} with the following table:
(3.2)⊕0αβγ00000αα0α0βββ00γγβα0
where 0:=(0,0),   α:=(0,1),   β:=(1,0), and γ:=(1,1). It is easy to see that (M(X),⊕,0) is a Q-algebra.Problems
(1) Find necessary conditions forM(X) to be a QS-algebra if (X,*,0) is a QS-algebra.
(2) Given a homomorphismf:X→Y of Q-algebras, construct a homomorphism f̂:M(X)→M(Y) of Q-algebras which is an extension of f.
(3) GivenQ-algebras X,Y, are the mirror algebras M(M(X×Y)) and M(X)×M(Y) isomorphic?



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*Source: 219496-2011-01-03.xml*