# (n

Authors: N. O. Alshehri; Hee Sik Kim; J. Neggers
Journal: The Scientific World Journal
 (2014)

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## Abstract

We define a ranked trigroupoid as a natural followup on the idea of a ranked bigroupoid.  We consider the idea of a derivation on such a trigroupoid as representing a two-step process on a pair of ranked bigroupoids where the mappingd is a self-derivation at each step. Following up on this idea we obtain several results and conclusions of interest.  We also discuss the notion of a couplet (D,d) on X, consisting of a two-step derivation d and its square D=d∘d, for example, whose defining property leads to further observations on the underlying ranked trigroupoids also.


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## Body

## 1. Introduction

The notion of derivations arising in analytic theory is extremely helpful in exploring the structures and properties of algebraic systems. Several authors [1, 2] studied derivations in rings and near rings. Jun and Xin [3] applied the notion of derivation in ring and near ring theory to BCI-algebras. In [4], the concept of derivation for lattices was introduced and some of its properties are investigated. For more details, the reader is referred to [3, 5–7].Iséki and Tanaka introduced two classes of abstract algebras:BCK-algebras and BCI-algebras [8, 9]. 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 [10]. Kim and Neggers [11] introduced the notion of Bin(X) and obtained a semigroup structure. Bell and Kappe [1] studied rings in which derivations satisfy certain algebraic conditions. Alshehri [12] applied the notion of derivations in incline algebras.The present authors [13] introduced the notion of ranked bigroupoids and discussed (X,*,ω)-self-(co)derivations. In addition, they defined rankomorphisms and (X,*,ω)-scalars for ranked bigroupoids and obtained some properties of these as well. Recently, Jun et al. [14] obtained further results on derivations of ranked bigroupoids, and Jun et al. [15] introduced the notion of generalized coderivations in ranked bigroupoids and showed that new generalized coderivations of ranked bigroupoids are obtained by combining a generalized self-coderivation with a rankomorphism.In this paper, we extend the theory of derivations on a ranked bigroupoid to that of a type of derivation on ranked trigroupoids, that is, two-step derivations on ranked trigroupoids(X,*,•,◊) considered as a couple of ranked bigroupoids (X,*,•) and (X,•,◊) with d:X→X such a two-step derivation on (X,*,•,◊) if it is a self-derivation on both (X,*,•) and (X,•,◊). The role of the operation • in this definition is the more interesting one since it acts as the minor operation in (X,*,•) and the major operation in (X,•,◊). From the results obtained below it is clear that it is indeed possible to obtain meaningful insights, especially via the notion of a couplet (D,d) on a ranked trigroupoid consisting of a pair of mappings D, d:X→X satisfying a natural condition (6) stated below which arises in a rather natural way from the context and is seen to be of interest in this study and presumably of any followup as well.


## 2. Preliminaries

And-algebra [10] is a nonempty set X with a constant 0 and a binary operation “*” satisfying the following axioms:(A)
x*x=0,(B)
0*x=0,(C)
x*y=0 and y*x=0 imply x=y for all x,y∈X.ABCK-algebra is a d-algebra X satisfying the following additional axioms:(D)
((x*y)*(x*z))*(z*y)=0,(E)
(x*(x*y))*y=0 for all x,y,z∈X.Given a nonempty setX, we let Bin(X) denote the collection of all groupoids (X,*), where *:X×X→X is a map and where *(x,y) is written in the usual product form. Given elements (X,*) and (X,•) of Bin(X), define a product “□” on these groupoids as follows:
(1)(X,*)□(X,•)=(X,□),
where
(2)x□y=(x*y)•(y*x)
for any x,y∈X. Using that notion, Kim and Neggers proved the following theorem.Theorem 1 (see [11]).
(
 Bin
(
X
)
,
□
) is a semigroup; that is, the operation “□” as defined in general is associative. Furthermore, the left-zero-semigroup is the identity for this operation.Aranked bigroupoid is an algebraic system (X,*,•) where X is a nonempty set and “*” and “•” are binary operations defined on X. We may consider the first binary operation * as themajor operation, and the second binary operation • as theminor operation.Example 2 (see [16]).
AK-algebra is defined as an algebraic system (G,•,⊙) where (G,•) is a group and where x⊙y:=x•y-1. Hence every K-algebra is a ranked bigroupoid.Example 3 (see [13]).
We construct a ranked bigroupoid from anyBCK-algebra. In fact, given a BCK-algebra (X,*,0), if we define a binary operation “∧” on X by x∧y:=x*(x*y) for any x,y∈X, then (X,*,∧) is a ranked bigroupoid.
We introduce the notion of “ranked bigroupoids” to distinguish two bigroupoids(X,*,•) and (X,•,*). Even though (X,*,•)=(X,•,*) in the sense of bigroupoids, the same is not true in the sense of ranked bigroupoids. This is analogous to the situation for sets, where {x,y}={y,x} but 〈x,y〉≠〈y,x〉 in general.
Given an element(X,*)∈Bin(X), (X,*) has a natural associated ranked bigroupoid (X,*,*); that is, the major operation and the minor operation coincide.
Given a ranked bigroupoid(X,*,ω), a map d:X→X is called an (X,*,ω)-self-derivation if for all x,y∈X,
(3)d(x*y)=(d(x)*y)ω(x*d(y)).
In the same setting, a map d:X→X is called an (X,*,ω)-self-coderivation if for all x,y∈X,
(4)d(x*y)=(x*d(y))ω(d(x)*y).
Note that if(X,ω) is a commutative groupoid, then (X,*,ω)-self-derivations are (X,*,ω)-self-coderivations. A map d:X→X is called anabelian-(X,*,ω)-self-derivation if it is both an (X,*,ω)-self-derivation and an (X,*,ω)-self-coderivation.


## 3. Two-Step Derivations and Couplets on Trigroupoids

An algebraic system(X,*,•,◊) is said to be aranked trigroupoid if algebraic systems (X,*,•) and (X,•,◊) are ranked bigroupoids. Atwo-step derivation on a ranked trigroupoid (X,*,•,◊) is a mapping d:X→X such that d is both an (X,*,•)-self-derivation and an (X,•,◊)-self-derivation.Obviously, if one considers rankedn-groupoids (X,*1,*2,…,*n), then one may consider (n-1)-step derivations d:X→X for which one has d as an (X,*k,*k+1)-self-derivation for k=1,…,n-1.In this paper we will mostly be interested in the case of two-step-derivations on ranked trigroupoids and some related pairs of maps(D,d) which we callcouplets. For ranked n-groupoids where n≥4, we obtain triplets, quadruplets, and so forth, as the appropriate generalizations.Letd:X→X be a two-step derivation on a ranked trigroupoid (X,*,•,◊). Then, for any x,y∈X, we have
(5)d2(x*y)=d(d(x*y))=d[(d(x)*y)•(x*d(y))]=[d(d(x)*y)•(x*d(y))]m◊[(d(x)*y)•d(x*d(y))]=[{(d2(x)*y)•(d(x)*d(y))}•(x*d(y))]m◊[(d(x)*y)•[(d(x)*d(y))•(x*d2(y))]].If we letD:=d2, then it follows that
(6)D(x*y)=[{(D(x)*y)•(d(x)*d(y))}•(x*d(y))]m◊[(d(x)*y)•[(d(x)*d(y))•(x*D(y))]].We call(D,d) acouplet on a ranked trigroupoid (X,*,•,◊) if it satisfies condition (6), and D(0)=0 if X contains a constant 0.Example 4.
LetR be the set of all real numbers and let (R,·,+,-) be the ranked trigroupoid where ·,+,and- are usual multiplication, addition, and subtraction, respectively. If we let (D,d) be a couplet on the ranked trigroupoid (R,·,+,-), then
(7)D(x·y)=[D(x)y+d(x)d(y)+xdy]-[d(x)y+d(x)d(y)+xD(y)]=[D(x)y-xD(y)]+[xd(y)-d(x)y].
If we lety:=x in (7), then
(8)D(x2)=[D(x)x-xD(x)]+[xd(x)-d(x)x]=0.
ThusD(x2)=0 for all x∈R, whence D(t)=0 for all t≥0. If we let y:=1 in (7), then D(x)=[D(x)-xD(1)]+[xd(1)-d(x)]=D(x)+xd(1)-d(x). It follows that, for any x∈R,
(9)d(x)=xd(1),
which implies that xd(y)-d(x)y=xyd(1)-xd(1)y=0. Hence, by (7), we have D(xy)=D(x)y-xD(y)+xd(y)-d(x)y=D(x)y-xD(y); that is,
(10)D(x·y)=D(x)y-xD(y).
If we letx<0 and y:=-1 in (9), then we have 0=D(-x)=D(x)·(-1)-xD(-1). Hence D(x)=-xD(-1) for all x<0.In Example4, if d is a two-step derivation on (R,·,+,-), then d(xy)=d(x)y+xd(y) and d(x+y)=(d(x)+y)-(x+d(y)) for any x,y∈R. It follows that d(2x)=d(x+x)=(d(x)+x)-(x+d(x))=0, which proves that d(x)=d(2·(x/2))=0 for all x∈R.Example 5.
LetX:=[0,∞) and let (X,·,+,+) be a ranked trigroupoid where ·, + are usual multiplication and addition, respectively. Define a map φ:X→X satisfying φ(a+b)=φ(a)+φ(b) for all a,b∈X. For k>0, k≠1, λ∈R, we define a map D:X→X by D(x):=φ(a)ka+λ where x=ka and D(0)=0. Then D is a (X,·,+)-self-derivation. In fact, if x=ka, y=kb for some a,b∈X, then
(11)D(x·y)=D(ka·kb)=φ(a+b)ka+b+λ=[φ(a)+φ(b)]ka+b+λ=[φ(a)ka+λ]kb+[φ(b)kb+λ]ka=D(x)y+xD(y).
Assume that (D,d) is a couplet on the ranked trigroupoid (X,·,+,+) for some self-(X,+,+)-derivation d. Then,
(12)D(x·y)=D(x)y+d(x)d(y)+xd(y)+d(x)y+d(x)d(y)+xD(y).
Since D(x·y)=D(x)y+xD(y), we obtain
(13)0=d(x)[y+d(y)]+[x+d(x)]d(y).Proposition 6.
There is no two-step derivation on the ranked trigroupoid(R,·,-,+).Proof.
Assume thatd:R→R is a two-step derivation on (R,·,-,+). Then, for any x,y∈R,
(14)d(x·y)=d(x)y-xd(y),(15)d(x-y)=(d(x)-y)+(x-d(y)).
If we lety∶=-x in (15), then
(16)d(2x)=(d(x)-(-x))+(x-d(-x))=2x+(d(x)-d(-x)),(17)d(-x2)=d(x)·(-x)-x·d(-x)=-x·d(x)-x·d(-x).
If we let x:=0 in (16), then d(0)=d(2·0)=2·0+(d(0)-d(-0))=0. On the other hand, if we let x:=1 and y:=0 in (15), then d(1-0)=(d(1)-0)+(1-d(0)), proving that d(0)=1 is a contradiction. Hence there is no two-step derivation on the ranked trigroupoid (R,·,-,+).Proposition 7.
There is no two-step derivation on the ranked trigroupoid(R,·,+,+).Proof.
Assume thatd:R→R is a two-step derivation on (R,·,+,+). Then, for any x,y∈R,
(18)d(x·y)=d(x)y+xd(y),(19)d(x+y)=(d(x)+y)+(x+d(y)).
If we let y:=0 in (19), then d(x+0)=d(x)+0+x+d(0) and hence d(0)=-x for all x∈R. If we let y:=0,x:=0 in (18), then -x=d(0·0)=d(0)·0+0·d(0)=0, for any x∈R, is a contradiction.Proposition 8.
Let(D,d) be a couplet on a ranked trigroupoid (R,·,-,+). If d(1)≠1/2 and λ:=(D(1)-d(1))/(2d(1)-1), then D(xy)=D(x)y+xD(y)-2λ2xy for all x,y∈R. In particular, if D(1)=d(1), then D is a (R,·,+)-self-derivation.Proof.
If(D,d) is a couplet on a ranked trigroupoid (R,·,-,+), then, for any x,y∈R, we have
(20)D(xy)=(D(x)y-d(x)d(y))-xd(y)+d(x)y-(d(x)d(y)-xD(y))=D(x)y+xD(y)-2d(x)d(y)+(d(x)y-xd(y)).
Since d(1)≠1/2, if we let y:=1 in (20), then D(x)=D(x)+xD(1)-2d(x)d(1)+d(x)-xd(1). It follows that
(21)0=xD(1)-2d(x)d(1)+d(x)-xd(1)=x[D(1)-d(1)]-[2d(1)-1]d(x).
Hence d(x)=(D(1)-d(1))/(2d(1)-1)x=λx for any  x∈R. If we change d(x) into λx for any x∈R, then we obtain
(22)D(xy)=D(x)y+xD(y)-2(λx)(λy)+(λx)y-x(λy)=D(x)y+xD(y)-2λ2xy
for all x,y∈R. In particular, if D(1)=d(1), then λ=0 and hence D(xy)=D(x)y+xD(y); that is, D is a (R,·,+)-self-derivation.


## 4. Frame Algebras andfr(3)-Algebras

A groupoid(X,*,0) is said to be aframe algebra if it satisfies the axioms (A),(B), and(F)
x*0=x,for allx∈X.Example 9.
(
1
) Every BCK-algebra is a frame algebra.
(
2
) Every lattice implication algebra (see [17, 18]) is a frame algebra.
The collection of frame algebras includes the collection ofBCK-algebras and it is a variety. In a frame algebra the element 0 is unique. Indeed, if 01 and 02 are both zeros, then x≠01, 02 yields 01=x*x=02.Proposition 10.
The collection of all frame algebras(X,*i,0) forms a subsemigroup of the semigroup (Bin(X),□).Proof.
Given frame algebras(X,*,0),(X,•,0), if we let (X,□):=(X,*)□(X,•), then x□y=(x*y)•(y*x) for all x,y∈X. It follows that x□x=(x*x)•(x*x)=0•0=0, 0x=(0*x)•(x*0)=0•x=0, and x□0=(x*0)•(0*x)=x•0=x, proving that (X,□) is a frame algebra. This proves the proposition.Given groupoids(X,*),(X,•)∈Bin(X), we define
(23)(X,*)S(X,•)≔{(X,∇)∣∀x,y∈X,mmmmmx∇y∈{x*y,x•y}}.Proposition 11.
Let(X,*,0) and (X,•,0) be frame algebras. If (X,∇)∈(X,*)S(X,•), then (X,∇) is a frame algebra.Proof.
If(X,∇)∈(X,*)S(X,•), then x∇y∈{x*y,x•y} for all x,y∈X. It follows that x∇x∈{x*x,x•x}={0} implies that x∇x=0. Moreover, 0∇x∈{0*x,0•x}={0} shows that 0∇x=0, and x∇0∈{x*0,x•0}={x} shows that x∇0=x, proving that (X,∇) is a frame algebra.A ranked trigroupoid(X,*,•,◊) is called anfr(3)-algebra if(G)
(X,*,01),(X,•,02),(X,◊,03) are frame algebras,(H)
01=02=03.Theorem 12.
Let(X,*,•,◊) be anfr(3)-algebra. If d:X→X is a two-step derivation on X, then (i)
d(0)=0,(ii)
d(x)=d(x)•x=d(x)◊x for all x∈X.Proof.
(i) Ifd is a two-step derivation on (X,*,•,◊), then for any x,y∈x, we have
(24)d(x*y)=(d(x)*y)•(x*d(y)),(25)d(x•y)=(d(x)•y)◊(x•d(y)).
It follows that d(0)=d(0*y)=(d(0)*y)•(0*d(y))=(d(0)*y)•0=d(0)*y; that is, d(0)=d(0)*y, for all y∈X. If we let y:=d(0), then by applying (A) we obtain d(0)=d(0)*d(0)=0.
(ii) Givenx∈X, we have d(x)=d(x*0)=(d(x)*0)•(x*d(0))=d(x)•x and d(x)=d(x•0)=(d(x)•0)◊(x•d(0))=d(x)◊x.Given a two-step derivationd on a trigroupoid (X,*,•,◊), we denote its kernel by Ker⁡(d):={x∈X∣d(x)=0}.Proposition 13.
Let(X,*,•,◊) be an fr(3)-algebra. If d:X→X is a two-step derivation on X, then (i)
x*d(x), x•d(x)∈Ker(d),(ii)
x∈Ker(d) implies that x*y, x•y∈Ker(d),(iii)
Ker(d)⊆Ker(d2),for all x,y∈X.Proof.
(i) If we lety:=d(x) in (24) and (25), respectively, then d(x*d(x))=(d(x)*d(x))•(x*d(d(x)))=0•(x*d2(x))=0 an d(x•d(x))=(d(x)•d(x))◊(x•d(d(x)))=0◊(x•d2(x))=0, proving that x*d(x), x•d(x)∈Ker⁡(d) for any x∈X.
(ii) Ifx∈Ker⁡(d), then d(x*y)=(d(x)*y)•(x*d(y))=(0*y)•(x*d(y))=0 and d(x•y)=(d(x)•y)◊(x•d(y))=0 for any y∈X, proving that x*y,   x•y∈Ker⁡(d).
(iii) Ifx∈Ker⁡(d), then d2(x)=d(d(x))=d(0)=0 by Theorem 12, which shows that x∈Ker⁡(d2).Proposition 14.
Let(X,*,•,◊) be an fr(3)-algebra. If d:X→X is a two-step derivation on X, then
(26)x∈Ker(d2)impliesx*y∈Ker(d2)
for all y∈X.Proof.
Ifx∈Ker⁡(d2), then d(d(x))=0 and hence
(27)d2(x*y)=d(d(x*y))=d[(d(x)*y)•(x*d(y))]=[d(d(x)*y)•(x*d(y))]◊[(d(x)*y)•d(x*d(y))]=[{(d(d(x)*y))•(d(x)*d(y))}•(x*d(y))]◊[(d(x)*y)•d(x*d(y))]=0◊[(d(x)*y)•d(x*d(y))]=0,
for all y∈X, which proves that x*y∈Ker⁡(d2).Note thatd2(x•y) may not be computable unless the behavior of d(u◊v) is specified, since d2(x•y)=d((d(x)•y)◊(x•d(y)))=d(u◊v) for some u,v∈X.Proposition 15.
Let(X,*,•,◊) be an fr(3)-algebra. If (D,d) is a couplet of (X,*,•,◊), then Ker(D)*X⊆Ker(D).Proof.
If(D,d) is a couplet of (X,*,•,◊) and if x∈Ker⁡(D), then D(x)*y=0*y=0 for any y∈X. It follows from (6) that D(x*y)=0◊[(d(x)*y)•[(d(x)*d(y))•(x*D(y))]]=0, proving that x*y∈Ker⁡(D).Let(X,≤) be a poset with minimal element 0. Define a binary operation “*” on X by
(28)x*y={0,ifx≤y,x,otherwise.
Then (X,*,0) is a BCK-algebra, called astandardBCK-algebra inherited from the poset (X,≤).Proposition 16.
Let(X,*,0) be a standard BCK-algebra. Let (X,*,•,◊) be anfr(3)-algebra and let (D,d) be a couplet of (X,*,•,◊). If D(x*y)≠0 for some x,y∈X, then D(x*y)=D(x)*y=D(x) and x*y=x.Proof.
LetD(x*y)≠0. We claim that D(x)*y≠0. Suppose that D(x)*y=0. Since (X,*,•,◊) is afr(3)-algebra, by applying (6), we obtain that D(x*y)=0 is a contradiction. Since (X,*,0) is a standard BCK-algebra, we obtain D(x)*y=D(x). We claim that x*y=x. If x*y=0, then D(x*y)=D(0)=0 is a contradiction. It follows that D(x*y)=D(x)=D(x)*y, proving the proposition.


## 5. Classification of Linear Ranked Trigroupoids

LetX:=R be the real field and let (X,*,•,◊) be a ranked trigroupoid, where (X,*),(X,•),and(X,◊) are linear groupoids; that is, x*y:=A+Bx+Cy, x•y:=a+bx+cy, and x◊y:=α+βx+γy for any x,y∈X, where A,B,C,a,b,c,α,β,γ∈X (fixed).Thus, ifd:X→X is a two-step derivation on (X,*,•,◊), then for any x,y∈X, we have
(29)d(x*y)=(d(x)*d(y))•(x*d(y))=(A+Bd(x)+Cy)•(A+Bx+Cd(y))=a+(b+c)A+B(cx+bd(x))+C(by+cd(y))
and also in a similar manner we obtain
(30)d(x•y)=α+(β+γ)a+b(γx+βd(x))+c(βy+γd(y)).Ifd(x)=0 for all x∈X, then (29) and (30) become
(31)a+(b+c)A+Bcx+Cby=0,α+(β+γ)a+bγx+cβy=0
for any x,y∈X. If we let x:=0, y:=0 in (31), then we obtain a+(b+c)A=0,  α+(β+γ)a=0. Hence, (31) reduces to
(32)Bcx+Cby=0,(33)bγx+cβy=0
for all x,y∈X. If we let x:=1,  y:=0 and let x:=0,  y:=1 in (32) and (33), respectively, then we obtain
(34)Bc=0,Cb=0,bγ=0,cβ=0.
From this information, we obtain the following propositions.Proposition 17.
LetX:=R be the real field and let (X,*,•,◊) be a ranked trigroupoid, where (X,*),(X,•),(X,◊) are linear groupoids; that is, x*y:=A+Bx+Cy, x•y:=a+bx+cy, and x◊y:=α+βx+γy for any x,y∈X, where A,B,C,a,b,c,α,β,γ∈X (fixed). Let d:X→X be a two-step derivation such that d(x)=0 for all x∈X. (1)
Ifbc≠0 and b+c≠0, then x*y=-a/(b+c), x•y:=a+bx+cy, x◊y=0;(2)
Ifbc≠0 and b+c=0, then x*y=A,x•y=b(x-y), x◊y=0.Proof.
(i) Ifbc≠0, then it follows from (34) that we obtain B=C=0, β=γ=α=0. If b+c≠0, then we have x*y=-a/(b+c) and x◊y=0.
(ii) Ifbc≠0 and b+c=0, then 0=a+(b+c)A=a and A is arbitrary, and hence we obtain the result.Proposition 18.
LetX:=R be the real field and let (X,*,•,◊) be a ranked trigroupoid, where (X,*),(X,•),(X,◊) are linear groupoids; that is, x*y:=A+Bx+Cy, x•y:=a+bx+cy, and x◊y:=α+βx+γy for any x,y∈X, where A,B,C,a,b,c,α,β,γ∈X (fixed). Let d:X→X be a two-step derivation such that d(x)=0 for all x∈X:(i)
ifb=c=0 and b+c≠0, then x*y=A+Bx+Cy, x•y:=0, x◊y=βx+γy;(ii)
ifb≠0, c=β=0, then x*y=-a/b+Bx, x•y=a+bx, x◊y=0;(iii)
ifb≠0, c=0, β≠0, then x*y=α/bβ+Bx, x•y=-α/β+bx, x◊y=α+βy;(iv)
ifb=0, c≠0, γ=0, then x*y=-a/c+Cy, x•y=a+cy, x◊y=0;(v)
ifb=0, c≠0, γ≠0, then x*y=α/cγ+Cy, x•y=-α/γ+cy, x◊y=α+γy.Proof.
The proof is similar to Proposition17 and we omit it.In Propositions17 and 18, we observe that there are 6 different types of linear ranked trigroupoids in the special case of d(x)=0 for all x∈X, and most of them are classified by the properties of b,c in x•y=a+bx+cy.


## 6. Conclusion

The notion of two-step derivations is a generalization of derivations. This leads to the study of trigroupoids, and we explore some relations with several algebras, for example,BCK-algebras, frame algebras, and so forth. The classification of linear ranked trigroupoids then explains a number of concrete algebraic structures with derivations.



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*Source: 726470-2014-07-08.xml*