# On Generalized Derivations ofBCI-Algebras  and Their Properties

Authors: L. Kamali Ardekani; B. Davvaz
Journal: Journal of Mathematics
 (2014)

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## Abstract

We introduce the concept of(f,g)-derivations of BCI-algebras and we investigate some fundamental properties and establish some results on (f,g)-derivations. Also, we treat to generalization of right derivation and left derivation of BCI-algebras and consider some related properties.


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## Body

## 1. Basic Facts aboutBCI-Algebras

In 1966, Iséki introduced the concept ofBCI-algebra, which is a generalization of theBCK-algebra, as an algebraic counterpart of theBCI-logic [1]; also see [2–6]. In this section, we summarize some basic concepts which will be used throughout the paper. For more details, we refer to the references in [7–12]. Let us recall the definition.ABCI-algebra X is an abstract algebra (X,*,0) of type (2,0), satisfying the following conditions, for all x,y,z∈X:(BCI1)
((x*y)*(x*z))*(z*y)=0;(BCI2)
(x*(x*y))*y=0;(BCI3)
x*x=0;(BCI4)
x*y=0 and y*x=0 imply that x=y. A nonempty subset S of aBCI-algebra X is called asubalgebra of X if x*y∈S, for all x,y∈S. In anyBCI-algebra X, one can define a partial order “≤” by putting x≤y if and only if x*y=0. ABCI-algebra X satisfying 0≤x, for all x∈X, is called aBCK-algebra. In any BCK/BCI-algebra X, the following properties are valid, for all x,y,z∈X: (1)
x*0=x;(2)
(x*y)*z=(x*z)*y;(3)
x≤y implies that x*z≤y*z, z*y≤z*x;(4)
(x*z)*(y*z)≤x*y;(5)
x*(x*(x*y))=x*y;(6)
0*(x*y)=(0*x)*(0*y);(7)
x*0=0 implies that x=0. Let X be aBCI-algebra; the set X+={x∈X:0≤x} is a subalgebra and is called theBCK-part of X. ABCI-algebra X is calledproper if X-X+≠∅. If X+={0}, then X is called a p-semisimple BCI-algebra. In anyBCI-algebra X, the following properties are equivalent, for all x,y∈X: (1) X is p-semisimple, (2) 0*(0*x)=x, (24) x*y=0⇒x=y, and (4) y*(y*x)=x. In any p-semisimpleBCI-algebra X, the following properties are valid, for all x,y,z∈X: (1) (x*z)*(y*z)=x*y, (2) x*(0*y)=y*(0*x), (3) x*y=x*z implies that y=z, and (4) y*x=z*x implies that y=z. For aBCI-algebra X, the set G(X)={x∈X:0*x=x} is called the BCI-Gpart of X. Note that G(X)∩X+={0}.LetX be a p-semisimpleBCI-algebra. We define addition “+” as x+y=x*(0*y), for all x,y∈X. Then, (X,+) is an abelian group with identity 0 and x-y=x*y. Conversely, let (X,+) be an abelian group with identity 0 and x*y=x-y. Then, X is a p-semisimpleBCI-algebra and x+y=x*(0*y), for all x,y∈X (see [2]).LetX be aBCI-algebra; we define the binary operation ∧ as x∧y=y*(y*x), for all x,y∈X. In particular, we denote ax:=x∧0=0*(0*x). An element a∈X is said to be aninitial element (p-atom) of X, if x≤a implies x=a. We denote by Lp(X) the set of all initial elements (p-atoms) of X, indeed Lp(X)={a∈X:x*a=0⇒x=a,∀x∈X}, and we call it thecenter of X. Note that Lp(X)={x∈X:ax=x}, which is the p-semisimple part of X and X is a p-semisimpleBCI-algebra if and only if Lp(X)=X. Let X be aBCI-algebra with Lp(X) as its center and a∈Lp(X). Then, the set V(a)={x∈X:a≤x} is called thebranch of X with respect to a.For anyBCI-algebra X the following results are valid.(1)
Ifx∈V(a) and y∈V(b), then x*y∈V(a*b), for all a,b∈Lp(X).(2)
Ifx≤y, then x,y are contained in the same branch of X.(3)
Ifx,y∈V(a), for some a∈Lp(X), then x*y,y*x∈X+.(4)
Ifa,b∈Lp(X), then a*y=a*b, for all y∈V(b).(5)
0*x∈Lp(X), for all x∈X.(6)
ax∈Lp(X), for all x∈X. Indeed, 0*(0*ax)=ax, for all x∈X which implies that ax*y∈Lp(X), for all y∈X.(7)
G(X)⊆Lp(X).(8)
x*(x*a)=a and a*x∈Lp(X), for all x∈X and a∈Lp(X). A self-map f of aBCI-algebra X (i.e., a mapping of X into itself) is called anendomorphism of X if f(x*y)=f(x)*f(y), for all x,y∈X. Note that f(0)=0. A subset I of aBCI-algebra X is called anideal of X, if it satisfies (1)
0∈I,(2)
x*y∈I and y∈I imply that x∈I, for all x,y∈X. Let f be an endomorphism of aBCI-algebra X and let Lp(X) be its center. Then, according to [13], we have the following results: (1)
f(a)∈Lp(X), for all a∈Lp(X),(2)
fx*fy∈Lp(X) and fx*y=fx*fy, for all x,y∈X, where fx=0*(0*f(x)),(3)
f(a)=0*(0*f(x)), for all x∈V(a). ABCI-algebra X is calledcommutative if x≤y implies x=x∧y=y*(y*x). It is calledbranchwise commutative, if x∧y=y∧x, for all x,y∈V(a) and all a∈Lp(X). Note that aBCI-algebra X is commutative if and only if it is branchwise commutative.


## 2.(f,g)-Derivations ofBCI-Algebras

Recently greater interest has been developed in the derivation ofBCI-algebras, introduced by Jun and Xin [14]. The notion was further explored in the form of f-derivations ofBCI-algebras by Zhan and Liu [15]. We recall the following definition from [14]. Let X be aBCI-algebra. Aleft-right derivation (briefly, (l,r)-derivation) of X is a self-map d of X satisfying the identity
(1)d(x*y)=(d(x)*y)∧(x*d(y)),∀x,y∈X.
If d satisfies the identity d(x*y)=(x*d(y))∧(d(x)*y), for all x,y∈X, then d is called aright-left derivation (briefly, (r,l)-derivation) of X. Moreover, if d is both an (l,r)- and (r,l)-derivation, then it is called aderivation.According to [15], aleft-rightf-derivation (briefly, (l,r)-f-derivation) of X is a self-map d of X satisfying the identity
(2)d(x*y)=(d(x)*f(y))∧(f(x)*d(y)),∀x,y∈X,
where f is an endomorphism of X. If d satisfies the identity d(x*y)=(f(x)*d(y))∧(d(x)*f(y)), for all x,y∈X, then d is called aright-leftf-derivation (briefly, (r,l)-f-derivation) of X. Moreover, if d is both an (l,r)- and (r,l)-f-derivation, then it is called an f-derivation.Now, we introduce the notion of(f,g)-derivation ofBCI-algebras.Definition 1.
LetX be aBCI-algebra. Aleft-right(f,g)-derivation (briefly, (l,r)-(f,g)-derivation) of X is a self-map d of X satisfying the identity
(3)d(x*y)=(d(x)*f(y))∧(g(x)*d(y)),∀x,y∈X,
where f, g are endomorphisms of X. If d satisfies the identity d(x*y)=(f(x)*d(y))∧(d(x)*g(y)), for all x,y∈X, then d is called aright-left(f,g)-derivation (briefly, (r,l)-(f,g)-derivation) of X. Moreover, if d is both an (l,r)- and (r,l)-(f,g)-derivation, it is called an (f,g)-derivation.Example 2.
LetX={0,a,b} and a binary operation * is defined as follows:
(4)*0ab00baaa0bbba0
Then, it forms a p-semisimpleBCI-algebra (see [16]). Define maps d:X→X and f,g:X→X by d(x)=0, for all x∈X, f=0 and
(5)g(x)={0ifx=0bifx=aaifx=b.
Then, f and g are endomorphisms. It is easy to check that d is both (l,r)- and (r,l)-(f,g)-derivation of X. So, d is an (f,g)-derivation.
Now, we considerh:X→X by h=I. Then, h is an endomorphism and d is not an (l,r)-(h,g)-derivation, since d(a*b)=0 but (d(a)*h(b))∧(g(a)*d(b))=a∧b=a. Also, d is not an (r,l)-(h,g)-derivation, since d(a*b)=0 but (h(a)*d(b))∧(d(a)*g(b))=a∧b=a.Example 3.
LetX={0,a,b,c,d,e} be aBCI-algebra with the following Cayley table:
(6)*0abcde000bbbbaa0bbbbbbb0000ccba000ddbaa0aeebaaa0
Define maps d:X→X and f,g:X→X by g(x)=0, for all x∈X:
(7)d(x)={bifx=0,a0otherwise,f(x)={0ifx=0,abotherwise.
Then, f and g are endomorphisms. d is both derivation and f-derivation of X (see Example 2.2 of [15]). Also, it is easy to check that d is both (l,r)- and (r,l)-(f,g)-derivation of X. So, d is an (f,g)-derivation.Example 4.
Let bothX and d:X→X be as in Example 3. Define f,g:X→X by f=0 and g=I. Then, f,g are endomorphisms. d is not a f-derivation of X (see Example 2.3 of [15]). Also, d is not an (l,r)-(f,g)-derivation, since d(b*c)=d(0)=b but (d(b)*f(c))∧(g(b)*d(c))=0∧b=0.  d is not an (r,l)-(f,g)-derivation, since d(b*c)=d(0)=b but (f(b)*d(c))∧(d(b)*g(c))=0∧(0*c)=0∧b=0.Theorem 5.
Letf, g be endomorphisms ofBCI-algebra X and let d be a self-map of X defined by d(x)=fx, for all x∈X. Then, d is an (l,r)-(f,g)-derivation of X.Proof.
Suppose thatx,y∈X. Then, we have
(8)d(x*y)=fx*y=0*(0*f(x*y))=0*(0*(f(x)*f(y)))=(0*(0*f(x)))*(0*(0*f(y)))=fx*(0*(0*f(y)))=(0*(0*fx))*(0*(0*f(y)))=0*(0*(fx*f(y)))=0*(0*((0*(0*f(x)))*f(y)))=0*(0*((0*f(y))*(0*f(x))))=0*(0*(0*f(y*x)))=0*f(y*x)=0*(f(y)*f(x))=(0*f(y))*(0*f(x))=(0*(0*f(x)))*f(y)=fx*f(y).
Since fx∈Lp(X), fx*f(y)∈Lp(X) and fx*f(y)=(fx*f(y))∧(g(x)*fy). Hence, by (8), we get
(9)d(x*y)=(fx*f(y))∧(g(x)*fy)=(d(x)*f(y))∧(g(x)*d(y)).
So, d is an (l,r)-(f,g)-derivation of X.Theorem 6.
Letd be a self-map of aBCI-algebra X. Then, the following properties hold. (1)
Ifd is an (l,r)-(f,g)-derivation of X, then d(x)=d(x)∧g(x), for all x∈X.(2)
Ifd is an (r,l)-(f,g)-derivation of X, then d(x)=f(x)∧d(x), for all x∈X if and only if d(0)=0.Proof.
(1) Suppose thatx∈X. Then, we have
(10)d(x)=d(x*0)=(d(x)*f(0))∧(g(x)*d(0))=d(x)∧(g(x)*d(0))=(g(x)*d(0))*((g(x)*d(0))*d(x))=(g(x)*d(0))*((g(x)*d(x))*d(0))≤g(x)*(g(x)*d(x))=d(x)∧g(x).
It is clear that d(x)∧g(x)≤d(x). So, d(x)=d(x)∧g(x).
(2) Suppose thatd is an (r,l)-(f,g)-derivation of X. If d(x)=f(x)∧d(x), for all x∈X, then d(0)=f(0)∧d(0)=0∧d(0)=0. Conversely, let d(0)=0. Then, d(x)=d(x*0)=(f(x)*d(0))∧(d(x)*g(0))=f(x)∧d(x).Corollary 7.
Letd be an (l,r)-(f,g)-derivation of aBCI-algebra X. Then, d(0)∈Lp(X).Proof.
By Theorem6(1), d(0)=d(0)∧g(0)=0*(0*d(0)). So, d(0)∈Lp(X).Theorem 8.
LetX be a p-semisimpleBCI-algebra and let d be an (l,r)-(f,g)-derivation X. Then, (1)
d(a)=d(0)*(0*f(a))=d(0)+f(a), for all a∈Lp(X);(2)
d(a)∈Lp(X), for all a∈Lp(X);(3)
d(a+b)=d(a)+d(b)-d(0), for all a,b∈Lp(X).Proof.
(1) Leta∈Lp(X). Then, f(a)∈Lp(X). Hence, by Corollary 7, we have
(11)d(a)=d(0*(0*a))=(d(0)*f(0*a))∧(g(0)*d(0*a))=(d(0)*f(0*a))∧(0*d(0*a))=(0*d(0*a))*((d(0)*f(0*a))(0*d(0*a))kkkkkkkkkkkkkkkk*(d(0)*f(0*a)))=(0*d(0*a))*((0*(d(0)*f(0*a)))kkkkkkkkkkkkkkkkk*d(0*a)(0*(d(0)*f(0*a))))=0*(0*(d(0)*f(0*a)))=0*(0*(d(0)*(0*f(a))))=(0*(0*d(0)))*(0*(0*(0*f(a))))=(0*(0*d(0)))*(0*f(a))=d(0)*(0*f(a))=d(0)+f(a).
The proof of (2) and (3) follows directly from (1).Theorem 9.
Letd be an (r,l)-(f,g)-derivation ofBCI-algebra X. Then, (1)
d(a)∈G(X), for all a∈G(X);(2)
d(a)∈Lp(X), for all a∈Lp(X);(3)
d(a)=f(a)*d(0)=f(a)+d(0), for all a∈Lp(X);(4)
d(a+b)=d(a)+d(b)-d(0), for all a,b∈Lp(X).Proof.
(1) Suppose thata∈G(X). Then, we have
(12)d(a)=d(0*a)=(f(0)*d(a))∧(d(0)*g(a))=(0*d(a))∧(d(0)*g(a))=0*d(a).
So, d(a)∈G(X).
(2) Suppose thata∈Lp(X). Then, we have
(13)d(a)=d(0*(0*a))=(f(0)*d(0*a))∧(d(0)*g(0*a))=0*d(0*a)∈Lp(X).
(3) Suppose thata∈Lp(X). Then, we have
(14)d(a)=d(a*0)=(f(a)*d(0))∧(d(a)*g(0))=f(a)*d(0)=f(a)*(0*d(0))=f(a)+d(0).
(4) The proof follows from (3).Definition 10.
An(f,g)-derivation d of aBCI-algebra X is calledregular if d(0)=0. If d(0)≠0, then d is calledirregular.Theorem 11.
LetX be a commutativeBCI-algebra and let d be a regular (r,l)-(f,g)-derivation of X. Then, both f(x) and d(x) belong to the same branch, for all x∈X.Proof.
Suppose thatx∈X. Then, we have
(15)0=d(0)=d(ax*x)=(f(ax)*d(x))∧(d(ax)*g(x))=(fx*d(x))∧(d(ax)*g(x))=fx*d(x).
Hence, fx≤d(x) and so d(x)∈V(fx). Also, we have f(x)∈V(fx), since fx≤f(x). This completes the proof.Theorem 12.
Letd be a regular (r,l)-(f,g)-derivation ofBCI-algebra X. Then, (1)
d(x)≤f(x), for all x∈X;(2)
d(x)*f(y)≤f(x)*d(y), for all x,y∈X.Proof.
(1) By Theorem6 (2), d(x)=f(x)∧d(x)≤f(x), for all x∈X.
(2) By part (1),d(x)*f(y)≤f(x)*f(y)≤f(x)*d(y), for all x∈X.Theorem 13.
Every(r,l)-(f,g)-derivation ((l,r)-(f,g)-derivation) of aBCK-algebra X is regular.Proof.
Suppose thatX is aBCK-algebra and d is an (r,l)-(f,g)-derivation of X. Then, for all x∈X, we have
(16)d(0)=d(0*x)=(f(0)*d(x))∧(d(0)*g(x))=(0*d(x))∧(d(0)*g(x))=0∧(d(0)*g(x))=0.
So, d is regular.
Now, suppose thatd is an (l,r)-(f,g)-derivation of X. Then, d(0)=d(0*x)=(d(0)*f(x))∧(g(0)*d(x))=(d(0)*f(x))∧0=0.Theorem 14.
Letf be an epimorphism and let g be an endomorphism of aBCI-algebra X. Also, let d be an (r,l)-(f,g)-derivation of X and a∈X such that a*d(x)=0 and f(a)*d(x)=0, for all x∈X. Then, d is regular. Moreover, X is aBCK-algebra.Proof.
For allx∈X, we have
(17)0=a*d(a*x)=a*((f(a)*d(x))∧(d(a)*g(x)))=a*(0∧(d(a)*g(x)))=a.
Hence, d(0)=d(0*0)=(f(0)*d(0))∧d(0)=0∧d(0)=0.
By Theorem12(1), 0*f(x)≤0*d(x)=0. So, (0*f(x))*0=0, for all x∈X. Then, 0*f(x)=0, for all x∈X. Therefore, f(X) is aBCK-algebra. This implies that X is aBCK-algebra, since f is an epimorphism.Theorem 15.
Letf be an epimorphism and let g be an endomorphism of aBCI-algebra X. Also, let d be an (f,g)-derivation of X and a∈X such that d(x)*a=0 and d(x)*f(a)=0, for all x∈X. Then, d is regular. Moreover, X is aBCK-algebra.Proof.
For allx∈X, we have
(18)0=d(x*a)*a=((d(x)*f(a))∧(g(x)*d(a)))*a=(0∧(g(x)*d(a)))*a=0*a.
So,
(19)d(0)=d(0*a)=(d(0)*f(a))∧(g(0)*d(a))=0∧(0*d(a))=0.
By Theorem12(1), we get
(20)0*f(x)≤0*d(x)=(d(x)*a)*d(x)=(d(x)*d(x))*a=0*a=0.
Thus, (0*f(x))*0=0, for all x∈X. So, 0*f(x)=0, for all x∈X. Hence, f(X) is aBCK-algebra. This implies that X is aBCK-algebra, since f is an epimorphism.Theorem 16.
LetX be a p-semisimpleBCI-algebra, and let d and d′ be (l,r)-(f,g)-derivations (resp., (r,l)-(f,g)-derivations) of X. Also, let f∘f=f. Then, d∘d′ is also an (l,r)-(f,g)-derivation (resp., (r,l)-(f,g)-derivation) of X.Proof.
Suppose thatd and d′ are (l,r)-(f,g)-derivations of X. Then, for all x,y∈X,
(21)(d∘d′)(x*y)=d(d′(x*y))=d((d′(x)*f(y))∧(g(x)*d′(y)))=d(d′(x)*f(y))=(d(d′(x))*f(f(y)))∧(g(d′(x))*d(f(y)))=(d∘d′)(x)*f(y)=((d∘d′)(x)*f(y))∧(g(x)*(d∘d′)(y)).
So, d∘d′ is an (l,r)-(f,g)-derivation of X. Now, let d, d′be (r,l)-(f,g)-derivations of X. Then, for all x,y∈X, we have
(22)(d∘d′)(x*y)=d((f(x)*d′(y))∧(d′(x)*g(y)))=d(f(x)*d′(y))=(f(f(x))*d(d′(y)))∧(d(f(x))*g(d′(y)))=f(x)*d∘d′(y)=(f(x)*(d∘d′)(y))∧((d∘d′)(x)*g(y)).
So, d∘d′ is an (r,l)-(f,g)-derivation of X.Theorem 17.
LetX be a p-semisimpleBCI-algebra, and let d and d′ be, respectively, (r,l)-(f,g)-derivation and (l,r)-(f,g)-derivation of X such that f∘d=d∘f, d′∘f=f∘d′. Then, d∘d′=d′∘d.Proof.
For allx,y∈X, we have
(23)(d∘d′)(x*y)=d((d′(x)*f(y))∧(g(x)*d′(y)))=d(d′(x)*f(y))=(f(d′(x))*d(f(y)))∧(d(d′(x))*g(f(y)))=f(d′(x))*d(f(y))=f∘d′(x)*d∘f(y).
Also, for all x,y∈X,
(24)(d′∘d)(x*y)=d′((f(x)*d(y))∧(d(x)*g(y)))=d′(f(x)*d(y))=(d′(f(x))*f(d(y)))∧(g(f(x))*d′(d(y)))=d′(f(x))*f(d(y))=d′∘f(x)*f∘d(y)=f∘d′(x)*d∘f(y).
By using (23) and (24), we obtain d∘d′(x*y)=d′∘d(x*y), for all x,y∈X. By putting y=0, we get d∘d′(x)=d′∘d(x), for all x∈X.Definition 18 (see [16]).
LetX be aBCI-algebra and let d, d′ be two self-maps of X. We define d•d′:X→X by (d•d′)(x)=d(x)*d′(x), for all x∈X.Theorem 19.
LetX be a p-semisimpleBCI-algebra and let d, d′ be (f,g)-derivations of X. Then, (1)
(f∘d′)•(d∘f)=(d∘f)•(f∘d′),(2)
(d∘d′)•(f∘f)=(f∘f)•(d∘d′).Proof.
(1) Sinced is an (r,l)-(f,g)-derivation and d′ is an (l,r)-(f,g)-derivation of X, for all x,y∈X, we have
(25)(d∘d′)(x*y)=d((d′(x)*f(y))∧(g(x)*d′(y)))=d(d′(x)*f(y))=(f(d′(x))*d(f(y)))∧(d(d′(x))*g(f(y)))=f∘d′(x)*d∘f(y).
Also, since d is an (l,r)-(f,g)-derivation and d′ is an (r,l)-(f,g)-derivation of X, for all x,y∈X, we have
(26)(d∘d′)(x*y)=d((f(x)*d′(y))∧(d′(x)*g(y)))=d(f(x)*d′(y))=(d(f(x))*f(d′(y)))∧(g(f(x))*d(d′(y)))=d(f(x))*f(d′(y))=d∘f(x)*f∘d′(y).
From (25) and (26), we get f∘d′(x)*d∘f(y)=d∘f(x)*f∘d′(y), for all x,y∈X. By putting x=y, (f∘d′•d∘f)(x)=(d∘f•f∘d′)(x), for all x∈X.
(2) Sinced and d′ are (l,r)-(f,g)-derivations, then for all x,y∈X, we have
(27)(d∘d′)(x*y)=d((d′(x)*f(y))∧(g(x)*d′(y)))=d(d′(x)*f(y))=(d(d′(x))*f(f(y)))∧(g(d′(x))*d(f(y)))=d∘d′(x)*f∘f(y).
On the other hand, since d and d′ are (r,l)-(f,g)-derivations, then for all x,y∈X, we have
(28)(d∘d′)(x*y)=d((f(x)*d′(y))∧(d′(x)*g(y)))=d(f(x)*d′(y))=(f(f(x))*d(d′(y)))∧(d(f(x))*g(d′(y)))=f∘f(x)*d∘d′(y).
By (27) and (28), for all x,y∈X,
(29)d∘d′(x)*f∘f(y)=f∘f(x)*d∘d′(y).
By putting x=y, we get (d∘d′•f∘f)(x)=(f∘f•d∘d′)(x), for all x∈X.Theorem 20.
LetX be a commutative p-semisimpleBCI-algebra and let d, d′ be (f,g)-derivations of X. Then, d=d′ if d(x) and d′(x) lie in the same branch.Proof.
For allx,y∈X, we have
(30)(d∧d′)(x*y)=d(x*y)∧d′(x*y)=d(x*y).
On the other hand,
(31)(d∧d′)(x*y)=d(x*y)∧d′(x*y)=d′(x*y)∧d(x*y)=d′(x*y).
By (30) and (31), d(x*y)=d′(x*y), for all x,y∈X. By putting y=0, we obtain d(x)=d′(x), for all x∈X.We denote byDer(X) the set of all (f,g)-derivations on X.Definition 21 (see [16]).
Letd,d′∈Der(X). The binary operation ∧ on Der(X) is defined as (d∧d′)(x)=d(x)∧d′(x), for all x∈X.Theorem 22.
LetX be a p-semisimpleBCI-algebra and let d, d′ be (l,r)-(f,g)-derivations (resp., (r,l)-(f,g)-derivations) of X. Then, d∧d′ is also an (l,r)-(f,g)-derivation (resp., (r,l)-(f,g)-derivation) of X.Proof.
Suppose thatd, d′ are (l,r)-(f,g)-derivations of X and x,y∈X. Then, we have
(32)(d∧d′)(x*y)=d(x*y)∧d′(x*y)=d(x*y)=(d(x)*f(y))∧(g(x)*d(y))=d(x)*f(y)=(d(x)∧d′(x))*f(y)=(d∧d′)(x)*f(y)=((d∧d′)(x)*f(y))∧(g(x)*(d∧d′)(y)).
So, d∧d′ is an (l,r)-(f,g)-derivation of X.
Suppose thatd, d′ are (r,l)-(f,g)-derivations of X and x,y∈X. Then,
(33)(d∧d′)(x*y)=d(x*y)∧d′(x*y)=d(x*y)=f(x)*d(y)=f(x)*(d(y)∧d′(y))=f(x)*(d∧d′)(y)=(f(x)*(d∧d′)(y))∧((d∧d′)(x)*g(y)).
So, d∧d′ is an (r,l)-(f,g)-derivation of X.Theorem 23.
LetX be a p-semisimpleBCI-algebra. Then, (Der(X),∧) is a semigroup.Proof.
By Theorem22, d∧d′∈Der(X). Suppose that d,d′,d′′∈Der(X). Then,
(34)(d∧(d′∧d′′))(x*y)=d(x*y)∧(d′∧d′′)(x*y)=d(x*y).
Also, we have
(35)((d∧d′)∧d′′)(x*y)=(d∧d′)(x*y)=d(x*y)∧d′(x*y)=d(x*y).
Therefore, d∧(d′∧d′′)=(d∧d′)∧d′′. This completes the proof.At the end of this section, we classify(f,g)-derivations onBCK-algebras of order 3.There is only oneBCK-algebra of order 2. It is called B2-1 and it is as follows:
(36)*0a000aa0
There are only two endomorphisms on B2-1. They are as d1=0 and d2=I. We have Table 1, and so
(37)∧d1d2d1d1d1d2d1d2
There are only threeBCK-algebras of order 3. In the following, we classify (f,g)-derivations on them.Table 1
f
g
(
f
,
g
)-derivations
d
1
d
1
d
1
d
2
d
1
d
2
d
1
d
1
d
2
d
1, d2LetB3-1={0,a,b}. Consider the following table:
(38)*0ab0000aa00bba0
There are only two endomorphisms on B3-1. They are as d1=0 and d2=I. Set
(39)d3={0ifx=0,aaifx=b;
then we have  Table 2, and so
(40)∧d1d2d3d1d1d1d1d2d1d2d3d3d1d3d3
Let B3-2={0,a,b}. Consider the following table:
(41)*0ab0000aa00bbb0
There are only four endomorphisms on B3-2. They are as d1=0, d2=I,
(42)d3={0ifx=0,aaifx=b,d4={0ifx=0,abifx=b.
Set
(43)d5={0ifx=0,baifx=a;
then we have  Table 3, and so
(44)∧d1d2d3d4d5d1d1d1d1d1d1d2d1d2d3d4d5d3d1d3d3d1d1d4d1d4d3d4d1d5d1d5d1d1d5
Let B3-3={0,a,b}. Consider the following table:
(45)*0ab0000aa0abbb0
There are only seven endomorphisms on B3-3. They are as d1=0, d2=I,
(46)d3={0ifx=0,aaifx=b,d4={0ifx=0,abifx=b,d5={0ifx=0,baifx=a,d6={0ifx=0,bbifx=a,d7={0ifx=0aifx=bbifx=a.
Obviously, if we set f=d1=0, then for all g=di, 1≤i≤7, we have only one (f,g)-derivation and it is d1=0. Also, we have Table 4, and so
(47)∧d1d2d3d4d5d6d7d1d1d1d1d1d1d1d1d2d1d2d1d4d5d1d1d3d1d1d3d1d1d1d3d4d1d4d1d4d1d1d1d5d1d5d1d1d5d1d1d6d1d1d1d1d1d6d6d7d1d1d3d1d1d6d7Table 2
f
g
(
f
,
g
)-derivations
d
1
d
1
d
1
d
2
d
1
d
2
d
1
d
1
d
2
d
1, d2, d3Table 3
f
g
(
f
,
g
)-derivations
f
g
(
f
,
g
)-derivations
d
1
d
1
d
1
d
3
d
1
d
1
d
2
d
1
d
2
d
1
d
3
d
1
d
3
d
1, d3
d
4
d
1
d
4
d
1
d
2
d
1
d
1
d
4
d
1
d
1
d
2
d
1, d2, d4, d5
d
2
d
1, d4
d
3
d
1
d
3
d
1, d3
d
4
d
1, d4
d
4
d
1, d4Table 4
f
g
(
f
,
g
)-derivations
f
g
(
f
,
g
)-derivations
f
g
(
f
,
g
)-derivations
d
2
d
1
d
1
d
4
d
1
d
1
d
6
d
1
d
1
d
2
d
1, d2, d4, d5
d
2
d
1, d4
d
2
d
1
d
3
d
1
d
3
d
1
d
3
d
1
d
4
d
1, d4
d
4
d
1, d4
d
4
d
1
d
5
d
1, d5
d
5
d
1
d
5
d
1
d
6
d
1
d
6
d
1
d
6
d
1, d6
d
7
d
1
d
7
d
1
d
7
d
1, d6
d
3
d
1
d
1
d
5
d
1
d
1
d
7
d
1
d
1
d
2
d
1
d
2
d
1, d5
d
2
d
1
d
3
d
1, d3
d
3
d
1
d
3
d
1, d3
d
4
d
1
d
4
d
1
d
4
d
1
d
5
d
1
d
5
d
1, d5
d
5
d
1
d
6
d
1
d
6
d
1
d
6
d
1, d6
d
7
d
1, d3
d
7
d
1
d
7
d
1, d3, d6, d7


## 3.(F,G)-Derivations ofBCI-Algebras

LetX be aBCI-algebra. Aright derivation of X is a self-map D of X satisfying the identity
(48)D(x*y)=(x*D(y))∧(y*D(x)),∀x,y∈X.
If D satisfies the identity D(x*y)=(D(x)*y)∧(D(y)*x), for all x,y∈X, then D is called aleft derivation of X.LetX be aBCI-algebra. ArightF-derivation of X is a self-map D of X satisfying the identity
(49)D(x*y)=(F(x)*D(y))∧(F(y)*D(x)),∀x,y∈X,
where F is an endomorphism of X. If D satisfies the identity D(x*y)=(D(x)*F(y))∧(D(y)*F(x)), for all x,y∈X, then D is called aleftF-derivation of X. Moreover, if D is both right and left F-derivation, then D is called an F-derivation of X; see [13].The notion of left(F,G)-derivation ofBCI-algebras is introduced in [17]. In this section, we introduce the notion of right (F,G)-derivation and give some examples and propositions to explain the theory of left (F,G)-derivation and right (F,G)-derivation inBCI-algebras.Definition 24.
LetX be aBCI-algebra. Aright(F,G)-derivation of X is a self-map D of X satisfying the identity
(50)D(x*y)=(F(x)*D(y))∧(G(y)*D(x)),∀x,y∈X,
where F and G are endomorphisms of X. If D satisfies the identity D(x*y)=(D(x)*F(y))∧(D(y)*G(x)), for all x,y∈X, then D is called aleft(F,G)-derivation of X. Moreover, if D is both right and left (F,G)-derivation, then it is said that D is an (F,G)-derivation of X.Example 25.
LetX, d, f, and g be as in Example 3. Define D=d, F=f, and G=g=0. It is easy to see that D is both right (F,G)-derivation and left (F,G)-derivation. So, D is an (F,G)-derivation.Example 26.
LetX and d be as in Example 3. Define D=d, F=0, and G=I. D is not a right (F,G)-derivation, since D(b*c)=b but (F(b)*D(c))∧(G(c)*D(b))=0∧c=0. Also, D is not a left (F,G)-derivation, since D(b*c)=b but (D(b)*F(c))∧(D(c)*G(b))=0∧b=0.Example 27.
LetX={0,a,b} and the binary operation * is defined as follows:
(51)*0ab0000aa0abbb0
Then, (X,*) is a commutativeBCK-algebra. Define maps D,F,G:X→X by D(x)=0, for all x∈X, F=I and
(52)G(x)={0ifx=0bifx=aaifx=b.
Then, F and G are endomorphisms. It is easy to check that D is a left (F,G)-derivation. D is not a right (F,G)-derivation, since D(b*a)=0 but (F(b)*D(a))∧(G(a)*D(b))=b*0=b.Definition 28.
An(F,G)-derivation D of aBCI-algebra X is calledregular if D(0)=0. If D(0)≠0, then D is calledirregular.Theorem 29.
Every right(F,G)-derivation of aBCK-algebra is regular.Proof.
Suppose thatX is aBCK-algebra and D is a right (F,G)-derivation of X. Then,
(53)D(0)=D(0*0)=(F(0)*D(0))∧(G(0)*D(0))=0.
So, D is regular.Theorem 30.
LetD be a right (F,G)-derivation of aBCI-algebra X. Then, D(x)∈G(X), for all x∈G(X).Proof.
Suppose thatx∈G(X). Then,
(54)D(x)=D(0*x)=(F(0)*D(x))∧(G(x)*D(0))≤F(0)*D(x)=0*D(x)∈Lp(X).
Thus, D(x)=0*D(x). So, D(x)∈G(X).Theorem 31.
LetD be a right (F,G)-derivation of aBCI-algebra X. Then, the following hold: (1)
D(x)∈Lp(X), for all x∈X;(2)
F(y)*(F(y)*D(x))=D(x), for all x,y∈X;(3)
D(x)*F(y)=0*(F(y)*D(x)), for all x,y∈X;(4)
D(x)*F(y)∈Lp(X), for all x,y∈X.Proof.
(1) For allx∈X, we have
(55)D(x)=D(x*0)=(F(x)*D(0))∧(G(0)*D(x))=(0*D(x))*((0*D(x))*(F(x)*D(0)))=(0*D(x))*((0*(F(x)*D(0)))*D(x))≤0*(0*(F(x)*D(0))).
So, D(x)∈Lp(X).
(2), (3), and (4) are clear.Theorem 32.
LetX be a p-semisimpleBCI-algebra. Then (1)
ifD is an (l,r)-(F,G)-derivation of X, then D is a left (F,G)-derivation of X;(2)
ifD is an (r,l)-(F,G)-derivation of X, then D is a right (F,G)-derivation of X.Proof.
(1) Suppose thatx,y∈X and D is an (l,r)-(F,G)-derivation. Then,
(56)D(x*y)=(D(x)*F(y))∧(G(x)*D(y))≤D(x)*F(y).
So, D(x*y)=D(x)*F(y)=(D(x)*F(y))∧(D(y)*G(x)), which implies that D is a left (F,G)-derivation of X.
(2) Suppose thatx,y∈X and D is an (r,l)-(F,G)-derivation. Then,
(57)D(x*y)=(F(x)*D(y))∧(D(x)*G(y))≤F(x)*D(y).
So, D(x*y)=F(x)*D(y)=(F(x)*D(y))∧(G(y)*D(x)), which implies that D is a right (F,G)-derivation of X.Theorem 33.
LetD be a self-map and let V(x0) be any branch of aBCI-algebra X. If, for any x∈V(x0), D(x)=F(x0), then D is a regular left (F,G)-derivation.Proof.
Suppose thatx,y∈X, x∈V(x0), and y∈V(y0); it is not necessary that x0≠y0. Then, x*y∈V(x0*y0). By using hypothesis, we get
(58)D(x*y)=F(x0*y0).
Also, by using hypothesis, we have
(59)(D(x)*F(y))∧(D(y)*G(x))=(F(x0)*F(y))∧(F(y0)*G(x))≤F(x0)*F(y)=F(x0*y)=F(x0*y0).
From x0,y0∈Lp(X), we get F(x0*y0)∈Lp(X). By using (58) and (59), D(x*y)=(D(x)*F(y))∧(D(y)*G(x)). So, D is a left (F,G)-derivation. It is clear that D is regular.Theorem 34.
A self-mapD of aBCI-algebra X defined by D(x)=0*(0*F(x))=Fx, for all x∈X, is a left (F,G)-derivation.Proof.
Suppose thatx,y∈X. We have D(x*y)=Fx*y=Fx*Fy and
(60)(D(x)*F(y))∧(D(y)*G(x))=(Fx*F(y))∧(Fy*G(x))≤Fx*F(y).
So,
(61)(D(x)*F(y))∧(D(y)*G(x))*(Fx*Fy)≤(Fx*F(y))*(Fx*Fy)≤Fy*F(y)=0.
Hence, (D(x)*F(y))∧(D(y)*G(x))≤Fx*Fy. Therefore, (D(x)*F(y))∧(D(y)*G(x))=Fx*Fy=D(x*y), which implies that D is a left (F,G)-derivation.Theorem 35.
LetD be a left (F,G)-derivation of a commutativeBCI-algebra X. Then, x≤y implies that D(x) and D(y) belong to the same branch of X.Proof.
Suppose thatx,y∈X and x≤y. Since X is commutative, y*(y*x)=x. Since D is a left (F,G)-derivation, we have
(62)D(x)=D(y*(y*x))=(D(y)*F(y*x))∧(D(y*x)*G(y))≤D(y)*F(y*x)=D(y)*(F(y)*F(x)).
Since x≤y, 0=F(0)=F(x*y)=F(x)*F(y). So, F(x) and F(y) are contained in the same branch. Hence, F(y)*F(x)∈X+. By (62), we have
(63)D(x)*D(y)≤(D(y)*(F(y)*F(x)))*D(y)=(D(y)*D(y))*(F(y)*F(x))=0*(F(y)*F(x))=0.
Thus, D(x)*D(y)=0. So, D(x)≤D(y). Hence, D(x) and D(y) are contained in the same branch of X.Theorem 36.
LetD be an (F,G)-derivation of aBCI-algebra X. Also, let V(x0) and V(y0) be two arbitrary branches of X such that F(x0)∈V(x0) and, for all y∈V(y0), F(y)∈V(y0). Then, D(x)=y0 implies D(y)∈V(x0).Proof.
Suppose thatx∈V(x0), y∈V(y0), and D(x)=y0. Since D is a left (F,G)-derivation, we have
(64)D(x*y)=(D(x)*F(y))∧(D(y)*G(x))≤D(x)*F(y)=y0*F(y).
On the other hand, y0*F(y)=0, since F(y)∈V(y0). Therefore, D(x*y)=0. Since D is a right (F,G)-derivation, 0=D(x*y)=(F(x)*D(y))∧(G(y)*D(x))≤F(x)*D(y). Hence, 0*F(x)≤(F(x)*D(y))*F(x)=0*D(y). So, 0*(0*D(y))≤0*(0*F(x))=F(x0). Since x0∈Lp(X), F(x0)∈Lp(X). Therefore, F(x0)=0*(0*D(y))≤D(y). Hence, F(x0) and D(y) belong to the same branch of X. Then, D(y)∈V(x0), since F(x0)∈V(x0).



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