On Generalized Derivations of BCI-Algebras and Their Properties

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.

1. Basic Facts about BCI-Algebras

In 1966, Iséki introduced the concept of BCI-algebra, which is a generalization of the BCK-algebra, as an algebraic counterpart of the BCI-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.

A BCI-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 a BCI-algebra 𝑋 is called a subalgebra of 𝑋 if 𝑥∗𝑦∈𝑆, for all 𝑥,𝑦∈𝑆. In any BCI-algebra 𝑋, one can define a partial order “≤” by putting 𝑥≤𝑦 if and only if 𝑥∗𝑦=0. A BCI-algebra 𝑋 satisfying 0≤𝑥, for all 𝑥∈𝑋, is called a BCK-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 a BCI-algebra; the set 𝑋₊={𝑥∈𝑋∶0≤𝑥} is a subalgebra and is called the BCK-part of 𝑋. A BCI-algebra 𝑋 is called proper if 𝑋−𝑋₊≠∅. If 𝑋₊={0}, then 𝑋 is called a 𝑝-semisimple BCI-algebra. In any BCI-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-semisimple BCI-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 a BCI-algebra 𝑋, the set 𝐺(𝑋)={𝑥∈𝑋∶0∗𝑥=𝑥} is called the 𝐵𝐶𝐼−𝐺 part of 𝑋. Note that 𝐺(𝑋)∩𝑋₊={0}.

Let 𝑋 be a 𝑝-semisimple BCI-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-semisimple BCI-algebra and 𝑥+𝑦=𝑥∗(0∗𝑦), for all 𝑥,𝑦∈𝑋 (see [2]).

Let 𝑋 be a BCI-algebra; we define the binary operation ∧ as 𝑥∧𝑦=𝑦∗(𝑦∗𝑥), for all 𝑥,𝑦∈𝑋. In particular, we denote 𝑎_𝑥∶=𝑥∧0=0∗(0∗𝑥). An element 𝑎∈𝑋 is said to be an initial element (𝑝-atom) of 𝑋, if 𝑥≤𝑎 implies 𝑥=𝑎. We denote by 𝐿_𝑝(𝑋) the set of all initial elements (𝑝-atoms) of 𝑋, indeed 𝐿_𝑝(𝑋)={𝑎∈𝑋∶𝑥∗𝑎=0⇒𝑥=𝑎,∀𝑥∈𝑋}, and we call it the center of 𝑋. Note that 𝐿_𝑝(𝑋)={𝑥∈𝑋∶𝑎_𝑥=𝑥}, which is the 𝑝-semisimple part of 𝑋 and 𝑋 is a 𝑝-semisimple BCI-algebra if and only if 𝐿_𝑝(𝑋)=𝑋. Let 𝑋 be a BCI-algebra with 𝐿_𝑝(𝑋) as its center and 𝑎∈𝐿_𝑝(𝑋). Then, the set 𝑉(𝑎)={𝑥∈𝑋∶𝑎≤𝑥} is called the branch of 𝑋 with respect to 𝑎.

For any BCI-algebra 𝑋 the following results are valid.(1)If 𝑥∈𝑉(𝑎) and 𝑦∈𝑉(𝑏), then 𝑥∗𝑦∈𝑉(𝑎∗𝑏), for all 𝑎,𝑏∈𝐿_𝑝(𝑋).(2)If 𝑥≤𝑦, then 𝑥,𝑦 are contained in the same branch of 𝑋.(3)If 𝑥,𝑦∈𝑉(𝑎), for some 𝑎∈𝐿_𝑝(𝑋), then 𝑥∗𝑦,𝑦∗𝑥∈𝑋₊.(4)If 𝑎,𝑏∈𝐿_𝑝(𝑋), then 𝑎∗𝑦=𝑎∗𝑏, for all 𝑦∈𝑉(𝑏).(5)0∗𝑥∈𝐿_𝑝(𝑋), for all 𝑥∈𝑋.(6)𝑎_𝑥∈𝐿_𝑝(𝑋), for all 𝑥∈𝑋. Indeed, 0∗(0∗𝑎_𝑥)=𝑎_𝑥, for all 𝑥∈𝑋 which implies that 𝑎_𝑥∗𝑦∈𝐿_𝑝(𝑋), for all 𝑦∈𝑋.(7)𝐺(𝑋)⊆𝐿_𝑝(𝑋).(8)𝑥∗(𝑥∗𝑎)=𝑎 and 𝑎∗𝑥∈𝐿_𝑝(𝑋), for all 𝑥∈𝑋 and 𝑎∈𝐿_𝑝(𝑋). A self-map 𝑓 of a BCI-algebra 𝑋 (i.e., a mapping of 𝑋 into itself) is called an endomorphism of 𝑋 if 𝑓(𝑥∗𝑦)=𝑓(𝑥)∗𝑓(𝑦), for all 𝑥,𝑦∈𝑋. Note that 𝑓(0)=0. A subset 𝐼 of a BCI-algebra 𝑋 is called an ideal 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 a BCI-algebra 𝑋 and let 𝐿_𝑝(𝑋) be its center. Then, according to [13], we have the following results: (1)𝑓(𝑎)∈𝐿_𝑝(𝑋), for all 𝑎∈𝐿_𝑝(𝑋),(2)𝑓_𝑥∗𝑓_𝑦∈𝐿_𝑝(𝑋) and 𝑓_𝑥∗𝑦=𝑓_𝑥∗𝑓_𝑦, for all 𝑥,𝑦∈𝑋, where 𝑓_𝑥=0∗(0∗𝑓(𝑥)),(3)𝑓(𝑎)=0∗(0∗𝑓(𝑥)), for all 𝑥∈𝑉(𝑎). A BCI-algebra 𝑋 is called commutative if 𝑥≤𝑦 implies 𝑥=𝑥∧𝑦=𝑦∗(𝑦∗𝑥). It is called branchwise commutative, if 𝑥∧𝑦=𝑦∧𝑥, for all 𝑥,𝑦∈𝑉(𝑎) and all 𝑎∈𝐿_𝑝(𝑋). Note that a BCI-algebra 𝑋 is commutative if and only if it is branchwise commutative.

2. (𝑓,𝑔)-Derivations of BCI-Algebras

Recently greater interest has been developed in the derivation of BCI-algebras, introduced by Jun and Xin [14]. The notion was further explored in the form of 𝑓-derivations of BCI-algebras by Zhan and Liu [15]. We recall the following definition from [14]. Let 𝑋 be a BCI-algebra. A left-right derivation (briefly, (𝑙,𝑟)-derivation) of 𝑋 is a self-map 𝑑 of 𝑋 satisfying the identity 𝑑𝑥∗𝑦=𝑑(𝑥)∗𝑦∧𝑥∗𝑑𝑦,∀𝑥,𝑦∈𝑋. (1) If 𝑑 satisfies the identity 𝑑(𝑥∗𝑦)=(𝑥∗𝑑(𝑦))∧(𝑑(𝑥)∗𝑦), for all 𝑥,𝑦∈𝑋, then 𝑑 is called a right-left derivation (briefly, (𝑟,𝑙)-derivation) of 𝑋. Moreover, if 𝑑 is both an (𝑙,𝑟)- and (𝑟,𝑙)-derivation, then it is called a derivation.

According to [15], a left-right 𝑓-derivation (briefly, (𝑙,𝑟)−𝑓-derivation) of 𝑋 is a self-map 𝑑 of 𝑋 satisfying the identity 𝑑=𝑑(𝑥)𝑦∧𝑓(𝑥)𝑦𝑥∗𝑦∗𝑓∗𝑑,∀𝑥,𝑦∈𝑋, (2) 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 a right-left 𝑓-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 of BCI-algebras.

Definition 1. Let 𝑋 be a BCI-algebra. A left-right (𝑓,𝑔)-derivation (briefly, (𝑙,𝑟)−(𝑓,𝑔)-derivation) of 𝑋 is a self-map 𝑑 of 𝑋 satisfying the identity 𝑑=𝑑(𝑥)𝑦∧𝑔(𝑥)𝑦𝑥∗𝑦∗𝑓∗𝑑,∀𝑥,𝑦∈𝑋, (3) 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 a right-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. Let 𝑋={0,𝑎,𝑏} and a binary operation ∗ is defined as follows: ∗0𝑎𝑏0𝑎0𝑏𝑎𝑏𝑎0𝑏𝑏𝑎0 (4) Then, it forms a 𝑝-semisimple BCI-algebra (see [16]). Define maps 𝑑∶𝑋→𝑋 and 𝑓,𝑔∶𝑋→𝑋 by 𝑑(𝑥)=0, for all 𝑥∈𝑋, 𝑓=0 and 𝑔(𝑥)=0if𝑥=0𝑏if𝑥=𝑎𝑎if𝑥=𝑏. (5) 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 consider ℎ∶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. Let 𝑋={0,𝑎,𝑏,𝑐,𝑑,𝑒} be a BCI-algebra with the following Cayley table: ∗0𝑎𝑏𝑐𝑑𝑒0𝑎00𝑏𝑏𝑏𝑏𝑏𝑎0𝑏𝑏𝑏𝑏𝑐𝑏𝑏0000𝑑𝑐𝑏𝑎000𝑒𝑑𝑏𝑎𝑎0𝑎𝑒𝑏𝑎𝑎𝑎0 (6) Define maps 𝑑∶𝑋→𝑋 and 𝑓,𝑔∶𝑋→𝑋 by 𝑔(𝑥)=0, for all 𝑥∈𝑋: 𝑑(𝑥)=𝑓(𝑥)=𝑏if𝑥=0,𝑎0otherwise,0if𝑥=0,𝑎𝑏otherwise. (7) 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 both 𝑋 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. Let 𝑓, 𝑔 be endomorphisms of BCI-algebra 𝑋 and let 𝑑 be a self-map of 𝑋 defined by 𝑑(𝑥)=𝑓_𝑥, for all 𝑥∈𝑋. Then, 𝑑 is an (𝑙,𝑟)-(𝑓,𝑔)-derivation of 𝑋.

Proof. Suppose that 𝑥,𝑦∈𝑋. Then, we have 𝑑𝑥∗𝑦=𝑓_𝑥∗𝑦𝑓(𝑥)𝑦=(𝑥)∗𝑦=0∗0∗𝑓𝑥∗𝑦=0∗0∗∗𝑓0∗0∗𝑓0∗0∗𝑓=𝑓_𝑥∗𝑦=0∗0∗𝑓0∗0∗𝑓_𝑥∗𝑦𝑓0∗0∗𝑓=0∗0∗_𝑥𝑦(𝑥)𝑦𝑦∗(𝑥)𝑓𝑦(𝑥)=𝑦∗(𝑥)=(𝑥𝑦∗𝑓=0∗0∗0∗0∗𝑓∗𝑓=0∗0∗0∗𝑓0∗𝑓=0∗0∗0∗𝑓𝑦∗𝑥=0∗𝑓𝑦∗𝑥=0∗∗𝑓0∗𝑓0∗𝑓0∗0∗𝑓)∗𝑓=𝑓_𝑥∗𝑓𝑦. (8) Since 𝑓_𝑥∈𝐿_𝑝(𝑋), 𝑓_𝑥∗𝑓(𝑦)∈𝐿_𝑝(𝑋) and 𝑓_𝑥∗𝑓(𝑦)=(𝑓_𝑥∗𝑓(𝑦))∧(𝑔(𝑥)∗𝑓_𝑦). Hence, by (8), we get 𝑑𝑥∗𝑦=𝑓_𝑥∗𝑓𝑦∧𝑔(𝑥)∗𝑓_𝑦=𝑑(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑𝑦. (9) So, 𝑑 is an (𝑙,𝑟)-(𝑓,𝑔)-derivation of 𝑋.

Theorem 6. Let 𝑑 be a self-map of a BCI-algebra 𝑋. Then, the following properties hold. (1)If 𝑑 is an (𝑙,𝑟)-(𝑓,𝑔)-derivation of 𝑋, then 𝑑(𝑥)=𝑑(𝑥)∧𝑔(𝑥), for all 𝑥∈𝑋.(2)If 𝑑 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 that 𝑥∈𝑋. Then, we have 𝑑(𝑥)()=𝑑(𝑥)(0)∧𝑔(𝑥)(0)(𝑥)∧𝑔(𝑥)(0)=𝑔(𝑥)(0)∗𝑔(𝑥)(0)(𝑥)=𝑔(𝑥)(0)∗𝑔(𝑥)(𝑥)(0)(𝑥)∗𝑔(𝑥)(𝑥)(𝑥)(𝑥).=𝑑𝑥∗0∗𝑓∗𝑑=𝑑∗𝑑∗𝑑∗𝑑∗𝑑∗𝑑∗𝑑∗𝑑≤𝑔∗𝑑=𝑑∧𝑔 (10) It is clear that 𝑑(𝑥)∧𝑔(𝑥)≤𝑑(𝑥). So, 𝑑(𝑥)=𝑑(𝑥)∧𝑔(𝑥).
(2) Suppose that 𝑑 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. Let 𝑑 be an (𝑙,𝑟)-(𝑓,𝑔)-derivation of a BCI-algebra 𝑋. Then, 𝑑(0)∈𝐿_𝑝(𝑋).

Proof. By Theorem 6(1), 𝑑(0)=𝑑(0)∧𝑔(0)=0∗(0∗𝑑(0)). So, 𝑑(0)∈𝐿_𝑝(𝑋).

Theorem 8. Let 𝑋 be a 𝑝-semisimple BCI-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∈𝐿_𝑝(𝑋);(2)𝑑(𝑎)∈𝐿_𝑝(𝑋), for all 𝑎∈𝐿_𝑝(𝑋);(3)𝑑(𝑎+𝑏)=𝑑(𝑎)+𝑑(𝑏)−𝑑(0), for all 𝑎,𝑏∈𝐿_𝑝(𝑋).

Proof. (1) Let 𝑎∈𝐿_𝑝(𝑋). Then, 𝑓(𝑎)∈𝐿_𝑝(𝑋). Hence, by Corollary 7, we have 𝑑(𝑎)((=𝑑(0)()∧𝑔(0)()=𝑑(0)()∧((=((∗((∗𝑑(0)()=((∗𝑑(0)()(𝑑(0)()𝑑(0)∗(𝑎)=(((0∗(𝑎=(((0∗(𝑎)(0)∗(𝑎)(0)(𝑎).=𝑑0∗0∗𝑎))∗𝑓0∗𝑎∗𝑑0∗𝑎∗𝑓0∗𝑎0∗𝑑0∗𝑎))0∗𝑑0∗𝑎))0∗𝑑0∗𝑎))∗𝑓0∗𝑎0∗𝑑0∗𝑎))0∗∗𝑓0∗𝑎∗𝑑0∗𝑎)=0∗0∗∗𝑓0∗𝑎=0∗0∗0∗𝑓0∗0∗𝑑)))0∗0∗0∗𝑓)0∗0∗𝑑)))0∗𝑓=𝑑0∗𝑓=𝑑+𝑓 (11) The proof of (2) and (3) follows directly from (1).

Theorem 9. Let 𝑑 be an (𝑟,𝑙)-(𝑓,𝑔)-derivation of BCI-algebra 𝑋. Then, (1)𝑑(𝑎)∈𝐺(𝑋), for all 𝑎∈𝐺(𝑋);(2)𝑑(𝑎)∈𝐿_𝑝(𝑋), for all 𝑎∈𝐿_𝑝(𝑋);(3)𝑑(𝑎)=𝑓(𝑎)∗𝑑(0)=𝑓(𝑎)+𝑑(0), for all 𝑎∈𝐿_𝑝(𝑋);(4)𝑑(𝑎+𝑏)=𝑑(𝑎)+𝑑(𝑏)−𝑑(0), for all 𝑎,𝑏∈𝐿_𝑝(𝑋).

Proof. (1) Suppose that 𝑎∈𝐺(𝑋). Then, we have 𝑑(𝑎)()=𝑓(0)(𝑎)∧𝑑(0)(𝑎)=((𝑎∧𝑑(0)(𝑎)(𝑎).=𝑑0∗𝑎∗𝑑∗𝑔0∗𝑑))∗𝑔=0∗𝑑 (12) So, 𝑑(𝑎)∈𝐺(𝑋).
(2) Suppose that 𝑎∈𝐿_𝑝(𝑋). Then, we have 𝑑(𝑎)((=𝑓(0)()∧𝑑(0)()()=𝑑0∗0∗𝑎))∗𝑑0∗𝑎∗𝑔0∗𝑎=0∗𝑑0∗𝑎∈𝐿_𝑝(𝑋). (13)
(3) Suppose that 𝑎∈𝐿_𝑝(𝑋). Then, we have 𝑑(𝑎)()=𝑓(𝑎)(0)∧𝑑(𝑎)(0)(𝑎)(0)(𝑎)∗((0(𝑎)(0).=𝑑𝑎∗0∗𝑑∗𝑔=𝑓∗𝑑=𝑓0∗𝑑))=𝑓+𝑑 (14)
(4) The proof follows from (3).

Definition 10. An (𝑓,𝑔)-derivation 𝑑 of a BCI-algebra 𝑋 is called regular if 𝑑(0)=0. If 𝑑(0)≠0, then 𝑑 is called irregular.

Theorem 11. Let 𝑋 be a commutative BCI-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 that 𝑥∈𝑋. Then, we have 0=𝑑(0)=𝑑𝑎_𝑥∗𝑥=𝑓𝑎_𝑥∗𝑑(𝑥)∧𝑑𝑎_𝑥(𝑥)=𝑓∗𝑔_𝑥∗𝑑(𝑥)∧𝑑𝑎_𝑥∗𝑔(𝑥)=𝑓_𝑥∗𝑑(𝑥). (15) Hence, 𝑓_𝑥≤𝑑(𝑥) and so 𝑑(𝑥)∈𝑉(𝑓_𝑥). Also, we have 𝑓(𝑥)∈𝑉(𝑓_𝑥), since 𝑓_𝑥≤𝑓(𝑥). This completes the proof.

Theorem 12. Let 𝑑 be a regular (𝑟,𝑙)-(𝑓,𝑔)-derivation of BCI-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 Theorem 6 (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 a BCK-algebra 𝑋 is regular.

Proof. Suppose that 𝑋 is a BCK-algebra and 𝑑 is an (𝑟,𝑙)-(𝑓,𝑔)-derivation of 𝑋. Then, for all 𝑥∈𝑋, we have 𝑑(0)()=𝑓(0)(𝑥)∧𝑑(0)(𝑥)=((𝑥∧𝑑(0)(𝑥)𝑑(0)(𝑥)=𝑑0∗𝑥∗𝑑∗𝑔0∗𝑑))∗𝑔=0∧∗𝑔=0. (16) So, 𝑑 is regular.
Now, suppose that 𝑑 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. Let 𝑓 be an epimorphism and let 𝑔 be an endomorphism of a BCI-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 a BCK-algebra.

Proof. For all 𝑥∈𝑋, we have ()𝑓(𝑎)(𝑥)∧𝑑(𝑎)(𝑥)𝑑(𝑎)(𝑥)0=𝑎∗𝑑𝑎∗𝑥=𝑎∗∗𝑑∗𝑔=𝑎∗0∧∗𝑔=𝑎. (17) Hence, 𝑑(0)=𝑑(0∗0)=(𝑓(0)∗𝑑(0))∧𝑑(0)=0∧𝑑(0)=0.
By Theorem 12(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 a BCK-algebra. This implies that 𝑋 is a BCK-algebra, since 𝑓 is an epimorphism.

Theorem 15. Let 𝑓 be an epimorphism and let 𝑔 be an endomorphism of a BCI-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 a BCK-algebra.

Proof. For all 𝑥∈𝑋, we have ()𝑑(𝑥)(𝑎)∧𝑔(𝑥)(𝑎)=𝑔(𝑥)(𝑎)0=𝑑𝑥∗𝑎∗𝑎=∗𝑓∗𝑑∗𝑎0∧∗𝑑∗𝑎=0∗𝑎. (18) So, 𝑑(0)()=𝑑(0)(𝑎)∧𝑔(0)(𝑎)((𝑎=𝑑0∗𝑎∗𝑓∗𝑑=0∧0∗𝑑))=0. (19)
By Theorem 12(1), we get (𝑥)(𝑥)=(𝑑(𝑥))(𝑥)=(𝑑(𝑥)(𝑥0∗𝑓≤0∗𝑑∗𝑎∗𝑑∗𝑑))∗𝑎=0∗𝑎=0. (20) Thus, (0∗𝑓(𝑥))∗0=0, for all 𝑥∈𝑋. So, 0∗𝑓(𝑥)=0, for all 𝑥∈𝑋. Hence, 𝑓(𝑋) is a BCK-algebra. This implies that 𝑋 is a BCK-algebra, since 𝑓 is an epimorphism.

Theorem 16. Let 𝑋 be a 𝑝-semisimple BCI-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 that 𝑑 and 𝑑^′ are (𝑙,𝑟)-(𝑓,𝑔)-derivations of 𝑋. Then, for all 𝑥,𝑦∈𝑋, 𝑑∘𝑑^′𝑑𝑥∗𝑦=𝑑^′𝑑𝑥∗𝑦=𝑑^′(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑^′𝑦=𝑑𝑑^′(𝑥)𝑦=𝑑𝑑∗𝑓_′(𝑥)∗𝑓𝑓𝑦∧𝑔𝑑^′(𝑥)𝑓𝑦=∗𝑑𝑑∘𝑑^′(𝑥)𝑦=∗𝑓𝑑∘𝑑^′(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑∘𝑑^′𝑦. (21) 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 𝑑∘𝑑^′𝑓(𝑥)𝑥∗𝑦=𝑑∗𝑑^′𝑦∧𝑑^′(𝑥)𝑦𝑓(𝑥)∗𝑔=𝑑∗𝑑^′𝑦=𝑓𝑓(𝑥)∗𝑑𝑑^′𝑦∧𝑑𝑓(𝑥)∗𝑔𝑑^′𝑦=𝑓(𝑥)∗𝑑∘𝑑^′𝑦=𝑓(𝑥)∗𝑑∘𝑑^′𝑦∧𝑑∘𝑑^′(𝑥)∗𝑔𝑦. (22) So, 𝑑∘𝑑^′ is an (𝑟,𝑙)-(𝑓,𝑔)-derivation of 𝑋.

Theorem 17. Let 𝑋 be a 𝑝-semisimple BCI-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 all 𝑥,𝑦∈𝑋, we have 𝑑∘𝑑^′𝑑𝑥∗𝑦=𝑑^′(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑^′𝑦=𝑑𝑑^′(𝑥)𝑦=𝑓𝑑∗𝑓_′(𝑥)∗𝑑𝑓𝑦∧𝑑𝑑^′(𝑥𝑓𝑦𝑑)∗𝑔=𝑓^′(𝑥)∗𝑑𝑓𝑦=𝑓∘𝑑^′(𝑥)∗𝑑∘𝑓𝑦. (23) Also, for all 𝑥,𝑦∈𝑋, 𝑑^′∘𝑑𝑥∗𝑦=𝑑^′𝑓(𝑥)∗𝑑𝑦∧𝑑(𝑥)∗𝑔𝑦=𝑑^′𝑓(𝑥)𝑦=𝑑∗𝑑_′𝑓(𝑥)∗𝑓𝑑𝑦∧𝑔𝑓(𝑥)∗𝑑^′𝑑𝑦=𝑑^′𝑓(𝑥)∗𝑓𝑑𝑦=𝑑^′∘𝑓(𝑥)∗𝑓∘𝑑𝑦=𝑓∘𝑑^′(𝑥)∗𝑑∘𝑓𝑦. (24) 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]). Let 𝑋 be a BCI-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. Let 𝑋 be a 𝑝-semisimple BCI-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) Since 𝑑 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 𝑑∘𝑑^′𝑑𝑥∗𝑦=𝑑^′(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑^′𝑦=𝑑𝑑^′(𝑥)𝑦=𝑓𝑑∗𝑓_′(𝑥)∗𝑑𝑓𝑦∧𝑑𝑑^′(𝑥)∗𝑔𝑓𝑦=𝑓∘𝑑^′(𝑥)∗𝑑∘𝑓𝑦. (25) 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 𝑑∘𝑑^′𝑓(𝑥)𝑥∗𝑦=𝑑∗𝑑^′𝑦∧𝑑^′(𝑥)𝑦𝑓(𝑥)∗𝑔=𝑑∗𝑑^′𝑦=𝑑𝑓(𝑥)∗𝑓𝑑^′𝑦∧𝑔𝑓(𝑥)∗𝑑𝑑^′𝑦=𝑑𝑓(𝑥)∗𝑓𝑑^′𝑦=𝑑∘𝑓(𝑥)∗𝑓∘𝑑^′𝑦. (26) 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) Since 𝑑 and 𝑑^′ are (𝑙,𝑟)-(𝑓,𝑔)-derivations, then for all 𝑥,𝑦∈𝑋, we have 𝑑∘𝑑^′𝑑𝑥∗𝑦=𝑑^′(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑^′𝑦=𝑑𝑑^′(𝑥)𝑦=𝑑𝑑∗𝑓_′(𝑥)∗𝑓𝑓𝑦∧𝑔𝑑^′(𝑥)∗𝑑𝑓𝑦=𝑑∘𝑑^′(𝑥)∗𝑓∘𝑓𝑦. (27) On the other hand, since 𝑑 and 𝑑^′ are (𝑟,𝑙)-(𝑓,𝑔)-derivations, then for all 𝑥,𝑦∈𝑋, we have 𝑑∘𝑑^′𝑓(𝑥)𝑥∗𝑦=𝑑∗𝑑^′𝑦∧𝑑^′(𝑥)𝑦𝑓(𝑥)∗𝑔=𝑑∗𝑑^′𝑦=𝑓𝑓(𝑥)∗𝑑𝑑^′𝑦∧𝑑𝑓(𝑥)∗𝑔𝑑^′𝑦=𝑓∘𝑓(𝑥)∗𝑑∘𝑑^′𝑦. (28) By (27) and (28), for all 𝑥,𝑦∈𝑋, 𝑑∘𝑑^′(𝑥)∗𝑓∘𝑓𝑦=𝑓∘𝑓(𝑥)∗𝑑∘𝑑^′𝑦. (29) By putting 𝑥=𝑦, we get (𝑑∘𝑑^′•𝑓∘𝑓)(𝑥)=(𝑓∘𝑓•𝑑∘𝑑^′)(𝑥), for all 𝑥∈𝑋.

Theorem 20. Let 𝑋 be a commutative 𝑝-semisimple BCI-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 all 𝑥,𝑦∈𝑋, we have 𝑑∧𝑑^′𝑥∗𝑦=𝑑𝑥∗𝑦∧𝑑^′𝑥∗𝑦=𝑑𝑥∗𝑦. (30) On the other hand, 𝑑∧𝑑^′𝑥∗𝑦=𝑑𝑥∗𝑦∧𝑑^′𝑥∗𝑦=𝑑^′𝑥∗𝑦∧𝑑𝑥∗𝑦=𝑑^′𝑥∗𝑦. (31) 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 by Der(𝑋) the set of all (𝑓,𝑔)-derivations on 𝑋.

Definition 21 (see [16]). Let 𝑑,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. Let 𝑋 be a 𝑝-semisimple BCI-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 that 𝑑, 𝑑^′ are (𝑙,𝑟)-(𝑓,𝑔)-derivations of 𝑋 and 𝑥,𝑦∈𝑋. Then, we have 𝑑∧𝑑^′𝑥∗𝑦=𝑑𝑥∗𝑦∧𝑑^′=𝑑(𝑥)𝑦∧𝑔(𝑥)𝑦(𝑥)𝑦=𝑑(𝑥)𝑥∗𝑦=𝑑𝑥∗𝑦∗𝑓∗𝑑=𝑑∗𝑓∧𝑑^′(𝑥)∗𝑓𝑦=𝑑∧𝑑^′(𝑥)𝑦=∗𝑓𝑑∧𝑑^′(𝑥)∗𝑓𝑦∧𝑔(𝑥)∗𝑑∧𝑑^′𝑦. (32) So, 𝑑∧𝑑^′ is an (𝑙,𝑟)-(𝑓,𝑔)-derivation of 𝑋.
Suppose that 𝑑, 𝑑^′ are (𝑟,𝑙)-(𝑓,𝑔)-derivations of 𝑋 and 𝑥,𝑦∈𝑋. Then, 𝑑∧𝑑^′𝑥∗𝑦=𝑑𝑥∗𝑦∧𝑑^′(𝑥)𝑦(𝑥)∗𝑑𝑦𝑥∗𝑦=𝑑𝑥∗𝑦=𝑓∗𝑑=𝑓∧𝑑^′𝑦=𝑓(𝑥)∗𝑑∧𝑑^′𝑦=𝑓(𝑥)∗𝑑∧𝑑^′𝑦∧𝑑∧𝑑^′(𝑥)∗𝑔𝑦. (33) So, 𝑑∧𝑑^′ is an (𝑟,𝑙)-(𝑓,𝑔)-derivation of 𝑋.

Theorem 23. Let 𝑋 be a 𝑝-semisimple BCI-algebra. Then, (𝐷𝑒𝑟(𝑋),∧) is a semigroup.

Proof. By Theorem 22, 𝑑∧𝑑^′∈Der(𝑋). Suppose that 𝑑,𝑑^′,𝑑^′′∈Der(𝑋). Then, 𝑑∧𝑑^′∧𝑑^′′𝑥∗𝑦=𝑑𝑥∗𝑦∧𝑑^′∧𝑑^′′.𝑥∗𝑦=𝑑𝑥∗𝑦 (34) Also, we have 𝑑∧𝑑^′∧𝑑^′′𝑥∗𝑦=𝑑∧𝑑^′𝑥∗𝑦=𝑑𝑥∗𝑦∧𝑑^′.𝑥∗𝑦=𝑑𝑥∗𝑦 (35) Therefore, 𝑑∧(𝑑^′∧𝑑^′′)=(𝑑∧𝑑^′)∧𝑑^′′. This completes the proof.

At the end of this section, we classify (𝑓,𝑔)-derivations on BCK-algebras of order 3.

There is only one BCK-algebra of order 2. It is called 𝐵₂₋₁ and it is as follows: ∗0𝑎0𝑎00𝑎0 (36) There are only two endomorphisms on 𝐵₂₋₁. They are as 𝑑₁=0 and 𝑑₂=𝐼. We have Table 1, and so ∧𝑑₁𝑑₂𝑑₁𝑑₁𝑑₁𝑑₂𝑑₁𝑑₂ (37) There are only three BCK-algebras of order 3. In the following, we classify (𝑓,𝑔)-derivations on them.

Table 1 

Let 𝐵₃₋₁={0,𝑎,𝑏}. Consider the following table: ∗0𝑎𝑏0𝑎000𝑏𝑎00𝑏𝑎0 (38) There are only two endomorphisms on 𝐵₃₋₁. They are as 𝑑₁=0 and 𝑑₂=𝐼. Set 𝑑₃=0if𝑥=0,𝑎𝑎if𝑥=𝑏; (39) then we have Table 2, and so ∧𝑑₁𝑑₂𝑑₃𝑑₁𝑑₁𝑑₁𝑑₁𝑑₂𝑑₁𝑑₂𝑑₃𝑑₃𝑑₁𝑑₃𝑑₃ (40) Let 𝐵₃₋₂={0,𝑎,𝑏}. Consider the following table: ∗0𝑎𝑏0𝑎000𝑏𝑎00𝑏𝑏0 (41) There are only four endomorphisms on 𝐵₃₋₂. They are as 𝑑₁=0, 𝑑₂=𝐼, 𝑑₃=𝑑0if𝑥=0,𝑎𝑎if𝑥=𝑏,₄=0if𝑥=0,𝑎𝑏if𝑥=𝑏. (42) Set 𝑑₅=0if𝑥=0,𝑏𝑎if𝑥=𝑎; (43) then we have Table 3, and so ∧𝑑₁𝑑₂𝑑₃𝑑₄𝑑₅𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₂𝑑₁𝑑₂𝑑₃𝑑₄𝑑₅𝑑₃𝑑₁𝑑₃𝑑₃𝑑₁𝑑₁𝑑₄𝑑₁𝑑₄𝑑₃𝑑₄𝑑₁𝑑₅𝑑₁𝑑₅𝑑₁𝑑₁𝑑₅ (44) Let 𝐵₃₋₃={0,𝑎,𝑏}. Consider the following table: ∗0𝑎𝑏0𝑎000𝑏𝑎0𝑎𝑏𝑏0 (45) There are only seven endomorphisms on 𝐵₃₋₃. They are as 𝑑₁=0, 𝑑₂=𝐼, 𝑑₃=𝑑0if𝑥=0,𝑎𝑎if𝑥=𝑏,₄=𝑑0if𝑥=0,𝑎𝑏if𝑥=𝑏,₅=𝑑0if𝑥=0,𝑏𝑎if𝑥=𝑎,₆=𝑑0if𝑥=0,𝑏𝑏if𝑥=𝑎,₇=0if𝑥=0𝑎if𝑥=𝑏𝑏if𝑥=𝑎. (46) Obviously, if we set 𝑓=𝑑₁=0, then for all 𝑔=𝑑_𝑖, 1≤𝑖≤7, we have only one (𝑓,𝑔)-derivation and it is 𝑑₁=0. Also, we have Table 4, and so ∧𝑑₁𝑑₂𝑑₃𝑑₄𝑑₅𝑑₆𝑑₇𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₂𝑑₁𝑑₂𝑑₁𝑑₄𝑑₅𝑑₁𝑑₁𝑑₃𝑑₁𝑑₁𝑑₃𝑑₁𝑑₁𝑑₁𝑑₃𝑑₄𝑑₁𝑑₄𝑑₁𝑑₄𝑑₁𝑑₁𝑑₁𝑑₅𝑑₁𝑑₅𝑑₁𝑑₁𝑑₅𝑑₁𝑑₁𝑑₆𝑑₁𝑑₁𝑑₁𝑑₁𝑑₁𝑑₆𝑑₆𝑑₇𝑑₁𝑑₁𝑑₃𝑑₁𝑑₁𝑑₆𝑑₇ (47)

Table 2 

Table 3 

Table 4 

3. (𝐹,𝐺)-Derivations of BCI-Algebras

Let 𝑋 be a BCI-algebra. A right derivation of 𝑋 is a self-map 𝐷 of 𝑋 satisfying the identity 𝐷=𝑦∧(𝑥)𝑥∗𝑦𝑥∗𝐷𝑦∗𝐷,∀𝑥,𝑦∈𝑋. (48) If 𝐷 satisfies the identity 𝐷(𝑥∗𝑦)=(𝐷(𝑥)∗𝑦)∧(𝐷(𝑦)∗𝑥), for all 𝑥,𝑦∈𝑋, then 𝐷 is called a left derivation of 𝑋.

Let 𝑋 be a BCI-algebra. A right 𝐹-derivation of 𝑋 is a self-map 𝐷 of 𝑋 satisfying the identity 𝐷=𝐹(𝑥)𝑦∧𝐹𝑦(𝑥)𝑥∗𝑦∗𝐷∗𝐷,∀𝑥,𝑦∈𝑋, (49) 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 a left 𝐹-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 of BCI-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 in BCI-algebras.

Definition 24. Let 𝑋 be a BCI-algebra. A right (𝐹,𝐺)-derivation of 𝑋 is a self-map 𝐷 of 𝑋 satisfying the identity 𝐷=𝐹(𝑥)𝑦∧𝐺𝑦(𝑥)𝑥∗𝑦∗𝐷∗𝐷,∀𝑥,𝑦∈𝑋, (50) 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 a left (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. Let 𝑋, 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. Let 𝑋 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. Let 𝑋={0,𝑎,𝑏} and the binary operation ∗ is defined as follows: ∗0𝑎𝑏0𝑎000𝑏𝑎0𝑎𝑏𝑏0 (51) Then, (𝑋,∗) is a commutative BCK-algebra. Define maps 𝐷,𝐹,𝐺∶𝑋→𝑋 by 𝐷(𝑥)=0, for all 𝑥∈𝑋, 𝐹=𝐼 and 𝐺(𝑥)=0if𝑥=0𝑏if𝑥=𝑎𝑎if𝑥=𝑏. (52) 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 a BCI-algebra 𝑋 is called regular if 𝐷(0)=0. If 𝐷(0)≠0, then 𝐷 is called irregular.

Theorem 29. Every right (𝐹,𝐺)-derivation of a BCK-algebra is regular.

Proof. Suppose that 𝑋 is a BCK-algebra and 𝐷 is a right (𝐹,𝐺)-derivation of 𝑋. Then, 𝐷(0)=𝐷(0∗0)=(𝐹(0)∗𝐷(0))∧(𝐺(0)∗𝐷(0))=0. (53) So, 𝐷 is regular.

Theorem 30. Let 𝐷 be a right (𝐹,𝐺)-derivation of a BCI-algebra 𝑋. Then, 𝐷(𝑥)∈𝐺(𝑋), for all 𝑥∈𝐺(𝑋).

Proof. Suppose that 𝑥∈𝐺(𝑋). Then, 𝐷(𝑥)()=(𝐹(0)(𝑥∧(𝐺(𝑥)(0(0)(𝑥)(𝑥)=𝐷0∗𝑥∗𝐷))∗𝐷))≤𝐹∗𝐷=0∗𝐷∈𝐿_𝑝(𝑋). (54) Thus, 𝐷(𝑥)=0∗𝐷(𝑥). So, 𝐷(𝑥)∈𝐺(𝑋).

Theorem 31. Let 𝐷 be a right (𝐹,𝐺)-derivation of a BCI-algebra 𝑋. Then, the following hold: (1)𝐷(𝑥)∈𝐿_𝑝(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)∈𝐿_𝑝(𝑋), for all 𝑥,𝑦∈𝑋.

Proof. (1) For all 𝑥∈𝑋, we have 𝐷(𝑥)()=(𝐹(𝑥)(0∧(𝐺(0)(𝑥=((𝑥∗(𝑥∗(𝐹(𝑥)(0=((𝑥∗(𝐹(𝑥)(0(𝑥((𝐹(𝑥)(0.=𝐷𝑥∗0∗𝐷))∗𝐷))0∗𝐷))((0∗𝐷))∗𝐷)))0∗𝐷))((0∗∗𝐷)))∗𝐷))≤0∗0∗∗𝐷))) (55) So, 𝐷(𝑥)∈𝐿_𝑝(𝑋).
(2), (3), and (4) are clear.

Theorem 32. Let 𝑋 be a 𝑝-semisimple BCI-algebra. Then (1)if 𝐷 is an (𝑙,𝑟)-(𝐹,𝐺)-derivation of 𝑋, then 𝐷 is a left (𝐹,𝐺)-derivation of 𝑋;(2)if 𝐷 is an (𝑟,𝑙)-(𝐹,𝐺)-derivation of 𝑋, then 𝐷 is a right (𝐹,𝐺)-derivation of 𝑋.

Proof. (1) Suppose that 𝑥,𝑦∈𝑋 and 𝐷 is an (𝑙,𝑟)-(𝐹,𝐺)-derivation. Then, 𝐷=𝐷(𝑥)𝑦∧𝐺(𝑥)𝑦(𝑥)𝑦.𝑥∗𝑦∗𝐹∗𝐷≤𝐷∗𝐹 (56) 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 that 𝑥,𝑦∈𝑋 and 𝐷 is an (𝑟,𝑙)-(𝐹,𝐺)-derivation. Then, 𝐷=𝐹(𝑥)𝑦∧𝐷(𝑥)𝑦(𝑥)𝑦.𝑥∗𝑦∗𝐷∗𝐺≤𝐹∗𝐷 (57) 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. Let 𝐷 be a self-map and let 𝑉(𝑥₀) be any branch of a BCI-algebra 𝑋. If, for any 𝑥∈𝑉(𝑥₀), 𝐷(𝑥)=𝐹(𝑥₀), then 𝐷 is a regular left (𝐹,𝐺)-derivation.

Proof. Suppose that 𝑥,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 𝐷𝑥∗𝑦=𝐹𝑥₀∗𝑦₀. (58) Also, by using hypothesis, we have 𝐷(𝑥)𝑦∧𝐷𝑦(𝑥)=𝐹𝑥∗𝐹∗𝐺₀∗𝐹𝑦∧𝐹𝑦₀(𝑥)𝑥∗𝐺≤𝐹₀∗𝐹𝑦=𝐹𝑥₀𝑥∗𝑦=𝐹₀∗𝑦₀. (59) From 𝑥₀,𝑦₀∈𝐿_𝑝(𝑋), we get 𝐹(𝑥₀∗𝑦₀)∈𝐿_𝑝(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-map 𝐷 of a BCI-algebra 𝑋 defined by 𝐷(𝑥)=0∗(0∗𝐹(𝑥))=𝐹_𝑥, for all 𝑥∈𝑋, is a left (𝐹,𝐺)-derivation.

Proof. Suppose that 𝑥,𝑦∈𝑋. We have 𝐷(𝑥∗𝑦)=𝐹_𝑥∗𝑦=𝐹_𝑥∗𝐹_𝑦 and 𝐷(𝑥)𝑦∧𝐷𝑦(𝑥)=𝐹∗𝐹∗𝐺_𝑥∗𝐹𝑦∧𝐹_𝑦∗𝐺(𝑥)≤𝐹_𝑥∗𝐹𝑦. (60) So, 𝐷(𝑥)∗𝐹𝑦∧𝐷𝑦∗𝐺(𝑥)∗𝐹_𝑥∗𝐹_𝑦≤𝐹_𝑥∗𝐹𝑦∗𝐹_𝑥∗𝐹_𝑦≤𝐹_𝑦∗𝐹𝑦=0. (61) Hence, (𝐷(𝑥)∗𝐹(𝑦))∧(𝐷(𝑦)∗𝐺(𝑥))≤𝐹_𝑥∗𝐹_𝑦. Therefore, (𝐷(𝑥)∗𝐹(𝑦))∧(𝐷(𝑦)∗𝐺(𝑥))=𝐹_𝑥∗𝐹_𝑦=𝐷(𝑥∗𝑦), which implies that 𝐷 is a left (𝐹,𝐺)-derivation.

Theorem 35. Let 𝐷 be a left (𝐹,𝐺)-derivation of a commutative BCI-algebra 𝑋. Then, 𝑥≤𝑦 implies that 𝐷(𝑥) and 𝐷(𝑦) belong to the same branch of 𝑋.

Proof. Suppose that 𝑥,𝑦∈𝑋 and 𝑥≤𝑦. Since 𝑋 is commutative, 𝑦∗(𝑦∗𝑥)=𝑥. Since 𝐷 is a left (𝐹,𝐺)-derivation, we have 𝐷(𝑥)=𝐷𝑦∧𝐷𝑦𝑦𝑦∗𝐹𝑦(𝑥).=𝐷𝑦∗𝑦∗𝑥∗𝐹𝑦∗𝑥𝑦∗𝑥∗𝐺≤𝐷∗𝐹𝑦∗𝑥=𝐷∗𝐹 (62) 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 𝐷(𝑥)𝑦≤𝐷𝑦∗𝐹𝑦(𝑥)𝑦=𝐷𝑦𝑦∗𝐹𝑦(𝑥)𝐹𝑦(𝑥)∗𝐷∗𝐹∗𝐷∗𝐷∗𝐹=0∗∗𝐹=0. (63) 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. Let 𝐷 be an (𝐹,𝐺)-derivation of a BCI-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 that 𝑥∈𝑉(𝑥₀), 𝑦∈𝑉(𝑦₀), and 𝐷(𝑥)=𝑦₀. Since 𝐷 is a left (𝐹,𝐺)-derivation, we have 𝐷=𝐷(𝑥)𝑦∧𝐷𝑦(𝑥)(𝑥)𝑦𝑥∗𝑦∗𝐹∗𝐺≤𝐷∗𝐹=𝑦₀∗𝐹𝑦. (64) 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∈𝐿_𝑝(𝑋), 𝐹(𝑥₀)∈𝐿_𝑝(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).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.