Pseudo BCI-algebras with derivations

TL;DR

This paper introduces and analyzes implicative derivations on pseudo-BCI algebras, characterizing their properties, invariance, and algebraic structures, especially in the context of p-semisimple algebras.

Contribution

It defines new types of derivations on pseudo-BCI algebras and explores their properties, including characterization, invariance, and algebraic structure, with a focus on p-semisimple cases.

Findings

Regular derivations of type II are characterized by invariance of deductive systems.

The set of implicative derivations forms a commutative monoid under composition.

In p-semisimple pseudo-BCI algebras, certain derivations coincide and are characterized by identity.

Abstract

In this paper we define two types of implicative derivations on pseudo-BCI algebras, we investigate their properties and we give a characterization of regular implicative derivations of type II. We also define the notion of a $d$-invariant deductive system of a pseudo-BCI algebra $A$ proving that $d$ is a regular derivation of type II if and only if every deductive system on $A$ is $d$-invariant. It is proved that a pseudo-BCI algebra is $p$-semisimple if and only if the only regular derivation of type II is the identity map. Another main result consists of proving that the set of all implicative derivations of a $p$-semisimple pseudo-BCI algebra forms a commutative monoid with respect to function composition. Two types of symmetric derivations on pseudo-BCI algebras are also introduced and it is proved that in the case of $p$-semisimple pseudo-BCI algebras the sets of type II implicative derivations and type II symmetric derivations are equal.

Equations

$$\hskip 28.45274pt\begin{array}[]{c|ccccc}\rightarrow&a&b&c&d&1\\ \hline\cr a&1&1&1&d&1\\ b&b&1&1&d&1\\ c&b&b&1&d&1\\ d&d&d&d&1&d\\ 1&a&b&c&d&1\end{array}\hskip 28.45274pt\begin{array}[]{c|ccccc}\rightsquigarrow&a&b&c&d&1\\ \hline\cr a&1&1&1&d&1\\ b&c&1&1&d&1\\ c&a&b&1&d&1\\ d&d&d&d&1&d\\ 1&a&b&c&d&1\end{array}.$$

$$\begin{array}[]{c|cccccc}x&a&b&c&d&1\\ \hline\cr d_{1}(x)&a&b&c&d&1\\ d_{2}(x)&d&d&d&1&d\\ d_{3}(x)&1&1&1&d&1\end{array}.$$

$$\hskip 28.45274pt\begin{array}[]{c|cccccc}\rightarrow&a&b&c&d&e&1\\ \hline\cr a&1&d&e&b&c&a\\ b&c&1&a&e&d&b\\ c&e&a&1&c&b&d\\ d&b&e&d&1&a&c\\ e&d&c&b&a&1&e\\ 1&a&b&c&d&e&1\end{array}\hskip 28.45274pt\begin{array}[]{c|cccccc}\rightsquigarrow&a&b&c&d&e&1\\ \hline\cr a&1&c&b&e&d&a\\ b&d&1&e&a&c&b\\ c&b&e&1&c&a&d\\ d&e&a&d&1&b&c\\ e&c&d&a&b&1&e\\ 1&a&b&c&d&e&1\end{array}.$$

$$\hskip 28.45274pt\begin{array}[]{c|cccccc}\rightarrow&a&b&x&y&g&1\\ \hline\cr a&1&b&g&y&g&1\\ b&a&1&x&g&g&1\\ x&g&x&1&a&1&g\\ y&y&g&b&1&1&g\\ g&y&x&b&a&1&g\\ 1&a&b&x&y&g&1\end{array}\hskip 28.45274pt\begin{array}[]{c|cccccc}\rightsquigarrow&a&b&x&y&g&1\\ \hline\cr a&1&b&x&g&g&1\\ b&a&1&g&y&g&1\\ x&x&g&1&b&1&g\\ y&g&y&a&1&1&g\\ g&x&y&a&b&1&g\\ 1&a&b&x&y&g&1\end{array}.$$

$$\begin{array}[]{c|cccccc}x&a&b&x&y&g&1\\ \hline\cr d_{1}(x)&a&b&x&y&g&1\\ d_{2}(x)&g&g&1&1&1&g\\ d_{3}(x)&1&1&g&g&g&1\end{array}.$$

$$\hskip 28.45274pt\begin{array}[]{c|ccc}\rightarrow&a&b&1\\ \hline\cr a&1&a&b\\ b&b&1&a\\ 1&a&b&1\end{array}\hskip 28.45274pt\begin{array}[]{c|ccc}x&a&b&1\\ \hline\cr d_{1}(x)&a&b&1\\ d_{2}(x)&b&1&a\\ d_{3}(x)&1&a&b\end{array}.$$

$$\hskip 28.45274pt\begin{array}[]{c|ccccc}\rightarrow&0&a&b&c&1\\ \hline\cr 0&1&1&1&1&1\\ a&0&1&b&1&1\\ b&a&a&1&1&1\\ c&0&a&b&1&1\\ 1&0&a&b&c&1\end{array}\hskip 28.45274pt\begin{array}[]{c|ccccc}\rightsquigarrow&0&a&b&c&1\\ \hline\cr 0&1&1&1&1&1\\ a&b&1&b&1&1\\ b&0&a&1&1&1\\ c&0&a&b&1&1\\ 1&0&a&b&c&1\end{array}.$$

$$\begin{array}[]{c|cccccc}x&0&a&b&c&1\\ \hline\cr d_{1}(x)&0&a&b&c&1\\ d_{2}(x)&0&a&b&1&1\\ d_{3}(x)&0&1&1&c&1\\ d_{4}(x)&0&1&1&1&1\\ d_{5}(x)&1&1&1&c&1\\ d_{6}(x)&1&1&1&1&1\end{array}.$$

Full text

Keywords: pseudo-BCI algebra, commutative pseudo-BCI algebra, $p$-semisimple pseudo-BCI algebra, implicative derivation, regular derivation, isotone derivation, invariant deductive system, symmetric derivation

AMS Mathematics Subject Classification (2010): 06D35, 06F05, 03F50

1. Introduction

The notion of derivations from the analytic theory was introduced in 1957 by Posner ([41]) to a prime ring $(R,+,\cdot)$ as a map $d:R\longrightarrow R$ satisfying the conditions $d(x+y)=d(x)+d(y)$ and $d(x\cdot y)=d(x)\cdot y+x\cdot d(y)$, for all $x,y\in R$. Since the derivation proved to be helpful for studying the properties of algebraic systems, this notion has been defined and studied by many authors for the cases of lattices ([44], [23], [45], [46]) and algebras of fuzzy logic: MV-algebras ([3]), [47], [22]), BCI-algebras ([30], [1], [2]), commutative residuated lattice ([25]), BCC-algebras ([42], [4]), BE-algebras ([32])), basic algebras ([34]) and pseudo-MV algebras ([43]).

The aim of this paper is to introduce the concept of derivations on pseudo-BCI algebras and to investigate their properties. We define the type I and type II implicative derivations on pseudo-BCI algebras, we introduce the notion of a regular implicative derivation and we give a characterization of regular implicative derivations of type II. The notion of isotone implicative derivations is also defined, and it is proved that any regular implicative derivation of type II is isotone. For an implicative derivation $d$ on a pseudo-BCK algebra $A$ we define the notion of a $d$-invariant deductive system of $A$ proving that $d$ is a regular implicative derivation of type II if and only if every deductive system on $A$ is $d$-invariant. We investigate the particular case of implicative derivations on the $p$-semisimple pseudo-BCI algebras and we prove that a pseudo-BCI algebra is $p$-semisimple if and only if the only regular derivation of type II is the identity map. Another main result consists of proving that the set of all implicative derivations of a $p$-semisimple pseudo-BCI algebra forms a commutative monoid with respect to function composition. Two types of symmetric derivations on pseudo-BCI algebras are also introduced and it is proved that in the case of $p$-semisimple pseudo-BCI algebras the sets of type II implicative derivations and type II symmetric derivations are equal.

2. Preliminaries

Pseudo-BCK algebras were introduced by G. Georgescu and A. Iorgulescu in [21] as algebras with ”two differences”, a left- and right-difference, and with a constant element [math] as the least element. Nowadays pseudo-BCK algebras are used in a dual form, with two implications, $\rightarrow$ and $\rightsquigarrow$ and with one constant element $1$, that is the greatest element. Thus such pseudo-BCK algebras are in the ”negative cone” and are also called ”left-ones”. Pseudo-BCK algebras were intensively studied in [27], [28], [26], [35], [9]. Pseudo-BCI algebras were defined by [11] as generalizations of pseudo-BCK algebra and BCI-algebras, and they form an important tool for an algebraic axiomatization of implicational fragment of non-classical logic ([15]). In this section we recall some basic notions and results regarding pseudo-BCI algebras from [11], [12]-[19], [7]-[8], [20].

Definition 2.1**.**

$\rm($[8]$\rm)$ A pseudo-BCI algebra is a structure ${\mathcal{A}}=(A,\rightarrow,\rightsquigarrow,1)$ of type $(2,2,0)$ satisfying the following axioms, for all $x,y,z\in A:$

$(psBCI_{1})$ $(x\rightarrow y)\rightsquigarrow[(y\rightarrow z)\rightsquigarrow(x\rightarrow z)]=1;$

$(psBCI_{2})$ $(x\rightsquigarrow y)\rightarrow[(y\rightsquigarrow z)\rightarrow(x\rightsquigarrow z)]=1;$

$(psBCI_{3})$ $1\rightarrow x=x;$

$(psBCI_{4})$ $1\rightsquigarrow x=x;$

$(psBCI_{5})$ $(x\rightarrow y=1$ and $y\rightarrow x=1)$ implies $x=y$.

It is proved in [8, Lemma 2.1] that $x\rightarrow y=1$ iff $x\rightsquigarrow y=1$, so that axiom $(psBCI_{5})$ is equivalent to the following axiom:

$(psBCI^{\prime}_{5})$ $(x\rightsquigarrow y=1$ and $y\rightsquigarrow x=1)$ implies $x=y$.

Every pseudo-BCI algebra satisfying $x\rightarrow y=x\rightsquigarrow y$ for all $x,y\in A$ is a BCI-algebra, and every pseudo-BCI algebra satisfying $x\leq 1$ for all $x\in A$ is a pseudo-BCK algebra. A pseudo-BCI algebra is said to be proper if it is not a BCI-algebra and it is not a pseudo-BCK algebra. In a pseudo-BCI algebra $(A,\rightarrow,\rightsquigarrow,1)$, one can define a binary relation $``\leq"$ by

                  $x\leq y$ iff $x\rightarrow y=1$ iff $x\rightsquigarrow y=1$, for all $x,y\in A$.

We will refer to $(A,\rightarrow,\rightsquigarrow,1)$ by its universe $A$.

Lemma 2.2**.**

$\rm($[20]$\rm)$* Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra. Then the following hold for all $x,y,z\in A$:

$(1)$ $x\rightarrow x=x\rightsquigarrow x=1;$

$(2)$ $x\leq(x\rightarrow y)\rightsquigarrow y$ and $x\leq(x\rightsquigarrow y)\rightarrow y;$

$(3)$ $x\rightarrow y=1$ iff $x\rightsquigarrow y=1;$

$(4)$ $x\leq y\rightarrow z$ iff $y\leq x\rightsquigarrow z;$

$(5)$ $x\leq y$ implies $y\rightarrow z\leq x\rightarrow z$ and $y\rightsquigarrow z\leq x\rightsquigarrow z;$

$(6)$ $x\leq y$ implies $z\rightarrow x\leq z\rightarrow y$ and $z\rightsquigarrow x\leq z\rightsquigarrow y;$

$(7)$ $x\rightarrow y\leq(z\rightarrow x)\rightarrow(z\rightarrow y)$ and $x\rightsquigarrow y\leq(z\rightsquigarrow x)\rightsquigarrow(z\rightsquigarrow y);$

$(8)$ $x\rightarrow(y\rightsquigarrow z)=y\rightsquigarrow(x\rightarrow z);$

$(9)$ if $1\leq x$, then $x=1;$

$(10)$ if $x\leq y$ and $y\leq z$, then $x\leq z;$

$(11)$ $x\rightarrow 1=x\rightsquigarrow 1;$

$(12)$ $(x\rightarrow y)\rightarrow 1=(x\rightarrow 1)\rightsquigarrow(y\rightarrow 1)$ and $(x\rightsquigarrow y)\rightsquigarrow 1=(x\rightsquigarrow 1)\rightarrow(y\rightsquigarrow 1)$.*

For any pseudo-BCI algebra $A$ the set ${\rm K}\,(A)=\{x\in A\mid x\leq 1\}$ is a subalgebra of $A$ called the pseudo-BCK part of $A$, since $({\rm K}\,(A),\rightarrow,\rightsquigarrow,1)$ is a pseudo-BCK algebra. An element $a$ of a pseudo-BCI algebra $A$ is called an atom if $a\leq x$ implies $x=a$, for all $x\in A$. Denote by ${\rm At}\,(A)$ the set of all atoms of $A$ and it is proved in [14] that ${\rm At}\,(A)=\{x\in A\mid x=(x\rightarrow 1)\rightarrow 1\}$. For any $a\in A$ we denote $V(a)=\{x\in X\mid x\leq a\}$ and it is obvious that $V(a)\neq\emptyset$, since $a\leq a$ gives $a\in V(a)$. If $a\in{\rm At}\,(A)$, then the set $V(a)$ is called a branch of $A$ determined by the element $a$. According to [13] we have: $(1)$ branches determined by different elements are disjoints; $(2)$ a pseudo-BCI algebra is a set-theoretic union of branches; $(3)$ comparable elements are in the same branch; $(4)$ the elements $x$ and $y$ belong to the same branch if and only if $x\rightarrow y\in V(1)$ or equivalently $x\rightsquigarrow y\in V(1)$. A pseudo-BCI algebra $A$ is $p$-semisimple if $x\leq 1$ implies $x=1$, for all $x\in A$, that is ${\rm K}\,(A)=\{1\}$.

Proposition 2.3**.**

$\rm($[13*]**,[17]$\rm)$ In any pseudo-BCI algebra $A$ the following are equivalent, for all $x,y\in A$:

$(a)$ $A$ is $p$-semisimple;

$(b)$ if $x\leq y$, then $x=y;$

$(c)$ $(x\rightarrow y)\rightsquigarrow y=(x\rightsquigarrow y)\rightarrow y=x;$

$(d)$ $(x\rightarrow 1)\rightsquigarrow 1=(x\rightsquigarrow 1)\rightarrow 1=x;$

$(e)$ $(x\rightarrow 1)\rightsquigarrow y=(y\rightsquigarrow 1)\rightarrow x;$

$(f)$ $x\rightarrow a=y\rightarrow a$ implies $x=y;$

$(g)$ $x\rightsquigarrow a=y\rightsquigarrow a$ implies $x=y;$

$(h)$ $A={\rm At}\,(A);$

$(i)$ $(A,\cdot,^{-1},1)$ is a group, where $x\cdot y=(x\rightarrow 1)\rightsquigarrow y=(y\rightsquigarrow 1)\rightarrow x$, $x^{-1}=x\rightarrow 1=x\rightsquigarrow 1$, $x\rightarrow y=y\cdot x^{-1}$ and $x\rightsquigarrow y=x^{-1}\cdot y$, for all $x,y\in A$.*

Proposition 2.4**.**

$\rm($[14]$\rm)$* In any pseudo-BCI algebra $A$ the following are equivalent, for all $a,x,y\in A$:

$(a)$ $a\in{\rm At}\,(A);$

$(b)$ $(a\rightarrow x)\rightsquigarrow x=(a\rightsquigarrow x)\rightarrow x=a;$

$(c)$ $x\rightarrow a=(a\rightarrow x)\rightsquigarrow 1;$

$(d)$ $x\rightsquigarrow a=(a\rightsquigarrow x)\rightarrow 1;$

$(e)$ $x\rightarrow a=(a\rightarrow y)\rightsquigarrow(x\rightarrow y);$

$(f)$ $x\rightsquigarrow a=(a\rightsquigarrow y)\rightarrow(x\rightsquigarrow y);$

$(g)$ $x\rightarrow a=((x\rightarrow a)\rightsquigarrow y)\rightarrow y;$

$(h)$ $x\rightsquigarrow a=((x\rightsquigarrow a)\rightarrow y)\rightsquigarrow y;$

$(i)$ $x\rightarrow a=(a\rightarrow 1)\rightsquigarrow(x\rightarrow 1);$

$(j)$ $x\rightsquigarrow a=(a\rightsquigarrow 1)\rightarrow(x\rightsquigarrow 1);$

$(k)$ $(a\rightarrow 1)\rightsquigarrow 1=(a\rightsquigarrow 1)\rightarrow 1=a$.*

Corollary 2.5**.**

If $a\in{\rm At}\,(A)$, then $x\rightarrow a,x\rightsquigarrow a\in{\rm At}\,(A)$, for all $x\in A$.

Proof.

It follows taking $y:=1$ in Proposition 2.4$(g)$,$(h)$ and applying $(k)$. ∎

Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra. Denote: $x\Cup_{1}y=(x\rightarrow y)\rightsquigarrow y$ and $x\Cup_{2}y=(x\rightsquigarrow y)\rightarrow y$, for all $x,y\in A$.

Lemma 2.6**.**

*In any pseudo-BCI algebra $A$ the following hold for all $x,x_{1},x_{2},y\in A:$

$(1)$ $1\Cup_{1}x=1\Cup_{2}x=1;$

$(2)$ $x\leq y$ iff $x\Cup_{1}y=y$ iff $x\Cup_{2}y=y;$

$(3)$ $x\Cup_{1}x=x\Cup_{2}x=x;$

$(4)$ if $A$ is $p$-semisimple, then $x\Cup_{1}1=x\Cup_{2}1=x;$

$(5)$ $x\leq x\Cup_{1}y$ and $x\leq x\Cup_{2}y;$

$(6)$ $x_{1}\leq x_{2}$ implies $x_{1}\Cup_{1}y\leq x_{2}\Cup_{1}y$ and $x_{1}\Cup_{2}y\leq x_{2}\Cup_{2}y;$

$(7)$ $x\Cup_{1}y\rightarrow y=x\rightarrow y$ and $x\Cup_{2}y\rightsquigarrow y=x\rightsquigarrow y$.*

Proof.

The proof is straightforward. ∎

A pseudo-BCK algebra $A$ is said to be commutative if it satisfies the following identities for all $x,y\in A$:

$(comm_{1})$ $x\Cup_{1}y=y\Cup_{1}x$,

$(comm_{2})$ $x\Cup_{2}y=y\Cup_{2}x$.

Note that if a pseudo-BCI algebra $A$ satisfies $(comm_{1})$ and $(comm_{2})$ for all $x,y\in A$, then $A$ is a pseudo-BCK algebra. Indeed, $x\leq(x\rightarrow 1)\rightsquigarrow 1=(1\rightarrow x)\rightsquigarrow x=1$, for all $x\in A$. A pseudo-BCI algebra satisfying $(comm_{1})$ and $(comm_{2})$ for all $x$ and $y$ belonging to the same branch is called branchwise commutative, and it is called commutative if it satisfies the quasi-identities

$(comm_{3})$ $x\Cup_{1}y=x\Cup_{2}y=x$, whenever $y\leq x$.

It was proved in [12] that a pseudo-BCI algebra is commutative if and only if it is branchwise commutative. Moreover, each branch of a commutative pseudo-BCI algebra is a semilattice with the join $\vee$ defined by $x\vee y=x\Cup_{1}y=x\Cup_{2}y$. It is known that any $p$-semisimple pseudo-BCI algebra is commutative ([12]).

A subset $D$ of a pseudo-BCI algebra $A$ is called a deductive system of $A$ if it satisfies the following axioms: $(ds_{1})$ $1\in D,$ $(ds_{2})$ $x\in D$ and $x\rightarrow y\in D$ imply $y\in D$.

A subset $D$ of $A$ is a deductive system if and only if it satisfies $(ds_{1})$ and the axiom:

$(ds^{\prime}_{2})$ $x\in D$ and $x\rightsquigarrow y\in D$ imply $y\in D$.

Denote by ${\mathcal{D}S}(A)$ the set of all deductive systems of $A$. A deductive system $D$ of $A$ is proper if $D\neq A$. A deductive system $D$ of a pseudo-BCK algebra $A$ is said to be compatible if it satisfies the condition: $(ds_{3})$ for all $x,y\in A$, $x\rightarrow y\in D$ iff $x\rightsquigarrow y\in D$.

Denote by ${\mathcal{D}S_{c}}(A)$ the set of all compatible deductive systems of $A$. A deductive system is closed if it is subalgebra of $A$. Denote by $\mathcal{CON}(A)$ the set of all congruences of $A$. We say that $\theta\in\mathcal{CON}(A)$ is a relative congruence of $A$ if the quotient algebra $(A/\theta,\rightarrow,\rightsquigarrow,[1]_{\theta})$ is a pseudo-BCI algebra. It was proved in [16] that the relative congruences of $A$ correspond one-to-one to closed compatible deductive systems of $A$. For details regarding the deductive systems and congruence relations on a pseudo-BCI algebra we refer the reader to [16].

3. Implicative derivation operators on pseudo-BCI algebras

In this section we define the type I and type II implicative derivations on pseudo-BCI algebras, and we investigate their properties. The notions of regular, isotone and idempotent implicative derivations are introduced, and a characterization of regular implicative derivations of type II is given. It is also proved that any regular implicative derivation of type II is isotone. If $d$ is a regular implicative derivation of type II on a pseudo-BCI algebra $A$ such that its kernel coincides with the pseudo-BCK part of $A$, then it is proved that $d$ is idempotent. Finally, given an implicative derivation $d$ on $A$, we define the notion of a $d$-invariant deductive system of $A$, and we prove that $d$ is a regular implicative derivation of type II if and only if every deductive system of $A$ is $d$-invariant.

Definition 3.1**.**

Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra. A mapping $d:A\longrightarrow A$ is called an implicative derivation operator of type I or a type I implicative derivation operator or a type I implicative derivation on $A$ if it satisfies the following conditions for all $x,y\in A:$

$(idop_{1})$ $d(x\rightarrow y)=(x\rightarrow d(y))\Cup_{2}(d(x)\rightarrow y)$.

$(idop_{2})$ $d(x\rightsquigarrow y)=(x\rightsquigarrow d(y))\Cup_{1}(d(x)\rightsquigarrow y)$.

Definition 3.2**.**

Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra. A mapping $d:A\longrightarrow A$ is called an implicative derivation operator of type II or a type II implicative derivation operator or a type II implicative derivation on $A$ if it satisfies the following conditions for all $x,y\in A:$

$(idop_{3})$ $d(x\rightarrow y)=(d(x)\rightarrow y)\Cup_{2}(x\rightarrow d(y))$,

$(idop_{4})$ $d(x\rightsquigarrow y)=(d(x)\rightsquigarrow y)\Cup_{1}(x\rightsquigarrow d(y))$.

Let $A$ be a pseudo-BCI algebra. Denote:

                  $\mathcal{IDOP}^{(I)}(A)$ the set of all implicative derivation operators of type I on $A$,

                  $\mathcal{IDOP}^{(II)}(A)$ the set of all implicative derivation operators of type II on $A$,

                  $\mathcal{IDOP}(A)=\mathcal{IDOP}^{(I)}(A)\cap\mathcal{IDOP}^{(II)}(A)$.

A map $d\in\mathcal{IDOP}(A)$ is called an implicative derivation on $A$.

In what follows we will denote $dx$ instead of $d(x)$.

Example 3.3**.**

Let $A$ be a pseudo-BCI algebra. If ${\rm Id}\,_{A}:A\longrightarrow A$, defined by ${\rm Id}\,_{A}(x)=x$ for all $x\in A$, then ${\rm Id}\,_{A}\in\mathcal{IDOP}(A)$.

Example 3.4**.**

Consider the structure $(A,\rightarrow,\rightsquigarrow,1)$, where the operations $\rightarrow$ and $\rightsquigarrow$ on $A=\{a,b,c,d,1\}$ are defined as follows:

[TABLE]

Then $(A,\rightarrow,\rightsquigarrow,1)$ is a pseudo-BCI algebra ([18]). Consider the maps $d_{1},d_{2},d_{3}:A\longrightarrow A$ given in the table below:

[TABLE]

One can check that $\mathcal{IDOP}^{(I)}(A)=\mathcal{IDOP}^{(II)}(A)=\{d_{1},d_{2},d_{3}\}$.

Proposition 3.5**.**

*Let $A$ be a pseudo-BCI algebra. Then the following hold for all $x\in A$:

$(1)$ if $d\in\mathcal{IDOP}^{(I)}(A)$, then $dx=dx\Cup_{1}x=dx\Cup_{2}x;$

$(2)$ if $d\in\mathcal{IDOP}^{(II)}(A)$, then $dx=x\Cup_{1}dx=x\Cup_{2}dx$ iff $d1=1$.*

Proof.

$(1)$ Consider $d\in\mathcal{IDOP}^{(I)}(A)$ and $x\in A$. Applying Lema 2.2$(8)$,$(7)$ we get:

                  $dx=d(1\rightarrow x)=(1\rightarrow dx)\Cup_{2}(d1\rightarrow x)=dx\Cup_{2}(d1\rightarrow x)$

                       $=(dx\rightsquigarrow(d1\rightarrow x))\rightarrow(d1\rightarrow x)$

                       $=(d1\rightarrow(dx\rightsquigarrow x))\rightarrow(d1\rightarrow x)$

                       $\geq(dx\rightsquigarrow x)\rightarrow x=dx\Cup_{2}x$.

On the other hand $dx\leq dx\Cup_{2}x$, hence $dx=dx\Cup_{2}x$.

Similarly we have:

                  $dx=d(1\rightsquigarrow x)=(1\rightsquigarrow dx)\Cup_{1}(d1\rightsquigarrow x)=dx\Cup_{1}(d1\rightsquigarrow x)$

                       $=(dx\rightarrow(d1\rightsquigarrow x))\rightsquigarrow(d1\rightsquigarrow x)$

                       $=(d1\rightsquigarrow(dx\rightarrow x))\rightsquigarrow(d1\rightsquigarrow x)$

                       $\geq(dx\rightarrow x)\rightsquigarrow x=dx\Cup_{1}x$.

Since $dx\leq dx\Cup_{1}x$, it follows that $dx=dx\Cup_{1}x$.

$(2)$ Let $d\in\mathcal{IDOP}^{(II)}(A)$ and let $x\in A$. Then:

$dx=d(1\rightarrow x)=(d1\rightarrow x)\Cup_{2}(1\rightarrow dx)=(d1\rightarrow x)\Cup_{2}dx$ and $dx=d(1\rightsquigarrow x)=(d1\rightsquigarrow x)\Cup_{1}(1\rightsquigarrow dx)=(d1\rightsquigarrow x)\Cup_{1}dx$. If $d1=1$, then $dx=x\Cup_{1}dx=x\Cup_{2}dx$. Conversely, if $dx=x\Cup_{1}dx=x\Cup_{2}dx$ for all $x\in A$, then $d1=1\Cup_{1}d1=1\Cup_{2}d1=1$. ∎

For any $x\in A$, define $\varphi:A\longrightarrow A$ by $\varphi(x)=\varphi_{x}=x\Cup_{1}1$. By Lemma 2.2$(11)$ we have $\varphi_{x}=x\Cup_{1}1=x\Cup_{2}1=(x\rightarrow 1)\rightarrow 1=(x\rightsquigarrow 1)\rightsquigarrow 1$.

Lemma 3.6**.**

*In any pseudo-BCI algevra $A$ the following hold for any $x\in A$:

$(1)$ $\varphi_{x}\in{\rm At}\,(A);$

$(2)$ $\varphi_{x}\rightarrow x,\varphi_{x}\rightsquigarrow x\in{\rm K}\,(A);$

$(3)$ $x\in{\rm K}\,(A)$ implies $\varphi_{x}=1$.*

Proof.

$(1)$ Since by Lemma 2.6$(7)$, $(\varphi_{x}\rightarrow 1)\rightsquigarrow 1=(x\Cup_{1}1\rightarrow 1)\rightsquigarrow 1=(x\rightarrow 1)\rightsquigarrow 1=\varphi_{x}$ and $(\varphi_{x}\rightsquigarrow 1)\rightarrow 1=(x\Cup_{2}1\rightsquigarrow 1)\rightarrow 1=(x\rightsquigarrow 1)\rightarrow 1=\varphi_{x}$, applying Proposition 2.4$(b)$ it follows that $\varphi_{x}\in{\rm At}\,(A)$.

$(2)$ From $x\leq\varphi_{x}$ we have $\varphi_{x}\rightarrow 1\leq x\rightarrow 1$, that is $(\varphi_{x}\rightarrow 1)\rightsquigarrow(x\rightarrow 1)=1$. Applying Lemma 2.2$(12)$ we get $(\varphi_{x}\rightarrow x)\rightarrow 1=1$, hence $\varphi_{x}\rightarrow x\leq 1$, that is $\varphi_{x}\rightarrow x\in{\rm K}\,(A)$. Similarly, $\varphi_{x}\rightsquigarrow x\in{\rm K}\,(A)$.

$(3)$ It is obvious. ∎

Proposition 3.7**.**

Let $A$ be a pseudo-BCI algebra and let $d_{\varphi}:A\longrightarrow A$, defined by $dx=\varphi_{x}$ for any $x\in A$. Then $d_{\varphi}\in\mathcal{IDOP}^{(I)}(A)$.

Proof.

Let $A$ be a pseudo-BCI algebra and let $x,y\in A$. By Lemma 3.6, $\varphi_{y}\in{\rm At}\,(A)$ and by Corollary 2.5, $x\rightarrow\varphi_{y},x\rightsquigarrow\varphi_{y}\in{\rm At}\,(A)$. Applying Lemma 2.2$(12)$ and Proposition 2.4$(b)$ we get:

          $d(x\rightarrow y)=\varphi_{x\rightarrow y}=((x\rightarrow y)\rightarrow 1)\rightsquigarrow 1=((x\rightarrow 1)\rightsquigarrow(y\rightarrow 1))\rightsquigarrow 1$

                        $=((x\rightarrow 1)\rightsquigarrow 1)\rightarrow((y\rightarrow 1)\rightsquigarrow 1)=((x\rightarrow 1)\rightsquigarrow 1)\rightarrow\varphi_{y}$

                        $=((x\rightarrow 1)\rightsquigarrow 1)\rightarrow((\varphi_{y}\rightarrow 1)\rightsquigarrow 1)=((x\rightarrow 1)\rightsquigarrow(\varphi_{y}\rightarrow 1))\rightsquigarrow 1$

                        $=((x\rightarrow\varphi_{y})\rightarrow 1)\rightsquigarrow 1=x\rightarrow\varphi_{y}=((x\rightarrow\varphi_{y})\rightsquigarrow(\varphi_{x}\rightarrow y))\rightarrow(\varphi_{x}\rightarrow y)$

                        $=(x\rightarrow\varphi_{y})\Cup_{2}(\varphi_{x}\rightarrow y)=(x\rightarrow dy)\Cup_{2}(dx\rightarrow y)$.

Similarly we have:

          $d(x\rightsquigarrow y)=\varphi_{x\rightsquigarrow y}=((x\rightsquigarrow y)\rightarrow 1)\rightsquigarrow 1=((x\rightsquigarrow y)\rightsquigarrow 1)\rightarrow 1$

                        $=((x\rightsquigarrow 1)\rightarrow(y\rightsquigarrow 1))\rightarrow 1=((x\rightsquigarrow 1)\rightarrow 1)\rightsquigarrow((y\rightsquigarrow 1)\rightarrow 1)$

                        $=((x\rightsquigarrow 1)\rightarrow 1)\rightsquigarrow\varphi_{y}=((x\rightsquigarrow 1)\rightarrow 1)\rightsquigarrow((\varphi_{y}\rightsquigarrow 1)\rightarrow 1)$

                        $=((x\rightsquigarrow 1)\rightarrow(\varphi_{y}\rightsquigarrow 1))\rightarrow 1=((x\rightsquigarrow\varphi_{y})\rightsquigarrow 1)\rightarrow 1=x\rightsquigarrow\varphi_{y}$

                        $=((x\rightsquigarrow\varphi_{y})\rightarrow(\varphi_{x}\rightsquigarrow y))\rightsquigarrow(\varphi_{x}\rightsquigarrow y)=(x\rightsquigarrow\varphi_{y})\Cup_{1}(\varphi_{x}\rightsquigarrow y)$

                        $=(x\rightsquigarrow dy)\Cup_{1}(dx\rightsquigarrow y)$.

We conclude that $d_{\varphi}\in\mathcal{IDOP}^{(I)}(A)$. ∎

Theorem 3.8**.**

Let $A$ be a commutative pseudo-BCI algebra, and let $d_{\varphi}:A\longrightarrow A$, defined by $d_{\varphi}x=\varphi_{x}$ for any $x\in A$. Then $d_{\varphi}\in\mathcal{IDOP}(A)$.

Proof.

By Proposition 3.7, we have $d_{\varphi}\in\mathcal{IDOP}^{(I)}(A)$, and let $x,y\in A$. As we mentioned in the proof of Proposition 3.7, $\varphi_{x},\varphi_{y},x\rightarrow\varphi_{y},x\rightsquigarrow\varphi_{y}\in{\rm At}\,(A)$. Then we have:

                  $x\rightarrow\varphi_{y}=((x\rightarrow\varphi_{y})\rightarrow 1)\rightsquigarrow 1=((x\rightarrow 1)\rightsquigarrow(\varphi_{y}\rightarrow 1))\rightsquigarrow 1$

                              $=((x\rightarrow 1)\rightsquigarrow 1)\rightarrow((\varphi_{y}\rightarrow 1)\rightsquigarrow 1)=\varphi_{x}\rightarrow\varphi_{y}$,

hence $x\rightarrow\varphi_{y}\in V(\varphi_{x}\rightarrow\varphi_{y})$. Moreover:

                  $\varphi_{x}\rightarrow y\leq((\varphi_{x}\rightarrow y)\rightarrow 1)\rightsquigarrow 1=((\varphi_{x}\rightarrow 1)\rightsquigarrow(y\rightarrow 1))\rightsquigarrow 1$

                              $=((\varphi_{x}\rightarrow 1)\rightsquigarrow 1)\rightarrow((y\rightarrow 1)\rightsquigarrow 1)=\varphi_{x}\rightarrow\varphi_{y}$,

that is $\varphi_{x}\rightarrow y\in V(\varphi_{x}\rightarrow\varphi_{y})$. Since $A$ is commutative, it follows that $A$ is branch commutative, and in the similar way to the proof of Proposition 3.7 we get:

          $(d_{\varphi}x\rightarrow y)\Cup_{2}(x\rightarrow d_{\varphi}y)=(\varphi_{x}\rightarrow y)\Cup_{2}(x\rightarrow\varphi_{y})=(x\rightarrow\varphi_{y})\Cup_{2}(\varphi_{x}\rightarrow y)$

                                              $=\varphi_{x\rightarrow y}=d_{\varphi}(x\rightarrow y)$.

Similarly we have:

          $(d_{\varphi}x\rightsquigarrow y)\Cup_{1}(x\rightsquigarrow d_{\varphi}y)=(\varphi_{x}\rightsquigarrow y)\Cup_{1}(x\rightsquigarrow\varphi_{y})=(x\rightsquigarrow\varphi_{y})\Cup_{1}(\varphi_{x}\rightsquigarrow y)$

                                              $=\varphi_{x\rightsquigarrow y}=d_{\varphi}(x\rightsquigarrow y)$,

hence $d_{\varphi}\in\mathcal{IDOP}^{(II)}(A)$. We conclude that $d_{\varphi}\in\mathcal{IDOP}(A)$. ∎

Definition 3.9**.**

Let $A$ be a pseudo-BCI algebra $A$. A type I or a type II derivation $d$ on $A$ is said to be:

$(1)$ regular if $d1=1;$

$(2)$ isotone if $x\leq y$ implies $dx\leq dy;$

$(3)$ idempotent if $d^{2}=d$, where $d^{2}=d\circ d$.

Example 3.10**.**

Consider the pseudo-BCI algebra $A$ and its derivations from Example 3.4. Then:

$(1)$ $d_{\varphi}=d_{3}\in\mathcal{IDOP}(A)$.

$(2)$ $d_{1}$ and $d_{3}$ are regular implicative derivations, while $d_{2}$ is not regular.

$(3)$ $d_{1}$, $d_{2}$, $d_{3}$ are isotone.

$(4)$ $d_{1}$ and $d_{3}$ are idempotent.

Remark 3.11**.**

In Theorem 3.8 the pseudo-BCI algebra $A$ need not be commutative. Indeed, the pseudo-BCI algebra from Example 3.4 is not commutative, but $d_{\varphi}=d_{3}\in\mathcal{IDOP}(A)$.

Let $A$ be a pseudo-BCI algebra. Denote:

          $\mathcal{RIDOP}^{(I)}(A)$ the set of all regular implicative derivation operators of type I on $A$,

          $\mathcal{RIDOP}^{(II)}(A)$ the set of all regular implicative derivation operators of type II on $A$,

          $\mathcal{RIDOP}(A)=\mathcal{RIDOP}^{(I)}(A)\cap\mathcal{RIDOP}^{(II)}(A)$.

Proposition 3.12**.**

*Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{RIDOP}^{(II)}(A)$. Then the following hold for all $x,y\in A$:

$(1)$ $x\leq dx;$

$(2)$ $dx\rightarrow y\leq dx\rightarrow dy\leq x\rightarrow dy=d(x\rightarrow y)$ and $dx\rightsquigarrow y\leq dx\rightsquigarrow dy\leq x\rightsquigarrow dy=d(x\rightsquigarrow y);$

$(3)$ $d$ is isotone;

$(4)$ ${\rm Ker}\,(d)=\{x\in A\mid dx=1\}$ is a subalgebra of $A;$

$(5)$ ${\rm Ker}\,(d)\subseteq{\rm K}\,(A);$

$(6)$ $d({\rm K}\,(A))\subseteq{\rm K}\,(A);$

$(7)$ $\varphi_{x}$ and $dx$ belong to the same branch of $A;$

$(8)$ $\varphi_{x}\rightarrow dx,\varphi_{x}\rightsquigarrow dx\in{\rm K}\,(A)$.*

Proof.

$(1)$ By Lemma 2.6$(5)$ and Proposition 3.5$(2)$, $x\leq x\Cup_{1}dx=dx$.

$(2)$ From $y\leq dy$ we have $dx\rightarrow y\leq dx\rightarrow dy$, and by $x\leq dx$ we get $dx\rightarrow dy\leq x\rightarrow dy$. It follows that:

                  $dx\rightarrow y\leq dx\rightarrow dy\leq x\rightarrow dy=1\rightarrow(x\rightarrow dy)$

                              $=((dx\rightarrow y)\rightsquigarrow(x\rightarrow dy))\rightarrow(x\rightarrow dy)$

                              $=(dx\rightarrow y)\Cup_{2}(x\rightarrow dy)=d(x\rightarrow y)$.

Similarly, $dx\rightsquigarrow y\leq dx\rightsquigarrow dy\leq x\rightsquigarrow dy=d(x\rightsquigarrow y)$.

$(3)$ Let $x,y\in A$ such that $x\leq y$, that is $x\rightarrow y=1$. Then we have $dx\rightarrow y\leq(dx\rightarrow y)\Cup_{2}(x\rightarrow dy)=d(x\rightarrow y)=d1=1$. It follows that $dx\leq y\leq dy$, hence $d$ is isotone.

$(4)$ Since $d1=1$, it follows that $1\in{\rm Ker}\,(d)$. Let $x,y\in{\rm Ker}\,(d)$, that is $dx=dy=1$. Then we have $1=dx\rightarrow dy\leq d(x\rightarrow y)$, hence $d(x\rightarrow y)=1$, that is $x\rightarrow y\in{\rm Ker}\,(d)$. Similarly, $x\rightsquigarrow y\in{\rm Ker}\,(d)$, thus ${\rm Ker}\,(d)$ is a subalgebra of $A$.

$(5)$ For any $x\in{\rm Ker}\,(d)$ we have $x\leq dx=1$, that is $x\in{\rm K}\,(A)$. Hence ${\rm Ker}\,(d)\subseteq{\rm K}\,(A)$.

$(6)$ If $x\in d({\rm K}\,(A))$, there exists $x^{\prime}\in{\rm K}\,(A)$ such that $x=dx^{\prime}$. Since $x^{\prime}\leq 1$ and $x^{\prime}\leq dx^{\prime}=x$, applying Lemma 2.2 we get $x=1\rightarrow x\leq x^{\prime}\rightarrow x=1$, that is $x\in{\rm K}\,(A)$. It follows that $d({\rm K}\,(A))\subseteq{\rm K}\,(A)$.

$(7)$ From $x\leq dx$, it follows that $x$ and $dx$ are in a branch $V_{1}$ of $A$, while from $x\leq\varphi_{x}$ we get that $x$ and $\varphi_{x}$ are in a branch $V_{2}$ of $A$. Since $x\in V_{1}\cap V_{2}$, it follows that $V_{1}$ and $V_{2}$ coincide, hence $x$, $\varphi_{x}$ and $dx$ belong to the same branch of $A$.

$(8)$ According to $(7)$, there exists $a\in{\rm At}\,(A)$ such that $\varphi_{x},dx\in V(a)$. From $\varphi_{x}\leq a$, $dx\leq a$ we get $\varphi_{x}\rightarrow dx\leq\varphi_{x}\rightarrow a=1$, that is $\varphi_{x}\rightarrow dx\in{\rm K}\,(A)$. Similarly, $\varphi_{x}\rightsquigarrow dx\in{\rm K}\,(A)$. ∎

Proposition 3.13**.**

Let $A$ be a pseudo-BCI algebra and let $d_{1},d_{2}\in\mathcal{RIDOP}^{(II)}(A)$ such that $d_{2}$ is idempotent and $d_{1}\leq d_{2}$ (that is $d_{1}x\leq d_{2}x$, for all $x\in A$). Then $d_{2}\circ d_{1}=d_{2}$.

Proof.

Let $x\in A$. By Proposition 3.12$(1)$,$(3)$, $d_{2}x\leq d_{2}d_{1}x=(d_{2}\circ d_{1})(x)$, so $d_{2}\leq d_{2}\circ d_{1}$. Moreover, since $d_{1}x\leq d_{2}x$ we have $d_{2}d_{1}x\leq d_{2}d_{2}x=d^{2}_{2}x=d_{2}x$, that is $d_{2}\circ d_{1}\leq d_{2}$. Hence $d_{2}\circ d_{1}=d_{2}$. ∎

Proposition 3.14**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{RIDOP}^{(II)}(A)$. Then ${\rm Ker}\,(d)={\rm K}\,(A)$ if and only if $d=d_{\varphi}$.

Proof.

Assume that ${\rm Ker}\,(d)={\rm K}\,(A)$ and let $x\in A$. Since by Lemma 3.6, $\varphi_{x}\rightarrow x\in{\rm K}\,(A)$, we have $d(\varphi_{x}\rightarrow x)=1$. Applying Proposition 3.12$(2)$ we get $1=d(\varphi_{x}\rightarrow x)=\varphi_{x}\rightarrow dx$, hence $\varphi_{x}\leq dx$. On the other hand $\varphi_{x}\in{\rm At}\,(A)$, thus $dx=\varphi_{x}$. Conversely, assume that $dx=\varphi_{x}$ for all $x\in A$. Since for any $x\in{\rm K}\,(A)$, $dx=\varphi_{x}=1$, we have $x\in{\rm Ker}\,(d)$, so that ${\rm K}\,(A)\subseteq{\rm Ker}\,(d)$. By Proposition 3.12$(5)$, ${\rm Ker}\,(d)\subseteq{\rm K}\,(A)$, hence ${\rm Ker}\,(d)={\rm K}\,(A)$. ∎

Corollary 3.15**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{RIDOP}^{(II)}(A)$ such that ${\rm Ker}\,(d)={\rm K}\,(A)$. Then $d$ is idempotent.

Proof.

According to Proposition 3.14, for any $x\in A$ we have: $d^{2}x=ddx=d_{\varphi_{dx}}=(dx\rightarrow 1)\rightsquigarrow 1=(\varphi_{x}\rightarrow 1)\rightsquigarrow 1=\varphi_{x}=dx$, hence $d^{2}=d$. ∎

Corollary 3.16**.**

Let $A$ be a pseudo-BCI algebra and let $d_{1},d_{2}\in\mathcal{RIDOP}^{(II)}(A)$ such that $d_{1}\leq d_{2}$ and ${\rm Ker}\,(d_{1})={\rm K}\,(A)$. Then $d_{2}\circ d_{1}=d_{2}$.

Proof.

Since $d_{1}\leq d_{2}$, we have ${\rm Ker}\,(d_{1})\subseteq{\rm Ker}\,(d_{2})$, that is ${\rm Ker}\,(d_{1})={\rm Ker}\,(d_{2})={\rm K}\,(A)$. According to Corollary 3.15, $d_{1}$ and $d_{2}$ are idempotent, and applying Proposition 3.13 it follows that $d_{2}\circ d_{1}=d_{2}$. ∎

Example 3.17**.**

With the notations from Example 3.4, we have:

$(1)$ ${\rm K}\,(A)=\{a,b,c,1\};$

$(2)$ ${\rm Ker}\,(d_{3})=\{a,b,c,1\}={\rm K}\,(A)$ and $d_{3}^{2}=d_{3};$

$(3)$ $d_{1}({\rm K}\,(A))={\rm K}\,(A)$ and $d_{3}({\rm K}\,(A))=\{1\}\subseteq{\rm K}\,(A)$.

Proposition 3.18**.**

*Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}(A)$. If there exists $a\in A$ such that $a\leq dx$ for all $x\in A$, then:

$(1)$ $d\in\mathcal{RIDOP}(A);$

$(2)$ $A$ is a pseudo-BCK algebra.*

Proof.

$(1)$ From $a\leq dx$, we have $a\rightarrow dx=a\rightsquigarrow dx=1$ for all $x\in A$, so $a\rightarrow d(a\rightarrow x)=1$ and $a\rightarrow d1=a\rightsquigarrow d1=1$. Then we get $1=a\rightarrow d(a\rightarrow x)=a\rightarrow(a\rightarrow dx)\Cup_{2}(da\rightarrow x)=a\rightarrow(1\Cup_{2}(da\rightarrow x))=a\rightarrow 1$ and similarly, $a\rightsquigarrow 1=1$. It follows that $d1=d(a\rightarrow 1)=(da\rightarrow 1)\Cup_{1}(a\rightarrow d1)=(da\rightarrow 1)\Cup_{1}1=((da\rightarrow 1)\rightarrow 1)\rightsquigarrow 1=1$ (from $1\in{\rm At}\,(A)$ we have $da\rightarrow 1\in{\rm At}\,(A)$, hence $(da\rightarrow 1)\rightarrow 1)\rightsquigarrow 1=1$). Hence $d1=1$, so $d\in\mathcal{RIDOP}(A)$.

$(2)$ Since $d$ is regular, we have $x\leq dx$, thus $dx\rightarrow 1\leq x\rightarrow 1$ for all $x\in A$. Then we have: $x\rightarrow 1\geq dx\rightarrow 1=dx\rightarrow(a\rightsquigarrow dx)=a\rightsquigarrow(dx\rightarrow dx)=a\rightsquigarrow 1=1$, hence $x\rightarrow 1=1$. We conclude that $A$ is a pseudo-BCK algebra. ∎

Lemma 3.19**.**

*Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(I)}(A)$. Then:

$(1)$ $d1\in{\rm At}\,(A);$

$(2)$ $da=(a\rightarrow 1)\rightarrow d1=(a\rightarrow 1)\rightsquigarrow d1=a\cdot d1$, for all $a\in{\rm At}\,(A);$

$(3)$ $d(dx\rightarrow x)=d(dx\rightsquigarrow x)=1$, for all $x\in A$.*

Proof.

$(1)$ By Proposition 3.5$(1)$, $d1=d1\Cup_{1}1=d1\Cup_{2}1=(d1\rightarrow 1)\rightsquigarrow 1=(d1\rightsquigarrow 1)\rightarrow 1$, hence by Proposition 2.4$(k)$, $d1\in{\rm At}\,(A)$.

$(2)$ Applying Lemma 2.2$(8)$,$(7)$,$(2)$ we have:

                  $da=d((a\rightarrow 1)\rightsquigarrow 1)=((a\rightarrow 1)\rightsquigarrow d1)\Cup_{1}(d(a\rightarrow 1)\rightsquigarrow 1)$

                       $=(((a\rightarrow 1)\rightsquigarrow d1)\rightarrow(d(a\rightarrow 1)\rightsquigarrow 1))\rightsquigarrow(d(a\rightarrow 1)\rightsquigarrow 1)$

                       $=(d(a\rightarrow 1)\rightsquigarrow(((a\rightarrow 1)\rightsquigarrow d1)\rightarrow 1))\rightsquigarrow(d(a\rightarrow 1)\rightsquigarrow 1)$

                       $\geq(((a\rightarrow 1)\rightsquigarrow d1)\rightarrow 1)\rightsquigarrow 1\geq(a\rightarrow 1)\rightsquigarrow d1$.

Since by $(1)$, $d1\in{\rm At}\,(A)$, according to Corollary 2.5, $(a\rightarrow 1)\rightsquigarrow d1\in{\rm At}\,(A)$. It follows that $da=(a\rightarrow 1)\rightsquigarrow d1$. Similarly, $da=(a\rightarrow 1)\rightarrow d1$. Obviously $da=a\cdot d1$.

$(3)$ For any $x\in A$ we have $d(dx\rightarrow x)=(dx\rightarrow dx)\Cup_{2}(x\rightarrow ddx)=1\Cup_{2}(x\rightarrow ddx)=1$ and similarly, $d(dx\rightsquigarrow x)=1$. ∎

Lemma 3.20**.**

*Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(II)}(A)$. Then:

$(1)$ $d1\rightarrow x\leq dx$ and $d1\rightsquigarrow x\leq dx$, for all $x\in A;$

$(2)$ $da=d1\rightarrow a=d1\rightsquigarrow a=d1\cdot a$, for all $a\in{\rm At}\,(A)$.*

Proof.

$(1)$ For all $x\in A$ we have $dx=d(1\rightarrow x)=(d1\rightarrow x)\Cup_{2}dx\geq d1\rightarrow x$ and $dx=d(1\rightsquigarrow x)=(d1\rightsquigarrow x)\Cup_{1}dx\geq d1\rightsquigarrow x$.

$(2)$ From $d1\rightarrow a\leq da$, since $a\in{\rm At}\,(A)$ we get $d1\rightarrow a\in{\rm At}\,(A)$, hence $da=d1\rightarrow a$. Similarly, $da=d1\rightsquigarrow a$. Since $d1=d1\rightarrow 1$, we have $da=d1\rightsquigarrow a=(d1\rightarrow 1)\rightsquigarrow a=d1\cdot a$. ∎

Proposition 3.21**.**

*Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(I)}(A)\cup\mathcal{IDOP}^{(II)}(A)$. Then:

$(1)$ $d({\rm At}\,(A))\subseteq{\rm At}\,(A);$

$(2)$ if $d1\in{\rm K}\,(A)$, then $d1=1;$

$(3)$ $d(a\cdot b)=da\cdot(d1\rightarrow 1)\cdot db$, for all $a,b\in{\rm At}\,(A);$

$(4)$ $d_{\mid{\rm At}\,(A)}=Id_{{\rm At}\,(A)}$ iff $d1=1$.*

Proof.

$(1)$ Let $a\in{\rm At}\,(A)$ and let $d\in\mathcal{IDOP}^{(I)}(A)$, by Lemma 3.19$(1)$, $d1\in{\rm At}\,(A)$, so $(a\rightarrow 1)\rightarrow d1\in{\rm At}\,(A)$. Hence by Lemma 3.19$(2)$, $da\in{\rm At}\,(A)$. If $d\in\mathcal{IDOP}^{(II)}(A)$, the assertion follows applying Lemma 3.20$(2)$.

$(2)$ Since $d1\leq 1$ and by $(1)$, $d1\in{\rm At}\,(A)$, we get $d1=1$.

$(3)$ Let $a,b\in{\rm At}\,(A)$, that is $a\cdot b=(a\rightarrow 1)\rightsquigarrow b\in{\rm At}\,(A)$. If $d\in\mathcal{IDOP}^{(I)}(A)$, by Lemma 3.19$(2)$ we have:

                  $a\cdot b=(a\cdot b)\cdot d1=(a\cdot d1)\cdot(d1)^{-1}\cdot(b\cdot d1)=da\cdot(d1\rightarrow 1)\cdot db$.

For the case of $d\in\mathcal{IDOP}^{(II)}(A)$, applying Lemma 3.20$(2)$ we get:

                  $a\cdot b=d1\cdot(a\cdot b)=(d1\cdot a)\cdot(d1)^{-1}\cdot(d1\cdot b)=da\cdot(d1\rightarrow 1)\cdot db$.

$(4)$ If $d_{\mid{\rm At}\,(A)}=Id_{{\rm At}\,(A)}$, then $d1=1$. Conversely, if $d\in\mathcal{IDOP}^{(I)}(A)\cup\mathcal{IDOP}^{(II)}(A)$ such that $d1=1$, then for all $x\in{\rm At}\,(A)$ we have $dx=x\cdot d1=x\cdot 1=x$ and $dx=d1\cdot x=1\cdot x=x$, respectively. Hence $d_{\mid{\rm At}\,(A)}=Id_{{\rm At}\,(A)}$. ∎

Example 3.22**.**

Let $A$ be the pseudo-BCI algebra and its derivations from Example 3.4. We can see that ${\rm At}\,(A)=\{d,1\}$, $\mathcal{RIDOP}^{(II)}(A)=\{d_{1},d_{3}\}$ and $d_{{1}_{\mid{\rm At}\,(A)}}=d_{{3}_{\mid{\rm At}\,(A)}}=Id_{{\rm At}\,(A)}$. Moreover, for each $d\in\{d_{1},d_{2},d_{3}\}$ we have $d({\rm At}\,(A))=\{d,1\}={\rm At}\,(A)$.

Theorem 3.23**.**

Let $A$ be a pseudo-BCI algebra and let $d:A\longrightarrow A$. Then $d\in\mathcal{RIDOP}^{(II)}(A)$ if and only if $d1=1$ and $d(x\rightarrow y)=x\rightarrow dy$, $d(x\rightsquigarrow y)=x\rightsquigarrow dy$, for all $x,y\in A$.

Proof.

Let $d\in\mathcal{RIDOP}^{(II)}(A)$, and let $x,y\in A$. Clearly $d1=1$, and according to Proposition 3.12$(2)$, $d(x\rightarrow y)=x\rightarrow dy$ and $d(x\rightsquigarrow y)=x\rightsquigarrow dy$. Conversely, let $d:A\longrightarrow A$ such that $d1=1$ and $d(x\rightarrow y)=x\rightarrow dy$, $d(x\rightsquigarrow y)=x\rightsquigarrow dy$, for all $x,y\in A$. Taking $x:=d1$, $y:=1$ we get $d1=d(d1\rightarrow 1)=d1\rightarrow d1=1$. For $y:=x$, we have $x\rightarrow dx=d(x\rightarrow x)=d1=1$, so $x\leq dx$ for all $x\in A$. From $x\leq dx$ and $y\leq dy$ we get $dx\rightarrow y\leq x\rightarrow y\leq x\rightarrow dy$ and $dx\rightsquigarrow y\leq x\rightsquigarrow y\leq x\rightsquigarrow dy$. It follows that $d(x\rightarrow y)=x\rightarrow dy=1\rightarrow(x\rightarrow dy)=((dx\rightarrow y)\rightsquigarrow(x\rightarrow dy))\rightarrow(x\rightarrow dy)=(dx\rightarrow y)\Cup_{2}(x\rightarrow dy)$. Similarly, $d(x\rightsquigarrow y)=(dx\rightsquigarrow y)\Cup_{1}(x\rightsquigarrow dy)$, hence $d\in\mathcal{RIDOP}^{(II)}(A)$. ∎

Proposition 3.24**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(I)}(A)$ such that ${\rm Im}\,(d)\subseteq{\rm At}\,(A)$. Then $d(x\rightarrow y)=x\rightarrow dy$ and $d(x\rightsquigarrow y)=x\rightsquigarrow dy$, for all $x,y\in A$.

Proof.

Let $x,y\in A$. Since $dy\in{\rm At}\,(A)$, according to Corollary 2.5, $x\rightarrow dy\in{\rm At}\,(A)$. From $x\rightarrow dy\leq(x\rightarrow dy)\Cup_{2}(dx\rightarrow y)=d(x\rightarrow y)$, it follows that $d(x\rightarrow y)=x\rightarrow dy$. Similarly, $d(x\rightsquigarrow y)=x\rightsquigarrow dy$. ∎

Remark 3.25**.**

If $d\in\mathcal{RIDOP}^{(I)}(A)\cup\mathcal{RIDOP}^{(II)}(A)$ such that $d(x\rightarrow y)=dx\rightarrow y$ or $d(x\rightsquigarrow y)=dx\rightsquigarrow y$ for all $x,y\in A$, then $d={\rm Id}\,_{A}$. Indeed, $dx=d(1\rightarrow x)=d1\rightarrow x=1\rightarrow x=x$ for all $x\in A$, that is $d={\rm Id}\,_{A}$. Similarly for $d(x\rightsquigarrow y)=dx\rightsquigarrow y$.

Definition 3.26**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}(A)$. Then $D\in\mathcal{DS}(A)$ is said to be $d$-invariant if $d(D)\subseteq D$.

Proposition 3.27**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(I)}(A)\cup\mathcal{IDOP}^{(II)}(A)$. If every deductive system of $A$ is $d$-invariant, then $d$ is regular.

Proof.

Let $d\in\mathcal{IDOP}^{(I)}(A)\cup\mathcal{IDOP}^{(II)}(A)$ and assume that every $D\in\mathcal{DS}(A)$ is $d$-invariant. Since $\{1\}\in\mathcal{DS}(A)$, it follows that $d(\{1\})\subseteq\{1\}$, hence $d(1)=1$, that is $d$ is regular. ∎

Proposition 3.28**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{RIDOP}^{(II)}(A)$. Then every deductive system of $A$ is $d$-invariant.

Proof.

Assume that $d\in\mathcal{RIDOP}^{(II)}(A)$. Let $D\in\mathcal{DS}(A)$ and let $y\in d(D)$, that is there exists $x\in D$ such that $y=dx$. Since by Proposition 3.12, $x\rightarrow y=x\rightarrow dx=1\in D$, it follows that $y\in D$. Hence $d(D)\subseteq D$, that is $D$ is $d$-invariant. ∎

Corollary 3.29**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(II)}(A)$. Then $d\in\mathcal{RIDOP}^{(II)}(A)$ if and only if every deductive system of $A$ is $d$-invariant.

Example 3.30**.**

Let $A$ be the pseudo-BCI algebra and its derivations from Example 3.4. We can see that $\mathcal{DS}(A)=\{\{1\},\{c,1\},\{a,b,c,1\},A\}$ and consider $d_{2}\in\mathcal{IDOP}^{(II)}(A)$, but $d_{2}\notin\mathcal{RIDOP}^{(II)}(A)$. For $D=\{c,1\}\in\mathcal{DS}(A)$, one can easily check that $d_{2}(D)=\{d\}\nsubseteq D$, hence $D$ is not $d_{2}$-invariant. On the other hand, for $d_{1},d_{3}\subseteq\mathcal{RIDOP}^{(II)}(A)$ we have $d_{1}(D)\subseteq D$ and $d_{3}(D)\subseteq D$, so that $D$ is $d_{1}$-invariant and $d_{3}$-invariant.

Example 3.31**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{IDOP}^{(II)}(A)$. According to [16, Prop. 12], ${\rm K}\,(A)$ is a closed compatible deductive system of $A$. By Proposition 3.12, $d({\rm K}\,(A))\subseteq{\rm K}\,(A)$, hence ${\rm K}\,(A)$ is $d$-invariant.

4. Implicative derivation operators on $p$-semisimple pseudo-BCI algebras

In this section we investigate the particular case of implicative derivations on the $p$-semisimple pseudo-BCI algebras, and we prove that a pseudo-BCI algebra $A$ is $p$-semisimple if and only if the only regular derivation of type II on $A$ is the identity map. It is also proved that a pseudo-BCI algebra $A$ is $p$-semisimple if and only if the kernel of any regular implicative derivation $d$ of type II on $A$ is the set $\{1\}$. As a corollary it is proved that, for any pseudo-BCI algebra $A$, the only regular implicative derivation of type II on $A/{\rm K}\,(A)$ is the identity map. It is also proved that the set of all implicative derivations of a $p$-semisimple pseudo-BCI algebra forms a commutative monoid with respect to function composition.

Theorem 4.1**.**

*If $A$ is a pseudo-BCI algebra, the following are equivalent:

$(a)$ $A$ is $p$-semisimple;

$(b)$ ${\rm Ker}\,(d)=\{1\}$ for all $d\in\mathcal{RIDOP}^{(II)}(A);$

$(c)$ $\mathcal{RIDOP}^{(II)}(A)=\{{\rm Id}\,_{A}\}$.*

Proof.

$(a)\Rightarrow(b)$ Let $A$ be a $p$-semisimple pseudo-BCI algebra, that is ${\rm K}\,(A)=\{1\}$, and let $d\in\mathcal{RIDOP}^{(II)}(A)$. According to Proposition 3.12$(5)$, ${\rm Ker}\,(d)\subseteq{\rm K}\,(A)$, hence ${\rm Ker}\,(d)=\{1\}$.

$(b)\Rightarrow(a)$ Assume that ${\rm Ker}\,(d)=\{1\}$ for all $d\in\mathcal{RIDOP}^{(II)}(A)$. Consider $d_{\varphi}:A\longrightarrow A$, defined by $d_{\varphi}x=\varphi_{x}$ for any $x\in A$. Since any $p$-semisimple pseudo-BCI algebra is commutative, applying Theorem 3.8 it follows that $d_{\varphi}\in\mathcal{IDOP}^{(II)}(A)$. Taking into consideration that $d_{\varphi}1=1$, we have $d_{\varphi}\in\mathcal{RIDOP}^{(II)}(A)$, so that ${\rm Ker}\,(d_{\varphi})=\{1\}$. Moreover, for any $x\in{\rm K}\,(A)$, $d_{\varphi}x=(x\rightarrow 1)\rightsquigarrow 1=1$, that is $x\in{\rm Ker}\,(d_{\varphi})$. Thus ${\rm K}\,(A)\subseteq{\rm Ker}\,(d_{\varphi})$, and applying Proposition 3.12$(5)$ we get ${\rm K}\,(A)={\rm Ker}\,(d_{\varphi})=\{1\}$, hence $A$ is $p$-semisimple.

$(b)\Rightarrow(c)$ If $d\in\mathcal{RIDOP}^{(II)}(A)$, then by $(b)$ we have ${\rm Ker}\,(d)=\{1\}$. Since $(b)\Rightarrow(a)$ and by Proposition 2.3, ${\rm At}\,(A)=A$, applying Proposition 3.21 we get $d={\rm Id}\,_{A}$.

$(c)\Rightarrow(b)$ Let $d\in\mathcal{RIDOP}^{(II)}(A)$, so by $(c)$ we have $d={\rm Id}\,_{A}$, that is ${\rm Ker}\,(d)=\{1\}$. ∎

Corollary 4.2**.**

If $A$ is a pseudo-BCI algebra, then $\mathcal{RIDOP}^{(II)}({\rm At}\,(A))=\{{\rm Id}\,_{A}\}$.

Proof.

According to [13, Th. 4.13], ${\rm At}\,(A)$ is a $p$-semisimple pseudo-BCI subalgebra of $A$, and by Theorem 4.1 it follows that $\mathcal{RIDOP}^{(II)}({\rm At}\,(A))=\{{\rm Id}\,_{A}\}$. ∎

Corollary 4.3**.**

If $A$ is a pseudo-BCI algebra, then $\mathcal{RIDOP}^{(II)}(A/{\rm K}\,(A))=\{{\rm Id}\,_{A}\}$.

Proof.

According to [16, Prop.12, Th. 6], ${\rm K}\,(A)$ is a closed compatible deductive system of $A$ and $A/{\rm K}\,(A)$ is a $p$-semisimple pseudo-BCI algebra. Applying Theorem 4.1, it follows that

$\mathcal{RIDOP}^{(II)}(A/{\rm K}\,(A))=\{{\rm Id}\,_{A}\}$. ∎

Example 4.4**.**

Consider the structure $(A,\rightarrow,\rightsquigarrow,1)$, where the operations $\rightarrow$ and $\rightsquigarrow$ on $A=\{a,b,c,d,e,1\}$ are defined as follows:

[TABLE]

Then $(A,\rightarrow,\rightsquigarrow,1)$ is a $p$-semisimple pseudo-BCI algebra ([16]), and one can check that:

$(1)$ ${\rm At}\,(A)=A$, ${\rm K}\,(A)=\{1\}$ and $A/{\rm K}\,(A)=A;$

$(2)$ $\mathcal{RIDOP}^{(II)}(A)=\mathcal{RIDOP}^{(II)}({\rm At}\,(A))=\mathcal{RIDOP}^{(II)}(A/{\rm K}\,(A))=\{{\rm Id}\,_{A}\}$.

Example 4.5**.**

Let $A$ be the pseudo-BCI algebra and its derivations from Example 3.4. One can easily check that:

$(1)$ ${\rm At}\,(A)=\{d,1\};$

$(2)$ ${\rm At}\,(A)$ is a $p$-semisimple subalgebra of $A;$

$(3)$ $\mathcal{RIDOP}^{(II)}({\rm At}\,(A))=Id_{{\rm At}\,(A)}$.

Remark 4.6**.**

The notion of a $\rightarrow$medial pseudo-BCI algebra was defined in [31] as a pseudo-BCI algebra $A$ satisfying the identity $(u\rightsquigarrow v)\rightarrow(x\rightsquigarrow y)=(u\rightsquigarrow x)\rightarrow(v\rightsquigarrow y)$, for all $x,y,u,v\in A$. It was proved in [13] that a $\rightarrow$medial pseudo-BCI algebra is a $p$-semisimple BCI-algebra, so that by Theorem 4.1, $\mathcal{RIDOP}^{(II)}(A)=\{{\rm Id}\,_{A}\}$. Similarly for the case of a $\rightsquigarrow$medial pseudo-BCI algebra which is a pseudo-BCI algebra $A$ satisfying the identity $(u\rightarrow v)\rightsquigarrow(x\rightarrow y)=(u\rightarrow x)\rightsquigarrow(v\rightarrow y)$, for all $x,y,u,v\in A$ ([36]).

Proposition 4.7**.**

Let $A$ be a $p$-semisimple pseudo-BCI algebra and let $d_{1},d_{2}\in\mathcal{IDOP}^{(I)}(A)$. Then $d_{1}\circ d_{2}\in\mathcal{IDOP}^{(I)}(A)$.

Proof.

According to Proposition 2.3$(c)$, $x\Cup_{1}y=x\Cup_{2}y=x$ for all $x,y\in A$.

If $d_{1},d_{2}\in\mathcal{IDOP}^{(I)}(A)$ we have:

                  $(d_{1}\circ d_{2})(x\rightarrow y)=d_{1}d_{2}(x\rightarrow y)=d_{1}((x\rightarrow d_{2}y)\Cup_{2}(d_{2}x\rightarrow y))$

                                          $=d_{1}(x\rightarrow d_{2}y)=(x\rightarrow d_{1}d_{2}y)\Cup_{2}(d_{1}x\rightarrow d_{2}y)$

                                          $=x\rightarrow d_{1}d_{2}y=(x\rightarrow d_{1}d_{2}y)\Cup_{2}(d_{1}d_{2}x\rightarrow y)$

                                          $=(x\rightarrow(d_{1}\circ d_{2})(y))\Cup_{2}((d_{1}\circ d_{2})(x)\rightarrow y)$.

Similarly, $(d_{1}\circ d_{2})(x\rightsquigarrow y)=(x\rightsquigarrow(d_{1}\circ d_{2})(y))\Cup_{1}((d_{1}\circ d_{2})(x)\rightsquigarrow y)$, hence $d_{1}\circ d_{2}\in\mathcal{IDOP}^{(I)}(A)$. ∎

Proposition 4.8**.**

Let $A$ be a $p$-semisimple pseudo-BCI algebra and let $d_{1},d_{2}\in\mathcal{IDOP}^{(II)}(A)$. Then $d_{1}\circ d_{2}\in\mathcal{IDOP}^{(II)}(A)$.

Proof.

Let $d_{1},d_{2}\in\mathcal{IDOP}^{(II)}(A)$. Similar to Proposition 4.7 we get:

                  $(d_{1}\circ d_{2})(x\rightarrow y)=d_{1}d_{2}(x\rightarrow y)=d_{1}((d_{2}x\rightarrow y)\Cup_{2}(x\rightarrow d_{2}y))$

                                          $=d_{1}(d_{2}x\rightarrow y)=(d_{1}d_{2}x\rightarrow y)\Cup_{2}(d_{2}x\rightarrow d_{1}y)$

                                          $=d_{1}d_{2}x\rightarrow y=(d_{1}d_{2}x\rightarrow y)\Cup_{2}(x\rightarrow d_{1}d_{2}y)$

                                          $=((d_{1}\circ d_{2})(x)\rightarrow y)\Cup_{2}(x\rightarrow(d_{1}\circ d_{2})(y))$.

Similarly, $(d_{1}\circ d_{2})(x\rightsquigarrow y)=((d_{1}\circ d_{2})(x)\rightsquigarrow y)\Cup_{1}(x\rightsquigarrow(d_{1}\circ d_{2})(y))$, thus $d_{1}\circ d_{2}\in\mathcal{IDOP}^{(II)}(A)$. ∎

Proposition 4.9**.**

Let $A$ be a $p$-semisimple pseudo-BCI algebra and let $d_{1},d_{2}\in\mathcal{IDOP}(A)$. Then $d_{1}\circ d_{2}=d_{2}\circ d_{1}$.

Proof.

Let $d_{1},d_{2}\in\mathcal{IDOP}(A)$. Using the definitions of type I and type II derivations we get:

                  $(d_{1}\circ d_{2})(x)=(d_{1}\circ d_{2})(1\rightarrow x)=d_{1}((1\rightarrow d_{2}x)\Cup_{2}(d_{2}1\rightarrow x))$

                                    $=d_{1}(1\rightarrow d_{2}x)=(d_{1}1\rightarrow d_{2}x)\Cup_{2}(1\rightarrow d_{1}d_{2}x)=d_{1}1\rightarrow d_{2}x$.

                  $(d_{2}\circ d_{1})(x)=(d_{2}\circ d_{1})(1\rightarrow x)=d_{2}((d_{1}1\rightarrow x)\Cup_{2}(1\rightarrow d_{1}x))$

                                    $=d_{2}(d_{1}1\rightarrow x)=(d_{1}1\rightarrow d_{2}x)\Cup_{2}(d_{2}d_{1}1\rightarrow x)=d_{1}1\rightarrow d_{2}x$.

Hence $(d_{1}\circ d_{2})(x)=(d_{2}\circ d_{1})(x)$ for all $x\in A$. It follows that $d_{1}\circ d_{2}=d_{2}\circ d_{1}$. ∎

Theorem 4.10**.**

If $A$ is a $p$-semisimple pseudo-BCI algebra then $(\mathcal{IDOP}(A),\circ,{\rm Id}\,_{A})$ is a commutative monoid.

Proof.

It follows by Propositions 4.7, 4.8, 4.9 taking into consideration that the composition of functions is always associative. ∎

Remark 4.11**.**

In Theorem 4.10 the pseudo-BCI algebra $A$ need not be $p$-semisimple. Indeed, consider the pseudo-BCI algebra $A$ and its derivations from Example 3.4. Since $a\leq b$ and $a\neq b$, the condition $(b)$ from Proposition 2.3 is not satisfied, so $A$ is not $p$-semisimple. On the other hand we have $d_{1}\circ d_{2}=d_{2}\circ d_{1}=d_{2}$, $d_{1}\circ d_{3}=d_{3}\circ d_{1}=d_{3}$, $d_{2}\circ d_{3}=d_{3}\circ d_{2}=d_{2}$, hence $(\mathcal{IDOP}(A),\circ,{\rm Id}\,_{A})$ is a commutative monoid.

Proposition 4.12**.**

Let $A$ be a $p$-semisimple pseudo-BCI algebra and let $d_{1},d_{2}\in\mathcal{IDOP}(A)$. Define $(d_{1}\rightarrow d_{2})(x)=d_{1}x\rightarrow d_{2}x$ and $(d_{1}\rightsquigarrow d_{2})(x)=d_{1}x\rightsquigarrow d_{2}x$, for all $x\in A$. Then $d_{1}\rightarrow d_{2}=d_{2}\rightarrow d_{1}$ and $d_{1}\rightsquigarrow d_{2}=d_{2}\rightsquigarrow d_{1}$.

Proof.

Let $d_{1},d_{2}\in\mathcal{IDOP}(A)$. We use again the fact that in any $p$-semisimple pseudo-BCI algebra $A$ we have $x\Cup_{1}y=x\Cup_{2}y=x$ for all $x,y\in A$. Applying the definitions of type I and type II derivations we get:

                  $(d_{1}\circ d_{2})(1)=(d_{1}\circ d_{2})(x\rightarrow x)=d_{1}((x\rightarrow d_{2}x)\Cup_{2}(d_{2}x\rightarrow x))$

                                    $=d_{1}(x\rightarrow d_{2}x)=(d_{1}x\rightarrow d_{2}x)\Cup_{2}(x\rightarrow d_{1}d_{2}x)=d_{1}x\rightarrow d_{2}x$.

                  $(d_{1}\circ d_{2})(x)=(d_{1}\circ d_{2})(x\rightarrow x)=d_{1}((d_{2}x\rightarrow x)\Cup_{2}(x\rightarrow d_{2}x))$

                                    $=d_{1}(d_{2}x\rightarrow x)=(d_{2}x\rightarrow d_{1}x)\Cup_{2}(d_{1}d_{2}x\rightarrow x)=d_{2}x\rightarrow d_{1}x$.

Hence $(d_{1}\rightarrow d_{2})(x)=(d_{2}\rightarrow d_{1})(x)=(d_{1}\circ d_{2})(x)$ for all $x\in A$. It follows that $d_{1}\rightarrow d_{2}=d_{2}\rightarrow d_{1}$. Similarly, $d_{1}\rightsquigarrow d_{2}=d_{2}\rightsquigarrow d_{1}$. ∎

5. Symmetric derivation operators on pseudo-BCI algebras

In this section we give a generalization of the concept of left derivation introduced in [2] for BCI-algebras. Two types of symmetric derivations on pseudo-BCI algebras are defined and studied, and the relationship between implicative and symmetric derivations is investigated. We prove that for the case of a $p$-semisimple pseudo-BCI algebra the sets of type II implicative derivations and type II symmetric derivations are equal, while for a $p$-semisimple BCI-algebra this result is also valid for the sets of type I implicative derivations and type I symmetric derivations. Finally, we show that a type I or type II symmetric derivation $d$ on a pseudo-BCI algebra $A$ is regular if and only if every deductive system of $A$ is $d$-invariant.

Definition 5.1**.**

Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra. A mapping $d:A\longrightarrow A$ is called a symmetric derivation operator of type I or a type I symmetric derivation operator or a type I symmetric derivation on $A$ if it satisfies the following conditions for all $x,y\in A:$

$(sdop_{1})$ $d(x\rightarrow y)=(x\rightarrow d(y))\Cup_{2}(y\rightarrow d(x))$.

$(sdop_{2})$ $d(x\rightsquigarrow y)=(x\rightsquigarrow d(y))\Cup_{1}(y\rightsquigarrow d(x))$.

Definition 5.2**.**

Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra. A mapping $d:A\longrightarrow A$ is called a symmetric derivation operator of type II or a type II symmetric derivation operator or a type II symmetric derivation on $A$ if it satisfies the following conditions for all $x,y\in A:$

$(sdop_{3})$ $d(x\rightarrow y)=(d(x)\rightarrow y)\Cup_{2}(d(y)\rightarrow x)$.

$(sdop_{4})$ $d(x\rightsquigarrow y)=(d(x)\rightsquigarrow y)\Cup_{1}(d(y)\rightsquigarrow x)$.

Let $A$ be a pseudo-BCI algebra. Denote:

                  $\mathcal{SDOP}^{(I)}(A)$ the set of all symmetric derivation operators of type I on $A$,

                  $\mathcal{SDOP}^{(II)}(A)$ the set of all symmetric derivation operators of type II on $A$,

                  $\mathcal{SDOP}(A)=\mathcal{SDOP}^{(I)}(A)\cap\mathcal{SDOP}^{(II)}(A)$.

A map $d\in\mathcal{SDOP}(A)$ is called a symmetric derivation on $A$.

In what follows we will denote $dx$ instead of $d(x)$. Similar to the case of implicative derivations we can define the regular symmetric derivations on pseudo-BCI algebras. Denote by $\mathcal{RSDOP}^{(I)}(A)$, $\mathcal{RSDOP}^{(II)}(A)$ and $\mathcal{RSDOP}(A)$ the set of all regular symmetric derivations from $\mathcal{SDOP}^{(I)}(A)$, $\mathcal{SDOP}^{(II)}(A)$ and $\mathcal{SDOP}(A)$, respectively.

Proposition 5.3**.**

*Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra and let $d\in\mathcal{SDOP}^{(I)}(A)$. The following hold for all $x,y\in A:$

$(1)$ $d1=x\rightarrow dx=x\rightsquigarrow dx;$

$(2)$ $dx=dx\Cup_{1}d1=dx\Cup_{2}d1;$

If $d$ is regular, then:

$(3)$ $x\leq dx;$

$(4)$ $dx\in{\rm At}\,(A);$

$(5)$ $dx=dx\Cup_{1}y=dx\Cup_{2}y$.*

Proof.

$(1)$ We have $d1=d(x\rightarrow x)=(x\rightarrow dx)\Cup_{2}(x\rightarrow dx)=x\rightarrow dx$ and similarly, $d1=x\rightsquigarrow dx$.

$(2)$ Using Lemmas 2.2$(7)$,$(8)$ and 2.6$(5)$ we get:

                  $dx=d(1\rightsquigarrow x)=dx\Cup_{1}(x\rightsquigarrow d1)=(dx\rightarrow(x\rightsquigarrow d1))\rightsquigarrow(x\rightsquigarrow d1)$

                                        $=(x\rightsquigarrow(dx\rightarrow d1))\rightsquigarrow(x\rightsquigarrow d1)$

                                        $\geq(dx\rightarrow d1)\rightsquigarrow d1=dx\Cup_{1}d1\geq dx$.

Hence $dx=dx\Cup_{1}d1$ and similarly, $dx=dx\Cup_{2}d1$.

$(3)$ It follows from $(1)$, since $d1=1$.

$(4)$ Since $d1=1$, applying $(2)$ we get $dx=(dx\rightarrow 1)\rightsquigarrow 1=(dx\rightsquigarrow 1)\rightarrow 1$, hence by Proposition 2.4$(k)$, $dx\in{\rm At}\,(A)$.

$(5)$ It follows by Proposition 2.4$(b)$, since $dx\in{\rm At}\,(A)$. ∎

Proposition 5.4**.**

*Let $(A,\rightarrow,\rightsquigarrow,1)$ be a pseudo-BCI algebra and let $d\in\mathcal{SDOP}^{(II)}(A)$. The following hold for all $x,y\in A:$

$(1)$ $d1=dx\rightarrow x=dx\rightsquigarrow x;$

$(2)$ $dx=\varphi_{dx\Cup_{1}x}=\varphi_{dx\Cup_{2}x};$

$(3)$ $dx=dx\Cup_{1}x=dx\Cup_{2}x;$

$(4)$ $dx\in{\rm At}\,(A);$

$(5)$ if $d$ is regular, then $d={\rm Id}\,_{A}$.*

Proof.

$(1)$ We have $d1=d(x\rightarrow x)=(dx\rightarrow x)\Cup_{2}(dx\rightarrow x)=dx\rightarrow x$ and similarly, $dx=dx\rightsquigarrow x$.

$(2)$ Applying Lemma 2.2$(7)$,$(8)$ and $(1)$ we get:

                  $dx=d(1\rightsquigarrow x)=(d1\rightsquigarrow x)\Cup_{1}(dx\rightsquigarrow 1)=((d1\rightsquigarrow x)\rightarrow(dx\rightsquigarrow 1))\rightsquigarrow(dx\rightsquigarrow 1)$

                       $=(dx\rightsquigarrow((d1\rightsquigarrow x)\rightarrow 1))\rightsquigarrow(dx\rightsquigarrow 1)$

                       $\geq((d1\rightsquigarrow x)\rightarrow 1)\rightsquigarrow 1=(((dx\rightarrow x)\rightsquigarrow x)\rightarrow 1)\rightsquigarrow 1$

                       $=(dx\Cup_{1}x\rightarrow 1)\rightsquigarrow 1=\varphi_{dx\Cup_{1}x}$.

On the other hand, by Lemmas 2.2$(2)$ and 2.6$(5)$ we have:

                  $\varphi_{dx\Cup_{1}x}=(dx\Cup_{1}x\rightarrow 1)\rightsquigarrow 1\geq dx\Cup_{1}x\geq dx$.

We conclude that $dx=\varphi_{dx\Cup_{1}x}$. Similarly, $dx=\varphi_{dx\Cup_{2}x}$.

$(3)$ By $(2)$, $dx=\varphi_{dx\Cup_{1}x}=(dx\Cup_{1}x)\rightarrow 1)\rightsquigarrow 1\geq dx\Cup_{1}x\geq dx$, so that $dx=dx\Cup_{1}x$.

Similarly, $dx=dx\Cup_{2}x$.

$(4)$ It follows by $(2)$, since $\varphi_{x}\in{\rm At}\,(A)$, for all $x\in A;$

$(5)$ Since $d1=1$, by $(3)$ and $(1)$ we get: $dx=dx\Cup_{1}x=(dx\rightarrow x)\rightsquigarrow x=d1\rightsquigarrow x=1\rightsquigarrow x=x$, for all $x\in A$. Hence $d={\rm Id}\,_{A}$. ∎

Corollary 5.5**.**

*If $d\in\mathcal{SDOP}^{(II)}(A)$, then the following hold for all $x,y\in A:$

$(1)$ ${\rm Im}\,(d)\subseteq{\rm At}\,(A);$

$(2)$ $dx=dx\Cup_{1}y=dx\Cup_{2}y;$

$(3)$ $x\rightarrow dy,x\rightsquigarrow dy\in{\rm At}\,(A)$.*

Proposition 5.6**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{SDOP}^{(I)}(A)\cup\mathcal{SDOP}^{(II)}(A)$. Then $d(x\cdot y)=dx\cdot y=x\cdot dy$, for all $x,y\in{\rm At}\,(A)$.

Proof.

Let $d\in\mathcal{SDOP}^{(I)}(A)$ and let $x,y\in{\rm At}\,(A)$. Since $1,x,y\in{\rm At}\,(A)$, by Corollary 2.5 and Proposition 5.4, $x\rightarrow 1,y\rightsquigarrow 1,dx,dy,d(x\rightarrow 1),d(y\rightsquigarrow 1),d1,d1\rightarrow x,d1\rightsquigarrow x\in{\rm At}\,(A)$. From $x\cdot y=(x\rightarrow 1)\rightsquigarrow y=(y\rightsquigarrow 1)\rightarrow x$, using Proposition 2.4$(b)$ we get:

                  $d(x\cdot y)=d((x\rightarrow 1)\rightsquigarrow y)=((x\rightarrow 1)\rightsquigarrow dy)\Cup_{1}(y\rightsquigarrow d(x\rightarrow 1))$

                              $=(x\rightarrow 1)\rightsquigarrow dy=x\cdot dy$

                  $d(x\cdot y)=d((y\rightsquigarrow 1)\rightarrow x)=((y\rightsquigarrow 1)\rightarrow dx)\Cup_{2}(x\rightarrow d(y\rightsquigarrow 1))$

                              $=(y\rightsquigarrow 1)\rightarrow dx=dx\cdot y$.

Similarly, for $d\in\mathcal{SDOP}^{(II)}(A)$ we have:

                  $d(x\cdot y)=d((x\rightarrow 1)\rightsquigarrow y)=(d(x\rightarrow 1)\rightsquigarrow y)\Cup_{1}(dy\rightsquigarrow(x\rightarrow 1))$

                              $=d(x\rightarrow 1)\rightsquigarrow y=((dx\rightarrow 1)\Cup_{2}(d1\rightarrow x))\rightsquigarrow y$

                              $=(dx\rightarrow 1)\rightsquigarrow y=dx\cdot y$.

                  $d(x\cdot y)=d((y\rightsquigarrow 1)\rightarrow x)=(d(y\rightsquigarrow 1)\rightarrow x)\Cup_{2}(dx\rightarrow(y\rightsquigarrow 1))$

                              $=d(y\rightsquigarrow 1)\rightarrow x=((dy\rightsquigarrow 1)\Cup_{1}(d1\rightsquigarrow y))\rightarrow x$

                              $=(dy\rightsquigarrow 1)\rightarrow x=x\cdot dy$.

Hence, in both cases $d(x\cdot y)=dx\cdot y=x\cdot dy$. ∎

Proposition 5.7**.**

Let $A$ be a pseudo-BCI algebra and let $d_{\varphi}:A\longrightarrow A$, defined by $dx=\varphi_{x}$ for any $x\in A$. Then $d_{\varphi}\in\mathcal{SDOP}^{(I)}(A)$.

Proof.

Similar to the proof of Proposition 3.7, since $x\rightarrow\varphi_{y}\in{\rm At}\,(A)$, we get:

          $d(x\rightarrow y)=\varphi_{x\rightarrow y}=x\rightarrow\varphi_{y}=((x\rightarrow\varphi_{y})\rightsquigarrow(y\rightarrow\varphi_{x}))\rightarrow(y\rightarrow\varphi_{x})$

                        $=(x\rightarrow\varphi_{y})\Cup_{2}(y\rightarrow\varphi_{x})=(x\rightarrow dy)\Cup_{2}(y\rightarrow dx)$.

Similarly we have:

          $d(x\rightsquigarrow y)=\varphi_{x\rightsquigarrow y}=x\rightsquigarrow\varphi_{y}=((x\rightsquigarrow\varphi_{y})\rightarrow(y\rightsquigarrow\varphi_{x}))\rightsquigarrow(y\rightsquigarrow\varphi_{x})$

                        $=(x\rightsquigarrow\varphi_{y})\Cup_{1}(y\rightsquigarrow\varphi_{x})=(x\rightsquigarrow dy)\Cup_{1}(y\rightsquigarrow dx)$.

We conclude that $d_{\varphi}\in\mathcal{SDOP}^{(I)}(A)$. ∎

Corollary 5.8**.**

$\mathcal{IDOP}^{(I)}(A)\cap\mathcal{SDOP}^{(I)}(A)\neq\emptyset$.

Proof.

If $d_{\varphi}:A\longrightarrow A$, defined by $dx=\varphi_{x}$, for all $x\in A$, then by Propositions 3.7 and 5.7, $d_{\varphi}\in\mathcal{IDOP}^{(I)}(A)\cap\mathcal{SDOP}^{(I)}(A)$. ∎

Example 5.9**.**

Consider the pseudo-BCI algebra $(A,\rightarrow,\rightsquigarrow,1)$ and the maps $d_{1},d_{2},d_{3}:A\longrightarrow A$ from Example 3.4. One can check that $d_{\varphi}=d_{3}$, $\mathcal{SDOP}^{(I)}(A)=\{d_{2},d_{3}\}$ and $\mathcal{SDOP}^{(II)}(A)=\{d_{3}\}$.

Example 5.10**.**

Consider the structure $(A,\rightarrow,\rightsquigarrow,1)$, where the operations $\rightarrow$ and $\rightsquigarrow$ on $A=\{a,b,x,y,g,1\}$ are defined as follows:

[TABLE]

Then $(A,\rightarrow,\rightsquigarrow,1)$ is a pseudo-BCI algebra ([20]). Consider the maps $d_{1},d_{2},d_{3}:A\longrightarrow A$ given in the table below:

[TABLE]

One can see that $\mathcal{SDOP}^{(I)}(A)=\{d_{2},d_{3}\}$ and $\mathcal{SDOP}^{(II)}(A)=\emptyset$. We also mention that $d_{\varphi}=d_{3}$ and $\mathcal{IDOP}^{(I)}(A)=\mathcal{IDOP}^{(II)}(A)=\{d_{1},d_{2},d_{3}\}$.

Proposition 5.11**.**

*Let $A$ be a $p$-semisimple pseudo-BCI algebra and let $d\in\mathcal{SDOP}^{(II)}(A)$. Then the following hold for all $x,y\in A:$

$(1)$ $d(x\rightarrow y)=dx\rightarrow y$ and $d(x\rightsquigarrow y)=dx\rightsquigarrow y;$

$(2)$ $x\rightarrow dx=y\rightarrow dy$ and $x\rightsquigarrow dx=y\rightsquigarrow dy;$

$(3)$ $x\rightarrow dx=dy\rightarrow y$ and $x\rightsquigarrow dx=dy\rightsquigarrow y$.*

Proof.

$(1)$ It follows by Proposition 2.3$(c)$ since $A$ is $p$-semisimple.

$(2)$ By $(psBCI_{1})$ we have:

                  $(x\rightarrow dx)\rightsquigarrow((dx\rightarrow y)\rightsquigarrow(x\rightarrow y))=(y\rightarrow dy)\rightsquigarrow((dy\rightarrow x)\rightsquigarrow(y\rightarrow x))(=1)$.

It follows by $(1)$ that:

                  $(x\rightarrow dx)\rightsquigarrow(d(x\rightarrow y)\rightsquigarrow(x\rightarrow y))=(y\rightarrow dy)\rightsquigarrow(d(y\rightarrow x)\rightsquigarrow(y\rightarrow x))$.

Applying Proposition 5.4$(1)$ we have:

                  $d(x\rightarrow y)\rightsquigarrow(x\rightarrow y)=d(y\rightarrow x)\rightsquigarrow(y\rightarrow x)(=d1)$.

Hence $(x\rightarrow dx)\rightsquigarrow(d(x\rightsquigarrow y)\rightarrow(x\rightarrow y))=(y\rightarrow dy)\rightsquigarrow(d(x\rightarrow y)\rightsquigarrow(x\rightarrow y))$.

Finally applying Proposition 2.3$(f)$ we get $x\rightarrow dx=y\rightarrow dy$. Similarly, $x\rightsquigarrow dx=y\rightsquigarrow dy$.

$(3)$ Since by Proposition 5.4$(1)$, $d1=dy\rightarrow y$, applying $(2)$ we get $x\rightarrow dx=1\rightarrow d1=d1=dy\rightarrow y$, that is $x\rightarrow dx=dy\rightarrow y$. Similarly, $x\rightsquigarrow dx=dy\rightsquigarrow y$. ∎

Proposition 5.12**.**

If $A$ is a $p$-semisimple pseudo-BCI algebra, then $\mathcal{SDOP}^{(II)}(A)=\mathcal{IDOP}^{(II)}(A)$.

Proof.

Let $d\in\mathcal{SDOP}^{(II)}(A)$. By Proposition 5.11$(1)$, since $A$ is $p$-semisimple we have $d(x\rightarrow y)=dx\rightarrow y=(dx\rightarrow y)\Cup_{2}(x\rightarrow dy)$, that is $d\in\mathcal{IDOP}^{(II)}(A)$. Hence $\mathcal{SDOP}^{(II)}(A)\subseteq\mathcal{IDOP}^{(II)}(A)$. Conversely, if $d\in\mathcal{IDOP}^{(II)}(A)$, then $d(x\rightarrow y)=(dx\rightarrow y)\Cup_{2}(x\rightarrow dy)=dx\rightarrow y=(dx\rightarrow y)\Cup_{2}(dy\rightarrow x)$, that is $d\in\mathcal{SDOP}^{(II)}(A)$. Thus $\mathcal{IDOP}^{(II)}(A)\subseteq\mathcal{SDOP}^{(II)}(A)$. We conclude that $\mathcal{SDOP}^{(II)}(A)=\mathcal{IDOP}^{(II)}(A)$. ∎

Proposition 5.13**.**

If $A$ is a $p$-semisimple BCI-algebra, then $\mathcal{SDOP}^{(I)}(A)=\mathcal{IDOP}^{(I)}(A)$.

Proof.

Similar to [2, Th. 3.13], based on the fact that any $p$-semisimple BCI-algebra is medial. ∎

Example 5.14**.**

Consider the structure $(A,\rightarrow,1)$ and the maps $d_{1},d_{2},d_{3}:A\longrightarrow A$, defined in the tables below:

[TABLE]

Then $(A,\rightarrow,1)$ is a $p$-semisimple BCI-algebra ([5]). One can check that $\mathcal{SDOP}^{(I)}(A)=\mathcal{IDOP}^{(I)}(A)=\{d_{1},d_{2},d_{3}\}$ and $\mathcal{SDOP}^{(II)}(A)=\mathcal{IDOP}^{(II)}(A)=\{d_{1}\}$.

Theorem 5.15**.**

Let $A$ be a pseudo-BCI algebra and let $d\in\mathcal{SDOP}^{(I)}(A)\cup\mathcal{SDOP}^{(II)}(A)$. Then $d$ is regular if and only if every deductive system of $A$ is $d$-invariant.

Proof.

Let $d\in\mathcal{SDOP}^{(I)}(A)\cup\mathcal{SDOP}^{(II)}(A)$ and assume that every $D\in\mathcal{DS}(A)$ is $d$-invariant. Since $\{1\}\in\mathcal{DS}(A)$, it follows that $d(\{1\})\subseteq\{1\}$, hence $d(1)=1$, that is $d$ is regular. Conversely, let $D\in\mathcal{DS}(A)$ and let $y\in d(D)$, that is there exists $x\in D$ such that $y=dx$.

If $d\in\mathcal{SDOP}^{(I)}(A)$, then by Proposition 5.3$(3)$ we have $x\rightarrow y=x\rightarrow dx=1\in D$, hence $y\in D$. For $d\in\mathcal{SDOP}^{(II)}(A)$, by Proposition 5.4$(5)$ we get $x\rightarrow y=x\rightarrow dx=x\rightarrow x=1$, that is $y\in D$. It follows that $d(D)\subseteq D$, hence $D$ is $d$-invariant. ∎

6. Conclusions and future work

In this paper we introduce and study two concepts of implicative derivation operators on pseudo-BCI algebras: type I implicative derivation defined by conditions $(idop_{1})$, $(idop_{2})$ and type II implicative derivation defined by conditions $(idop_{3})$, $(idop_{4})$. These conditions were required by the proof of Proposition 3.5, which is crucial for the results of this paper. For the particular case of the pseudo-BCK algebras the above mentioned result is also valid for another two types of implicative derivations:

$-$ type III implicative derivation defined by the following conditions, for all $x,y\in A:$

$(idop_{5})$ $d(x\rightarrow y)=(x\rightarrow dy)\Cup_{1}(dx\rightarrow y)$

$(idop_{6})$ $d(x\rightsquigarrow y)=(x\rightsquigarrow dy)\Cup_{2}(dx\rightsquigarrow y)$,

$-$ type IV implicative derivation defined by the following conditions, for all $x,y\in A:$

$(idop_{7})$ $d(x\rightarrow y)=(dx\rightarrow y)\Cup_{1}(x\rightarrow dy)$

$(idop_{8})$ $d(x\rightsquigarrow y)=(dx\rightsquigarrow y)\Cup_{2}(x\rightsquigarrow dy)$.

In what follows we give an example of these derivations, but the investigation of type III and type IV implicative derivations on pseudo-BCK algebras is the topic of another work ([10]).

Consider the structure $(A,\rightarrow,\rightsquigarrow,1)$, where the operations $\rightarrow$ and $\rightsquigarrow$ on $A=\{0,a,b,c,1\}$ are defined as follows:

[TABLE]

Then $(A,\rightarrow,\rightsquigarrow,1)$ is a pseudo-BCK algebra ([9]). Consider the maps $d_{i}:A\longrightarrow A$, $i=1,\cdots,6$, given in the table below:

[TABLE]

One can check that $\mathcal{IDOP}^{(I)}(A)=\mathcal{IDOP}^{(II)}(A)=\mathcal{IDOP}^{(IV)}(A)=\{d_{1},d_{2},d_{5},d_{6}\}$ and $\mathcal{IDOP}^{(III)}(A)=\{d_{1},d_{2},d_{3},d_{4},d_{5},d_{6}\}$.

We also introduce two types of symmetric derivations on pseudo-BCI algebras and study the relationship between implicative and symmetric derivations.

The results presented in this paper could be extended to other “pseudo” algebras such as pseudo-BCH algebras, pseudo-BE algebras, pseudo-CI algebras, non-commutative residuated lattices. As a generalization of the concept of derivations, the $f$-derivation has been defined for lattices ([6]), BCI-algebras ([24], [39], [48]), BE-algebras ([33]). If $(A,*,0)$ is a BCI-algebra and $f$ is an endomorphism of $A$, then a map $d:A\longrightarrow A$ is called a left-right $f$-derivation if $d(x*y)=(dx*f(y))\wedge(f(x)*dy)$ and a right-left $f$-derivation if $d(x*y)=(f(x)*dy)\wedge(dx*f(y))$ for all $x,y\in A$, where $x\wedge y=y*(y*x)$. As another direction of research one could define and study the concept of $f$-derivations on pseudo-BCI algebras and other “pseudo” algebras.

The concept of generalized derivations on BCI-algebras defined and investigated in [40], [38], [37] could be also introduced and studied for the case of pseudo-BCI algebras.

The BCI-algebras with product (BCI(P)-algebras) $(A,\odot,\rightarrow,1)$ were originally introduced by $\rm Is\acute{e}ki$ ([29]) as BCI-algebras with condition (S). The concept of multiplicative derivations on BCI(P)-algebras could be defined and studied as another topic of future research.