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*Axioms*
**2018**,
*7*(2),
26;
doi:10.3390/axioms7020026

Article

Quotient Structures of BCK/BCI-Algebras Induced by Quasi-Valuation Maps

^{1}

Department of Mathematics, Jeju National University, Jeju 63243, Korea

^{2}

Department of Mathematical Sciences, Shahid Beheshti University, Tehran 1983969411, Iran

^{3}

Department of Mathematics Education, Gyeongsang National University, Jinju 52828, Korea

^{*}

Author to whom correspondence should be addressed.

Received: 4 February 2018 / Accepted: 11 April 2018 / Published: 23 April 2018

## Abstract

**:**

Relations between I-quasi-valuation maps and ideals in $BCK/BCI$-algebras are investigated. Using the notion of an I-quasi-valuation map of a $BCK/BCI$-algebra, the quasi-metric space is induced, and several properties are investigated. Relations between the I-quasi-valuation map and the I-valuation map are considered, and conditions for an I-quasi-valuation map to be an I-valuation map are provided. A congruence relation is introduced by using the I-valuation map, and then the quotient structures are established and related properties are investigated. Isomorphic quotient $BCK/BCI$-algebras are discussed.

Keywords:

ideal; I-quasi-valuation map; I-valuation map; quasi-metricMSC:

06F35; 03G25; 03C05## 1. Introduction

$BCK/BCI$-algebras are an important class of logical algebras introduced by Imai and Iséki (see [1,2,3,4]), and have been extensively investigated by several researchers. It is known that the class of $BCK$-algebras is a proper subclass of $BCI$-algebras. Song et al. [5] introduced the notion of quasi-valuation maps based on a subalgebra and an ideal in $BCK/BCI$-algebras, and then they investigated several properties. They provided relations between a quasi-valuation map based on a subalgebra and a quasi-valuation map based on an ideal, and gave a condition for a quasi-valuation map based on an ideal to be a quasi-valuation map based on a subalgebra in $BCI$-algebras. Using the notion of a quasi-valuation map based on an ideal, they constructed (pseudo) metric spaces, and showed that the binary operation ∗ in $BCK$-algebras is uniformly continuous.

In this paper, we discuss relations between I-quasi-valuation maps and ideals in $BCK/BCI$-algebras. Using the notion of an I-quasi-valuation map of a $BCK/BCI$-algebra, we induce the quasi-metric space, and investigate several properties. We discuss relations between the I-quasi-valuation map and the I-valuation map. We provide conditions for an I-quasi-valuation map to be an I-valuation map. We use I-quasi-valuation maps to introduce a congruence relation, and then we construct the quotient structures and investigate related properties. We establish isomorphic quotient $BCK/BCI$-algebras.

## 2. Preliminaries

By a $BCI$-algebra, we mean a nonempty set X with a binary operation ∗ and a special element 0 satisfying the following axioms:

- (I)
- $(\forall x,y,z\in X)$$\left(\right((x\ast y)\ast (x\ast z))\ast (z\ast y)=0),$
- (II)
- $(\forall x,y\in X)$$\left(\right(x\ast (x\ast y))\ast y=0),$
- (III)
- $(\forall x\in X)$$(x\ast x=0),$
- (IV)
- $(\forall x,y\in X)$$(x\ast y=0,y\ast x=0\phantom{\rule{0.166667em}{0ex}}\Rightarrow \phantom{\rule{0.166667em}{0ex}}x=y).$

If a BCI-algebra X satisfies the following identity:

- (V)
- $(\forall x\in X)(0\ast x=0),$

$$\begin{array}{c}(\forall x\in X)(x\ast 0=x),\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y,z\in X)(x\ast y=0\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}(x\ast z)\ast (y\ast z)=0,\phantom{\rule{3.33333pt}{0ex}}(z\ast y)\ast (z\ast x)=0),\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y,z\in X)\left(\right(x\ast y)\ast z=(x\ast z)\ast y),\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y,z\in X)\left(\right((x\ast z)\ast (y\ast z))\ast (x\ast y)=0).\hfill \end{array}$$

Any $BCI$-algebra X satisfies the following condition:

$$\begin{array}{c}\hfill (\forall x,y\in X)(0\ast (x\ast y)=(0\ast x)\ast (0\ast y\left)\right).\end{array}$$

We can define a partial ordering ≤ on X as follows:

$$\begin{array}{c}\hfill (\forall x,y\in X)\left(x\le y\phantom{\rule{3.33333pt}{0ex}}\u27fa\phantom{\rule{3.33333pt}{0ex}}x\ast y=0\right).\end{array}$$

A nonempty subset S of a $BCK/BCI$-algebra X is called a subalgebra of X if $x\ast y\in S$ for all $x,y\in S.$ A subset I of a $BCK/BCI$-algebra X is called an ideal of X if it satisfies the following conditions:

$$\begin{array}{c}0\in I,\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y\in X)\left(x\ast y\in I,\phantom{\rule{3.33333pt}{0ex}}y\in I\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}x\in I\right).\hfill \end{array}$$

An ideal I of a $BCI$-algebra X is said to be closed if

$$\begin{array}{c}\hfill (\forall x\in X)(x\in I\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}0\ast x\in I).\end{array}$$

## 3. Quasi-Valuation Maps on BCK/BCI-Algebras

In what follows, let X denote a $BCK/BCI$-algebra unless otherwise specified.

**Definition**

**1**

**([5]).**

By a quasi-valuation map of X based on an ideal (briefly I-quasi-valuation map of X), we mean a mapping $f:X\to \mathbb{R}$ which satisfies the conditions

$$\begin{array}{c}f\left(0\right)=0,\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y\in X)\left(f\left(x\right)\ge f(x\ast y)+f\left(y\right)\right).\hfill \end{array}$$

The I-quasi-valuation map f is called an I-valuation map of X if

$$\begin{array}{c}\hfill (\forall x\in X)\left(f\right(x)=0\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}x=0).\end{array}$$

**Lemma**

**1**

**([5]).**

For any I-quasi-valuation map f of $X,$ we have the following assertions:

- (1)
- f is order reversing.
- (2)
- $f(x\ast y)+f(y\ast x)\le 0$ for all $x,y\in X.$
- (3)
- $f(x\ast y)\ge f(x\ast z)+f(z\ast y)$ for all $x,y,z\in X.$

**Corollary**

**1.**

Every quasi-valuation map f of a $BCK$-algebra X satisfies:

$$\begin{array}{c}\hfill (\forall x\in X)\left(f\right(x)\le 0).\end{array}$$

**Theorem**

**1.**

For any ideal I of X, define a map
where t is a negative number in $\mathbb{R}$. Then, ${f}_{I}$ is an I-quasi-valuation map of X. Moreover, ${f}_{I}$ is an I-valuation map of X if and only if I is the trivial ideal of X (i.e., $I=\left\{0\right\}$).

$$\begin{array}{c}\hfill {f}_{I}:X\to \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}x\mapsto \left\{\begin{array}{cc}0\hfill & \mathrm{if}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.166667em}{0ex}}x\in I,\hfill \\ t\hfill & \mathrm{otherwise}\hfill \end{array}\right.,\end{array}$$

**Proof.**

Straightforward. ☐

**Theorem**

**2.**

If f is an I-quasi-valuation map of X, then the set
is an ideal of X.

$$\begin{array}{c}\hfill {A}_{f}:=\{x\in X\mid f\left(x\right)\ge 0\}\end{array}$$

**Proof.**

Obviously $0\in {A}_{f}$. Let $x,y\in X$ be such that $x\ast y\in {A}_{f}$ and $y\in {A}_{f}$. Then, $f(x\ast y)\ge 0$ and $f\left(y\right)\ge 0$. It follows from (10) that $f\left(x\right)\ge f(x\ast y)+f\left(y\right)\ge 0$ and so that $x\in {A}_{f}$. Therefore ${A}_{f}$ is an ideal of X. ☐

Note that if an ideal of a $BCI$-algebra X is of finite order, then it is a closed ideal of X, and every ideal of a $BCK$-algebra X is a closed ideal of X (see [6]). Hence, we have the following corollary.

**Corollary**

**2.**

Let X be a finite $BCI$-algebra or a $BCK$-algebra. If f is an I-quasi-valuation map of X, then the set ${A}_{f}$ is a closed ideal of X.

**Theorem**

**3.**

If I is an ideal of X, then ${A}_{{f}_{I}}=I$.

**Proof.**

We get ${A}_{{f}_{I}}=\{x\in X\mid {f}_{I}\left(x\right)\ge 0\}=\{x\in X\mid x\in I\}=I$. ☐

**Definition**

**2.**

A real-valued function d on $X\times X$ is called a quasi-metric if it satisfies:

$$\begin{array}{c}(\forall x,y\in X)\left(d(x,y)\le 0,\phantom{\rule{3.33333pt}{0ex}}d(x,x)=0\right),\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y\in X)\left(d(x,y)=d(y,x)\right),\hfill \end{array}$$

$$\begin{array}{c}(\forall x,y,z\in X)\left(d(x,z)\ge d(x,y)+d(y,z)\right).\hfill \end{array}$$

The pair $(X,d)$ is called the quasi-metric space.

Given a real-valued function f on X, define a mapping

$$\begin{array}{c}\hfill {d}_{f}:X\times X\to \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}(x,y)\mapsto f(x\ast y)+f(y\ast x).\end{array}$$

**Theorem**

**4.**

If a real-valued function f on X is an I-quasi-valuation map of X, then ${d}_{f}$ is a quasi-metric on $X\times X$.

The pair $(X,{d}_{f})$ is called the quasi-metric space induced by f.

**Proof.**

Using Lemma 1(2), we have ${d}_{f}(x,y)=f(x\ast y)+f(y\ast x)\le 0$ for all $(x,y)\in X\times X$. Obviously, ${d}_{f}(x,x)=0$ and ${d}_{f}(x,y)={d}_{f}(y,x)$ for all $x,y\in X$. Using Lemma 1(3), we get
for all $x,y,z\in X$. Therefore ${d}_{f}$ is a quasi-metric on X. ☐

$$\begin{array}{cc}\hfill {d}_{f}(x,y)+{d}_{f}(y,z)& =\left(f\right(x\ast y)+f(y\ast x\left)\right)+\left(f\right(y\ast z)+f(z\ast y\left)\right)\hfill \\ & =\left(f\right(x\ast y)+f(y\ast z\left)\right)+\left(f\right(z\ast y)+f(y\ast x\left)\right)\hfill \\ & \le f(x\ast z)+f(z\ast x)={d}_{f}(x,z)\hfill \end{array}$$

**Proposition**

**1.**

Let f be an I-quasi-valuation map of a $BCK$-algebra X such that

$$\begin{array}{c}\hfill (\forall x\in X)(x\ne 0\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}f(x)\ne 0).\end{array}$$

Then, the quasi-metric space $(X,{d}_{f})$ induced by f satisfies:

$$\begin{array}{c}\hfill (\forall x,y\in X)({d}_{f}(x,y)=0\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}x=y).\end{array}$$

**Proof.**

Assume that ${d}_{f}(x,y)=0$ for $x,y\in X$. Then, $f(x\ast y)+f(y\ast x)=0$, and so $f(x\ast y)=0$ and $f(y\ast x)=0$ by Corollary 1. It follows from (15) that $x\ast y=0$ and $y\ast x=0$. Hence $x=y$. ☐

We provide conditions for an I-quasi-valuation map to be an I-valuation map.

**Theorem**

**5.**

Let f be an I-quasi-valuation map of a $BCI$-algebra X such that ${A}_{f}$ is a closed ideal of X. If the quasi-metric ${d}_{f}$ induced by f satisfies the condition (16), then f is an I-valuation map of X.

**Proof.**

Assume that f does not satisfy the condition (11). Then, there exists $x\in X$ such that $x\ne 0$ and $f\left(x\right)=0$. Thus, $x\in {A}_{f}$, and so $0\ast x\in {A}_{f}$ since ${A}_{f}$ is a closed ideal of X. Hence $f(0\ast x)\ge 0$, which implies that

$$\begin{array}{c}\hfill 0=f\left(0\right)\ge f(0\ast x)+f\left(x\right)=f(0\ast x)\ge 0.\end{array}$$

Thus, $f(0\ast x)=0$, and so ${d}_{f}(x,0)=f(x\ast 0)+f(0\ast x)=f\left(x\right)=0$. It follows from (16) that $x=0$. Therefore, f is an I-valuation map of X. ☐

Since every ideal is closed in a $BCK$-algebra, we have the following corollary.

**Corollary**

**3.**

Given an I-quasi-valuation map f of a $BCK$-algebra X, if the quasi-metric ${d}_{f}$ induced by f satisfies the condition (16), then f is an I-valuation map of X.

Consider the $BCI$-algebra $(\mathbb{Z},-,0)$ and define a map f on $\mathbb{Z}$ as follows:
where k is a negative integer. For any $x\in \mathbb{Z}\backslash \left\{0\right\}$ and $y\in \mathbb{Z}$, we have ${f}_{k}\left(x\right)=k-x$ and

$$\begin{array}{c}\hfill {f}_{k}:\mathbb{Z}\to \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}x\mapsto \left\{\begin{array}{cc}0\hfill & \mathrm{if}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.166667em}{0ex}}x=0,\hfill \\ k-x\hfill & \mathrm{otherwise}\hfill \end{array}\right.,\end{array}$$

$$\begin{array}{c}\hfill {f}_{k}(x-y)+{f}_{k}\left(y\right)=\left\{\begin{array}{cc}k-x\hfill & \mathrm{if}\phantom{\rule{4.pt}{0ex}}\mathrm{either}\phantom{\rule{4.pt}{0ex}}y=0\phantom{\rule{4.pt}{0ex}}\mathrm{or}\phantom{\rule{4.pt}{0ex}}y=x,\hfill \\ 2k-x\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

It follows that ${f}_{k}\left(x\right)\ge {f}_{k}(x-y)+{f}_{k}\left(y\right)$ for all $x,y\in \mathbb{Z}$, and so ${f}_{k}$ is an I-quasi-valuation map of $(\mathbb{Z},-,0)$. It is clear that the set
is an ideal of $(\mathbb{Z},-,0)$ which is not closed. Using Theorem 4, we know that ${d}_{{f}_{k}}$ is a quasi-metric induced by ${f}_{k}$ and satisfies:

$$\begin{array}{c}\hfill {A}_{{f}_{k}}=\{x\in \mathbb{Z}\mid {f}_{k}\left(x\right)\ge 0\}=\{x\in \mathbb{Z}\mid x\le k\}\cup \left\{0\right\}\end{array}$$

$$\begin{array}{c}\hfill (\forall x,y\in X)({d}_{{f}_{k}}(x,y)=0\phantom{\rule{3.33333pt}{0ex}}\Rightarrow \phantom{\rule{3.33333pt}{0ex}}x=y).\end{array}$$

However, ${f}_{k}$ is not an I-valuation map of $(\mathbb{Z},-,0)$ since ${f}_{k}\left(k\right)=0$ and $k\ne 0$. This shows that if ${A}_{f}$ is not a closed ideal of X, then the conclusion of Theorem 5 is not true.

**Proposition**

**2.**

Given an I-quasi-valuation map f of X, the quasi-metric space $(X,{d}_{f})$ satisfies:

- (1)
- ${d}_{f}(x,y)\le min\{{d}_{f}(x\ast a,y\ast a),{d}_{f}(a\ast x),{d}_{f}(a\ast y)\}$,
- (2)
- ${d}_{f}(x\ast y,a\ast b)\ge {d}_{f}(x\ast y,a\ast y)+{d}_{f}(a\ast y,a\ast b)$,

**Proof.**

Let $x,y,a,b\in X$. Using (4), we have

$$\begin{array}{c}\hfill (y\ast a)\ast (x\ast a)\le y\ast x\text{\hspace{1em}}\mathrm{and}\text{\hspace{1em}}(x\ast a)\ast (y\ast a)\le x\ast y.\end{array}$$

Since f is order reversing, it follows that

$$\begin{array}{c}\hfill f(y\ast x)\le f\left(\right(y\ast a)\ast (x\ast a\left)\right)\text{\hspace{1em}}\mathrm{and}\text{\hspace{1em}}f(x\ast y)\le f\left(\right(x\ast a)\ast (y\ast a\left)\right).\end{array}$$

Thus,

$$\begin{array}{cc}\hfill {d}_{f}(x,y)& =f(x\ast y)+f(y\ast x)\hfill \\ & \le f\left(\right(y\ast a)\ast (x\ast a\left)\right)+f\left(\right(x\ast a)\ast (y\ast a\left)\right)\hfill \\ & ={d}_{f}(x\ast a,y\ast a).\hfill \end{array}$$

Similarly, we get

$$\begin{array}{c}\hfill {d}_{f}(x,y)\le {d}_{f}(a\ast x,a\ast y).\end{array}$$

Therefore, (1) is valid. Now, using Lemma 1(3) implies that
and
for all $x,y,a,b\in X$. Hence
for all $x,y,a,b\in X$. Therefore, (2) is valid. ☐

$$\begin{array}{c}\hfill f\left(\right(x\ast y)\ast (a\ast b\left)\right)\ge f\left(\right(x\ast y)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (a\ast b\left)\right)\end{array}$$

$$\begin{array}{c}\hfill f\left(\right(a\ast b)\ast (x\ast y\left)\right)\ge f\left(\right(a\ast b)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (x\ast y\left)\right)\end{array}$$

$$\begin{array}{cc}\hfill {d}_{f}(x\ast y,a\ast b)& =f\left(\right(x\ast y)\ast (a\ast b\left)\right)+f\left(\right(a\ast b)\ast (x\ast y\left)\right)\hfill \\ & \ge f\left(\right(x\ast y)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (a\ast b\left)\right)\hfill \\ & \phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}+f\left(\right(a\ast b)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (x\ast y\left)\right)\hfill \\ & \ge f\left(\right(x\ast y)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (x\ast y\left)\right)\hfill \\ & \phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}+f\left(\right(a\ast b)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (a\ast b\left)\right)\hfill \\ & ={d}_{f}(x\ast y,a\ast y)+{d}_{f}(a\ast y,a\ast b)\hfill \end{array}$$

**Definition**

**3.**

Let f be an I-quasi-valuation map of X. Define a relation ${\theta}_{f}$ on X by

$$\begin{array}{c}\hfill (\forall x,y\in X)\left((x,y)\in {\theta}_{f}\u27faf(x\ast y)+f(y\ast x)=0\right).\end{array}$$

**Proof.**

It is clear that ${\theta}_{f}$ is an equivalence relation on X. Let $x,$ $y,$ $u,$ $v\in X$ be such that $(x,y)\in {\theta}_{f}$ and $(u,v)\in {\theta}_{f}$. Then, $f(x\ast y)+f(y\ast x)=0$ and $f(u\ast v)+f(v\ast u)=0$. It follows from Proposition 2 that

$$\begin{array}{c}f\left(\right(x\ast u)\ast (y\ast v\left)\right)+f\left(\right(y\ast v)\ast (x\ast u\left)\right)\hfill \\ ={d}_{f}(x\ast u,y\ast v)\ge {d}_{f}(x,y)\hfill \\ =f(x\ast y)+f(y\ast x)=0.\hfill \end{array}$$

Hence, $f\left(\right(x\ast u)\ast (y\ast v\left)\right)+f\left(\right(y\ast v)\ast (x\ast u\left)\right)=0$, and so $(x\ast u,y\ast v)\in {\theta}_{f}$. Therefore, ${\theta}_{f}$ is a congruence relation on X. ☐

**Definition**

**4.**

Let f be an I-quasi-valuation map of X and ${\theta}_{f}$ be a congruence relation on X induced by f. Given $x\in X$, the set
is called an equivalence class of x.

$$\begin{array}{c}\hfill {x}_{f}:=\{y\in X\mid (x,y)\in {\theta}_{f}\}\end{array}$$

Denote by ${X}_{f}$ the set of all equivalence classes; that is,

$$\begin{array}{c}\hfill {X}_{f}:=\{{x}_{f}\mid x\in X\}.\end{array}$$

**Theorem**

**7.**

Let f be an I-quasi-valuation map of X. Then, $({X}_{f},\odot ,{0}_{f})$ is a $BCK/BCI$-algebra where “⊙” is the binary operation on ${X}_{f}$ which is defined as follows:

$$\begin{array}{c}\hfill (\forall {x}_{f},{y}_{f}\in {X}_{f})\left({x}_{f}\odot {y}_{f}={(x\ast y)}_{f}\right).\end{array}$$

**Proof.**

Let X be a $BCI$-algebra. The operation ⊙ is well-defined since f is an I-quasi-valuation map of X. For any ${x}_{f},{y}_{f},{z}_{f}\in {X}_{f}$, we have
.

$$\begin{array}{c}(({x}_{f}\odot {y}_{f})\odot ({x}_{f}\odot {z}_{f}))\odot ({z}_{f}\odot {y}_{f})={(((x\ast y)\ast (x\ast z))\ast (z\ast y))}_{f}={0}_{f},\hfill \\ ({x}_{f}\odot ({x}_{f}\odot {y}_{f}))\odot {y}_{f}={((x\ast (x\ast y))\ast y)}_{f}={0}_{f},\hfill \\ {x}_{f}\odot {x}_{f}={(x\ast x)}_{f}={0}_{f}\hfill \end{array}$$

Assume that ${x}_{f}\odot {y}_{f}={0}_{f}$ and ${y}_{f}\odot {x}_{f}={0}_{f}$. Then, ${(x\ast y)}_{f}={0}_{f}$ and ${(y\ast x)}_{f}={0}_{f}$, which imply that $(x\ast y,0)\in {\theta}_{f}$ and $(y\ast x,0)\in {\theta}_{f}$. It follows from (1), (5), and (10) that
and

$$\begin{array}{cc}\hfill 0& =f\left(\right(x\ast y)\ast 0)+f(0\ast (x\ast y\left)\right)\hfill \\ & =f(x\ast y)+f\left(\right(0\ast x)\ast (0\ast y\left)\right)\hfill \\ & \le f(x\ast y)+f(0\ast x)-f(0\ast y)\hfill \end{array}$$

$$\begin{array}{cc}\hfill 0& =f\left(\right(y\ast x)\ast 0)+f(0\ast (y\ast x\left)\right)\hfill \\ & =f(y\ast x)+f\left(\right(0\ast y)\ast (0\ast x\left)\right)\hfill \\ & \le f(y\ast x)+f(0\ast y)-f(0\ast x).\hfill \end{array}$$

Hence, $f(x\ast y)+f(0\ast x)-f(0\ast y)=0$ and $f(y\ast x)+f(0\ast y)-f(0\ast x)=0$, which imply that $f(x\ast y)+f(y\ast x)=0$. Hence, $(x,y)\in {\theta}_{f}$; that is, ${x}_{f}={y}_{f}$. Therefore, $({X}_{f},\odot ,{0}_{f})$ is a $BCI$-algebra. Moreover, if X is a $BCK$-algebra, then $0\ast x=0$ for all $x\in X$. Hence, ${0}_{f}\odot {x}_{f}={(0\ast x)}_{f}={0}_{f}$ for all ${x}_{f}\in {X}_{f}$. Hence, $({X}_{f},\odot ,{0}_{f})$ is a $BCK$-algebra. ☐

The following example illustrates Theorem 7.

**Example**

**1.**

Let $X=\{0,a,b,c,d\}$ be a set with the ∗-operation given by Table 1.

Then, $(X;\ast ,0)$ is a BCK-algebra (see [7]), and a real-valued function f on X defined by
is an I-quasi-valuation map of X (see [5]). It is routine to verify that
and ${X}_{f}=\{{0}_{f},{a}_{f},{b}_{f},{d}_{f}\}$ is a $BCK$-algebra where ${0}_{f}=\{0,c\},$ ${a}_{f}=\left\{a\right\},$ ${b}_{f}=\left\{b\right\}$, and ${d}_{f}=\left\{d\right\}$.

$$f=\left(\begin{array}{ccccc}0& a& b& c& d\\ 0& -4& -9& 0& -11\end{array}\right)$$

$$\begin{array}{c}\hfill {\theta}_{f}=\{(0,0),(a,a),(b,b),(c,c),(d,d),(0,c),(c,0)\},\end{array}$$

**Proposition**

**3.**

Given an I-quasi-valuation map f of a $BCI$-algebra X, if ${A}_{f}$ is a closed ideal of X, then ${A}_{f}\subseteq {0}_{f}$.

**Proof.**

Let $x\in {A}_{f}$. Then, $0\ast x\in {A}_{f}$ since ${A}_{f}$ is a closed ideal, and so $f\left(x\right)\ge 0$ and $f(0\ast x)\ge 0$. It follows from (1) that
and so that $f(0\ast x)+f(x\ast 0)=0$ by using Lemma 1(2). Hence, $(0,x)\in {\theta}_{f}$; that is, $x\in {0}_{f}$. Therefore, ${A}_{f}\subseteq {0}_{f}$. ☐

$$\begin{array}{c}\hfill f(0\ast x)+f(x\ast 0)=f(0\ast x)+f\left(x\right)\ge 0,\end{array}$$

**Corollary**

**4.**

If f is an I-quasi-valuation map of a $BCK$-algebra X, then ${A}_{f}\subseteq {0}_{f}$.

**Proposition**

**4.**

Let f be an I-quasi-valuation map of a $BCI$-algebra such that

$$\begin{array}{c}\hfill (\forall x\in X)\left(f\right(x)\le 0).\end{array}$$

Then, ${0}_{f}\subseteq {A}_{f}$.

**Proof.**

Let $x\in {0}_{f}$. Then, $(0,x)\in {\theta}_{f}$, and s

$$\begin{array}{c}\hfill f(0\ast x)+f\left(x\right)=f(0\ast x)+f(x\ast 0)=0.\end{array}$$

It follows from (18) that $f(0\ast x)=0=f\left(x\right)$. Hence, $x\in {A}_{f}$, and therefore ${0}_{f}\subseteq {A}_{f}$. ☐

Let I be an ideal of X and let ${\eta}_{I}$ be a relation on X defined as follows:

$$\begin{array}{c}\hfill (\forall x,y\in X)((x,y)\in {\eta}_{I}\phantom{\rule{3.33333pt}{0ex}}\iff \phantom{\rule{3.33333pt}{0ex}}x\ast y\in I,\phantom{\rule{3.33333pt}{0ex}}y\ast x\in I).\end{array}$$

Then, ${\eta}_{I}$ is a congruence relation on X, which is called the ideal congruence relation on X induced by I (see [6]). Denote by $X/I$ the set of all equivalence classes; that is,
where ${\left[x\right]}_{I}=\{y\in X\mid (x,y)\in {\eta}_{I}\}$. If we define a binary operation ${\ast}_{I}$ on $X/I$ by ${\left[x\right]}_{I}{\ast}_{I}{\left[y\right]}_{I}={[x\ast y]}_{I}$ for all ${\left[x\right]}_{I},{\left[y\right]}_{I}\in X/I$, then $(X,{\ast}_{I},{\left[0\right]}_{I})$ is a $BCK/BCI$-algebra (see [6]).

$$\begin{array}{c}\hfill X/I:=\{{\left[x\right]}_{I}\mid x\in X\},\end{array}$$

**Proposition**

**5.**

If f is an I-quasi-valuation map of X, then ${\eta}_{{A}_{f}}\subseteq {\theta}_{f}$.

**Proof.**

Let $x,y\in X$ be such that $(x,y)\in {\eta}_{{A}_{f}}$. Then, $x\ast y\in {A}_{f}$ and $y\ast x\in {A}_{f}$, which imply that $f(x\ast y)\ge 0$ and $f(y\ast x)\ge 0$. Hence, $f(x\ast y)+f(y\ast x)\ge 0$, and so $f(x\ast y)+f(y\ast x)=0$ by using Lemma 1(2). Thus, $(x,y)\in {\theta}_{f}$. This completes the proof. ☐

**Proposition**

**6.**

If f is an I-quasi-valuation map of X such that ${A}_{f}=X$, then ${\theta}_{f}\subseteq {\eta}_{{A}_{f}}$.

**Proof.**

Let $x,y\in X$ be such that $(x,y)\in {\theta}_{f}$. Then, $f(x\ast y)+f(y\ast x)=0$, and so $f(x\ast y)=0$ and $f(y\ast x)=0$ by the condition ${A}_{f}=X$. It follows that $x\ast y\in {A}_{f}$ and $y\ast x\in {A}_{f}$. Hence, $(x,y)\in {\eta}_{{A}_{f}}$, and therefore ${\theta}_{f}\subseteq {\eta}_{{A}_{f}}$. ☐

**Theorem**

**8.**

If I is an ideal of X, then ${\eta}_{I}={\theta}_{{f}_{I}}$.

**Proof.**

Let $x,y\in X$ be such that $(x,y)\in {\eta}_{I}$. Then, $x\ast y\in I$ and $y\ast x\in I$. It follows that ${f}_{I}(x\ast y)=0$ and ${f}_{I}(y\ast x)=0$. Hence, ${f}_{I}(x\ast y)+{f}_{I}(y\ast x)=0$, and thus $(x,y)\in {\theta}_{{f}_{I}}$.

Conversely, let $(x,y)\in {\theta}_{{f}_{I}}$ for $x,y\in X$. Then, ${f}_{I}(x\ast y)+{f}_{I}(y\ast x)=0$, which implies that ${f}_{I}(x\ast y)=0$ and ${f}_{I}(y\ast x)=0$ since ${f}_{I}\left(x\right)\le 0$ for all $x\in X$. Hence, $x\ast y\in I$ and $y\ast x\in I$; that is, $(x,y)\in {\eta}_{I}$. This completes the proof. ☐

**Corollary**

**5.**

If f is an I-quasi-valuation map of X, then ${\eta}_{{A}_{f}}={\theta}_{{f}_{{A}_{f}}}$.

**Theorem**

**9.**

For any two different I-quasi-valuation maps f and g of X, if ${0}_{f}={0}_{g}$, then ${\theta}_{f}$ and ${\theta}_{g}$ coincide, and so ${X}_{f}={X}_{g}$.

**Proof.**

Let $x,y\in X$ be such that $(x,y)\in {\theta}_{f}$. Then, $(x\ast y,0)=(x\ast y,y\ast y)\in {\theta}_{f}$, and so $x\ast y\in {0}_{f}$. Similarly, we have $y\ast x\in {0}_{f}$. It follows from ${0}_{f}={0}_{g}$ that ${x}_{g}\odot {y}_{g}={(x\ast y)}_{g}={0}_{g}$ and ${y}_{g}\odot {x}_{g}={(y\ast x)}_{g}$ $={0}_{g}$. Hence, ${x}_{g}={y}_{g}$, and so $(x,y)\in {\theta}_{g}$. Similarly, we can verify that if $(x,y)\in {\theta}_{g}$, then $(x,y)\in {\theta}_{f}$. Therefore, ${\theta}_{f}$ and ${\theta}_{g}$ coincide and so ${X}_{f}={X}_{g}$. ☐

**Theorem**

**10.**

Let I be an ideal of X and let f be an I-quasi-valuation map of X such that ${0}_{f}\subseteq I$. If we denote
then the following assertions are valid.

$$\begin{array}{c}\hfill {I}_{f}:=\{{x}_{f}\mid x\in I\},\end{array}$$

- (1)
- $(\forall x\in X)(x\in I\phantom{\rule{3.33333pt}{0ex}}\iff \phantom{\rule{3.33333pt}{0ex}}{x}_{f}\in {I}_{f})$.
- (2)
- ${I}_{f}$ is an ideal of ${X}_{f}$.

**Proof.**

(1) It is clear that if $x\in I$, then ${x}_{f}\in {I}_{f}$. Let $x\in X$ be such that ${x}_{f}\in {I}_{f}$. Then, there exists $y\in I$ such that ${x}_{f}={y}_{f}$. Hence, $(x,y)\in {\theta}_{f}$, and so $(x\ast y,0)=(x\ast y,y\ast y)\in {\theta}_{f}$. It follows that $x\ast y\in {0}_{f}\subseteq I$ and so that $x\in I$.

(2) Clearly, ${0}_{f}\in {I}_{f}$ since $0\in I$. Let $x,y\in X$ be such that ${x}_{f}\odot {y}_{f}\in {I}_{f}$ and ${y}_{f}\in {I}_{f}$. Then, ${(x\ast y)}_{f}={x}_{f}\odot {y}_{f}\in {I}_{f}$, and so $x\ast y\in I$ and $y\in I$ by (1). Since I is an ideal of X, it follows that $x\in I$ and so that ${x}_{f}\in {I}_{f}$. Therefore, ${I}_{f}$ is an ideal of ${X}_{f}$. ☐

**Theorem**

**11.**

For any I-quasi-valuation map f of X, if ${J}^{\ast}$ is an ideal of ${X}_{f}$, then the set
is an ideal of X containing ${0}_{f}$.

$$\begin{array}{c}\hfill J:=\{x\in X\mid {x}_{f}\in {J}^{\ast}\}\end{array}$$

**Proof.**

It is obvious that $0\in {0}_{f}\subseteq J$. Let $x,y\in X$ be such that $x\ast y\in J$ and $y\in J$. Then, ${y}_{f}\in {J}^{\ast}$ and ${x}_{f}\odot {y}_{f}={(x\ast y)}_{f}\in {J}^{\ast}$. Since ${J}^{\ast}$ is an ideal of ${X}_{f}$, it follows that ${x}_{f}\in {J}^{\ast}$ (i.e., $x\in J$). Therefore, J is an ideal of X. ☐

Let $\mathcal{I}\left({X}_{f}\right)$ denote the set of all ideals of ${X}_{f}$, and let $\mathcal{I}(X,f)$ denote the set of all ideals of X containing ${0}_{f}$. Then, there exists a bijection between $\mathcal{I}\left({X}_{f}\right)$ and $\mathcal{I}(X,f)$; that is, $\psi :\mathcal{I}\left({X}_{f}\right)\to \mathcal{I}(X,f),\phantom{\rule{3.33333pt}{0ex}}I\mapsto {I}_{f}$ is a bijection.

**Proposition**

**7.**

Let $\phi :X\to Y$ be a homomorphism of $BCK/BCI$-algebras. If f is an I-quasi-valuation map of Y, then the composition $f\circ \phi $ of f and φ is an I-quasi-valuation map of X.

**Proof.**

We have $(f\circ \phi )\left(0\right)=f\left(\phi \right(0\left)\right)=f\left(0\right)=0$. For any $x,y\in X$, we get

$$\begin{array}{cc}\hfill (f\circ \phi )\left(x\right)& =f\left(\phi \right(x\left)\right)\hfill \\ & \ge f\left(\phi \right(x)\ast \phi (y\left)\right)+f\left(\phi \right(y\left)\right)\hfill \\ & =f\left(\phi \right(x\ast y\left)\right)+f\left(\phi \right(y\left)\right)\hfill \\ & =(f\circ \phi )(x\ast y)+(f\circ \phi )\left(y\right).\hfill \end{array}$$

Hence, $f\circ \phi $ is an I-quasi-valuation map of X. ☐

**Theorem**

**12.**

Let $\phi :X\to Y$ be an onto homomorphism of $BCK/BCI$-algebras. If f is an I-quasi-valuation map of Y, then ${X}_{f\circ \phi}$ and ${Y}_{f}$ are isomorphic.

**Proof.**

Define a map $\zeta :{X}_{f\circ \phi}\to {Y}_{f}$ by $\zeta \left({x}_{f\circ \phi}\right)=\phi {\left(x\right)}_{f}$ for all $x\in X$. If we let ${x}_{f\circ \phi}={a}_{f\circ \phi}$ for $a,x\in X$, then
which implies that $\zeta \left({x}_{f\circ \phi}\right)=\phi {\left(x\right)}_{f}=\phi {\left(a\right)}_{f}=\zeta \left({a}_{f\circ \phi}\right)$. Hence, $\zeta $ is well-defined. For any $a,x\in X$, we have

$$\begin{array}{cc}\hfill 0& =(f\circ \phi )(x\ast a)+(f\circ \phi )(a\ast x)\hfill \\ & =f\left(\phi \right(x\ast a\left)\right)+f\left(\phi \right(a\ast x\left)\right)\hfill \\ & =f\left(\phi \right(x)+\phi (a\left)\right)+f\left(\phi \right(a)\ast \phi (x\left)\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \zeta ({x}_{f\circ \phi}\odot {a}_{f\circ \phi})& =\zeta ({(x\ast a)}_{f\circ \phi})=\phi {(x\ast a)}_{f}\hfill \\ & ={(\phi \left(x\right)\ast \phi \left(a\right))}_{f}=\phi {\left(x\right)}_{f}\odot \phi {\left(a\right)}_{f}\hfill \\ & =\zeta \left({x}_{f\circ \phi}\right)\odot \zeta \left({a}_{f\circ \phi}\right).\hfill \end{array}$$

This shows that $\zeta $ is a homomorphism. For any ${y}_{f}$ in ${Y}_{f}$, there exists $x\in X$ such that $\phi \left(x\right)=y$, since $\phi $ is surjective. It follows that $\zeta \left({x}_{f\circ \phi}\right)=\phi {\left(x\right)}_{f}={y}_{f}$. Thus, $\zeta $ is surjective. Suppose that $\zeta \left({x}_{f\circ \phi}\right)=\zeta \left({a}_{f\circ \phi}\right)$ for any ${x}_{f\circ \phi},{a}_{f\circ \phi}\in {X}_{f\circ \phi}$. Then, $\phi {\left(x\right)}_{f}=\phi {\left(a\right)}_{f}$, and so

$$\begin{array}{cc}\hfill (f\circ \phi )(x\ast a)+(f\circ \phi )(a\ast x)& =f\left(\phi \right(x\ast a\left)\right)+f\left(\phi \right(a\ast x\left)\right)\hfill \\ & =f\left(\phi \right(x)\ast \phi (a\left)\right)+f\left(\phi \right(a)\ast \phi (x\left)\right)=0.\hfill \end{array}$$

Hence, ${x}_{f\circ \phi}={a}_{f\circ \phi}$. This shows that $\zeta $ is injective, and therefore ${X}_{f\circ \phi}$ and ${Y}_{f}$ are isomorphic. ☐

**Theorem**

**13.**

Given an I-quasi-valuation map f of X, the following assertions are valid.

- (1)
- The map $\pi :X\to {X}_{f},\phantom{\rule{3.33333pt}{0ex}}x\mapsto {x}_{f}$ is an onto homomorphism.
- (2)
- For each I-quasi-valuation map ${g}^{\ast}$ of ${X}_{f}$, there exist an I-quasi-valuation map g of X such that $g={g}^{\ast}\circ \pi $.
- (3)
- If ${A}_{f}=X$, then the map$$\begin{array}{c}\hfill {f}^{\ast}:{X}_{f}\to \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}{x}_{f}\mapsto f\left(x\right)\end{array}$$

**Proof.**

(1) and (2) are straightforward.

(3) Assume that ${x}_{f}={y}_{f}$ for $x,y\in X$. Then, $f(x\ast y)+f(y\ast x)=0$, which implies from the assumption that $f(x\ast y)=0=f(y\ast x)$. Since $x\ast (x\ast y)\le y$ for all $x,y\in X$, we get $f\left(y\right)\le f(x\ast (x\ast y\left)\right)$. It follows that

$$\begin{array}{cc}\hfill f\left(x\right)& \ge f(x\ast (x\ast y\left)\right)+f(x\ast y)\ge f(x\ast y)+f\left(y\right)\ge f\left(y\right).\hfill \end{array}$$

Similarly, we show that $f\left(x\right)\le f\left(y\right)$, and so $f\left(x\right)=f\left(y\right)$; that is, ${f}^{\ast}\left({x}_{f}\right)={f}^{\ast}\left({y}_{f}\right)$. Therefore, ${f}^{\ast}$ is well-defined. Now, we have ${f}^{\ast}\left({0}_{f}\right)=f\left(0\right)=0$ and

$$\begin{array}{c}\hfill {f}^{\ast}\left({x}_{f}\right)=f\left(x\right)\ge f(x\ast y)+f\left(y\right)={f}^{\ast}\left({(x\ast y)}_{f}\right)+{f}^{\ast}\left({y}_{f}\right)={f}^{\ast}({x}_{f}\odot {y}_{f})+{f}^{\ast}\left({y}_{f}\right).\end{array}$$

Therefore, ${f}^{\ast}$ is an I-quasi-valuation map of ${X}_{f}$. ☐

## 4. Conclusions

Quasi-valuation maps on $BCK/BCI$-algebras were studied by Song et al. in [5]. The aim of this paper was to study the quotient structures of $BCK/BCI$-algebras induced by quasi-valuation maps. We have described relations between I-quasi-valuation maps and ideals in $BCK/BCI$-algebras. We have induced the quasi-metric space by using an I-quasi-valuation map of a $BCK/BCI$-algebra, and have investigated several properties. We have considered relations between the I-quasi-valuation map and the I-valuation map, and have provided conditions for an I-quasi-valuation map to be an I-valuation map. We have used I-quasi-valuation maps to introduce a congruence relation, and then constructed the quotient structures with related properties. We have established isomorphic quotient $BCK/BCI$-algebras. In the future, from a purely mathematical standpoint, we will apply the concepts and results in this article to related algebraic structures, such as $BCC$-algebras (see [8]), pseudo $BCI$-algebras (see [9,10]), and so on. From an application standpoint, we will try to find the possibility of extending our proposed approach to some decision-making problem, mathematical programming, medical diagnosis, etc.

## Acknowledgments

The authors wish to thank the anonymous reviewers for their valuable suggestions.

## Author Contributions

All authors contributed equally and significantly to the study and preparation of the manuscript. They have read and approved the final article.

## Conflicts of Interest

The authors declare no conflict of interest.

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∗ | 0 | a | b | c | d |
---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 |

a | a | 0 | 0 | a | 0 |

b | b | b | 0 | b | 0 |

c | c | c | c | 0 | c |

d | d | d | d | d | 0 |

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