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

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.

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\Rightarrow x=y).$

If a BCI-algebra X satisfies the following identity:

(V)

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

then X is called a $BCK$-algebra. Any $BCK/BCI$-algebra X satisfies the following conditions:

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

(1)

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

(2)

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

(3)

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

(4)

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

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

(5)

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

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

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:

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

(6)

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

(7)

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

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

(8)

We refer the reader to the books [6,7] for further information regarding $BCK/BCI$-algebras.

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

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

(9)

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

(10)

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

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

(11)

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

$$\begin{array}{c}\hfill f_I:X\to \mathbb{R},x\mapsto \left\{\begin{array}{cc}0\hfill & \mathrm{if}x\in I,\hfill \\ t\hfill & \mathrm{otherwise}\hfill \end{array}\right.,\end{array}$$

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., $I=\left\{0\right\}$).

Proof. 

Straightforward. ☐

Theorem 2.

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

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

is an ideal of X.

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:

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

(12)

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

(13)

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

(14)

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

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

$$\begin{array}{c}\hfill d_f:X\times X\to \mathbb{R},(x,y)\mapsto f(x\ast y)+f(y\ast x).\end{array}$$

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

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

for all $x,y,z\in X$. Therefore $d_f$ is a quasi-metric on X. ☐

Proposition 1.

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

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

(15)

Then, the quasi-metric space $(X,d_f)$ induced by f satisfies:

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

(16)

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:

$$\begin{array}{c}\hfill f_k:\mathbb{Z}\to \mathbb{R},x\mapsto \left\{\begin{array}{cc}0\hfill & \mathrm{if}x=0,\hfill \\ k-x\hfill & \mathrm{otherwise}\hfill \end{array}\right.,\end{array}$$

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

$$\begin{array}{c}\hfill f_k(x-y)+f_k\left(y\right)=\left\{\begin{array}{cc}k-x\hfill & \mathrm{if}\mathrm{either}y=0\mathrm{or}y=x,\hfill \\ 2k-x\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

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

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

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:

$$\begin{array}{c}\hfill (\forall x,y\in X)(d_{f_k}(x,y)=0\Rightarrow x=y).\end{array}$$

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)

$d_f(x\ast y,a\ast b)\ge d_f(x\ast y,a\ast y)+d_f(a\ast y,a\ast b)$,

for all $x,y,a,b\in X$.

Proof. 

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

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

Since f is order reversing, it follows that

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

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

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

and

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

for all $x,y,a,b\in X$. Hence

$$\begin{array}{cc}\hfill d_f(x\ast y,a\ast b)& =f\left(\right(x\ast y)\ast (a\ast b\left)\right)+f\left(\right(a\ast b)\ast (x\ast y\left)\right)\hfill \\ & \ge f\left(\right(x\ast y)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (a\ast b\left)\right)\hfill \\ & +f\left(\right(a\ast b)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (x\ast y\left)\right)\hfill \\ & \ge f\left(\right(x\ast y)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (x\ast y\left)\right)\hfill \\ & +f\left(\right(a\ast b)\ast (a\ast y\left)\right)+f\left(\right(a\ast y)\ast (a\ast b\left)\right)\hfill \\ & =d_f(x\ast y,a\ast y)+d_f(a\ast y,a\ast b)\hfill \end{array}$$

for all $x,y,a,b\in X$. Therefore, (2) is valid. ☐

Definition 3.

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

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

(17)

Theorem 6.

The relation ${\theta }_f$ on X which is given in (17) is a congruence relation on X.

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

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

is called an equivalence class of x.

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

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

and

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

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. ∗-operation.

∗	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

.

Then, $(X;\ast ,0)$ is a BCK-algebra (see [7]), and a real-valued function f on X defined by

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

is an I-quasi-valuation map of X (see [5]). It is routine to verify that

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

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 $d_f=\left\{d\right\}$.

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

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

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, $A_f\subseteq 0_f$. ☐

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

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

(18)

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:

$$\begin{array}{c}\hfill (\forall x,y\in X)((x,y)\in {\eta }_I\iff x\ast y\in I,y\ast x\in I).\end{array}$$

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,

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

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]).

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

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

then the following assertions are valid.

(1)

$(\forall x\in X)(x\in I\iff x_f\in I_f)$.

(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

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

is an ideal of X containing $0_f$.

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, $\psi :\mathcal{I}\left(X_f\right)\to \mathcal{I}(X,f),I\mapsto I_f$ 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

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

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

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

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 $\pi :X\to X_f,x\mapsto x_f$ 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

$$\begin{array}{c}\hfill f^{\ast }:X_f\to \mathbb{R},x_f\mapsto f\left(x\right)\end{array}$$

is an I-quasi-valuation map of $X_f$.

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.