Neutrosophic Quadruple BCK/BCI-Algebras

Abstract

The notion of a neutrosophic quadruple $BCK/BCI$-number is considered, and a neutrosophic quadruple $BCK/BCI$-algebra, which consists of neutrosophic quadruple $BCK/BCI$-numbers, is constructed. Several properties are investigated, and a (positive implicative) ideal in a neutrosophic quadruple $BCK$-algebra and a closed ideal in a neutrosophic quadruple $BCI$-algebra are studied. Given subsets A and B of a $BCK/BCI$-algebra, the set $NQ(A,B)$, which consists of neutrosophic quadruple $BCK/BCI$-numbers with a condition, is established. Conditions for the set $NQ(A,B)$ to be a (positive implicative) ideal of a neutrosophic quadruple $BCK$-algebra are provided, and conditions for the set $NQ(A,B)$ to be a (closed) ideal of a neutrosophic quadruple $BCI$-algebra are given. An example to show that the set $\{\tilde{0}\}$ is not a positive implicative ideal in a neutrosophic quadruple $BCK$-algebra is provided, and conditions for the set $\{\tilde{0}\}$ to be a positive implicative ideal in a neutrosophic quadruple $BCK$-algebra are then discussed.

1. Introduction

The notion of a neutrosophic set was developed by Smarandache [1,2,3] and is a more general platform that extends the notions of classic sets, (intuitionistic) fuzzy sets, and interval valued (intuitionistic) fuzzy sets. Neutrosophic set theory is applied to a different field (see [4,5,6,7,8]). Neutrosophic algebraic structures in $BCK/BCI$-algebras are discussed in [9,10,11,12,13,14,15,16]. Neutrosophic quadruple algebraic structures and hyperstructures are discussed in [17,18].

In this paper, we will use neutrosophic quadruple numbers based on a set and construct neutrosophic quadruple $BCK/BCI$-algebras. We investigate several properties and consider ideals and positive implicative ideals in neutrosophic quadruple $BCK$-algebra, and closed ideals in neutrosophic quadruple $BCI$-algebra. Given subsets A and B of a neutrosophic quadruple $BCK/BCI$-algebra, we consider sets $NQ(A,B)$, which consist of neutrosophic quadruple $BCK/BCI$-numbers with a condition. We provide conditions for the set $NQ(A,B)$ to be a (positive implicative) ideal of a neutrosophic quadruple $BCK$-algebra and for the set $NQ(A,B)$ to be a (closed) ideal of a neutrosophic quadruple $BCI$-algebra. We give an example to show that the set $\{\tilde{0}\}$ is not a positive implicative ideal in a neutrosophic quadruple $BCK$-algebra, and we then consider conditions for the set $\{\tilde{0}\}$ to be a positive implicative ideal in a neutrosophic quadruple $BCK$-algebra.

2. Preliminaries

A $BCK/BCI$-algebra is an important class of logical algebras introduced by Iséki (see [19,20]).

By a $BCI$-algebra, we mean a set X with a special element 0 and a binary operation ∗ that satisfies the following conditions:

(I)

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

(II)

$(\forall x,y\in X)$$((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 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:

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

(1)

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

(2)

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

(3)

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

(4)

where $x\le y$ if and only if $x\ast y=0.$ Any $BCI$-algebra X satisfies the following conditions (see [21]):

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

(5)

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

(6)

A $BCK$-algebra X is said to be positive implicative if the following assertion is valid.

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

(7)

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

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

(8)

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

(9)

A subset I of a $BCI$-algebra X is called a closed ideal (see [21]) of X if it is an ideal of X which satisfies

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

(10)

A subset I of a $BCK$-algebra X is called a positive implicative ideal (see [22]) of X if it satisfies (8) and

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

(11)

Observe that every positive implicative ideal is an ideal, but the converse is not true (see [22]). Note also that a $BCK$-algebra X is positive implicative if and only if every ideal of X is positive implicative (see [22]).

We refer the reader to the books [21,22] for further information regarding $BCK/BCI$-algebras, and to the site “http://fs.gallup.unm.edu/neutrosophy.htm” for further information regarding neutrosophic set theory.

3. Neutrosophic Quadruple BCK/BCI-Algebras

We consider neutrosophic quadruple numbers based on a set instead of real or complex numbers.

Definition 1.

Let X be a set. A neutrosophic quadruple X-number is an ordered quadruple $(a,xT,yI,zF)$ where $a,x,y,z\in X$ and $T,$ $I,$ F have their usual neutrosophic logic meanings.

The set of all neutrosophic quadruple X-numbers is denoted by $NQ(X)$, that is,

$$\begin{array}{c}\hfill NQ(X):=\{(a,xT,yI,zF)\mid a,x,y,z\in X\},\end{array}$$

and it is called the neutrosophic quadruple set based on X. If X is a $BCK/BCI$-algebra, a neutrosophic quadruple X-number is called a neutrosophic quadruple $BCK/BCI$-number and we say that $NQ(X)$ is the neutrosophic quadruple $BCK/BCI$-set.

Let X be a $BCK/BCI$-algebra. We define a binary operation ⊙ on $NQ(X)$ by

$$\begin{array}{c}\hfill (a,xT,yI,zF)\odot (b,uT,vI,wF)=(a\ast b,(x\ast u)T,(y\ast v)I,(z\ast w)F)\end{array}$$

for all $(a,xT,yI,zF),$ $(b,uT,vI,wF)\in NQ(X)$. Given $a_1,a_2,a_3,a_4\in X$, the neutrosophic quadruple $BCK/BCI$-number $(a_1,a_{2}T,a_{3}I,a_{4}F)$ is denoted by $\tilde{a}$, that is,

$$\begin{array}{c}\hfill \tilde{a}=(a_1,a_{2}T,a_{3}I,a_{4}F),\end{array}$$

and the zero neutrosophic quadruple $BCK/BCI$-number $(0,0T,0I,0F)$ is denoted by $\tilde{0}$, that is,

$$\begin{array}{c}\hfill \tilde{0}=(0,0T,0I,0F).\end{array}$$

We define an order relation “≪” and the equality “=” on $NQ(X)$ as follows:

$$\begin{array}{c}\hfill \begin{array}{c}\tilde{x}\ll \tilde{y}\iff x_i\le y_i\mathrm{for}i=1,2,3,4\hfill \\ \tilde{x}=\tilde{y}\iff x_i=y_i\mathrm{for}i=1,2,3,4\hfill \end{array}\end{array}$$

for all $\tilde{x},\tilde{y}\in NQ(X)$. It is easy to verify that “≪” is an equivalence relation on $NQ(X)$.

Theorem 1.

If X is a $BCK/BCI$-algebra, then $(NQ(X);\odot ,\tilde{0})$ is a $BCK/BCI$-algebra.

Proof. 

Let X be a $BCI$-algebra. For any $\tilde{x},\tilde{y},\tilde{z}\in NQ(X)$, we have

$$\begin{array}{cc}\hfill (\tilde{x}\odot \tilde{y})\odot (\tilde{x}\odot \tilde{z})& =(x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)\hfill \\ \hfill & \odot (x_1\ast z_1,(x_2\ast z_2)T,(x_3\ast z_3)I,(x_4\ast z_4)F)\hfill \\ \hfill & =((x_1\ast y_1)\ast (x_1\ast z_1),((x_2\ast y_2)\ast (x_2\ast z_2))T,\hfill \\ \hfill & ((x_3\ast y_3)\ast (x_3\ast z_3))I,((x_4\ast y_4)\ast (x_4\ast z_4))T)\hfill \\ \hfill & \ll (z_1\ast y_1,(z_2\ast y_2)T,(z_3\ast y_3)I,(z_4\ast y_4)F)\hfill \\ \hfill & =\tilde{z}\odot \tilde{y}\hfill \end{array}$$

$$\begin{array}{cc}\hfill \tilde{x}\odot (\tilde{x}\odot \tilde{y})& =(x_1,x_{2}T,x_{3}I,x_{4}F)\odot (x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)\hfill \\ \hfill & =(x_1\ast (x_1\ast y_1),(x_2\ast (x_2\ast y_2))T,(x_3\ast (x_3\ast y_3))I,(x_4\ast (x_4\ast y_4))F)\hfill \\ \hfill & \ll (y_1,y_{2}T,y_{3}I,y_{4}F)\hfill \\ \hfill & =\tilde{y}\hfill \end{array}$$

$$\begin{array}{cc}\hfill \tilde{x}\odot \tilde{x}& =(x_1,x_{2}T,x_{3}I,x_{4}F)\odot (x_1,x_{2}T,x_{3}I,x_{4}F)\hfill \\ \hfill & =(x_1\ast x_1,(x_2\ast x_2)T,(x_3\ast x_3)I,(x_4\ast x_4)F)\hfill \\ \hfill & =(0,0T,0I,0F)=\tilde{0}.\hfill \end{array}$$

Assume that $\tilde{x}\odot \tilde{y}=\tilde{0}$ and $\tilde{y}\odot \tilde{x}=\tilde{0}$. Then

$$\begin{array}{c}\hfill (x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)=(0,0T,0I,0F)\end{array}$$

and

$$\begin{array}{c}\hfill (y_1\ast x_1,(y_2\ast x_2)T,(y_3\ast x_3)I,(y_4\ast x_4)F)=(0,0T,0I,0F).\end{array}$$

It follows that $x_1\ast y_1=0=y_1\ast x_1$, $x_2\ast y_2=0=y_2\ast x_2$, $x_3\ast y_3=0=y_3\ast x_3$ and $x_4\ast y_4=0=y_4\ast x_4$. Hence, $x_1=y_1,$ $x_2=y_2,$ $x_3=y_3$, and $x_4=y_4$, which implies that

$$\begin{array}{c}\hfill \tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)=(y_1,y_{2}T,y_{3}I,y_{4}F)=\tilde{y}.\end{array}$$

Therefore, we know that $(NQ(X);\odot ,\tilde{0})$ is a $BCI$-algebra. We call it the neutrosophic quadruple $BCI$-algebra. Moreover, if X is a $BCK$-algebra, then we have

$$\begin{array}{c}\hfill \tilde{0}\odot \tilde{x}=(0\ast x_1,(0\ast x_2)T,(0\ast x_3)I,(0\ast x_4)F)=(0,0T,0I,0F)=\tilde{0}.\end{array}$$

Hence, $(NQ(X);\odot ,\tilde{0})$ is a $BCK$-algebra. We call it the neutrosophic quadruple $BCK$-algebra. ☐

Example 1.

If $X=\{0,a\}$, then the neutrosophic quadruple set $NQ(X)$ is given as follows:

$$\begin{array}{c}\hfill NQ(X)=\{\tilde{0},\tilde{1},\tilde{2},\tilde{3},\tilde{4},\tilde{5},\tilde{6},\tilde{7},\tilde{8},\tilde{9},\widetilde{10},\widetilde{11},\widetilde{12},\widetilde{13},\widetilde{14},\widetilde{15}\}\end{array}$$

where$\tilde{0}=(0,0T,0I,0F)$, $\tilde{1}=(0,0T,0I,aF)$, $\tilde{2}=(0,0T,aI,0F)$, $\tilde{3}=(0,0T,aI,aF)$,$\tilde{4}=(0,aT,0I,0F)$, $\tilde{5}=(0,aT,0I,aF)$, $\tilde{6}=(0,aT,aI,0F)$, $\tilde{7}=(0,aT,aI,aF)$,$\tilde{8}=(a,0T,0I,0F)$, $\tilde{9}=(a,0T,0I,aF)$, $\widetilde{10}=(a,0T,aI,0F)$, $\widetilde{11}=(a,0T,aI,aF)$,$\widetilde{12}=(a,aT,0I,0F)$, $\widetilde{13}=(a,aT,0I,aF)$, $\widetilde{14}=(a,aT,aI,0F)$, and $\widetilde{15}=(a,aT,aI,aF)$.

Consider a $BCK$-algebra $X=\{0,a\}$ with the binary operation ∗, which is given in Table 1.

Table 1. Cayley table for the binary operation “∗”.

∗	0	a
0	0	0
a	a	0

Then $(NQ(X),\odot ,\tilde{0})$ is a $BCK$-algebra in which the operation ⊙ is given by Table 2.

Table 2. Cayley table for the binary operation “⊙”.

⊙	$\tilde{0}$	$\tilde{1}$	$\tilde{2}$	$\tilde{3}$	$\tilde{4}$	$\tilde{5}$	$\tilde{6}$	$\tilde{7}$	$\tilde{8}$	$\tilde{9}$	$\widetilde{10}$	$\widetilde{11}$	$\widetilde{12}$	$\widetilde{13}$	$\widetilde{14}$	$\widetilde{15}$
$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$
$\tilde{1}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$
$\tilde{2}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$
$\tilde{3}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$
$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$
$\tilde{5}$	$\tilde{5}$	$\tilde{4}$	$\tilde{5}$	$\tilde{4}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{5}$	$\tilde{4}$	$\tilde{5}$	$\tilde{4}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$
$\tilde{6}$	$\tilde{6}$	$\tilde{6}$	$\tilde{4}$	$\tilde{4}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$	$\tilde{6}$	$\tilde{6}$	$\tilde{4}$	$\tilde{4}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$
$\tilde{7}$	$\tilde{7}$	$\tilde{6}$	$\tilde{5}$	$\tilde{4}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$	$\tilde{7}$	$\tilde{6}$	$\tilde{5}$	$\tilde{4}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$
$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$
$\tilde{9}$	$\tilde{9}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{9}$	$\tilde{8}$	$\tilde{9}$	$\tilde{8}$	$\tilde{9}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$
$\widetilde{10}$	$\widetilde{10}$	$\widetilde{10}$	$\tilde{8}$	$\tilde{8}$	$\widetilde{10}$	$\widetilde{10}$	$\tilde{8}$	$\tilde{8}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{2}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$
$\widetilde{11}$	$\widetilde{11}$	$\widetilde{10}$	$\tilde{9}$	$\tilde{8}$	$\widetilde{11}$	$\widetilde{10}$	$\tilde{9}$	$\tilde{8}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$
$\widetilde{12}$	$\widetilde{12}$	$\widetilde{12}$	$\widetilde{12}$	$\widetilde{12}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{8}$	$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{4}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$	$\tilde{0}$
$\widetilde{13}$	$\widetilde{13}$	$\widetilde{12}$	$\widetilde{13}$	$\widetilde{12}$	$\tilde{9}$	$\tilde{8}$	$\tilde{9}$	$\tilde{8}$	$\tilde{5}$	$\tilde{4}$	$\tilde{5}$	$\tilde{4}$	$\tilde{1}$	$\tilde{0}$	$\tilde{1}$	$\tilde{0}$
$\widetilde{14}$	$\widetilde{14}$	$\widetilde{14}$	$\widetilde{12}$	$\widetilde{12}$	$\widetilde{10}$	$\widetilde{10}$	$\tilde{8}$	$\tilde{8}$	$\tilde{6}$	$\tilde{6}$	$\tilde{4}$	$\tilde{4}$	$\tilde{2}$	$\tilde{2}$	$\tilde{0}$	$\tilde{0}$
$\widetilde{15}$	$\widetilde{15}$	$\widetilde{14}$	$\widetilde{13}$	$\widetilde{12}$	$\widetilde{11}$	$\widetilde{10}$	$\tilde{9}$	$\tilde{8}$	$\tilde{7}$	$\tilde{6}$	$\tilde{5}$	$\tilde{4}$	$\tilde{3}$	$\tilde{2}$	$\tilde{1}$	$\tilde{0}$

Theorem 2.

The neutrosophic quadruple set $NQ(X)$ based on a positive implicative $BCK$-algebra X is a positive implicative $BCK$-algebra.

Proof. 

Let X be a positive implicative $BCK$-algebra. Then X is a $BCK$-algebra, so $(NQ(X);\odot ,\tilde{0})$ is a $BCK$-algebra by Theorem 1. Let $\tilde{x}$, $\tilde{y}$, $\tilde{z}\in NQ(X)$. Then

$$\begin{array}{c}\hfill (x_i\ast z_i)\ast (y_i\ast z_i)=(x_i\ast y_i)\ast z_i\end{array}$$

for all $i=1,2,3,4$ since $x_i,y_i,z_i\in X$ and X is a positive implicative $BCK$-algebra. Hence, $(\tilde{x}\odot \tilde{z})\odot (\tilde{y}\ast \tilde{z})=(\tilde{x}\odot \tilde{y})\odot \tilde{z}$; therefore, $NQ(X)$ based on a positive implicative $BCK$-algebra X is a positive implicative $BCK$-algebra. ☐

Proposition 1.

The neutrosophic quadruple set $NQ(X)$ based on a positive implicative $BCK$-algebra X satisfies the following assertions.

$$\begin{array}{c}\hfill (\forall \tilde{x},\tilde{y},\tilde{z}\in NQ(X))(\tilde{x}\odot \tilde{y}\ll \tilde{z}\Rightarrow \tilde{x}\odot \tilde{z}\ll \tilde{y}\odot \tilde{z})\end{array}$$

(12)

$$\begin{array}{c}\hfill (\forall \tilde{x},\tilde{y}\in NQ(X))(\tilde{x}\odot \tilde{y}\ll \tilde{y}\Rightarrow \tilde{x}\ll \tilde{y}).\end{array}$$

(13)

Proof. 

Let $\tilde{x},\tilde{y},\tilde{z}\in NQ(X)$. If $\tilde{x}\odot \tilde{y}\ll \tilde{z}$, then

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

so $\tilde{x}\odot \tilde{z}\ll \tilde{y}\odot \tilde{z}$. Assume that $\tilde{x}\odot \tilde{y}\ll \tilde{y}$. Using Equation (12) implies that

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{y}\ll \tilde{y}\odot \tilde{y}=\tilde{0},\end{array}$$

so $\tilde{x}\odot \tilde{y}=\tilde{0}$, i.e., $\tilde{x}\ll \tilde{y}$. ☐

Let X be a $BCK/BCI$-algebra. Given $a,b\in X$ and subsets A and B of X, consider the sets

$$\begin{array}{c}\hfill NQ(a,B):=\{(a,aT,yI,zF)\in NQ(X)\mid y,z\in B\}\end{array}$$

$$\begin{array}{c}\hfill NQ(A,b):=\{(a,xT,bI,bF)\in NQ(X)\mid a,x\in A\}\end{array}$$

$$\begin{array}{c}\hfill NQ(A,B):=\{(a,xT,yI,zF)\in NQ(X)\mid a,x\in A;y,z\in B\}\end{array}$$

$$\begin{array}{c}\hfill NQ(A^{\ast },B):=\bigcup _{a\in A}NQ(a,B)\end{array}$$

$$\begin{array}{c}\hfill NQ(A,B^{\ast }):=\bigcup _{b\in B}NQ(A,b)\end{array}$$

and

$$\begin{array}{c}\hfill NQ(A\cup B):=NQ(A,0)\cup NQ(0,B).\end{array}$$

The set $NQ(A,A)$ is denoted by $NQ(A)$.

Proposition 2.

Let X be a $BCK/BCI$-algebra. Given $a,b\in X$ and subsets A and B of X, we have

(1)

$NQ(A^{\ast },B)$ and $NQ(A,B^{\ast })$ are subsets of $NQ(A,B)$.

(1)

If $0\in A\cap B$ then $NQ(A\cup B)$ is a subset of $NQ(A,B)$.

Proof. 

Straightforward. ☐

Let X be a $BCK/BCI$-algebra. Given $a,b\in X$ and subalgebras A and B of X, $NQ(a,B)$ and $NQ(A,b)$ may not be subalgebras of $NQ(X)$ since

$$\begin{array}{c}\hfill (a,aT,x_{3}I,x_{4}F)\odot (a,aT,u_{3}I,v_{4}F)=(0,0T,(x_3\ast u_3)I,(x_4\ast v_4)F)\notin NQ(a,B)\end{array}$$

and

$$\begin{array}{c}\hfill (x_1,x_{2}T,bI,bF)\odot (u_1,u_{2}T,bI,bF)=(x_1\ast u_1,(x_2\ast u_2)T,0I,0F)\notin NQ(A,b)\end{array}$$

for $(a,$ $aT,$ $x_{3}I,$ $x_{4}F)\in NQ(a,B)$, $(a,$ $aT,$ $u_{3}I,$ $v_{4}F)\in NQ(a,B)$, $(x_1,$ $x_{2}T,$ $bI,$ $bF)\in NQ(A,b)$, and $(u_1,$ $u_{2}T,$ $bI,$ $bF)\in NQ(A,b)$.

Theorem 3.

If A and B are subalgebras of a $BCK/BCI$-algebra X, then the set $NQ(A,B)$ is a subalgebra of $NQ(X)$, which is called a neutrosophic quadruple subalgebra.

Proof. 

Assume that A and B are subalgebras of a $BCK/BCI$-algebra X. Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$ and $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$ be elements of $NQ(A,B)$. Then $x_1,$ $x_2,$ $y_1,$ $y_2\in A$ and $x_3,$ $x_4,$ $y_3,$ $y_4\in B$, which implies that $x_1\ast y_1\in A$, $x_2\ast y_2\in A$, $x_3\ast y_3\in B$, and $x_4\ast y_4\in B$. Hence,

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{y}=(x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)\in NQ(A,B),\end{array}$$

so $NQ(A,B)$ is a subalgebra of $NQ(X)$. ☐

Theorem 4.

If A and B are ideals of a $BCK/BCI$-algebra X, then the set $NQ(A,B)$ is an ideal of $NQ(X)$, which is called a neutrosophic quadruple ideal.

Proof. 

Assume that A and B are ideals of a $BCK/BCI$-algebra X. Obviously, $\tilde{0}\in NQ(A,B)$. Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$ and $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$ be elements of $NQ(X)$ such that $\tilde{x}\odot \tilde{y}\in NQ(A,B)$ and $\tilde{y}\in NQ(A,B)$. Then

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{y}=(x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)\in NQ(A,B),\end{array}$$

so $x_1\ast y_1\in A$, $x_2\ast y_2\in A$, $x_3\ast y_3\in B$ and $x_4\ast y_4\in B$. Since $\tilde{y}\in NQ(A,B)$, we have $y_1,y_2\in A$ and $y_3,y_4\in B$. Since A and B are ideals of X, it follows that $x_1,x_2\in A$ and $x_3,x_4\in B$. Hence, $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)\in NQ(A,B)$, so $NQ(A,B)$ is an ideal of $NQ(X)$. ☐

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

Corollary 1.

If A and B are ideals of a $BCK$-algebra X, then the set $NQ(A,B)$ is a subalgebra of $NQ(X)$.

The following example shows that Corollary 1 is not true in a $BCI$-algebra.

Example 2.

Consider a $BCI$-algebra $(\mathbb{Z},-,0)$. If we take $A=\mathbb{N}$ and $B=\mathbb{Z}$, then $NQ(A,B)$ is an ideal of $NQ(\mathbb{Z})$. However, it is not a subalgebra of $NQ(\mathbb{Z})$ since

$$\begin{array}{c}\hfill (2,3T,-5I,6F)\odot (3,5T,6I,-7F)=(-1,-2T,-11I,13F)\notin NQ(A,B)\end{array}$$

for $(2,3T,-5I,6F),$ $(3,5T,6I,-7F)\in NQ(A,B)$.

Theorem 5.

If A and B are closed ideals of a $BCI$-algebra X, then the set $NQ(A,B)$ is a closed ideal of $NQ(X)$.

Proof. 

If A and B are closed ideals of a $BCI$-algebra X, then the set $NQ(A,B)$ is an ideal of $NQ(X)$ by Theorem 4. Let $\tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)\in NQ(A,B)$. Then

$$\begin{array}{c}\hfill \tilde{0}\odot \tilde{x}=(0\ast x_1,(0\ast x_2)T,(0\ast x_3)I,(0\ast x_4)F)\in NQ(A,B)\end{array}$$

since $0\ast x_1,0\ast x_2\in A$ and $0\ast x_3,0\ast x_4\in B$. Therefore, $NQ(A,B)$ is a closed ideal of $NQ(X)$. ☐

Since every closed ideal of a $BCI$-algebra X is a subalgebra of X, we have the following corollary.

Corollary 2.

If A and B are closed ideals of a $BCI$-algebra X, then the set $NQ(A,B)$ is a subalgebra of $NQ(X)$.

In the following example, we know that there exist ideals A and B in a $BCI$-algebra X such that $NQ(A,B)$ is not a closed ideal of $NQ(X)$.

Example 3.

Consider $BCI$-algebras $(Y,\ast ,0)$ and $(\mathbb{Z},-,0)$. Then $X=Y\times \mathbb{Z}$ is a $BCI$-algebra (see [21]). Let $A=Y\times \mathbb{N}$ and $B=\{0\}\times \mathbb{N}$. Then A and B are ideals of X, so $NQ(A,B)$ is an ideal of $NQ(X)$ by Theorem 4. Let $((0,0),(0,1)T,(0,2)I,(0,3)F)\in NQ(A,B)$. Then

$$\begin{array}{c}((0,0),(0,0)T,(0,0)I,(0,0)F)\odot ((0,0),(0,1)T,(0,2)I,(0,3)F)\hfill \\ =((0,0),(0,-1)T,(0,-2)I,(0,-3)F)\notin NQ(A,B).\hfill \end{array}$$

Hence, $NQ(A,B)$ is not a closed ideal of $NQ(X)$.

We provide conditions wherethe set $NQ(A,B)$ is a closed ideal of $NQ(X)$.

Theorem 6.

Let A and B be ideals of a $BCI$-algebra X and let

$$\begin{array}{c}\hfill \Gamma :=\{\tilde{a}\in NQ(X)\mid (\forall \tilde{x}\in NQ(X))(\tilde{x}\ll \tilde{a}\Rightarrow \tilde{x}=\tilde{a})\}.\end{array}$$

Assume that, if $\Gamma \subseteq NQ(A,B)$, then $|\Gamma |<\infty$. Then $NQ(A,B)$ is a closed ideal of $NQ(X)$.

Proof. 

If A and B are ideals of X, then $NQ(A,B)$ is an ideal of $NQ(X)$ by Theorem 4. Let $\tilde{a}=(a_1,a_{2}T,a_{3}I,a_{4}F)\in NQ(A,B)$. For any $n\in \mathbb{N}$, denote $n(\tilde{a}):=\tilde{0}\odot {(\tilde{0}\odot \tilde{a})}^n$. Then $n(\tilde{a})\in \Gamma$ and

$$\begin{array}{cc}\hfill n(\tilde{a})& =(0\ast {(0\ast a_1)}^n,(0\ast {(0\ast a_2)}^n)T,(0\ast {(0\ast a_3)}^n)I,(0\ast {(0\ast a_4)}^n)F)\hfill \\ \hfill & =(0\ast (0\ast a_1^n),(0\ast (0\ast a_2^n))T,(0\ast (0\ast a_3^n))I,(0\ast (0\ast a_4^n))F)\hfill \\ \hfill & =\tilde{0}\odot (\tilde{0}\odot {\tilde{a}}^n).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill n(\tilde{a})\odot {\tilde{a}}^n& =(\tilde{0}\odot (\tilde{0}\odot {\tilde{a}}^n))\odot {\tilde{a}}^n\hfill \\ \hfill & =(\tilde{0}\odot {\tilde{a}}^n)\odot (\tilde{0}\odot {\tilde{a}}^n)\hfill \\ \hfill & =\tilde{0}\in NQ(A,B),\hfill \end{array}$$

so $n(\tilde{a})\in NQ(A,B)$, since $\tilde{a}\in NQ(A,B)$, and $NQ(A,B)$ is an ideal of $NQ(X)$. Since $|\Gamma |<\infty$, it follows that $k\in \mathbb{N}$ such that $n(\tilde{a})=(n+k)(\tilde{a})$, that is, $n(\tilde{a})=n(\tilde{a})\odot {(\tilde{0}\odot \tilde{a})}^k$, and thus

$$\begin{array}{cc}\hfill k(\tilde{a})& =\tilde{0}\odot {(\tilde{0}\odot \tilde{a})}^k\hfill \\ \hfill & =(n(\tilde{a})\odot {(\tilde{0}\odot \tilde{a})}^k)\odot n(\tilde{a})\hfill \\ \hfill & =n(\tilde{a})\odot n(\tilde{a})=\tilde{0},\hfill \end{array}$$

i.e., $(k-1)(\tilde{a})\odot (\tilde{0}\odot \tilde{a})=\tilde{0}$. Since $\tilde{0}\odot \tilde{a}\in \Gamma$, it follows that $\tilde{0}\odot \tilde{a}=(k-1)(\tilde{a})\in NQ(A,B)$. Therefore, $NQ(A,B)$ is a closed ideal of $NQ(X)$. ☐

Theorem 7.

Given two elements a and b in a $BCI$-algebra X, let

$$\begin{array}{c}\hfill A_a:=\{x\in X\mid a\ast x=a\}\mathrm{and}{B}_b:=\{x\in X\mid b\ast x=b\}.\end{array}$$

(14)

Then $NQ(A_a,B_b)$ is a closed ideal of $NQ(X)$.

Proof. 

Since $a\ast 0=a$ and $b\ast 0=b$, we have $0\in A_a\cap B_b$. Thus, $\tilde{0}\in NQ(A_a,B_b)$. If $x\in A_a$ and $y\in B_b$, then

$$\begin{array}{c}\hfill 0\ast x=(a\ast x)\ast a=a\ast a=0\mathrm{and}0\ast y=(b\ast y)\ast b=b\ast b=0.\end{array}$$

(15)

Let $x,y,c,d\in X$ be such that $x,y\ast x\in A_a$ and $c,d\ast c\in B_b$. Then

$$\begin{array}{c}\hfill (a\ast y)\ast a=0\ast y=(0\ast y)\ast 0=(0\ast y)\ast (0\ast x)=0\ast (y\ast x)=0\end{array}$$

and

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

that is, $a\ast y\le a$ and $b\ast d\le b$. On the other hand,

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

and

$$\begin{array}{c}\hfill b=b\ast (d\ast c)=(b\ast c)\ast (d\ast c)\le b\ast d.\end{array}$$

Thus, $a\ast y=a$ and $b\ast d=b$, i.e., $y\in A_a$ and $d\in B_b$. Hence, $A_a$ and $B_b$ are ideals of X, and $NQ(A_a,B_b)$ is therefore an ideal of $NQ(X)$ by Theorem 4. Let $\tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)\in NQ(A_a,B_b)$. Then $x_1,x_2\in A_a$, and $x_3,x_4\in B_b$. It follows from Equation (15) that $0\ast x_1=0\in A_a$, $0\ast x_2=0\in A_a$, $0\ast x_3=0\in B_b$, and $0\ast x_4=0\in B_b$. Hence,

$$\begin{array}{c}\hfill \tilde{0}\odot \tilde{x}=(0\ast x_1,(0\ast x_2)T,(0\ast x_3)I,(0\ast x_4)F)\in NQ(A_a,B_b).\end{array}$$

Therefore, $NQ(A_a,B_b)$ is a closed ideal of $NQ(X)$. ☐

Proposition 3.

Let A and B be ideals of a $BCK$-algebra X. Then

$$\begin{array}{c}\hfill NQ(A)\cap NQ(B)=\{\tilde{0}\}\iff (\forall \tilde{x}\in NQ(A))(\forall \tilde{y}\in NQ(B))(\tilde{x}\odot \tilde{y}=\tilde{x}).\end{array}$$

(16)

Proof. 

Note that $NQ(A)$ and $NQ(B)$ are ideals of $NQ(X)$. Assume that $NQ(A)\cap NQ(B)=\{\tilde{0}\}$. Let

$$\tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)\in NQ(A)\mathrm{and}\tilde{y}=(y_1,y_{2}T,y_{3}I,y_{4}F)\in NQ(B).$$

Since $\tilde{x}\odot (\tilde{x}\odot \tilde{y})\ll \tilde{x}$ and $\tilde{x}\odot (\tilde{x}\odot \tilde{y})\ll \tilde{y}$, it follows that $\tilde{x}\odot (\tilde{x}\odot \tilde{y})\in NQ(A)\cap NQ(B)=\{\tilde{0}\}$. Obviously, $(\tilde{x}\odot \tilde{y})\odot \tilde{x}\in \{\tilde{0}\}$. Hence, $\tilde{x}\odot \tilde{y}=\tilde{x}$.

Conversely, suppose that $\tilde{x}\odot \tilde{y}=\tilde{x}$ for all $\tilde{x}\in NQ(A)$ and $\tilde{y}\in NQ(B)$. If $\tilde{z}\in NQ(A)\cap NQ(B)$, then $\tilde{z}\in NQ(A)$ and $\tilde{z}\in NQ(B)$, which is implied from the hypothesis that $\tilde{z}=\tilde{z}\odot \tilde{z}=\tilde{0}$. Hence $NQ(A)\cap NQ(B)=\{\tilde{0}\}$. ☐

Theorem 8.

Let A and B be subsets of a $BCK$-algebra X such that

$$\begin{array}{c}\hfill (\forall a,b\in A\cap B)(K(a,b)\subseteq A\cap B)\end{array}$$

(17)

where $K(a,b):=\{x\in X\mid x\ast a\le b\}$. Then the set $NQ(A,B)$ is an ideal of $NQ(X)$.

Proof. 

If $x\in A\cap B$, then $0\in K(x,x)$ since $0\ast x\le x$. Hence, $0\in A\cap B$ by Equation (17), so it is clear that $\tilde{0}\in NQ(A,B)$. Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$ and $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$ be elements of $NQ(X)$ such that $\tilde{x}\odot \tilde{y}\in NQ(A,B)$ and $\tilde{y}\in NQ(A,B)$. Then

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{y}=(x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)\in NQ(A,B),\end{array}$$

so $x_1\ast y_1\in A$, $x_2\ast y_2\in A$, $x_3\ast y_3\in B$, and $x_4\ast y_4\in B$. Using (II), we have $x_1\in K(x_1\ast y_1,y_1)\subseteq A$, $x_2\in K(x_2\ast y_2,y_2)\subseteq A$, $x_3\in K(x_3\ast y_3,y_3)\subseteq B$, and $x_4\in K(x_4\ast y_4,y_4)\subseteq B$. This implies that $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)\in NQ(A,B)$. Therefore, $NQ(A,B)$ is an ideal of $NQ(X)$. ☐

Corollary 3.

Let A and B be subsets of a $BCK$-algebra X such that

$$\begin{array}{c}\hfill (\forall a,x,y\in X)(x,y\in A\cap B,(a\ast x)\ast y=0\Rightarrow a\in A\cap B).\end{array}$$

(18)

Then the set $NQ(A,B)$ is an ideal of $NQ(X)$.

Theorem 9.

Let A and B be nonempty subsets of a $BCK$-algebra X such that

$$\begin{array}{c}\hfill (\forall a,x,y\in X)(x,y\in A(\mathrm{or}B),a\ast x\le y\Rightarrow a\in A(\mathrm{or}B)).\end{array}$$

(19)

Then the set $NQ(A,B)$ is an ideal of $NQ(X)$.

Proof. 

Assume that the condition expressed by Equation (19) is valid for nonempty subsets A and B of X. Since $0\ast x\le x$ for any $x\in A(\mathrm{or}B)$, we have $0\in A(\mathrm{or}B)$ by Equation (19). Hence, it is clear that $\tilde{0}\in NQ(A,B)$. Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$ and $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$ be elements of $NQ(X)$ such that $\tilde{x}\odot \tilde{y}\in NQ(A,B)$ and $\tilde{y}\in NQ(A,B)$. Then

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{y}=(x_1\ast y_1,(x_2\ast y_2)T,(x_3\ast y_3)I,(x_4\ast y_4)F)\in NQ(A,B),\end{array}$$

so $x_1\ast y_1\in A$, $x_2\ast y_2\in A$, $x_3\ast y_3\in B$, and $x_4\ast y_4\in B$. Note that $x_i\ast (x_i\ast y_i)\le y_i$ for $i=1,2,3,4$. It follows from Equation (19) that $x_1,x_2\in A$ and $x_3,x_4\in B$. Hence,

$$\begin{array}{c}\hfill \tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)\in NQ(A,B);\end{array}$$

therefore, $NQ(A,B)$ is an ideal of $NQ(X)$. ☐

Theorem 10.

If A and B are positive implicative ideals of a $BCK$-algebra X, then the set $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$, which is called a positive implicative neutrosophic quadruple ideal.

Proof. 

Assume that A and B are positive implicative ideals of a $BCK$-algebra X. Obviously, $\tilde{0}\in NQ(A,B)$. Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$, $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$, and $\tilde{z}=(z_1,$ $z_{2}T,$ $z_{3}I,$ $z_{4}F)$ be elements of $NQ(X)$ such that $(\tilde{x}\odot \tilde{y})\odot \tilde{z}\in NQ(A,B)$ and $\tilde{y}\odot \tilde{z}\in NQ(A,B)$. Then

$$\begin{array}{cc}\hfill (\tilde{x}\odot \tilde{y})\odot \tilde{z}& =((x_1\ast y_1)\ast z_1,((x_2\ast y_2)\ast z_2)T,\hfill \\ \hfill & ((x_3\ast y_3)\ast z_3)I,((x_4\ast y_4)\ast z_4)F)\in NQ(A,B),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \tilde{y}\odot \tilde{z}=(y_1\ast z_1,(y_2\ast z_2)T,(y_3\ast z_3)I,(y_4\ast z_4)F)\in NQ(A,B),\end{array}$$

so $(x_1\ast y_1)\ast z_1\in A$, $(x_2\ast y_2)\ast z_2\in A$, $(x_3\ast y_3)\ast z_3\in B$, $(x_4\ast y_4)\ast z_4\in B$, $y_1\ast z_1\in A$, $y_2\ast z_2\in A$, $y_3\ast z_3\in B$, and $y_4\ast z_4\in B$. Since A and B are positive implicative ideals of X, it follows that $x_1\ast z_1,x_2\ast z_2\in A$ and $x_3\ast z_3,x_4\ast z_4\in B$. Hence,

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{z}=(x_1\ast z_1,(x_2\ast z_2)T,(x_3\ast z_3)I,(x_4\ast z_4)F)\in NQ(A,B),\end{array}$$

so $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$. ☐

Theorem 11.

Let A and B be ideals of a $BCK$-algebra X such that

$$\begin{array}{c}\hfill (\forall x,y,z\in X)((x\ast y)\ast z\in A(\mathrm{or}B)\Rightarrow (x\ast z)\ast (y\ast z)\in A(\mathrm{or}B)).\end{array}$$

(20)

Then $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$.

Proof. 

Since A and B are ideals of X, it follows from Theorem 4 that $NQ(A,B)$ is an ideal of $NQ(X)$. Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$, $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$, and $\tilde{z}=(z_1,$ $z_{2}T,$ $z_{3}I,$ $z_{4}F)$ be elements of $NQ(X)$ such that $(\tilde{x}\odot \tilde{y})\odot \tilde{z}\in NQ(A,B)$ and $\tilde{y}\odot \tilde{z}\in NQ(A,B)$. Then

$$\begin{array}{cc}\hfill (\tilde{x}\odot \tilde{y})\odot \tilde{z}& =((x_1\ast y_1)\ast z_1,((x_2\ast y_2)\ast z_2)T,\hfill \\ \hfill & ((x_3\ast y_3)\ast z_3)I,((x_4\ast y_4)\ast z_4)F)\in NQ(A,B),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \tilde{y}\odot \tilde{z}=(y_1\ast z_1,(y_2\ast z_2)T,(y_3\ast z_3)I,(y_4\ast z_4)F)\in NQ(A,B),\end{array}$$

so $(x_1\ast y_1)\ast z_1\in A$, $(x_2\ast y_2)\ast z_2\in A$, $(x_3\ast y_3)\ast z_3\in B$, $(x_4\ast y_4)\ast z_4\in B$, $y_1\ast z_1\in A$, $y_2\ast z_2\in A$, $y_3\ast z_3\in B$, and $y_4\ast z_4\in B$. It follows from Equation (20) that $(x_1\ast z_1)\ast (y_1\ast z_1)\in A$, $(x_2\ast z_2)\ast (y_2\ast z_2)\in A$, $(x_3\ast z_3)\ast (y_3\ast z_3)\in B$, and $(x_4\ast z_4)\ast (y_4\ast z_4)\in B$. Since A and B are ideals of X, we get $x_1\ast z_1\in A$, $x_2\ast z_2\in A$, $x_3\ast z_3\in B$, and $x_4\ast z_4\in B$. Hence,

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{z}=(x_1\ast z_1,(x_2\ast z_2)T,(x_3\ast z_3)I,(x_4\ast z_4)F)\in NQ(A,B).\end{array}$$

Therefore, $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$. ☐

Corollary 4.

Let A and B be ideals of a $BCK$-algebra X such that

$$\begin{array}{c}\hfill (\forall x,y\in X)((x\ast y)\ast y\in A(\mathrm{or}B)\Rightarrow x\ast y\in A(\mathrm{or}B)).\end{array}$$

(21)

Then $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$.

Proof. 

If the condition expressed in Equation (21) is valid, then the condition expressed in Equation (20) is true. Hence, $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$ by Theorem 11. ☐

Theorem 12.

Let A and B be subsets of a $BCK$-algebra X such that $0\in A\cap B$ and

$$\begin{array}{c}\hfill ((x\ast y)\ast y)\ast z\in A(\mathrm{or}B),z\in A(\mathrm{or}B)\Rightarrow x\ast y\in A(\mathrm{or}B)\end{array}$$

(22)

for all $x,y,z\in X$. Then $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$.

Proof. 

Since $0\in A\cap B$, it is clear that $\tilde{0}\in NQ(A,B)$. We first show that

$$\begin{array}{c}\hfill (\forall x,y\in X)(x\ast y\in A(\mathrm{or}B),y\in A(\mathrm{or}B)\Rightarrow x\in A(\mathrm{or}B)).\end{array}$$

(23)

Let $x,y\in X$ be such that $x\ast y\in A$ (or B) and $y\in A$ (or B). Then

$$\begin{array}{c}\hfill ((x\ast 0)\ast 0)\ast y=x\ast y\in A(\mathrm{or}B)\end{array}$$

by Equation (1), which, based on Equations (1) and (22), implies that $x=x\ast 0\in A$ (or B). Let $\tilde{x}=(x_1,$ $x_{2}T,$ $x_{3}I,$ $x_{4}F)$, $\tilde{y}=(y_1,$ $y_{2}T,$ $y_{3}I,$ $y_{4}F)$, and $\tilde{z}=(z_1,$ $z_{2}T,$ $z_{3}I,$ $z_{4}F)$ be elements of $NQ(X)$ such that $(\tilde{x}\odot \tilde{y})\odot \tilde{z}\in NQ(A,B)$ and $\tilde{y}\odot \tilde{z}\in NQ(A,B)$. Then

$$\begin{array}{cc}\hfill (\tilde{x}\odot \tilde{y})\odot \tilde{z}& =((x_1\ast y_1)\ast z_1,((x_2\ast y_2)\ast z_2)T,\hfill \\ \hfill & ((x_3\ast y_3)\ast z_3)I,((x_4\ast y_4)\ast z_4)F)\in NQ(A,B),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \tilde{y}\odot \tilde{z}=(y_1\ast z_1,(y_2\ast z_2)T,(y_3\ast z_3)I,(y_4\ast z_4)F)\in NQ(A,B),\end{array}$$

so $(x_1\ast y_1)\ast z_1\in A$, $(x_2\ast y_2)\ast z_2\in A$, $(x_3\ast y_3)\ast z_3\in B$, $(x_4\ast y_4)\ast z_4\in B$, $y_1\ast z_1\in A$, $y_2\ast z_2\in A$, $y_3\ast z_3\in B$, and $y_4\ast z_4\in B$. Note that

$$\begin{array}{c}\hfill (((x_i\ast z_i)\ast z_i)\ast (y_i\ast z_i))\ast ((x_i\ast y_i)\ast z_i)=0\in A(\mathrm{or}B)\end{array}$$

for $i=1,2,3,4$. Since $(x_i\ast y_i)\ast z_i\in A$ for $i=1,2$ and $(x_j\ast y_j)\ast z_j\in B$ for $j=3,4$, it follows from Equation (23) that $((x_i\ast z_i)\ast z_i)\ast (y_i\ast z_i)\in A$ for $i=1,2$, and $((x_j\ast z_j)\ast z_j)\ast (y_j\ast z_j)\in B$ for $j=3,4$. Moreover, since $y_i\ast z_i\in A$ for $i=1,2$, and $y_j\ast z_j\in B$ for $j=3,4$, we have $x_1\ast z_1\in A$, $x_2\ast z_2\in A$, $x_3\ast z_3\in B$, and $x_4\ast z_4\in B$ by Equation (22). Hence,

$$\begin{array}{c}\hfill \tilde{x}\odot \tilde{z}=(x_1\ast z_1,(x_2\ast z_2)T,(x_3\ast z_3)I,(x_4\ast z_4)F)\in NQ(A,B).\end{array}$$

Therefore, $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$. ☐

Theorem 13.

Let A and B be subsets of a $BCK$-algebra X such that $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$. Then the set

$$\begin{array}{c}\hfill {\Omega }_{\tilde{a}}:=\{\tilde{x}\in NQ(X)\mid \tilde{x}\odot \tilde{a}\in NQ(A,B)\}\end{array}$$

(24)

is an ideal of $NQ(X)$ for any $\tilde{a}\in NQ(X)$.

Proof. 

Obviously, $\tilde{0}\in {\Omega }_{\tilde{a}}$. Let $\tilde{x}$, $\tilde{y}\in NQ(X)$ be such that $\tilde{x}\odot \tilde{y}\in {\Omega }_{\tilde{a}}$ and $\tilde{y}\in {\Omega }_{\tilde{a}}$. Then $(\tilde{x}\odot \tilde{y})\odot \tilde{a}\in NQ(A,B)$ and $\tilde{y}\odot \tilde{a}\in NQ(A,B)$. Since $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$, it follows from Equation (11) that $\tilde{x}\odot \tilde{a}\in NQ(A,B)$ and therefore that $\tilde{x}\in {\Omega }_{\tilde{a}}$. Hence, ${\Omega }_{\tilde{a}}$ is an ideal of $NQ(X)$. ☐

Combining Theorems 12 and 13, we have the following corollary.

Corollary 5.

If A and B are subsets of a $BCK$-algebra X satisfying $0\in A\cap B$ and the condition expressed in Equation (22), then the set ${\Omega }_{\tilde{a}}$ in Equation (24) is an ideal of $NQ(X)$ for all $\tilde{a}\in NQ(X)$.

Theorem 14.

For any subsets A and B of a $BCK$-algebra X, if the set ${\Omega }_{\tilde{a}}$ in Equation (24) is an ideal of $NQ(X)$ for all $\tilde{a}\in NQ(X)$, then $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$.

Proof. 

Since $\tilde{0}\in {\Omega }_{\tilde{a}}$, we have $\tilde{0}=\tilde{0}\odot \tilde{a}\in NQ(A,B)$. Let $\tilde{x},$ $\tilde{y},$ $\tilde{z}\in NQ(X)$ be such that $(\tilde{x}\odot \tilde{y})\odot \tilde{z}\in NQ(A,B)$ and $\tilde{y}\odot \tilde{z}\in NQ(A,B)$. Then $\tilde{x}\odot \tilde{y}\in {\Omega }_{\tilde{z}}$ and $\tilde{y}\in {\Omega }_{\tilde{z}}$. Since ${\Omega }_{\tilde{z}}$ is an ideal of $NQ(X)$, it follows that $\tilde{x}\in {\Omega }_{\tilde{z}}$. Hence, $\tilde{x}\odot \tilde{z}\in NQ(A,B)$. Therefore, $NQ(A,B)$ is a positive implicative ideal of $NQ(X)$. ☐

Theorem 15.

For any ideals A and B of a $BCK$-algebra X and for any $\tilde{a}\in NQ(X)$, if the set ${\Omega }_{\tilde{a}}$ in Equation (24) is an ideal of $NQ(X)$, then $NQ(X)$ is a positive implicative $BCK$-algebra.

Proof. 

Let $\Omega$ be any ideal of $NQ(X)$. For any $\tilde{x},$ $\tilde{y},$ $\tilde{z}\in NQ(X)$, assume that $(\tilde{x}\odot \tilde{y})\odot \tilde{z}\in \Omega$ and $\tilde{y}\odot \tilde{z}\in \Omega$. Then $\tilde{x}\odot \tilde{y}\in {\Omega }_{\tilde{z}}$ and $\tilde{y}\in {\Omega }_{\tilde{z}}$. Since ${\Omega }_{\tilde{z}}$ is an ideal of $NQ(X)$, it follows that $\tilde{x}\in {\Omega }_{\tilde{z}}$. Hence, $\tilde{x}\odot \tilde{z}\in \Omega$, which shows that $\Omega$ is a positive implicative ideal of $NQ(X)$. Therefore, $NQ(X)$ is a positive implicative $BCK$-algebra. ☐

In general, the set $\{\tilde{0}\}$ is an ideal of any neutrosophic quadruple $BCK$-algebra $NQ(X)$, but it is not a positive implicative ideal of $NQ(X)$ as seen in the following example.

Example 4.

Consider a $BCK$-algebra $X=\{0,1,2\}$ with the binary operation ∗, which is given in Table 3.

Table 3. Cayley table for the binary operation “∗”.

∗	0	1	2
0	0	0	0
1	1	0	0
2	2	1	0

Then the neutrosophic quadruple $BCK$-algebra $NQ(X)$ has 81 elements. If we take $\tilde{a}=(2,2T,2I,2F)$ and $\tilde{b}=(1,1T,1I,1F)$ in $NQ(X)$, then

$$\begin{array}{cc}\hfill (\tilde{a}\odot \tilde{b})\odot \tilde{b}& =((2\ast 1)\ast 1,((2\ast 1)\ast 1)T,((2\ast 1)\ast 1)I,((2\ast 1)\ast 1)F)\hfill \\ \hfill & =(1\ast 1,(1\ast 1)T,(1\ast 1)I,(1\ast 1)F)=(0,0T,0I,0F)=\tilde{0},\hfill \end{array}$$

and $\tilde{b}\odot \tilde{b}=\tilde{0}$. However,

$$\begin{array}{c}\hfill \tilde{a}\odot \tilde{b}=(2\ast 1,(2\ast 1)T,(2\ast 1)I,(2\ast 1)F)=(1,1T,1I,1F)\ne \tilde{0}.\end{array}$$

Hence, $\{\tilde{0}\}$ is not a positive implicative ideal of $NQ(X)$.

We now provide conditions for the set $\{\tilde{0}\}$ to be a positive implicative ideal in the neutrosophic quadruple $BCK$-algebra.

Theorem 16.

Let $NQ(X)$ be a neutrosophic quadruple $BCK$-algebra. If the set

$$\begin{array}{c}\hfill \Omega (\tilde{a}):=\{\tilde{x}\in NQ(X)\mid \tilde{x}\ll \tilde{a}\}\end{array}$$

(25)

is an ideal of $NQ(X)$ for all $\tilde{a}\in NQ(X)$, then $\{\tilde{0}\}$ is a positive implicative ideal of $NQ(X)$.

Proof. 

We first show that

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

(26)

Assume that $(\tilde{x}\odot \tilde{y})\odot \tilde{y}=\tilde{0}$ for all $\tilde{x},\tilde{y}\in NQ(X)$. Then $\tilde{x}\odot \tilde{y}\ll \tilde{y}$, so $\tilde{x}\odot \tilde{y}\in \Omega (\tilde{y})$. Since $\tilde{y}\in \Omega (\tilde{y})$ and $\Omega (\tilde{y})$ is an ideal of $NQ(X)$, we have $\tilde{x}\in \Omega (\tilde{y})$. Thus, $\tilde{x}\ll \tilde{y}$, that is, $\tilde{x}\odot \tilde{y}=\tilde{0}$. Let $\tilde{u}:=(\tilde{x}\odot \tilde{y})\odot \tilde{y}$. Then

$$\begin{array}{c}\hfill ((\tilde{x}\odot \tilde{u})\odot \tilde{y})\odot \tilde{y}=((\tilde{x}\odot \tilde{y})\odot \tilde{y})\odot \tilde{u}=\tilde{0},\end{array}$$

which implies, based on Equations (3) and (26), that

$$\begin{array}{c}\hfill (\tilde{x}\odot \tilde{y})\odot ((\tilde{x}\odot \tilde{y})\odot \tilde{y})=(\tilde{x}\odot \tilde{y})\odot \tilde{u}=(\tilde{x}\odot \tilde{u})\odot \tilde{y}=\tilde{0},\end{array}$$

that is, $\tilde{x}\odot \tilde{y}\ll (\tilde{x}\odot \tilde{y})\odot \tilde{y}$. Since $(\tilde{x}\odot \tilde{y})\odot \tilde{y}\ll \tilde{x}\odot \tilde{y}$, it follows that

$$\begin{array}{c}\hfill (\tilde{x}\odot \tilde{y})\odot \tilde{y}=\tilde{x}\odot \tilde{y}.\end{array}$$

(27)

If we put $\tilde{y}=\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))$ in Equation (27), then

$$\begin{array}{cc}\hfill \tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))& =(\tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))))\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\hfill \\ \hfill & \ll (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\hfill \\ \hfill & \ll (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot \tilde{y})\hfill \\ \hfill & =(\tilde{y}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x})\hfill \\ \hfill & =((\tilde{y}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{y}\odot \tilde{x})\hfill \\ \hfill & \ll (\tilde{x}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}).\hfill \end{array}$$

On the other hand,

$$\begin{array}{c}((\tilde{x}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))))\hfill \\ =((\tilde{x}\odot (\tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))))\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}))\hfill \\ =((\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}))\hfill \\ \ll (\tilde{y}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\odot (\tilde{y}\odot \tilde{x}))=\tilde{0},\hfill \end{array}$$

so $((\tilde{x}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))))=\tilde{0}$, that is,

$$\begin{array}{c}\hfill ((\tilde{x}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}))\ll \tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))).\end{array}$$

Hence,

$$\begin{array}{c}\hfill \tilde{x}\odot (\tilde{x}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))=((\tilde{x}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x})).\end{array}$$

(28)

If we use $\tilde{y}\odot \tilde{x}$ instead of $\tilde{x}$ in Equation (28), then

$$\begin{array}{cc}\hfill \tilde{y}\odot \tilde{x}& =(\tilde{y}\odot \tilde{x})\odot \tilde{0}\hfill \\ \hfill & =(\tilde{y}\odot \tilde{x})\odot ((\tilde{y}\odot \tilde{x})\odot (\tilde{y}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))))\hfill \\ \hfill & =((\tilde{y}\odot \tilde{x})\odot ((\tilde{y}\odot \tilde{x})\odot \tilde{y}))\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\hfill \\ \hfill & =(\tilde{y}\odot \tilde{x})\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})),\hfill \end{array}$$

which, by taking $\tilde{x}=\tilde{y}\odot \tilde{x}$, implies that

$$\begin{array}{cc}\hfill \tilde{y}\odot (\tilde{y}\odot \tilde{x})& =(\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{y}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\hfill \\ \hfill & =(\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{y}\odot \tilde{x}).\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot \tilde{y})& =((\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot \tilde{y})\hfill \\ \hfill & \ll (\tilde{x}\odot (\tilde{y}\odot \tilde{x}))\odot (\tilde{x}\odot \tilde{y})\hfill \\ \hfill & =(\tilde{x}\odot (\tilde{x}\odot \tilde{y}))\odot (\tilde{y}\odot \tilde{x}),\hfill \end{array}$$

so,

$$\begin{array}{cc}\hfill \tilde{y}\odot \tilde{x}& =(\tilde{y}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\odot \tilde{0}\hfill \\ \hfill & =(\tilde{y}\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x})))\odot ((\tilde{y}\odot \tilde{x})\odot \tilde{y})\hfill \\ \hfill & \ll ((\tilde{y}\odot \tilde{x})\odot ((\tilde{y}\odot \tilde{x})\odot \tilde{y}))\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\hfill \\ \hfill & =(\tilde{y}\odot \tilde{x})\odot (\tilde{y}\odot (\tilde{y}\odot \tilde{x}))\hfill \\ \hfill & \ll (\tilde{y}\odot \tilde{x})\odot \tilde{x}.\hfill \end{array}$$

Since $(\tilde{y}\odot \tilde{x})\odot \tilde{x}\ll \tilde{y}\odot \tilde{x}$, it follows that

$$\begin{array}{c}\hfill (\tilde{y}\odot \tilde{x})\odot \tilde{x}=\tilde{y}\odot \tilde{x}.\end{array}$$

(29)

Based on Equation (29), it follows that

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

that is, $(\tilde{x}\odot \tilde{z})\ast (\tilde{y}\odot \tilde{z})\ll (\tilde{x}\odot \tilde{y})\odot \tilde{z}$. Note that

$$\begin{array}{c}((\tilde{x}\odot \tilde{y})\odot \tilde{z})\odot ((x\odot \tilde{z})\odot (\tilde{y}\odot \tilde{z}))\hfill \\ =((\tilde{x}\odot \tilde{y})\odot \tilde{z})\odot ((x\odot (\tilde{y}\odot \tilde{z}))\odot \tilde{z})\hfill \\ \ll (\tilde{x}\odot \tilde{y})\odot (\tilde{x}\odot (\tilde{y}\odot \tilde{z}))\hfill \\ \ll (\tilde{y}\odot \tilde{z})\odot \tilde{y}=\tilde{0},\hfill \end{array}$$

which shows that $(\tilde{x}\odot \tilde{y})\odot \tilde{z}\ll (\tilde{x}\odot \tilde{z})\odot (\tilde{y}\odot \tilde{z})$. Hence, $(\tilde{x}\odot \tilde{y})\odot \tilde{z}=(\tilde{x}\odot \tilde{z})\odot (\tilde{y}\odot \tilde{z})$. Therefore, $NQ(X)$ is a positive implicative, so $\{\tilde{0}\}$ is a positive implicative ideal of $NQ(X)$. ☐

4. Conclusions

We have considered a neutrosophic quadruple $BCK/BCI$-number on a set and established neutrosophic quadruple $BCK/BCI$-algebras, which consist of neutrosophic quadruple $BCK/BCI$-numbers. We have investigated several properties and considered ideal theory in a neutrosophic quadruple $BCK$-algebra and a closed ideal in a neutrosophic quadruple $BCI$-algebra. Using subsets A and B of a neutrosophic quadruple $BCK/BCI$-algebra, we have considered sets $NQ(A,B)$, which consist of neutrosophic quadruple $BCK/BCI$-numbers with a condition. We have provided conditions for the set $NQ(A,B)$ to be a (positive implicative) ideal of a neutrosophic quadruple $BCK$-algebra, and the set $NQ(A,B)$ to be a (closed) ideal of a neutrosophic quadruple $BCI$-algebra. We have provided an example to show that the set $\{\tilde{0}\}$ is not a positive implicative ideal in a neutrosophic quadruple $BCK$-algebra, and we have considered conditions for the set $\{\tilde{0}\}$ to be a positive implicative ideal in a neutrosophic quadruple $BCK$-algebra.