Implicative Neutrosophic Quadruple BCK-Algebras and Ideals

Abstract

A neutrosophic set is initiated by Smarandache, and it is a novel tool to deal with vagueness considering the truth, indeterminacy and falsity memberships satisfying the condition that their sum is less than 3. The concept of neutrosophic quadruple numbers was introduced by Florentin Smarandache. Using this idea, Jun et al. introduced the notion of neutrosophic quadruple $BCK/BCI$-numbers, and studied neutrosophic quadruple $BCK/BCI$-algebras. As a continuation of Jun et al.’s paper, the notion of implicative neutrosophic quadruple $BCK$-algebras is introduced, and several properties are investigated. Given a set Y, conditions for the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be a neutrosophic quadruple $BCI$-algebra are provided. Conditions for the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be an implicative neutrosophic quadruple $BCK$-algebra are provided. Given subsets I and J of a $BCK$-algebra Y, conditions for the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ to be an implicative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$ are discussed.

1. Introduction

$BCK/BCI$-algebras are an algebraic structure, which was introduced by Imai, Iséki and Tanaka in 1966, that describes fragments of the propositional calculus involving implication known as $BCK$ and $BCI$ logics. The notion of neutrosophic set, which is developed by Smarandache (see [1,2,3]), is a more general platform which extends the notions of (intuitionistic) fuzzy set, interval valued (intuitionistic) fuzzy set and classic set. Neutrosophic set theory has useful applications in several branches. Decision-making problems are some of the most widely used phenomena in our real-life applications or in various fields like science, engineering, operation research, and management sciences. Garg and Nancy [4] developed a nonlinear programming model based on the technique for order preference by similarity to ideal solution (TOPSIS), in order to solve decision-making problems in which criterion values and their importance are given in the form of interval neutrosophic numbers (INNs). Garg and Nancy [5] presented some new operational laws called logarithm operational laws with real number base for the single-valued neutrosophic (SVN) numbers, and applied it to multiattribute decision making. In algebraic structures of $BCK/BCI$-algebras and semigroup, neutrosophic set theory is discussed in the papers [6,7,8,9,10,11,12,13,14,15]. Smarandache [16] introduced the notion of neutrosophic quadruple numbers. Akinleye et al. [17] introduced the concept of neutrosophic quadruple algebraic structures. Jun et al. [18] studied the neutrosophic quadruple algebraic structures in $BCK/BCI$-algebras, and they introduced the notion of neutrosophic quadruple $BCK/BCI$-algebras.

In this article, we introduce the concept of implicative neutrosophic quadruple $BCK$-algebras, and investigate several properties. In the first and second sections, introduction and basic notions/results are displayed. In the third section, we discuss several properties and (implicative) neutrosophic quadruple ideals in (implicative) neutrosophic quadruple $BCK$-algebras. Given a set Y, we discuss conditions for the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be a neutrosophic quadruple $BCI$-algebra. We provide conditions for the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be an implicative neutrosophic quadruple $BCK$-algebra. Given subsets I and J of a $BCK$-algebra Y, we find conditions for the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ to be an implicative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$.

2. Preliminaries

A $BCI$-algebra (see [19]) is defined to be a set Y with a binary operation ∗ and a special element 0 which satisfies the following conditions:

(I)

$(\forall a,b,c\in Y)$$(((a\ast b)\ast (a\ast c))\ast (c\ast b)=0),$

(II)

$(\forall a\in Y)$$(a\ast 0=a),$

(III)

$(\forall u,v\in Y)$$(u\ast v=0,v\ast u=0\Rightarrow u=v).$

If a $BCI$-algebra Y satisfies the following identity:

(IV)

$(\forall a\in Y)$$(0\ast a=0),$

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

$$\begin{array}{c}\hfill a\ast 0=a,\end{array}$$

(1)

$$\begin{array}{c}\hfill a\le b\Rightarrow a\ast c\le b\ast c,c\ast b\le c\ast a,\end{array}$$

(2)

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

(3)

$$\begin{array}{c}\hfill (a\ast c)\ast (b\ast c)\le a\ast b\end{array}$$

(4)

for all $a,b,c\in Y$.

A $BCK$-algebra Y is said to be

• commutative if the following assertion is valid:

  $$\begin{array}{c}\hfill (\forall a,b\in Y)\left(a\ast (a\ast b)=b\ast (b\ast a)\right).\end{array}$$

  (5)
• implicative if the following assertion is valid:

  $$\begin{array}{c}\hfill (\forall a,b\in Y)\left(a\ast (b\ast a)=a\right).\end{array}$$

  (6)

  A subset I of a $BCK$-algebra Y is called
• an ideal of Y if it satisfies:

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

  (7)

  $$\begin{array}{c}(\forall a\in Y)\left(\forall b\in I\right)\left(a\ast b\in I\Rightarrow a\in I\right),\hfill \end{array}$$

  (8)
• a commutative ideal of Y if it satisfies Label (7) and

  $$\begin{array}{c}\hfill (\forall a,b\in Y)(\forall c\in I)\left((a\ast b)\ast c\in I\Rightarrow a\ast (b\ast (b\ast a))\in I\right).\end{array}$$

  (9)
• an implicative ideal (see [20]) of Y if it satisfies (7) and

  $$\begin{array}{c}\hfill (\forall a,b,c\in Y)(\left((a\ast (b\ast a))\ast c\in I,c\in I\Rightarrow a\in I\right).\end{array}$$

  (10)

We refer the reader to the books [19,20] for further information regarding $BCK/BCI$-algebras, and to [2,3] for further information regarding neutrosophic set theory.

Definition 1 

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

The set of all neutrosophic quadruple Y-numbers is denoted by ${\mathcal{N}}_q(Y)$, that is,

$$\begin{array}{c}\hfill {\mathcal{N}}_q(Y):=\{(a,xT,yI,zF)\mid a,x,y,z\in Y\},\end{array}$$

and it is called the neutrosophic quadruple set based on Y or neutrosophic quadruple Y-set.

Let Y be a set with a binary operation ∗ and a special number 0. We define a binary operation $\tilde{\ast }$ on ${\mathcal{N}}_q(Y)$ by

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

for all $(a,xT,yI,zF),$$(b,uT,vI,wF)\in {\mathcal{N}}_q(Y)$. Given $x_1,x_2,x_3,x_4\in Y$, the neutrosophic quadruple Y-number $(x_1,x_{2}T,x_{3}I,x_{4}F)$ is denoted by $\tilde{x}$, that is,

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

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

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

If Y has an order relation “≤”, then we define an order relation “≪” and the equality “=” on ${\mathcal{N}}_q(Y)$ as follows:

$$\begin{array}{c}\hfill \begin{array}{c}\tilde{a}\ll \tilde{b}\iff a_i\le b_i\mathrm{for}i=1,2,3,4,\hfill \\ \tilde{a}=\tilde{b}\iff a_i=b_i\mathrm{for}i=1,2,3,4\hfill \end{array}\end{array}$$

for all $\tilde{a},\tilde{b}\in {\mathcal{N}}_q(Y)$. It is easy to verify that, if “≤” is a partial order on Y, then “≪” is a partial order on ${\mathcal{N}}_q(Y)$.

Definition 2 

([18]). Given a set Y with a binary operation ∗ and a special number 0, the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ is called a neutrosophic quadruple $BCK/BCI$-algebra if $({\mathcal{N}}_q(Y);\tilde{\ast },\tilde{0})$ is a $BCK/BCI$-algebra.

3. Implicative Neutrosophic Quadruple Ideals

In this section, we first consider conditions for a the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be a neutrosophic quadruple $BCI$-algebra. We define the notion of (commutative, implicative) neutrosophic quadruple $BCK$-algebra and investigate related properties.

Theorem 1.

Given a set Y with a binary operation ∗ and a special number 0, if the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ has a binary operation “ $\tilde{\ast }$” and a partial ordering “≪” such that

(1)

$(\tilde{x}\tilde{\ast }\tilde{y})\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{z})\ll \tilde{z}\tilde{\ast }\tilde{y}$,

(2)

$\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y})\ll \tilde{y}$,

(3)

$\tilde{x}\tilde{\ast }\tilde{y}=\tilde{0}\iff \tilde{x}\ll \tilde{y}$

for all $\tilde{x},\tilde{y},\tilde{z}\in {\mathcal{N}}_q(Y)$, then ${\mathcal{N}}_q(Y)$ is a neutrosophic quadruple $BCI$-algebra.

Proof. 

Let $\tilde{x},\tilde{y},\tilde{z}\in {\mathcal{N}}_q(Y)$. Using conditions (1) and (3) of this theorem, we have

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

(11)

Assume that $\tilde{x}\tilde{\ast }\tilde{y}=\tilde{0}$ and $\tilde{y}\tilde{\ast }\tilde{x}=\tilde{0}$. Then, $\tilde{x}\ll \tilde{y}$ and $\tilde{y}\ll \tilde{x}$ by (3), which implies that $\tilde{x}=\tilde{y}$ by the anti-symmetry of ≪. By the condition (3) of this theorem and the reflexivity of ≪, we get $\tilde{x}\tilde{\ast }\tilde{x}=\tilde{0}$. Using conditions (2) and (3) of this theorem, we have

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

(12)

Putting $\tilde{y}=\tilde{0}$ in (12) implies that

$$\begin{array}{c}\hfill (\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{0}))\tilde{\ast }\tilde{0}=\tilde{0}.\end{array}$$

(13)

If we substitute $\tilde{x}\tilde{\ast }\tilde{0}$ and $\tilde{x}$ for $\tilde{y}$ and $\tilde{z}$, respectively, in (11), then

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

Hence, $\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{0})=\tilde{0}$. On the other hand, we get $(\tilde{x}\tilde{\ast }\tilde{0})\tilde{\ast }\tilde{x}=(\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{x}))\tilde{\ast }\tilde{x}=\tilde{0}$. It follows that $\tilde{x}\tilde{\ast }\tilde{0}=\tilde{x}$. Hence, $({\mathcal{N}}_q(Y);\tilde{\ast },\tilde{0})$ is a $BCI$-algebra, and therefore ${\mathcal{N}}_q(Y)$ is a neutrosophic quadruple $BCI$-algebra. □

Definition 3.

Given a set Y with a binary operation ∗ and a special number 0, the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ is called a (commutative, implicative) neutrosophic quadruple $BCK$-algebra if $({\mathcal{N}}_q(Y);\tilde{\ast },\tilde{0})$ is a (commutative, implicative) $BCK$-algebra.

Example 1.

Given a set $Y=\{0,a\}$, consider the neutrosophic quadruple Y-set as follows:

$$\begin{array}{c}\hfill {\mathcal{N}}_q(Y)=\{{\tilde{\alpha }}_0,{\tilde{\alpha }}_1,{\tilde{\alpha }}_2,{\tilde{\alpha }}_3,{\tilde{\alpha }}_4,{\tilde{\alpha }}_5,{\tilde{\alpha }}_6,{\tilde{\alpha }}_7,{\tilde{\alpha }}_8,{\tilde{\alpha }}_9,{\tilde{\alpha }}_{10},{\tilde{\alpha }}_{11},{\tilde{\alpha }}_{12},{\tilde{\alpha }}_{13},{\tilde{\alpha }}_{14},{\tilde{\alpha }}_{15}\},\end{array}$$

where

$${\tilde{\alpha }}_0=(0,0T,0I,0F),{\tilde{\alpha }}_1=(0,0T,0I,aF),{\tilde{\alpha }}_2=(0,0T,aI,0F),{\tilde{\alpha }}_3=(0,0T,aI,aF),$$

$${\tilde{\alpha }}_4=(0,aT,0I,0F),{\tilde{\alpha }}_5=(0,aT,0I,aF),{\tilde{\alpha }}_6=(0,aT,aI,0F),{\tilde{\alpha }}_7=(0,aT,aI,aF),$$

$${\tilde{\alpha }}_8=(a,0T,0I,0F),{\tilde{\alpha }}_9=(a,0T,0I,aF),{\tilde{\alpha }}_{10}=(a,0T,aI,0F),{\tilde{\alpha }}_{11}=(a,0T,aI,aF),$$

$${\tilde{\alpha }}_{12}=(a,aT,0I,0F),{\tilde{\alpha }}_{13}=(a,aT,0I,aF),{\tilde{\alpha }}_{14}=(a,aT,aI,0F),{\tilde{\alpha }}_{15}=(a,aT,aI,aF).$$

Define a binary operation “$\tilde{\ast }$” on ${\mathcal{N}}_q(Y)$ by Table 1.

Table 1. Binary operation “$\tilde{\ast }$”.

$\tilde{\ast }$	${\tilde{\mathit{\alpha }}}_0$	${\tilde{\mathit{\alpha }}}_1$	${\tilde{\mathit{\alpha }}}_2$	${\tilde{\mathit{\alpha }}}_3$	${\tilde{\mathit{\alpha }}}_4$	${\tilde{\mathit{\alpha }}}_5$	${\tilde{\mathit{\alpha }}}_6$	${\tilde{\mathit{\alpha }}}_7$	${\tilde{\mathit{\alpha }}}_8$	${\tilde{\mathit{\alpha }}}_9$	${\tilde{\mathit{\alpha }}}_{10}$	${\tilde{\mathit{\alpha }}}_{11}$	${\tilde{\mathit{\alpha }}}_{12}$	${\tilde{\mathit{\alpha }}}_{13}$	${\tilde{\mathit{\alpha }}}_{14}$	${\tilde{\mathit{\alpha }}}_{15}$
${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_1$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_3$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_5$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_6$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_7$	${\tilde{\alpha }}_7$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_7$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_9$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_{11}$	${\tilde{\alpha }}_{11}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_{11}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_{13}$	${\tilde{\alpha }}_{13}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{13}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_{14}$	${\tilde{\alpha }}_{14}$	${\tilde{\alpha }}_{14}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_0$	${\tilde{\alpha }}_0$
${\tilde{\alpha }}_{15}$	${\tilde{\alpha }}_{15}$	${\tilde{\alpha }}_{14}$	${\tilde{\alpha }}_{13}$	${\tilde{\alpha }}_{12}$	${\tilde{\alpha }}_{11}$	${\tilde{\alpha }}_{10}$	${\tilde{\alpha }}_9$	${\tilde{\alpha }}_8$	${\tilde{\alpha }}_7$	${\tilde{\alpha }}_6$	${\tilde{\alpha }}_5$	${\tilde{\alpha }}_4$	${\tilde{\alpha }}_3$	${\tilde{\alpha }}_2$	${\tilde{\alpha }}_1$	${\tilde{\alpha }}_0$

Then, ${\mathcal{N}}_q(Y)$ is a (commutative, implicative) neutrosophic quadruple $BCK$-algebra.

Lemma 1 

([18]). If Y is a $BCK/BCI$-algebra, then ${\mathcal{N}}_q(Y)$ is a neutrosophic quadruple $BCK/BCI$-algebra.

Theorem 2.

If Y is an implicative $BCK$-algebra, then the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ is an implicative neutrosophic quadruple $BCK$-algebra.

Proof. 

Let Y be an implicative $BCK$-algebra. Then, Y is a $BCK$-algebra, and so ${\mathcal{N}}_q(Y)$ is a neutrosophic quadruple $BCK$-algebra by Lemma 1. Let $\tilde{x}$, $\tilde{y}\in {\mathcal{N}}_q(Y)$. Then, $x_i\ast (y_i\ast x_i)=x_i$ for all $i=1,2,3,4$ since $x_i,y_i\in Y$ and Y is an implicative $BCK$-algebra. Hence,

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

and therefore ${\mathcal{N}}_q(Y)$ is an implicative neutrosophic quadruple $BCK$-algebra. □

Lemma 2 

([21]). If Y is a commutative $BCK$-algebra, then the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ is a commutative neutrosophic quadruple $BCK$-algebra.

Since every implicative $BCK$-algebra is a commutative $BCK$-algebra, we have the following corollary.

Corollary 1.

Every neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ based on an implicative $BCK$-algebra Y is a commutative neutrosophic quadruple $BCK$-algebra.

Proposition 1.

The neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ based on an implicative $BCK$-algebra Y satisfies the following assertions:

(1)

$(\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y}))\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y})=\tilde{y}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})$,

(2)

$(\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y}))\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})=\tilde{y}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})$,

(3)

$(\tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y}))\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y})=(\tilde{y}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x}))\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})$

for all $\tilde{x},\tilde{y}\in {\mathcal{N}}_q(Y)$.

Proof. 

Let Y be an implicative $BCK$-algebra. Then,

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

for all $x_i,y_i\in Y$ with $i=1,2,3,4$. Thus,

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

This proves (1). Similarly, we can prove (2) and (3). □

Theorem 3.

If the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ based on a $BCK$-algebra Y satisfies the condition (3) in Proposition 1, then it is an implicative neutrosophic quadruple $BCK$-algebra.

Proof. 

By Lemma 1, we know that ${\mathcal{N}}_q(Y)$ is a neutrosophic quadruple $BCK$-algebra. Let $\tilde{x},\tilde{y}\in {\mathcal{N}}_q(Y)$. Then,

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

(14)

for all $i=1,2,3,4$. If we substitute $x_i\ast y_i$ for $y_i$ in (14), then

$$\begin{array}{cc}\hfill x_i\ast y_i& =((x_i\ast y_i)\ast 0)\ast 0=((x_i\ast y_i)\ast ((x_i\ast y_i)\ast x_i))\ast ((x_i\ast y_i)\ast x_i)\hfill \\ & =(x_i\ast (x_i\ast (x_i\ast y_i)))\ast (x_i\ast (x_i\ast y_i))\hfill \\ & =(x_i\ast y_i)\ast (x_i\ast (x_i\ast y_i))\hfill \\ & =(x_i\ast (x_i\ast (x_i\ast y_i))\ast y_i\hfill \\ & =(x_i\ast y_i)\ast y_i.\hfill \end{array}$$

(15)

It follows from (15) and (3) in Proposition 1 that

$$\begin{array}{cc}\hfill x_i\ast (x_i\ast y_i)& =(x_i\ast (x_i\ast y_i))\ast (x_i\ast y_i)=(y_i\ast (y_i\ast x_i))\ast (y_i\ast x_i)=y_i\ast (y_i\ast x_i).\hfill \end{array}$$

(16)

Using (15) and (16), we have

$$\begin{array}{c}\hfill x_i\ast (x_i\ast (y_i\ast x_i))=(y_i\ast x_i)\ast ((y_i\ast x_i)\ast x_i)=((y_i\ast x_i)\ast x_i)\ast ((y_i\ast x_i)\ast x_i)=0.\end{array}$$

Obviously, $(x_i\ast (y_i\ast x_i))\ast x_i=0$. Hence, $x_i\ast (y_i\ast x_i)=x_i$, and thus

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

Hence, $({\mathcal{N}}_q(Y);\tilde{\ast },\tilde{0})$ is an implicative $BCK$-algebra, and therefore ${\mathcal{N}}_q(Y)$ is an implicative neutrosophic quadruple $BCK$-algebra. □

Given subsets I and J of a $BCK$-algebra Y, consider the set

$$\begin{array}{c}\hfill {\mathcal{N}}_q(I,J):=\{(a,xT,yI,zF)\in {\mathcal{N}}_q(Y)\mid a,x\in I;y,z\in J\},\end{array}$$

which is called the neutrosophic quadruple $(I,J)$-set. It is clear that the neutrosophic quadruple $(I,J)$-set is a subset of the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$.

Theorem 4.

If I and J are implicative ideals of a $BCK$-algebra Y, then the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is an implicative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$.

Proof. 

Assume that I and J are implicative ideals of a $BCK$-algebra Y. Obviously, $\tilde{0}\in {\mathcal{N}}_q(I,J)$. 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 ${\mathcal{N}}_q(Y)$ such that $(\tilde{x}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x}))\tilde{\ast }\tilde{z}\in {\mathcal{N}}_q(I,J)$ and $\tilde{z}\in {\mathcal{N}}_q(I,J)$. Then,

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

and so $(x_1\ast (y_1\ast x_1))\ast z_1\in I$, $(x_2\ast (y_2\ast x_2))\ast z_2\in I$, $(x_3\ast (y_3\ast x_3))\ast z_3\in J$ and $(x_4\ast (y_4\ast x_4))\ast z_4\in J$. Since $\tilde{z}\in {\mathcal{N}}_q(I,J)$, we have $z_1,z_2\in I$ and $z_3,z_4\in J$. Since I and I are implicative ideals of Y, it follows that $x_1,x_2\in I$ and $x_3,x_4\in J$. Hence, $\tilde{x}=(x_1,$$x_{2}T,$$x_{3}I,$$x_{4}F)\in {\mathcal{N}}_q(I,J)$, and therefore ${\mathcal{N}}_q(I,J)$ is an implicative ideal of $\mathcal{N}(Y)$. □

Lemma 3 

([21]). If I and J are commutative ideals of a $BCK$-algebra Y, then the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is a commutative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$.

Since every implicative ideal is a commutative ideal, we have the following corollary.

Corollary 2.

If I and J are implicative ideals of a $BCK$-algebra Y, then the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is a commutative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$.

The following example illustrates Theorem 4.

Example 2.

Consider a $BCK$-algebra $Y=\{0,a,b,c\}$ in which the binary operation ∗ is given by Table 2,

Table 2. Binary operation “∗”.

∗	0	a	b	c
0	0	0	0	0
a	a	0	a	0
b	b	b	0	0
c	c	b	a	0

Then, the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$ has 256 elements. Note that $I:=\{0,a\}$ and $J:=\{0,b\}$ are implicative ideals of Y. Hence, the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is given as follows:

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

and it is an implicative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$ where

$$\tilde{0}=(0,0T,0I,0F),\tilde{1}=(0,0T,0I,bF),\tilde{2}=(0,0T,bI,0F),\tilde{3}=(0,0T,bI,bF),$$

$$\tilde{4}=(0,aT,0I,0F),\tilde{5}=(0,aT,0I,bF),\tilde{6}=(0,aT,bI,0F),\tilde{7}=(0,aT,bI,bF),$$

$$\tilde{8}=(a,0T,0I,0F),\tilde{9}=(a,0T,0I,bF),\widetilde{10}=(a,0T,bI,0F),\widetilde{11}=(a,0T,bI,bF),$$

$$\widetilde{12}=(a,aT,0I,0F),\widetilde{13}=(a,aT,0I,bF),\widetilde{14}=(a,aT,bI,0F),\widetilde{15}=(a,aT,bI,bF).$$

Proposition 2.

If I and J are implicative ideals of a $BCK$-algebra Y, then the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ satisfies the following assertion:

$$\begin{array}{c}\hfill (\forall \tilde{x},\tilde{y}\in {\mathcal{N}}_q(Y))\left(\tilde{y}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})\in {\mathcal{N}}_q(I,J)\Rightarrow \tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y})\in {\mathcal{N}}_q(I,J)\right).\end{array}$$

(17)

Proof. 

Assume that I and J are implicative ideals of a $BCK$-algebra Y and $\tilde{y}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})\in {\mathcal{N}}_q(I,J)$ for all $\tilde{x},\tilde{y}\in {\mathcal{N}}_q(Y)$. Then,

$$\begin{array}{c}\hfill \tilde{y}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x})=\left(y_1\ast (y_1\ast x_1),(y_2\ast (y_2\ast x_2))T,(y_3\ast (y_3\ast x_3))I,(y_4\ast (y_4\ast x_4))F\right)\in {\mathcal{N}}_q(I,J),\end{array}$$

and so $y_1\ast (y_1\ast x_1)\in I$, $y_2\ast (y_2\ast x_2)\in I$, $y_3\ast (y_3\ast x_3)\in J$ and $y_4\ast (y_4\ast x_4)\in J$. Since $x_i\ast (x_i\ast y_i)\le x_i$ for $i=1,2,3,4$, we have

$$\begin{array}{c}\hfill y_i\ast x_i\le y_i\ast (x_i\ast (x_i\ast y_i)),\end{array}$$

which implies that

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

that is, $((x_i\ast (x_i\ast y_i))\ast (y_i\ast (x_i\ast (x_i\ast y_i))))\ast (y_i\ast (y_i\ast x_i))=0\in I\cap J$ for $i=1,2,3,4$. Since $y_i\ast (y_i\ast x_i)\in I$ for $i=1,2$, $y_j\ast (y_j\ast x_j)\in J$ for $j=3,4$, and I and J are implicative ideals of Y, it follows from (10) that $x_i\ast (x_i\ast y_i)\in I$ for $i=1,2$, and $x_j\ast (x_j\ast y_j)\in J$ for $j=3,4$. Hence,

$$\begin{array}{c}\hfill \tilde{x}\tilde{\ast }(\tilde{x}\tilde{\ast }\tilde{y})=\left(x_1\ast (x_1\ast y_1),(x_2\ast (x_2\ast y_2))T,(x_3\ast (x_3\ast y_3))I,(x_4\ast (x_4\ast y_4))F\right)\in {\mathcal{N}}_q(I,J).\end{array}$$

This completes the proof. □

Lemma 4 

([18]). If I and J are ideals of a $BCK$-algebra Y, then the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is an ideal of ${\mathcal{N}}_q(Y)$.

Theorem 5.

Let I and J be ideals of a $BCK$-algebra Y such that

$$\begin{array}{c}\hfill (x\ast y)\ast y\in I(\mathrm{resp}.,J)\Rightarrow x\ast y\in I(\mathrm{resp}.,J),\end{array}$$

(18)

$$\begin{array}{c}\hfill y\ast (y\ast x)\in I(\mathrm{resp}.,J)\Rightarrow x\ast (x\ast y)\in I(\mathrm{resp}.,J)\end{array}$$

(19)

for all $x,y\in Y$. Then, the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$.

Proof. 

If I and J are ideals of a $BCK$-algebra Y, then ${\mathcal{N}}_q(I,J)$ is an ideal of ${\mathcal{N}}_q(Y)$ by Lemma 4. Suppose $(\tilde{x}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x}))\tilde{\ast }\tilde{z}\in {\mathcal{N}}_q(I,J)$ and $\tilde{z}\in {\mathcal{N}}_q(I,J)$ for all $\tilde{x},\tilde{y},\tilde{z}\in {\mathcal{N}}_q(Y)$. Then,

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

and $\tilde{z}=\left(z_1,z_{2}T,z_{3}I,z_{4}F\right)\in {\mathcal{N}}_q(I,J)$. It follows that $z_1,z_2\in I$, $z_3,z_4\in J$, $(x_1\ast (y_1\ast x_1))\ast z_1\in I$, $(x_2\ast (y_2\ast x_2))\ast z_2\in I$, $(x_3\ast (y_3\ast x_3))\ast z_3\in J$ and $(x_4\ast (y_4\ast x_4))\ast z_4\in J$. Since I and J are ideals of Y, we have $x_1\ast (y_1\ast x_1)\in I$, $x_2\ast (y_2\ast x_2)\in I$, $x_3\ast (y_3\ast x_3)\in J$ and $x_4\ast (y_4\ast x_4)\in J$. Since $(y_i\ast (y_i\ast x_i))\ast (y_i\ast x_i)\le x_i\ast (y_i\ast x_i)$ for $i=1,2,3,4$, it follows that $(y_i\ast (y_i\ast x_i))\ast (y_i\ast x_i)\in I$ for $i=1,2$, and $(y_j\ast (y_j\ast x_j))\ast (y_j\ast x_j)\in J$ for $j=3,4$. Using (18), we obtain $y_i\ast (y_i\ast x_i)\in I$ for $i=1,2$, and $y_j\ast (y_j\ast x_j)\in J$ for $j=3,4$. It follows from (19) that

$$\begin{array}{c}\hfill x_i\ast (x_i\ast y_i)\in I\mathrm{for}i=1,2,\mathrm{and}{x}_j\ast (x_j\ast y_j)\in J\mathrm{for}j=3,4.\end{array}$$

(20)

Note that $(x_i\ast y_i)\ast z_i\le x_i\ast y_i\le x_i\ast (y_i\ast x_i)$ for $i=1,2,3,4$. Thus, $(x_i\ast y_i)\ast z_i\in I$ for $i=1,2$, and $(x_i\ast y_i)\ast z_i\in J$ for $j=3,4$. Since $z_1,z_2\in I$ and $z_3,z_4\in J$, we obtain $x_i\ast y_i\in I$ for $i=1,2$, and $x_j\ast y_j\in J$ for $j=3,4$, which imply from (20) that $x_1,x_2\in I$ and $x_3,x_4\in J$. Hence, $\tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)\in {\mathcal{N}}_q(I,J)$, and therefore ${\mathcal{N}}_q(I,J)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$. □

Theorem 6.

Let I and J be ideals of a $BCK$-algebra Y such that

$$\begin{array}{c}\hfill x\ast (y\ast x)\in I(\mathrm{resp}.,J)\Rightarrow x\in I(\mathrm{resp}.,J)\end{array}$$

(21)

for all $x,y\in Y$. Then, the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$.

Proof. 

If I and J are ideals of a $BCK$-algebra Y, then ${\mathcal{N}}_q(I,J)$ is an ideal of ${\mathcal{N}}_q(Y)$ by Lemma 4. Let $\tilde{x},\tilde{y},\tilde{z}\in {\mathcal{N}}_q(Y)$ be such that $(\tilde{x}\tilde{\ast }(\tilde{y}\tilde{\ast }\tilde{x}))\tilde{\ast }\tilde{z}\in {\mathcal{N}}_q(I,J)$ and $\tilde{z}\in {\mathcal{N}}_q(I,J)$. Then,

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

and $\tilde{z}=\left(z_1,z_{2}T,z_{3}I,z_F\right)\in {\mathcal{N}}_q(I,J)$. It follows that $z_i\in I$, $(x_i\ast (y_i\ast x_i))\ast z_i\in I$ for $i=1,2$ and $z_j\in J$, $(x_j\ast (y_j\ast x_j))\ast z_j\in J$ for $j=3,4$. Since I and J are ideals of Y, we have $x_i\ast (y_i\ast x_i)\in I$ for $i=1,2$ and $x_j\ast (y_j\ast x_j)\in J$ for $j=3,4$. Using (21), we get $x_1,x_2\in I$ and $x_3,x_4\in J$. Hence $\tilde{x}=(x_1,x_{2}T,x_{3}I,x_{4}F)\in {\mathcal{N}}_q(I,J)$, and therefore ${\mathcal{N}}_q(I,J)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$. □

Lemma 5 

([20]). If I is an implicative ideal of a $BCK$-algebra Y, then every ideal A containing I is implicative.

Theorem 7.

Let A, B, I and J be ideals of a $BCK$-algebra Y such that $A\subseteq I$ and $B\subseteq J$. If A and B are implicative ideals of Y, then the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$.

Proof. 

If A, B, I and J are ideals of Y, then ${\mathcal{N}}_q(A,B)$ and ${\mathcal{N}}_q(I,J)$ are ideals of ${\mathcal{N}}_q(Y)$ by Lemma 4 and ${\mathcal{N}}_q(A,B)\subseteq {\mathcal{N}}_q(I,J)$. Since A and B are implicative ideals of Y, it follows from Theorem 4 that ${\mathcal{N}}_q(A,B)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$. Therefore, the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ is an implicative ideal of ${\mathcal{N}}_q(Y)$ by Lemma 5. □

4. Conclusions

Based on the concept of neutrosophic quadruple numbers which is introduced by Florentin Smarandache, Jun et al. have introduced the notion of neutrosophic quadruple $BCK/BCI$-numbers, and have studied neutrosophic quadruple $BCK/BCI$-algebras. As a continuation of Jun et al.’s paper which has been published in Axioms, we have introduced the notion of implicative neutrosophic quadruple $BCK$-algebras and have investigated several properties. Given a set Y, we have provided conditions for the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be a neutrosophic quadruple $BCI$-algebra, and have considered conditions for the neutrosophic quadruple Y-set ${\mathcal{N}}_q(Y)$ to be an implicative neutrosophic quadruple $BCK$-algebra. Given subsets I and J of a $BCK$-algebra Y, we have discussed conditions for the neutrosophic quadruple $(I,J)$-set ${\mathcal{N}}_q(I,J)$ to be an implicative ideal of the neutrosophic quadruple $BCK$-algebra ${\mathcal{N}}_q(Y)$. In the forthcoming research and papers, we will continue these ideas and will define new notions. We will study several kinds of neutrosophic quadruple ideals in neutrosophic quadruple $BCK/BCI$-algebras.