Some Implicativities for Groupoids and BCK-Algebras

Abstract

In this paper, we generalize the notion of an implicativity discussed in $BCK$-algebras, and apply it to some groupoids and $BCK$-algebras. We obtain some relations among those axioms in the theory of groupoids.

1. Introduction

As a generalization of $BCK$-algebras, the notion of d-algebras was introduced by Neggers and Kim [1]. They discussed some relations between d-algebras and $BCK$-algebras as well as several other relations between d-algebras and oriented digraphs. Several properties on d-algebras, e.g., d-ideals, deformations, and companion d-algebras, were studied [2,3,4]. Recently, some notions of the graph theory were applied to the theory of groupoids [5].

The notion of an implicativity has a very important role in the study of $BCK$-algebras. An implicative $BCK$-algebra has some connections with distributive lattices, Boolean algebras, and semi-Brouwerian algebras.

In this paper, we generalize the notion of the implicativity, which is a useful tool for investigation of $BCK$-algebras by using the notion of a word in general algebraic structures, the most simple mathematical structure, i.e., in the theory of a groupoid. Moreover, we generalized the notion of the implicativity by using $Bin\left(X\right)$-product “□”, and obtain the notion of a weakly i-implicativity, and obtain several properties in $BCK$-algebras and other algebraic structures.

2. Preliminaries

A groupoid $(X,\ast )$ is said to be a left-zero-semigroup if $x\ast y:=x$ for all $x,y\in X$. Similarly, a groupoid $(X,\ast )$ is said to be a right-zero-semigroup if $x\ast y:=y$ for all $x,y\in X$ [6]. A groupoid $(X,\ast ,0)$ with constant 0 is said to be a d-algebra [1] if it satisfies the following conditions:

(I)

$x\ast x=0$,

(II)

$0\ast x=0$,

(III)

$x\ast y=0$ and $y\ast x=0$ imply $x=y$ for all $x,y\in X$.

For brevity, we call X a d-algebra. In a d-algebra X, we define a binary relation $‘‘\le "$ by $x\le y$ if and only if $x\ast y=0$. A d-algebra $(X,\ast ,0)$ is said to be an edge if $x\ast 0=x$ for all $x\in X$. Example 1 below is an edge d-algebra. For general references on d-algebras we refer to [2,3,4].

A $BCK$-algebra [7] is a d-algebra X satisfying the following additional axioms:

(IV)

$\left(\right(x\ast y)\ast (x\ast z\left)\right)\ast (z\ast y)=0$,

(V)

$(x\ast (x\ast y\left)\right)\ast y=0$ for all $x,y,z\in X$.

Theorem 1

([7]). If $(X,\ast ,0)$ is a $BCK$-algebra, then

$$(x\ast y)\ast z=(x\ast z)\ast y$$

for all $x,y,z\in X$.

Example 1.

Let $X:=\{0,a,b,c,d,1\}$ be a set with the following table:

$$\begin{array}{ccccccc}\ast & 0& a& b& c& d& 1\\ 0& 0& 0& 0& 0& 0& 0\\ a& a& 0& 0& a& 0& 0\\ b& b& a& 0& b& a& 0\\ c& c& c& b& 0& 0& 0\\ d& d& c& b& a& 0& 0\\ 1& 1& d& b& a& a& 0\end{array}$$

Then, $(X,\ast ,0)$ is an edge d-algebra which is not a $BCK$-algebra, since $(c\ast b)\ast d=b\ast d=a\ne 0=0\ast b=(c\ast d)\ast b$. For general references on $BCK$-algebras, we refer to [7,8,9].

Let $(X,\le )$ be a partially ordered set with minimal element 0, and let $(X,\ast )$ be its associated groupoid, i.e., * is a binary operation on X defined by

$$x\ast y:=\left\{\begin{array}{cc}0\hfill & \mathrm{if}x\le y,\hfill \\ x\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Then, $(X,\ast ,0)$ is a $BCK$-algebra, and we call it a standard $BCK$-algebra.

A $BCK$-algebra $(X,\ast ,0)$ is said to be implicative if $x=x\ast (y\ast x)$; commutative if $x\ast (x\ast y)=y\ast (y\ast x)$; positive implicative if $(x\ast y)\ast (y\ast z)=(x\ast y)\ast z$ for all $x,y\in X$ [7]. It is well known that a $BCK$-algebra is implicative if and only if it is both commutative and positive implicative. A group X is said to be Boolean if every element of X is its own inverse.

The notion of Smarandache algebras emerged and has been applied to several algebraic structures [10,11,12]. Two algebras $(X,\ast )$ and $(X,\circ )$ are said to be Smarandache disjoint [13,14] if we add some axioms of an algebra $(X,\ast )$ to an algebra $(X,\circ )$, then the algebra $(X,\circ )$ becomes a trivial algebra, i.e., $\left|X\right|=1$; or if we add some axioms of an algebra $(X,\circ )$ to an algebra $(X,\ast )$, then the algebra $(X,\circ )$ becomes a trivial algebra, i.e., $\left|X\right|=1$. Note that if we add an axiom $\left(A\right)$ of an algebra $(X,\ast )$ to another algebra $(X,\circ )$, then we replace the binary operation “∘” in $\left(A\right)$ by the binary operation “*”.

Let $Bin\left(X\right)$ be the collection of all groupoids $(X,\ast )$ defined on X. For any elements $(X,\ast )$ and $(X,•)$ in $Bin\left(X\right)$, we define a binary operation “□” on $Bin\left(X\right)$ by

$$(X,\ast )\square (X,•)=(X,\square ),$$

(1)

where

$$x\square y=(x\ast y)•(y\ast x)$$

(2)

for any $x,y\in X$. Using the notion, Kim and Neggers proved the following theorem.

Theorem 2

([6]). $\left(Bin\right(X),$$\square )$ is a semigroup, i.e., the operation “□" as defined in general is associative. Furthermore, the left zero semigroup is an identity for this operation.

3. (Weakly) Implicativity in Groupoids

By using the notion of words, we generalize the notion of an implicativity in groupoids. A groupoid (or a $BCK$-algebra) $(X,\ast )$ is said to be implicative if

$$x\ast (y\ast x)=x$$

for all $x,y\in X$.

Proposition 1.

If $(X,\ast )$ is a left-zero semigroup (respectively, a right-zero semigroup), i.e., $x\ast y=x$ (respectively, $x\ast y=y$) for all $x,y\in X$, then $(X,\ast )$ is implicative.

Proof. 

If $(X,\ast )$ is a left-zero semigroup, then $x\ast y=x$ for all $x,y\in X$. It follows that $x\ast (y\ast x)=x\ast y=x$, which proves that $(X,\ast )$ is implicative. Similarly, if $(X,\ast )$ is a right-zero semigroup, then it is also implicative. □

Proposition 2.

The class of implicative groupoids and the class of groups are Smarandache disjoint.

Proof. 

Assume $(X,•,e)$ is both a group and an implicative groupoid. Then, $e=e•(x•e)=x•e=x$ for all $x\in X$. This shows that $X=\left\{e\right\}$. □

Notice that the class of implicative groupoids is equationally defined and thus that it is a variety, i.e., it is closed under subgroups, epimorphic images, and direct products.

A groupoid $(X,\ast )$ is said to be weakly implicative if there exists a word $w\left(x\right)$ such that, for all $x,y\in X$,

$$x\ast (y\ast x)=w\left(x\right).$$

Note that $w\left(x\right)$ is an expression of “x”, e.g., $x\ast (x\ast x),x\ast x,\left(\right(x\ast x)\ast x)\ast x,\cdots$, and a zero element “0”, e.g., $x\ast (0\ast x),(0\ast x)\ast (x\ast 0),\cdots$, if necessary.

Proposition 3.

Let $(X,\ast ,0)$ be a weakly implicative groupoid with $w\left(x\right)=x\ast (0\ast x)$. If $(X,\ast ,0)$ is a $BCK$-algebra, then it is an implicative $BCK$-algebra.

Proof. 

Let $(X,\ast ,0)$ be a weakly implicative groupoid with $w\left(x\right):=x\ast (0\ast x)$. Since $(X,\ast ,0)$ is a $BCK$-algebra, we obtain $x\ast (y\ast x)=w\left(x\right)=x\ast (0\ast x)=x\ast 0=x$ for all $x,y\in X$. Hence, $(X,\ast ,0)$ is an implicative $BCK$-algebra. □

Corollary 1.

Let $(X,\ast ,0)$ be an edge d-algebra. If $(X,\ast ,0)$ is a weakly implicative with $w\left(x\right)=x\ast (0\ast x)$, then is an implicative edge d-algebra.

Proof. 

If $(X,\ast ,0)$ is an edge d-algebra, then $0\ast x=0$ and $x\ast 0=x$ for all $x\in X$. By Proposition 3, $(X,\ast ,0)$ is an implicative edge d-algebra. □

Let $(X,\ast )$ be a groupoid. Define a binary operation “•” on X by

$$x•y:=y\ast x$$

for all $x,y\in X$. We call $(X,•)$ an oppositie groupoid of a groupoid $(X,\ast )$.

Theorem 3.

The opposite groupoid of a $BCK$-algebra is weakly implicative.

Proof. 

Let $(X,\ast ,0)$ be a $BCK$-algebra and let $w\left(x\right):=0$ for all $x\in X$. Then, $x•(y•x)=(x\ast y)\ast x=(x\ast x)\ast y=0\ast y=0=w\left(x\right)$. Hence, $(X,•)$ is weakly implicative. □

Proposition 4.

There is no nontrivial implicative opposite groupoid derived from a $BCK$-algebra.

Proof. 

Let $(X,\ast ,0)$ be a $BCK$-algebra and let $\left|X\right|\ge 2$. Assume that $(X,•)$ is implicative. Then, $x=x•(y•x)=(x\ast y)\ast x=(x\ast x)\ast y=0\ast y=0$ for all $x\in X$, i.e., $X=\left\{0\right\}$, a contradiction. □

Theorem 4.

The class of weakly implicative groupoids and the class of groups are Smarandache disjoint.

Proof. 

Assume $(X,·,e)$ is both a group and a weakly groupoid. Then, there exists a word $w\left(x\right)$ such that $x·(y·x)=w\left(x\right)$ for all $x,y\in X$. It follows that $e·(x·e)=w\left(e\right)$ for all $x\in X$. Since $x=e·(x·e)$, we obtain $x=w\left(e\right)$, a constant. Hence, $X=\left\{w\right(e\left)\right\}$, i.e., $\left|X\right|=1$, a contradiction. □

4. Levels of Implicativities

Let $(X,\ast )$ be a groupoid and let $x,y\in X$. We define binary operations “${\square }_i$” on X by $x{\square }_{1}y:=(x\ast y)\ast (y\ast x)=x\square y$ and $x{\square }_{i+1}y:=\left(x{\square }_{i}y\right)\ast \left(y{\square }_{i}x\right)$ for all $x,y\in X$, where $i=1,2,3,\cdots$. Let $w\left(x\right)$ be a word of x. We define the following levels of implicativities as follows:

Level 0:

(i) $x\ast (y\ast x)=w\left(x\right)$ (weakly 0-implicative); (ii) $x\ast (y\ast x)=x$ (implicative).

Level 1:

(i) $x\ast \left(y{\square }_{1}x\right)=w\left(x\right)$ (weakly 1-implicative); (ii) $x\ast \left(y{\square }_{1}x\right)=x$ (1-implicative).

Level i:

(i) $x\ast \left(y{\square }_{i}x\right)=w\left(x\right)$ (weakly i-implicative); (ii) $x\ast \left(y{\square }_{i}x\right)=x$ (i-implicative).

Theorem 5.

Let $(X,·,e)$ be a group with $\left|X\right|\ge 2$. Then, X is weakly 1-implicative if and only if X is a Boolean group.

Proof. 

Let $(X,·,e)$ be a weakly 1-implicative groupoid. Then, $x·\left(y{\square }_{1}x\right)=w\left(x\right)$ for all $x,y\in X$. It follows that $x·\left(\right(y·x)·(x·y\left)\right)=w\left(x\right)$. If we let $x:=e$, then $e·\left(\right(y·e)·(e·y\left)\right)=w\left(e\right)$, and hence $y^2=w\left(e\right)$ for all $y\in X$. If we let $y:=e$, then $w\left(e\right)=e^2=e$. Hence $y^2=w\left(e\right)=e$ for all $y\in X$. Hence, $(X,·,e)$ is a Boolean group.

Assume $(X,·,e)$ is a Boolean group. Then, $x^2=e$ for all $x\in X$. It follows that, for any $x,y\in X$,

$$\begin{array}{ccc}\hfill x·\left(y{\square }_{1}x\right)& =& x·\left(\right(y·x)·(x·y\left)\right)\hfill \\ & =& xy{x}^{2}y\hfill \\ & =& x\hfill \\ & =& w\left(x\right).\hfill \end{array}$$

Hence, $(X,·,e)$ is a weakly 1-implicative groupoid. □

Theorem 6.

Let $(X,·,e)$ be a group. If $(X,·,e)$ is a weakly i-implicative groupoid, then it is i-implicative.

Proof. 

Given $x\in X$, we have $e{\square }_{1}x=(e·x)·(x·e)=x^2$, $x{\square }_{1}e=(x·e)·(e·x)=x^2$, $e{\square }_{2}x=\left(e{\square }_{1}x\right)·\left(x{\square }_{1}e\right)=x^2·x^2=x^4$, and $x{\square }_{2}e=x^4$. Similarly, we obtain $e{\square }_{i}x=x^{2^i}=x{\square }_{i}e$. Since X is a group and $w\left(x\right)$ is a word on x, we have $w\left(e\right)=e$. This shows that $e=w\left(e\right)=e·\left(y{\square }_{i}e\right)=e·y^{2^i}=y^2$ for all $y\in X$. Hence, $w\left(x\right)=x·\left(e{\square }_{i}x\right)=x·x^{2^i}=x·e^i=x$ for all $x\in X$, proving that $(X,·,e)$ is i-implicative. □

Proposition 5.

Let $(X,·,e)$ be a group. If $x^{2^i}=e$ for any $x\in X$, then X is i-implicative.

Proof. 

Given $x,y\in X$, we have $x·\left(y{\square }_{i}x\right)=x·x^{2^i}{y}^{2^i}=x$. Hence, X is i-implicative. □

Theorem 7.

Let $(X,\ast ,0)$ be a $BCK$-algebra. If it is weakly i-implicative, then it is i-implicative.

Proof. 

Suppose that $(X,\ast ,0)$ is weakly i-implicative. Then, there exists a mapping $H:X\times X\to X$ such that, for any $x,y\in X$, $x\ast \left(y{\square }_{i}x\right)=H\left(x\right)$. Since $(X,\ast ,0)$ is a $BCK$-algebra, we obtain $0{\square }_{1}x=(0\ast x)\ast (x\ast 0)=0,0{\square }_{2}x=\left(0{\square }_{1}x\right)\ast \left(x{\square }_{1}0\right)=0$. In this fashion, we obtain $0{\square }_{i}x=0$. Thus, $H\left(x\right)=x\ast \left(0{\square }_{i}x\right)=x\ast 0=x$, which proves that $x\ast \left(y{\square }_{i}x\right)=H\left(x\right)=x\ast \left(0{\square }_{i}x\right)=x$. Hence, $(X,\ast ,0)$ is i-implicative. □

Theorem 8.

Let $(X,\ast )$ be both a weakly 0-implicative groupoid and an 1-implicative groupoid. If $(X,\square ):=(X,\ast )\square (X,\ast )$, then $(X,\square )$ is weakly 0-implicative.

Proof. 

Since $(X,\square )=(X,\ast )\square (X,\ast )$, we have $x\square (y\square x)=(x\ast (y\square x\left)\right)\ast \left(\right(y\square x)\ast x)$ for any $x,y\in X$. It follows from $(X,\ast )$ is 1-implicative that $x=x\ast \left(y{\square }_{1}x\right)=x\ast (y\square x)$ for all $x,y\in X$. Let $z:=y\square x$. Since $(X,\ast )$ is weakly 0-implicative, we have $x\ast (z\ast x)=w\left(x\right)$ for some word $w\left(x\right)$. It follows that

$$\begin{array}{ccc}\hfill x\square (y\square x)& =& (x\ast (y\square x\left)\right)\ast \left(\right(y\square x)\ast x)\hfill \\ & =& x\ast \left(\right(y\square x)\ast x)\hfill \\ & =& x\ast (z\ast x)\hfill \\ & =& w\left(x\right),\hfill \end{array}$$

which proves that $(X,\square )$ is weakly 0-implicative. □

Corollary 2.

Let $(X,\ast )$ be both an implicative groupoid and a 1-implicative groupoid. If $(X,\square ):=(X,\ast )\square (X,\ast )$, then $(X,\square )$ is implicative.

Proof. 

Let $w\left(x\right):=x$ in Theorem 8. □

Let $(X,\ast )$ be a groupoid and let $(X,\square ):=(X,\ast )\square (X,\ast )$. If we assume that $x\square y:=x\ast y$ for any $x,y\in X$, then $x{\square }_{1}y=x\square y=x\ast y$ and hence $x{\square }_{2}y=\left(x{\square }_{1}y\right)\ast \left(y{\square }_{1}x\right)=(x\ast y)\ast (y\ast x)=x{\square }_{1}y=x\square y=x\ast y$. In this fashion, we obtain $x{\square }_{i}y=x\ast y$ for all $i=1,2,\cdots$.

Theorem 9.

Every implicative $BCK$-algebra $(X,\ast ,0)$ is an i-implicative $BCK$-algebra where $i=1,2,\cdots$.

Proof. 

Let $(X,\ast ,0)$ be an implicative $BCK$-algebra. Then, $x\ast (y\ast x)=x$ for any $x,y\in X$. It follows from Theorem 1 that

$$\begin{array}{ccc}\hfill y\square x& =& (y\ast x)\ast (x\ast y)\hfill \\ & =& (y\ast (x\ast y\left)\right)\ast x\hfill \\ & =& y\ast x,\hfill \end{array}$$

i.e., $y\square x=y\ast x$. This shows that $x\ast \left(y{\square }_{i}x\right)=x\ast (y\square x)=x\ast (y\ast x)=x$ for any $i=1,2,\cdots$. Hence, $(X,\ast ,0)$ is an i-implicative $BCK$-algebra. □

5. Weakly Implicative Groupoids with P$\left({\mathit{L}}_{\mathit{i}}\right)$

A groupoid $(X,\ast ,0)$ is said to have a condition $\left(L_i\right)$ if it satisfies the following condition, for any $x,y\in X$,

$$\begin{array}{cc}\hfill x{\square }_{i+1}y& =x{\square }_{i}y,\left(L_i\right);\hfill \end{array}$$

and a groupoid $(X,\ast ,0)$ is said to have a condition $\left(L_0\right)$ if it satisfies the following condition, for any $x,y\in X$,

$$\begin{array}{cc}\hfill x{\square }_{1}y& =x{\square }_{0}y,\left(L_0\right),\hfill \end{array}$$

i.e., $(x\ast y)\ast (y\ast x)=x\ast y$. Assume that a groupoid $(X,\ast )$ has the condition $\left(L_i\right)$. Then, $x{\square }_{i+2}y=\left(x{\square }_{i+1}y\right)\ast \left(y{\square }_{i+1}x\right)=\left(x{\square }_{i}y\right)\ast \left(y{\square }_{i}x\right)=x{\square }_{i+1}y$ for any $x,y\in X$. Similarly, $x{\square }_{i+3}y=x{\square }_{i+2}y=x{\square }_{i+1}y$. In this fashion, we have $x{\square }_{i+k}y=x{\square }_{i+k-1}y$ for any $k=1,2,\cdots$. Hence, $(X,\ast )$ satisfies the condition $\left(L_{i+k}\right)$.

Proposition 6.

If a groupoid $(X,\ast )$ is a weakly i-implicative groupoid with $\left(L_i\right)$, then it is a weakly $(i+k)$-implicative groupoid.

Proof. 

Let $(X,\ast )$ be a weakly i-implicative groupoid with $\left(L_i\right)$. Then, $x\ast \left(y{\square }_{i}x\right)=w\left(x\right)$ and $y{\square }_{i+k}x=y{\square }_{i}x$ for any $x,y\in X$, where $k=1,2,\cdots$. It follows that $x\ast \left(y{\square }_{i+k}x\right)=x\ast \left(y{\square }_{i}x\right)=w\left(x\right)$ for any $k=1,2,\cdots$. This proves that $(X,\ast )$ is a weakly $(i+k)$-implicative groupoid. □

Theorem 10.

Any standard $BCK$-algebra has the condition $\left(L_0\right)$.

Proof. 

Let $(X,\ast ,0)$ be a standard $BCK$-algebra. Given $x,y\in X$, we have 3 cases: (i) $x\ast y=0$; (ii) $y\ast x=0$; (iii) $x\ast y\ne 0,y\ast x\ne 0$. Case (i). If $x\ast y=0$, then $x\square y=(x\ast y)\ast (y\ast x)=0\ast (y\ast x)=0=x\ast y$. Case (ii). If $y\ast x=0$, then $x\square y=(x\ast y)\ast (y\ast x)=(x\ast y)\ast 0=x\ast y$. Case (iii). If $x\ast y\ne 0,y\ast x\ne 0$, then $x\ast y=x$ and $y\ast x=y$. It follows that $x\square y=(x\ast y)\ast (y\ast x)=x\ast y$. Hence, $x{\square }_{1}y=x{\square }_{0}y=x\ast y$. □

Note that nonstandard $BCK$-algebras need not have the condition $\left(L_0\right)$. Consider the following example.

Example 2.

Let $X:=\{0,1,2,3\}$ be a set with the following table:

$$\begin{array}{ccccc}\ast & 0& 1& 2& 3\\ 0& 0& 1& 2& 3\\ 1& 1& 0& 3& 2\\ 2& 2& 3& 0& 1\\ 3& 3& 2& 1& 0\end{array}$$

Then, $(X,\ast ,0)$ is a $BCK$-algebra ([7], p. 245). Since $2\ast 3=1$ and $(2\ast 3)\ast (3\ast 2)=1\ast 3=0$, we have $2\square 3\ne 2\ast 3$, i.e., $(X,\ast ,0)$ does not satisfy the condition $\left(L_0\right)$.

A groupoid $(X,\ast )$ is said to have a condition$\left(\alpha \right)$ if $X\times X=A\cup B\cup C$, where

$$\begin{array}{cc}\hfill A& =\left\{\right(x,y\left)\right|y\ast x=0\},\hfill \\ \hfill B& =\left\{\right(x,y\left)\right|x\ast y=0\},\hfill \\ \hfill C& =\left\{\right(x,y\left)\right|x\ast y=x,y\ast x=y\}.\hfill \end{array}$$

Theorem 11.

Let $(X,\ast ,0)$ be a groupoid with a condition $\left(\alpha \right)$. If $(X,\ast ,0)$ satisfies the following conditions: (i) $0\ast x=x$; (ii) $x\ast 0=x$; (iii) $x\ast x=0$; (iv) $y\ast x=0$ implies $x\ast y\in \{0,x\}$, then $(x\ast (x\ast y\left)\right)\ast y=0$ for all $x,y\in X$.

Proof. 

Case (i). If $(x,y)\in A$, then $y\ast x=0$. By (iv), we have $x\ast y\in \{0,x\}$. If $x\ast y=0$, then $(x\ast (x\ast y\left)\right)\ast y=(x\ast 0)\ast y=x\ast y=0$. If $x\ast y=x$, then $(x\ast (x\ast y\left)\right)\ast y=(x\ast x)\ast y=0\ast y=0$. Case (ii). If $(x,y)\in B$, then $x\ast y=0$ and hence $(x\ast (x\ast y\left)\right)\ast y=(x\ast 0)\ast y=x\ast y=0$. Case (iii). If $(x,y)\in C$, then $x\ast y=x$ and $y\ast x=y$. It follows that $(x\ast (x\ast y\left)\right)\ast y=(x\ast x)\ast y=0\ast y=0$. □

Theorem 12.

Let $(X,\ast ,0)$ be a groupoid with a condition $\left(\alpha \right)$. If $(X,\ast ,0)$ satisfies the following conditions: (i) $x\ast 0=x$; (ii) $0\ast (x\ast y)=y\ast x$ for all $x,y\in X$, then $(X,\ast ,0)$ satisfies the condition $\left(L_0\right)$.

Proof. 

Given $x,y\in X$, if $(x,y)\in A$, then $y\ast x=0$ and hence $x\square y=(x\ast y)\ast (y\ast x)=(x\ast y)\ast 0=x\ast y$. If $(x,y)\in B$, then $x\ast y=0$ and hence $x\square y=(x\ast y)\ast (y\ast x)=0\ast (y\ast x)=x\ast y$. If $(x,y)\in C$, then $x\ast y=x,y\ast x=y$ and hence $x\square y=(x\ast y)\ast (y\ast x)=x\ast y$, proving the theorem. □

Theorem 13.

Let K be a field and let $A,B,C\in K,\left|K\right|\ge 3$. Define a binary operation “*" on K by $x\ast y:=A+Bx+Cy$ for all $x,y\in K$. If $(K,\ast )$ is an implicative groupoid, then $x\ast y$ is one of the following:

(i)

$x\ast y=x$,

(ii)

$x\ast y=y$,

(iii)

$x\ast y=A-y$.

Proof. 

Since $(K,\ast )$ is an implicative groupoid, we have

$$\begin{array}{ccc}\hfill x& =& x\ast (y\ast x)\hfill \\ & =& A+Bx+C(A+Bx+Cy)\hfill \\ & =& A(1+C)+(B+C^2)x+BCy\hfill \end{array}$$

for any $x,y\in K$. It follows that $A(1+C)=0,B+C^2=1$, and $BC=0$. Case 1. Assume $B=0$. Since $B+C^2=1$, we obtain $C^2=1$, i.e., $C=\pm 1$. If $C=1$, then $A=0$, since $A(1+C)=0$. Hence, $x\ast y=y$. If $C=-1$, then A is arbitrary, since $A(1+C)=0$. Hence, $x\ast y=A-y$. Case 2. Assume $C=0$. Since $A(1+C)=0,B+C^2=1$, we obtain $A=0,B=1$, i.e., $x\ast y=x$. □

Theorem 14.

Let K be a field and let $A,B,C\in K,\left|K\right|\ge 3$. Define a binary operation “*" on K by $x\ast y:=A+Bx+Cy$ for all $x,y\in K$. If $(K,\ast )$ satisfies the condition $\left(L_0\right)$, then $x\ast y$ is one of the following:

(i)

$x\ast y=A$,

(ii)

$x\ast y=x$,

(iii)

$x\ast y=\frac{1}{2}(x+y)$,

(iv)

$x\ast y=A-\frac{1}{2}(x-y)$.

Proof. 

Since $x\ast y=A+Bx+Cy$ and $y\ast x=A+By+Cx$, we have

$$\begin{array}{ccc}\hfill (x\ast y)\ast (y\ast x)& =& (A+Bx+Cy)\ast (A+By+Cx)\hfill \\ & =& A+B(A+Bx+Cy)+C(A+By+Cx)\hfill \\ & =& A(1+B+C)+(B^2+C^2)x+2BCy\hfill \\ & =& x\ast y\hfill \\ & =& A+Bx+Cy\hfill \end{array}$$

for any $x,y\in K$. It follows that $A(1+B+C)=A,B^2+C^2=B$ and $2BC=C$. This shows that $C=0$ or $B=\frac{1}{2}$. Case 1. $C=0$. Since $B^2+C^2=B$, we obtain that either $B=0$ or $B=1$. If $B=0$, then $x\ast y=A$. If $B=1$, then $A=A(1+B+C)=2A$, i.e., $A=0$. Hence, $x\ast y=x$. Case 2. $B=\frac{1}{2}$. Since $B^2+C^2=B$, we obtain $C=\pm \frac{1}{2}$. If $C=\frac{1}{2}$, then $A=A(1+B+C)=2A$, i.e., $A=0$. Hence, $x\ast y=\frac{1}{2}(x+y)$. If $C=-\frac{1}{2}$, then $A=A(1+B+C)=A$, and hence A is arbitrary. Hence, $x\ast y=A-\frac{1}{2}(x-y)$. □

6. Conclusions

In this paper, we generalized the notion of an implicativity discussed mainly in $BCK$-algebras by using the notion of a word, and obtained several properties in groupoids and $BCK$-algebras. By using the notion of $Bin\left(X\right)$-product □, we generalized the notion of the implicativity in different directions, and obtained the notion of a weakly (i-)implicativity. We applied these notions to $BCK$-algebras and several groupoids, and investigated some relations among them. The notion of a weakly (i-)implicativity can be applied to positive implicative $BCK$-algebras, e.g., $x\ast y=\left(x{\square }_{i}y\right)\ast y$, and seek to find some relations with commutative $BCK$-algebras.

7. Future Research

Using the notions of the word and the $Bin\left(X\right)$-product, we will generalize the notions of the commutativity and the positive implicativity in $BCK$-algebras and groupoids, i.e., (weakly) i-commutative and (weakly) i-positive implicative $BCK$-algebras and groupoids. We will investigate some relations between (weakly) i-implicative $BCK$-algebras and (weakly) i-commutative and (weakly) i-positive implicative $BCK$-algebras. Moreover, we will generalize several equivalent conditions for positive implicative $BCK$-algebras, and investigate their relationships.