A Class of BCI-Algebra and Quasi-Hyper BCI-Algebra

Abstract

In this paper, we study the connection between generalized quasi-left alter $BCI$-algebra and commutative Clifford semigroup by introducing the concept of an adjoint semigroup. We introduce QM-$BCI$ algebra, in which every element is a quasi-minimal element, and prove that each QM-$BCI$ algebra is equivalent to generalized quasi-left alter $BCI$-algebra. Then, we introduce the notion of generalized quasi-left alter-hyper $BCI$-algebra and prove that every generalized quasi-left alter-hyper $BCI$-algebra is a generalized quasi-left alter $BCI$-algebra. Next, we propose a new notion of quasi-hyper $BCI$ algebra and discuss the relationship among them. Moreover, we study the subalgebras of quasi-hyper $BCI$ algebra and the relationships between $H_v$-group and quasi-hyper $BCI$-algebra, hypergroup and quasi-hyper $BCI$-algebra. Finally, we propose the concept of a generalized quasi-left alter quasi-hyper $BCI$ algebra and QM-quasi hyper $BCI$-algebra and discuss the relationships between them and related $BCI$-algebra.

1. Introduction

In 1966, Japanese mathematicians Imai and Is$\acute{e}$ki proposed the concepts of $BCK$/$BCI$-algebra based on logical algebra and the algebraic expression of combinators in combinatorial logic, which are the two kinds of algebraic structures closest to combinatorial logic, fuzzy logic, etc.) [1,2,3,4,5]. From the development of $BCI$-algebras, we know that $BCI$-algebras can be studied from several aspects:

(1) Find some special classes of $BCI$-algebra, such as associative $BCI$-algebra introduced by Qingping Hu and K. Is$\acute{e}$ki in 1980 [6]; generalized associative $BCI$-algebra proposed by Tiande Lei in 1985 [7]; generalized quasi-left alter $BCI$-algebra by X.H. Zhang in 1992 [8]; the mixed structure of BCI-algebra and semigroup [9].

(2) Starting from the operation of logical algebra, another new operation was derived, which is suitable for associative law. For example, the semigroup structure induced by a BCI-algebra [10]. W.P. Huang studied the adjoint semigroup of generalized associative $BCI$-algebra and proved that the adjoint semigroup of generalized associative $BCI$-algebra is an Abelian group [11]. In 1995, Zhang and Ye first revealed the internal relationship between BZ-algebras and general groups, which can be non-commutative [12].

F. Marty introduced the notion of hyperstructure (also called multialgebra) in 1934 [13], which was widely used in applied sciences (see [14,15,16,17]). Naturally, the idea of a hyperstructure is also applied to the study of non-classical logic algebras. Jun et al. introduced hyper BCK-algebra in 2000, and investigated hyper BCK-ideals and some related hyper algebras, such as hyper K-algebra and hyper $MV$-algebra (see [18,19,20,21,22,23]). In 2006, Jun and Borzooei et al. independently proposed the new concept of hyper BCC-algebra; Xin also introduced the definition of hyper BCI-algebra in 2006 and, since then, many research papers on hyper logical algebras have emerged (see [24,25,26,27,28,29,30,31]). In 2021, Y.D. Du and X.H. Zhang introduced the definition of hyper $BZ$-algebra and discuss the relationships between it and semihypergroups by an adjoint semigroup ([32]).

In sum, the study of both logical algebra and hyper logic algebra should start from their relationship with classical abstract algebra. Therefore, for this paper, we paid attention to the connection between quasi-hyper $BCI$-algebra and a semihypergroup.

The arrangement of the whole paper is as below. Firstly, we show a number of definitions, properties, and theorems in non-classical logic algebras and related hyper algebraic structures in Section 2. In Section 3, we study the adjoint semigroup of generalized quasi-left alter (hyper) $BCI$-algebra and QM-(hyper) $BCI$ algebra. We obtained some results. In Section 4, we propose a definition of quasi-hyper $BCI$ algebra, discuss some properties, investigate the relationship between quasi-hyper $BCI$-algebra and hyper $BCI$-algebra, weak hyper $BCI$-algebra. Moreover, we study relationships between quasi-hyper $BCI$-algebra, hypergroup and the $H_v$-group. Finally, we introduce the concepts of generalized quasi-left alter quasi-hyper $BCI$-algebra and QM-quasi-hyper $BCI$-algebra and discuss the relationships between them and the related $BCI$-algebra.

2. Preliminaries

Definition 1

([33]). Assume that S is a semigroup, $a\in S$. If $\exists x\in S$ s.t. $axa=a$, then a is called regular. S is regular, if each element of S is regular.

Definition 2

([33]). Assume that S be a semigroup. If $\forall a\in S$, there is a operation $a\mapsto a^{-1}$ on S, and it satisfies

$${\left(a^{-1}\right)}^{-1}=a,a{a}^{-1}a=a,a{a}^{-1}=a^{-1}a.$$

Then, it is a completely regular semigroup.

Definition 3

([33]). Assume that $(S,\mu {,}^{-1})$ be a completely regular semigroup. $\forall x,y$ in S

$$\left(x^{-1}x\right)\left(y^{-1}y\right)=\left(y^{-1}y\right)\left(x^{-1}x\right).$$

This is a Clifford semigroup.

Definition 4

([1,2]). Let $<X;\ast ,0>$ be a type (2,0) algebraic structure, if it satisfies: $\forall x,y,z\in X$,

• (BCI1) $\left(\right(x\ast z)\ast (y\ast z\left)\right)\ast (x\ast y)=0$;
• (BCI2) $(x\ast (x\ast y\left)\right)\ast y=0$;
• (BCI3) $x\ast 0=x$;
• (BCI4) $x\ast y=0$ and $y\ast x=0\Rightarrow x=y$.
• It is a $BCI$-algebra.

In $BCI$-algebra $<X;\ast ,0>$, if it satisfies:

• (BK) $0\ast x=0$, for all $x\in X$;
• This is a BCK-algebra.

In $BCK/BCI$-algebra, define ≤: $x\le y$ iff $x\ast y=0$.

Theorem 1

([1]). A type (2,0) algebraic structure $<X;\ast ,0>$ is a BCI-algebra iff $\forall x,y,z\in X,$ it satisfies:

• (BI1) $\left(\right(x\ast z)\ast (y\ast z\left)\right)\ast (x\ast y)=0$;
• (BI2) $(z\ast x)\ast y=(z\ast y)\ast x$;
• (BI3) $x\ast x=0$;
• (BI4) $x\ast y=0$ and $y\ast x=0\Rightarrow x=y$;
• (BI5) $(0\ast (0\ast x\left)\right)\ast x=0$.

Definition 5

([6]). $BCI$-algebra $<X;\ast ,0>$ is generalized associative, if $0\ast (0\ast x)=x$, $\forall x\in X$.

Theorem 2

([34]). In any $BCI$-algebra $<X;\ast ,0>$, the below conditions are equivalent:if $\forall x,y,z\in X$,

(1) 

X is generalized associative;

(2) 

$x\ast (0\ast y)=y\ast (0\ast x)$;

(3) 

$0\ast (x\ast y)=y\ast x$;

(4) 

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

(5) 

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

(6) 

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

(7) 

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

Proposition 1

([7]). In $BCI$-algebra$<X;\ast ,0>$, the below equations hold:

(1) 

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

(2) 

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

Definition 6

([7]). The necessary and sufficient condition for a type (2,0) algebraic structure $<X;\ast ,0>$ to be a quasi-alter BCK-algebra is that it meets: $\forall x,y\in X$, if $x=y$, $x\ast y=0$, otherwise, $x\ast y=x$.

Definition 7

([7]). In a $BCI$-algebra $<X;\ast ,0>$, if $\forall x,y\in X$, and $x\ne y$,

$$x\ast (x\ast y)=0\ast (0\ast y).$$

Then, it is said to be a generalized quasi-left alter $BCI$-algebra.

Theorem 3

([7]). Assume that $<X;\ast ,0>$ be a generalized quasi-left alter $BCI$-algebra. $\forall x\in X$, either $0\ast x=0$, or $0\ast (0\ast x)=x$.

Definition 8.

Assume that $<X;\ast ,0>$ is a $BCI$-algebra. Then,

$$BCK\left(X\right):=\{m\in X|0\ast m=0\}$$

is a $BCK$-part of X.

Definition 9.

Assume that $<X;\ast ,0>$ is a $BCI$-algebra. Then,

$$AG\left(X\right):=\{m\in X|0\ast (0\ast m)=0\}$$

is a generalized associative part of X.

Proposition 2

([7]). The $BCK$-part $B\left(X\right)$ of the generalized quasi-left alter $BCI$-algebra X is quasi-alter $BCK$-subalgebra. Let $G\left(X\right)=X-B\left(X\right)$ be the $BCK$-remainder of generalized quasi-left alter $BCI$-algebra X. We can see that $G\left(X\right)\cup \left\{0\right\}$ is generalized associative subalgebra.

Theorem 4

([7]). Assume that $<X;\ast ,0>$ is a generalized quasi-left alter $BCI$-algebra, $B\left(X\right)$, $G\left(X\right)$ are $BCK$-part and $BCK$-remainder of X, respectively. Then:

(1) 

$x\in B\left(X\right),y\in G\left(X\right)\cup \left\{0\right\}$imply $x\ast y=0\ast y$;

(2) 

$x\in G\left(X\right)\cup \left\{0\right\},y\in B\left(X\right)$ imply $x\ast y=x$.

Definition 10

([14]). A hypergroupoid $(H,\circ )$ is a semihypergroup, if $\forall x,y,z\in H$, we have $(x\circ y)\circ z=x\circ (y\circ z)$. That is,

$$\bigcup _{u\in x\circ y}u\circ z=\bigcup _{v\in y\circ z}x\circ v.$$

In semihypergroup $(H,\circ )$, for all $A,B,C\in P^{\ast }\left(H\right)$, $(A\circ B)\circ C=A\circ (B\circ C)$, where $P^{\ast }\left(H\right)$ represents the non-empty subset of H.

Definition 11

([14]). A semihypergroup $(H,\circ )$ is a hypergroup if $\forall a\in H$, $a\circ H=H\circ a=H$.

In a study of hyperstructure, $a\ll b$ represents $0\in a\circ b$. For each $S,B\subseteq H$, $S\ll B$ represents that, for all $s\in S$, there is $b\in B$, such that $s\ll b$.

Definition 12

([18]). Assume that $(H,\circ )$ is a hyper groupoid containing 0. If it meets these axioms: $\forall x,y,z\in H$,

• $\left(HBK1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y$;
• $\left(HBK2\right)$$(x\circ y)\circ z=(x\circ z)\circ y$
• $\left(HBK3\right)$$x\circ H\ll x$
• $\left(HBK4\right)$$x\ll y$ and $y\ll x\Rightarrow x=y.$
• We call this a hyper $BCK$-algebra.

Definition 13

([26]). Assume that $(H,\circ )$ is a hyper groupoid containing 0. If it meets these axioms: $\forall x,y,z\in H$,

• $\left(HBK1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y$;
• $\left(HBK2\right)$$(x\circ y)\circ z=(x\circ z)\circ y$;
• $\left(HBI3\right)$$x\ll x$;
• $\left(HBK4\right)$$x\ll y$ and $y\ll x\Rightarrow x=y$;
• $\left(HBI5\right)$$0\circ (0\circ x)\ll x$. Then, it is a hyper $BCI$-algebra.

Definition 14

([24]). Assume that $(H,\circ )$ is a hyper groupoid containing 0. If it meets these axioms:

• $\left(HBK1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y,\forall x,y,z\in H$;
• $\left(HBK2\right)$$(x\circ y)\circ z=(x\circ z)\circ y,\forall x,y,z\in H$;
• $\left(HBI3\right)$$x\ll x,\forall x\in H$;
• $\left(HBK4\right)$$x\ll y$ and $y\ll x\Rightarrow x=y,\forall x,y\in H$;
• $\left(HBI5\right)$$0\circ (0\circ x)\ll x$, $x\ne 0,\forall x\in H$.
• Then, it is a weak hyper $BCI$-algebra.

3. Generalized Quasi-Left Alter $\mathbf{BCI}$-Algebra

Firstly, we introduce the adjoint semigroup of $BCI$-algebra and provide some examples about the adjoint semigroup of $BCI$-algebra. Then, we discuss the relationship between the adjoint semigroup of a generalized quasi-left alter $BCI$-algebra and commutative Clifford semigroup.

Assume that $<X,\ast ,0>$ is a $BCI$-algebra. $\forall a,x\in X$, denote a map ${\rho }_a$:

$${\rho }_a:X\to X:x\mapsto x\ast a.$$

$\forall a,b\in X$, denote ${\rho }_a⊛{\rho }_b$: $X\to X$ as follows: $\forall x\in X$,

$$x\mapsto ({\rho }_a⊛{\rho }_b)\left(x\right)=(x\ast b)\ast a.$$

where ⊛ represents the composition operation of mappings.

Theorem 5.

Assume that $<X,\ast ,0>$ is a $BCI$-algebra. Denote $M\left(X\right)$ as a set of finite products ${\rho }_{a_1}⊛...⊛{\rho }_{a_s}$ ($a_1,...,a_s\in X$), where ⊛ represents the composition operation of mappings. Then, $\left(M\right(X),⊛)$ is a monoid and it is commutative.

Proof. 

$\forall a,b,c\in X$, and $\forall x\in X$, we have

$$({\rho }_a⊛{\rho }_b)⊛{\rho }_c\left(x\right)={\rho }_a⊛{\rho }_b(x\ast c)=((x\ast c)\ast b)\ast a,$$

$${\rho }_a⊛({\rho }_b⊛{\rho }_c)\left(x\right)={\rho }_a⊛({\rho }_b⊛{\rho }_c\left(x\right))={\rho }_a((x\ast c)\ast b)=((x\ast c)\ast b)\ast a,$$

Obviously, $({\rho }_a⊛{\rho }_b)⊛{\rho }_c\left(x\right)={\rho }_a⊛({\rho }_b⊛{\rho }_c)\left(x\right)$. So $\left(M\right(X),⊛)$ satisfies associative law.

$\forall x\in X$, ${\rho }_a\in M\left(X\right)$,

$${\rho }_0⊛{\rho }_a\left(x\right)={\rho }_0(x\ast a)=(x\ast a)\ast 0=x\ast a={\rho }_a\left(x\right),$$

$${\rho }_a⊛{\rho }_a\left(x\right)={\rho }_a(x\ast 0)=(x\ast 0)\ast a=x\ast a={\rho }_a\left(x\right).$$

Then, ${\rho }_0$ is the identity element in $M\left(X\right)$. Therefore, $\left(M\right(X),⊛)$ is monoid.

$\forall m,n\in X$, and $\forall x\in X$,

$${\rho }_m⊛{\rho }_n\left(x\right)=(x\ast n)\ast m=(x\ast m)\ast n={\rho }_n⊛{\rho }_m\left(x\right).$$

Therefore, $\left(M\right(X),⊛)$ is commutative. □

Example 1.

Assume that $X=\{0,1,2,3,4,5\}$. This operation on X is shown in

Table 1. $BCI$-algebra.

*	0	1	2	3	4	5
0	0	0	0	3	5	4
1	1	0	1	3	5	4
2	2	0	0	3	5	4
3	3	3	3	0	4	5
4	4	4	4	5	0	3
5	5	5	5	4	3	0

.

Clearly, $<X,\ast ,0>$ is a $BCI$-algebra.

That is, ${\rho }_0:X\to X:x\mapsto x\ast 0.$ So ${\rho }_0=\{0,1,2,3,4,5\}$;

${\rho }_1:X\to X:x\mapsto x\ast 1.$ So ${\rho }_1=\{0,0,0,3,4,5\}$;

${\rho }_2:X\to X:x\mapsto x\ast 2.$ So ${\rho }_2=\{0,1,0,3,4,5\}$;

${\rho }_3:X\to X:x\mapsto x\ast 3.$ So ${\rho }_3=\{3,3,3,0,5,4\}$;

${\rho }_4:X\to X:x\mapsto x\ast 4.$ So ${\rho }_4=\{5,5,5,4,0,3\}$;

${\rho }_5:X\to X:x\mapsto x\ast 5.$ So ${\rho }_5=\{4,4,4,5,3,0\}$.

We could confirm the following:

${\rho }_0⊛{\rho }_0={\rho }_0,{\rho }_0⊛{\rho }_1={\rho }_1,{\rho }_0⊛{\rho }_2={\rho }_2,{\rho }_0⊛{\rho }_3={\rho }_3,{\rho }_0⊛{\rho }_4={\rho }_4,{\rho }_0⊛{\rho }_5={\rho }_5;$

${\rho }_1⊛{\rho }_0={\rho }_1,{\rho }_1⊛{\rho }_1={\rho }_1,{\rho }_1⊛{\rho }_2={\rho }_1,{\rho }_1⊛{\rho }_3={\rho }_3,{\rho }_1⊛{\rho }_4={\rho }_4,{\rho }_1⊛{\rho }_5={\rho }_5$;

${\rho }_2⊛{\rho }_0={\rho }_2,{\rho }_2⊛{\rho }_1={\rho }_1,{\rho }_2⊛{\rho }_2={\rho }_2,{\rho }_2⊛{\rho }_3={\rho }_3$,${\rho }_2⊛{\rho }_4={\rho }_4$,${\rho }_2⊛{\rho }_5={\rho }_5$;

${\rho }_3⊛{\rho }_0={\rho }_3,{\rho }_3⊛{\rho }_1={\rho }_3,{\rho }_3⊛{\rho }_2={\rho }_3,{\rho }_3⊛{\rho }_3={\rho }_1,{\rho }_3⊛{\rho }_4={\rho }_5,{\rho }_3⊛{\rho }_5={\rho }_4$;

${\rho }_4⊛{\rho }_0={\rho }_4,{\rho }_4⊛{\rho }_1={\rho }_4,{\rho }_4⊛{\rho }_2={\rho }_4,{\rho }_4⊛{\rho }_3={\rho }_5$,${\rho }_4⊛{\rho }_4={\rho }_3$,${\rho }_4⊛{\rho }_5={\rho }_1$;

${\rho }_5⊛{\rho }_0={\rho }_5,{\rho }_5⊛{\rho }_1={\rho }_5,{\rho }_5⊛{\rho }_2={\rho }_5,{\rho }_5⊛{\rho }_3={\rho }_4,{\rho }_5⊛{\rho }_4={\rho }_1,{\rho }_5⊛{\rho }_5={\rho }_3$.

Then $M\left(X\right)=\{{\rho }_0,{\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4,{\rho }_5\},$ and $\left(M\right(X),⊛)$ is a monoid, and the operation ⊛ on it is shown in

Table 2. The adjoint semigroup of $BCI$-algebra.

⊛	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$	${\mathit{\rho }}_4$	${\mathit{\rho }}_5$
${\rho }_0$	${\rho }_0$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_5$
${\rho }_1$	${\rho }_1$	${\rho }_1$	${\rho }_1$	${\rho }_3$	${\rho }_4$	${\rho }_5$
${\rho }_2$	${\rho }_2$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_5$
${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_1$	${\rho }_5$	${\rho }_4$
${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_5$	${\rho }_3$	${\rho }_1$
${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_4$	${\rho }_1$	${\rho }_3$

. Thus $\left(M\right(X),⊛)$ is commutative monoid.

Theorem 6.

Assume that $<X,\ast ,0>$ is a generalized quasi-left alter $BCI$-algebra. Hence, $M\left(X\right)$ is a commutative Clifford semigroup.

Proof. 

Step 1:

Let $B\left(X\right)$ be a $BCK$-part of X and $L\left(X\right)=X-B\left(X\right)$ be $BCK$-remainder of X. $\forall x\in X$,

$${\rho }_0⊛{\rho }_0\left(x\right)=(x\ast 0)\ast 0=x\ast 0=x={\rho }_0\left(x\right).$$

Then, according to Definition 6, Proposition 2 and Theorem 4, $\forall x,y\in X-\left\{0\right\}$,

Case 1: $\forall x,y\in B\left(X\right)$, $\forall a\in X$, $a\ne x$ and $a\ne y$. That is,

$${\rho }_x⊛{\rho }_y\left(a\right)=(a\ast y)\ast x=a\ast x=a={\rho }_y\left(a\right)={\rho }_x\left(a\right).$$

$${\rho }_x⊛{\rho }_y\left(y\right)=(y\ast y)\ast x=0\ast x=0={\rho }_y\left(y\right).$$

$${\rho }_x⊛{\rho }_y\left(x\right)=(x\ast y)\ast x=x\ast x=0={\rho }_x\left(x\right).$$

Therefore, $\forall x,y\in B\left(X\right)$ and $x\ne y$, there is ${\rho }_x⊛{\rho }_y\in M\left(X\right)$.

Then:

(1) $({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(a\right)=((a\ast y)\ast y)\ast x=(a\ast y)\ast x=a\ast x=(a\ast y)\ast x={\rho }_x⊛{\rho }_y\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(y\right)=((y\ast y)\ast y)\ast x=(0\ast y)\ast x=0\ast x=(y\ast y)\ast x={\rho }_x⊛{\rho }_y\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(x\right)=((x\ast y)\ast y)\ast x=(x\ast y)\ast x=x\ast x=(x\ast y)\ast x={\rho }_x⊛{\rho }_y\left(x\right)$.

Therefore, for any $x,y\in B\left(X\right)$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_y={\rho }_x⊛{\rho }_y$.

(2) $({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(a\right)=((a\ast x)\ast y)\ast x=(a\ast y)\ast x=a\ast x=(a\ast y)\ast x={\rho }_x⊛{\rho }_y\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(y\right)=((y\ast x)\ast y)\ast x=(y\ast y)\ast x=0\ast x(y\ast y)\ast x={\rho }_x⊛{\rho }_y\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(x\right)=((x\ast x)\ast y)\ast x=(0\ast y)\ast x=0\ast x=0=x\ast x=(x\ast y)\ast x={\rho }_x⊛{\rho }_y\left(x\right)$.

Therefore, for any $x,y\in B\left(X\right)$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_x={\rho }_x⊛{\rho }_y$.

(3) $({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(a\right)=(((a\ast y)\ast x)\ast y)\ast x=((a\ast x)\ast y)\ast x=a\ast x=(a\ast y)\ast x={\rho }_x⊛{\rho }_y\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(y\right)=(((y\ast y)\ast x)\ast y)\ast x=((0\ast x)\ast y)\ast x=0\ast x=(y\ast y)\ast x={\rho }_x⊛{\rho }_y\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(x\right)=(((x\ast y)\ast x)\ast y)\ast x=((x\ast x)\ast y)\ast x=0\ast x=0=0\ast y=(x\ast x)\ast y=(x\ast y)\ast x={\rho }_x⊛{\rho }_y\left(x\right)$.

So, for any $x,y\in B\left(X\right)$, there is $({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)={\rho }_x⊛{\rho }_y$.

(4) $\forall z\in L\left(X\right)$, $b\in L\left(X\right)$ and $b\ne z$, and for any $m\in B\left(X\right)$, $m\ne x$ and $m\ne y$, there is

$({\rho }_x⊛{\rho }_y)⊛{\rho }_z\left(m\right)=((m\ast z)\ast y)\ast x=((0\ast z)\ast y)\ast x=0\ast z=m\ast z={\rho }_z\left(m\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_z\left(y\right)=((y\ast z)\ast y)\ast x=((0\ast z)\ast y)\ast x=0\ast z=y\ast z={\rho }_z\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_z\left(x\right)=((x\ast z)\ast y)\ast x=((0\ast z)\ast y)\ast x=0\ast z=x\ast z={\rho }_z\left(x\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_z\left(z\right)=((z\ast z)\ast y)\ast x=(0\ast y)\ast x=0\ast x=0=z\ast z={\rho }_z\left(z\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_z\left(b\right)=((b\ast z)\ast y)\ast x=((b\ast y)\ast z)\ast x=(b\ast z)\ast x=(b\ast x)\ast z=b\ast z={\rho }_z\left(b\right)$.

That is, for any $x,y\in B\left(X\right)$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_z={\rho }_z$.

Above all, let $\left|B\right(X\left)\right|=n$, $a_i\in B\left(X\right)$ and $i\in [1,n]$. Except for ${\rho }_{a_i}\in M\left(X\right)$, there are ${\displaystyle \prod _{i\in [1,n]}}{\rho }_{a_i}\in M\left(X\right)$. Additionally, for $k,l\in [1,n]$, there is $({\rho }_{a_k}⊛...⊛{\rho }_{a_l})⊛{\rho }_{a_k}={\rho }_{a_k}⊛...⊛{\rho }_{a_l}$, $({\rho }_{a_k}⊛...⊛{\rho }_{a_l})⊛{\rho }_{a_l}={\rho }_{a_k}⊛...⊛{\rho }_{a_l}$. For $j\in [1,n]$, $({\displaystyle \prod _{i=1}^n}{\rho }_{a_i})⊛{\rho }_{a_j}={\displaystyle \prod _{i=1}^n}{\rho }_{a_i}.$

Case 2: $\forall x\in B\left(X\right),y\in L\left(X\right)$, $a\in B\left(X\right)$ and $a\ne x$, $b\in L\left(X\right)$ and $b\ne y$. That is:

$${\rho }_x⊛{\rho }_y\left(a\right)=(a\ast y)\ast x=(0\ast y)\ast x=0\ast y=a\ast y={\rho }_y\left(a\right).$$

$${\rho }_x⊛{\rho }_y\left(y\right)=(y\ast y)\ast x=0\ast x=0={\rho }_y\left(y\right).$$

$${\rho }_x⊛{\rho }_y\left(x\right)=(x\ast y)\ast x=(0\ast y)\ast x=0\ast y=x\ast y={\rho }_y\left(x\right).$$

$${\rho }_x⊛{\rho }_y\left(b\right)=(b\ast y)\ast x=b\ast y={\rho }_y\left(b\right).$$

Therefore, ${\rho }_x⊛{\rho }_y={\rho }_y$.

Case 3: $\forall x\in L\left(X\right),y\in B\left(X\right)$, For any $a\in B\left(X\right)$ and $a\ne y$, $b\in L\left(X\right)$ and $b\ne x$. That is:

$${\rho }_x⊛{\rho }_y\left(a\right)=(a\ast y)\ast x=a\ast x=0\ast x=a\ast x={\rho }_x\left(a\right).$$

$${\rho }_x⊛{\rho }_y\left(y\right)=(y\ast y)\ast x=0\ast x=y\ast x={\rho }_x\left(y\right).$$

$${\rho }_x⊛{\rho }_y\left(x\right)=(x\ast y)\ast x=x\ast x=0={\rho }_x\left(x\right).$$

$${\rho }_x⊛{\rho }_y\left(b\right)=(b\ast y)\ast x=b\ast x={\rho }_x\left(b\right).$$

Therefore, ${\rho }_x⊛{\rho }_y={\rho }_x$.

Case 4: $\forall x,y\in L\left(X\right)$, For any $a\in B\left(X\right)$, $b\in L\left(X\right)$, $b\ne x$ and $b\ne y$. Let $z=y\ast (0\ast x)$. That is:

(1) $z\ne 0$, that is, $y\ne 0\ast x$,

$${\rho }_z\left(a\right)=a\ast z=0\ast z=0\ast (y\ast (0\ast x))=(0\ast x)\ast y=(0\ast y)\ast x=(a\ast y)\ast x={\rho }_x⊛{\rho }_y\left(a\right).$$

$${\rho }_z\left(y\right)=y\ast z=(y\ast 0)\ast (y\ast (0\ast x))=(0\ast x)\ast 0=(0\ast 0)\ast x=(y\ast y)\ast x={\rho }_x⊛{\rho }_y\left(y\right).$$

$${\rho }_z\left(x\right)=x\ast z=(0\ast (0\ast x))\ast (y\ast (0\ast x))=0\ast y=(x\ast x)\ast y=(x\ast y)\ast x={\rho }_x⊛{\rho }_y\left(x\right).$$

$$\begin{array}{ccc}{\rho }_z\left(b\right)\hfill & =\hfill & b\ast z=(0\ast (0\ast b\left)\right)\ast (y\ast (0\ast x\left)\right)\hfill \\ & =\hfill & (0\ast (y\ast (0\ast x)\left)\right)\ast (0\ast b)\hfill \\ & =\hfill & \left(\right(0\ast x)\ast y)\ast (0\ast b)\hfill \\ & =\hfill & \left(\right(0\ast x)\ast (0\ast b\left)\right)\ast y\hfill \\ & =\hfill & (b\ast x)\ast y=(b\ast y)\ast x={\rho }_x⊛{\rho }_y\left(b\right).\hfill \end{array}$$

Therefore, ${\rho }_x⊛{\rho }_y={\rho }_{y\ast (0\ast x)}$. According to Case 2 and Case 3, ${\rho }_x⊛{\rho }_y$ is a completely regular element.

(2) $z=0$, that is, $y=0\ast x$ and $x=0\ast y$.

$${\rho }_x⊛{\rho }_y\left(y\right)=(y\ast y)\ast x=0\ast x=y=y\ast 0={\rho }_0\left(y\right).$$

$${\rho }_x⊛{\rho }_y\left(x\right)=(x\ast y)\ast x=(x\ast x)\ast y=0\ast y=x=x\ast 0={\rho }_0\left(x\right).$$

$${\rho }_x⊛{\rho }_y\left(b\right)=(b\ast y)\ast x=(b\ast y)\ast (0\ast y)=b\ast 0={\rho }_0\left(b\right).$$

However,

$${\rho }_x⊛{\rho }_y\left(a\right)=(a\ast y)\ast x=(0\ast y)\ast x=x\ast x=0\ne {\rho }_0\left(a\right).$$

Therefore, ${\rho }_x\ast {\rho }_y\in M\left(X\right)$.

(i) Let $m\in B\left(X\right)$, and $m\ne a$, then:

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(a\right)=((a\ast m)\ast y)\ast x=(a\ast y)\ast x=(0\ast y)\ast x=(a\ast y)\ast x={\rho }_x⊛{\rho }_y\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(m\right)=((m\ast m)\ast y)\ast x=(0\ast y)\ast x=0$, ${\rho }_x⊛{\rho }_y\left(m\right)=(m\ast y)\ast x=(0\ast y)\ast x=0$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(y\right)=((y\ast m)\ast y)\ast x=(y\ast y)\ast x={\rho }_x⊛{\rho }_y\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(x\right)=((x\ast m)\ast y)\ast x=(x\ast y)\ast x={\rho }_x⊛{\rho }_y\left(x\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(b\right)=((b\ast m)\ast y)\ast x=(b\ast y)\ast x={\rho }_x⊛{\rho }_y\left(b\right)$.

Therefore, for $x,y\in L\left(X\right)$ and $y=0\ast x$, $m\in B\left(X\right)$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_m={\rho }_x⊛{\rho }_y$.

(ii) Let $m\in L\left(X\right)$, and $m\ne x$, $m\ne y$; then,

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(a\right)=((a\ast m)\ast y)\ast x=((0\ast m)\ast y)\ast x=((0\ast m)\ast (0\ast x))\ast x=(x\ast m)\ast x=(x\ast x)\ast m=0\ast m=a\ast m={\rho }_m\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(y\right)=((y\ast m)\ast y)\ast x=((y\ast y)\ast m)\ast x=(0\ast m)\ast x=(0\ast x)\ast m=y\ast m={\rho }_m\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(x\right)=((x\ast m)\ast y)\ast x=((x\ast m)\ast x)\ast y=((x\ast x)\ast m)\ast y=(0\ast m)\ast y=(0\ast y)\ast m=x\ast m={\rho }_m\left(x\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(m\right)=((m\ast m)\ast y)\ast x=(0\ast y)\ast x=x\ast x=0=m\ast m={\rho }_m\left(m\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_m\left(b\right)=((b\ast m)\ast y)\ast x=((b\ast y)\ast m)\ast x=((b\ast y)\ast x)\ast m=((b\ast y)\ast (0\ast y))\ast m=(b\ast 0)\ast m=b\ast m={\rho }_m\left(b\right)$.

Therefore, for $x,y\in L\left(X\right)$ and $y=0\ast x$, $m\in L\left(X\right)$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_m={\rho }_m$.

(iii) $({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(a\right)=((a\ast y)\ast y)\ast x=((0\ast y)\ast y)\ast x=(x\ast y)\ast x=(x\ast x)\ast y=0\ast y=a\ast y={\rho }_y\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(y\right)=((y\ast y)\ast y)\ast x=(0\ast y)\ast x=x\ast x=y\ast y={\rho }_y\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(x\right)=((x\ast y)\ast y)\ast x=((x\ast y)\ast x)\ast y=((x\ast x)\ast y)\ast y=(0\ast y)\ast y=x\ast y={\rho }_y\left(x\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_y\left(b\right)=((b\ast y)\ast y)\ast x=((b\ast y)\ast x)\ast y=((b\ast y)\ast (0\ast y))\ast y=(b\ast 0)\ast y=b\ast y={\rho }_y\left(b\right)$;

Therefore, for any $x,y\in L\left(X\right)$ and $y=0\ast x$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_y={\rho }_y$.

(iv) $({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(a\right)=((a\ast x)\ast y)\ast x=((0\ast x)\ast y)\ast x=(y\ast y)\ast x=0\ast x=a\ast x={\rho }_x\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(y\right)=((y\ast x)\ast y)\ast x=((y\ast y)\ast x)\ast x=(0\ast x)\ast x=y\ast x={\rho }_x\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(x\right)=((x\ast x)\ast y)\ast x=(0\ast y)\ast x=x\ast x={\rho }_x\left(x\right)$;

$({\rho }_x⊛{\rho }_y)⊛{\rho }_x\left(b\right)=((b\ast x)\ast y)\ast x=((b\ast x)\ast (0\ast x))\ast x=(b\ast 0)\ast x=b\ast x={\rho }_x\left(b\right)$.

Therefore, for any $x,y\in L\left(X\right)$ and $y=0\ast x$, there is $({\rho }_x⊛{\rho }_y)⊛{\rho }_x={\rho }_x$.

(v) $({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(a\right)=(((a\ast y)\ast x)\ast y)\ast x=(((0\ast y)\ast (0\ast y))\ast y)\ast x=(a\ast y)\ast x={\rho }_x⊛{\rho }_y\left(a\right)$;

$({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(y\right)=(((y\ast y)\ast x)\ast y)\ast x=((0\ast x)\ast y)\ast x=(y\ast y)\ast x={\rho }_x⊛{\rho }_y\left(y\right)$;

$({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(x\right)=(((x\ast y)\ast x)\ast y)\ast x=(((x\ast x)\ast y)\ast y)\ast x=((0\ast y)\ast y)\ast x=(x\ast y)\ast x={\rho }_x⊛{\rho }_y\left(x\right)$;

$({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)\left(b\right)=(((b\ast y)\ast x)\ast y)\ast x=((b\ast y)\ast (0\ast y)\ast y)\ast x=((b\ast 0)\ast y)\ast x=(b\ast y)\ast x={\rho }_x⊛{\rho }_y\left(b\right)$.

Therefore, for any $x,y\in L\left(X\right)$ and $y=0\ast x$, there is $({\rho }_x⊛{\rho }_y)⊛({\rho }_x⊛{\rho }_y)={\rho }_x⊛{\rho }_y$. At the same time, ${\rho }_x⊛{\rho }_y$ is idempotent.

In sum, all elements of $M\left(X\right)$ are aligned.

Step 2:

Case 1: According to Case1 in Step1, let $\left|B\right(X\left)\right|=n$, $x,a_i\in B\left(X\right)$ and $i\in [1,n]$, $\forall a\in X$. Then, ${\rho }_x⊛{\rho }_x\left(a\right)=(a\ast x)\ast x=a\ast x={\rho }_x$, ${\rho }_x$ is idempotent and a completely regular element. $({\displaystyle \prod _{i\in [1,n]}}{\rho }_{a_i})⊛({\displaystyle \prod _{i\in [1,n]}}{\rho }_{a_i})={\displaystyle \prod _{i\in [1,n]}}{\rho }_{a_i}$. That is, ${\displaystyle \prod _{i\in [1,n]}}{\rho }_{a_i}$ is idempotent and a completely regular element.

Case 2: $\forall x\in B\left(X\right),y\in L\left(X\right)$, $b\in L\left(X\right)$ and $b\ne y$. That is, there exists $z=0\ast y$ and $y=0\ast z$. Then,

(1)

$${\rho }_y⊛{\rho }_z⊛{\rho }_y\left(a\right)=((a\ast y)\ast z)\ast y=((0\ast y)\ast (0\ast y))\ast y=0\ast y=a\ast y={\rho }_y\left(a\right).$$

$${\rho }_z⊛{\rho }_y⊛{\rho }_z\left(a\right)=((a\ast z)\ast y)\ast z=((0\ast z)\ast (0\ast z))\ast z=0\ast z=a\ast z={\rho }_z\left(a\right).$$

$${\rho }_z⊛{\rho }_y\left(a\right)=(a\ast y)\ast z=(a\ast z)\ast y={\rho }_y⊛{\rho }_z\left(a\right).$$

(2)

$${\rho }_y⊛{\rho }_z⊛{\rho }_y\left(y\right)=((y\ast y)\ast z)\ast y=(0\ast z)\ast y=y\ast y=0=y\ast y={\rho }_y\left(y\right).$$

$${\rho }_z⊛{\rho }_y⊛{\rho }_z\left(y\right)=((y\ast z)\ast y)\ast z=((y\ast y)\ast z)\ast z=(0\ast z)\ast z=y\ast z={\rho }_z\left(y\right).$$

$${\rho }_z⊛{\rho }_y\left(y\right)=(y\ast y)\ast z=(y\ast z)\ast y={\rho }_y⊛{\rho }_z\left(y\right).$$

(3)

$${\rho }_y⊛{\rho }_z⊛{\rho }_y\left(z\right)=((z\ast y)\ast z)\ast y=((z\ast z)\ast y)\ast y=(0\ast y)\ast y=z\ast y={\rho }_y\left(z\right).$$

$${\rho }_z⊛{\rho }_y⊛{\rho }_z\left(z\right)=((z\ast z)\ast y)\ast z=(0\ast y)\ast z=z\ast z={\rho }_z\left(z\right).$$

$${\rho }_z⊛{\rho }_y\left(z\right)=(z\ast y)\ast z=(z\ast z)\ast y={\rho }_y⊛{\rho }_z\left(z\right).$$

(4)

$${\rho }_y⊛{\rho }_z⊛{\rho }_y\left(b\right)=((b\ast y)\ast z)\ast y=((b\ast y)\ast (0\ast y))\ast y=(b\ast 0)\ast y=b\ast y={\rho }_y\left(b\right).$$

$${\rho }_z⊛{\rho }_y⊛{\rho }_z\left(b\right)=((b\ast z)\ast y)\ast z=((b\ast z)\ast (0\ast z))\ast z=(b\ast 0)\ast z=b\ast z={\rho }_z\left(b\right).$$

$${\rho }_z⊛{\rho }_y\left(b\right)=(b\ast y)\ast z=(b\ast z)\ast y={\rho }_y⊛{\rho }_z\left(b\right).$$

Above all, ${\rho }_y$ is a completely regular element.

Case 3:

According to Case 4 in Step 1, $\forall x,y\in L\left(X\right)$, and $y\ne 0\ast x$, ${\rho }_x⊛{\rho }_y={\rho }_{y\ast (0\ast x)}$. According to Case 2, ${\rho }_x⊛{\rho }_y$ is a completely regular element. Additionally, $({\rho }_x⊛{\rho }_{0\ast x})⊛({\rho }_x⊛{\rho }_{0\ast x})={\rho }_x⊛{\rho }_{0\ast x}$. At the same time, ${\rho }_x⊛{\rho }_{0\ast x}$ is idempotent. Additionally, $({\rho }_y⊛{\rho }_{0\ast y})⊛({\rho }_y⊛{\rho }_{0\ast y})={\rho }_y⊛{\rho }_{0\ast y}$. At the same time, ${\rho }_y⊛{\rho }_{0\ast y}$ is idempotent.

Therefore, $M\left(X\right)$ is a completely regular semigroup. In addition, $M\left(X\right)$ is a commutative Clifford semigroup because of commutativity. □

Example 2.

Assume that $X=\{0,1,2,3,4,5\}$. The operation on X is shown in

Table 3. Generalized quasi-left alter $BCI$-algebra.

*	0	1	2	3	4	5
0	0	0	0	3	5	4
1	1	0	1	3	5	4
2	2	2	0	3	5	4
3	3	3	3	0	4	5
4	4	4	4	5	0	3
5	5	5	5	4	3	0

.

Then, $<X,\ast ,0>$ is a generalized quasi-left alter $BCI$-algebra and $M\left(X\right)=\{{\rho }_0,{\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4,$ ${\rho }_5,{\rho }_3^2\}$, where ${\rho }_3^2={\rho }_3\ast {\rho }_3$.

We can verify the following:

${\rho }_0⊛{\rho }_0={\rho }_0,{\rho }_0⊛{\rho }_1={\rho }_1,{\rho }_0⊛{\rho }_2={\rho }_2,{\rho }_0⊛{\rho }_3={\rho }_3,{\rho }_0⊛{\rho }_4={\rho }_4,{\rho }_0⊛{\rho }_5={\rho }_5,{\rho }_0⊛{\rho }_3^2={\rho }_3^2;$

${\rho }_1⊛{\rho }_0={\rho }_1,{\rho }_1⊛{\rho }_1={\rho }_1,{\rho }_1⊛{\rho }_2={\rho }_3^2,{\rho }_1⊛{\rho }_3={\rho }_3,{\rho }_1⊛{\rho }_4={\rho }_4,{\rho }_1⊛{\rho }_5={\rho }_5,{\rho }_1⊛{\rho }_3^2={\rho }_3^2$;

${\rho }_2⊛{\rho }_0={\rho }_2,{\rho }_2⊛{\rho }_1={\rho }_3^2,{\rho }_2⊛{\rho }_2={\rho }_2,{\rho }_2⊛{\rho }_3={\rho }_3,{\rho }_2⊛{\rho }_4={\rho }_4,{\rho }_2⊛{\rho }_5={\rho }_5,{\rho }_2⊛{\rho }_3^2={\rho }_3^2$;

${\rho }_3⊛{\rho }_0={\rho }_3,{\rho }_3⊛{\rho }_1={\rho }_3,{\rho }_3⊛{\rho }_2={\rho }_3,{\rho }_3⊛{\rho }_3={\rho }_3^2,{\rho }_2⊛{\rho }_4={\rho }_5,{\rho }_2⊛{\rho }_5={\rho }_4,{\rho }_3⊛{\rho }_3^2={\rho }_3$;

${\rho }_4⊛{\rho }_0={\rho }_4,{\rho }_4⊛{\rho }_1={\rho }_4,{\rho }_4⊛{\rho }_2={\rho }_4,{\rho }_4⊛{\rho }_3={\rho }_5,{\rho }_4⊛{\rho }_4={\rho }_3,{\rho }_4⊛{\rho }_5={\rho }_3^2,{\rho }_4⊛{\rho }_3^2={\rho }_4$;

${\rho }_5⊛{\rho }_0={\rho }_5,{\rho }_5⊛{\rho }_1={\rho }_5,{\rho }_5⊛{\rho }_2={\rho }_5,{\rho }_5⊛{\rho }_3={\rho }_4,{\rho }_5⊛{\rho }_4={\rho }_3^2,{\rho }_5⊛{\rho }_5={\rho }_3,{\rho }_5⊛{\rho }_3^2={\rho }_5$;

${\rho }_3^2⊛{\rho }_0={\rho }_3^2,{\rho }_3^2⊛{\rho }_1={\rho }_3^2,{\rho }_3^2⊛{\rho }_2={\rho }_3^2,{\rho }_3^2⊛{\rho }_3={\rho }_3,{\rho }_3^2⊛{\rho }_4={\rho }_4,{\rho }_3^2⊛{\rho }_5={\rho }_5,{\rho }_3^2⊛{\rho }_3^2={\rho }_3^2$.

Then $M\left(X\right)$ is a completely regular semigroup, and the operation ⊛ on it is shown in

Table 4. The adjoint semigroup of generalized quasi left alter $BCI$-algebra.

⊛	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$	${\mathit{\rho }}_4$	${\mathit{\rho }}_5$	${\mathit{\rho }}_3^2$
${\rho }_0$	${\rho }_0$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_5$	${\rho }_3^2$
${\rho }_1$	${\rho }_1$	${\rho }_1$	${\rho }_3^2$	${\rho }_3$	${\rho }_4$	${\rho }_5$	${\rho }_3^2$
${\rho }_2$	${\rho }_2$	${\rho }_3^2$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_5$	${\rho }_3^2$
${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_3^2$	${\rho }_5$	${\rho }_4$	${\rho }_3$
${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_5$	${\rho }_3$	${\rho }_3^2$	${\rho }_4$
${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_4$	${\rho }_3^2$	${\rho }_3$	${\rho }_5$
${\rho }_3^2$	${\rho }_3^2$	${\rho }_3^2$	${\rho }_3^2$	${\rho }_3$	${\rho }_4$	${\rho }_5$	${\rho }_3^2$

.

In the following, we introduce QM-$BCI$ algebra and discuss the relationship between QM-$BCI$ algebra and a generalized quasi-left alter $BCI$-algebra.

Definition 15.

Assume that $(X,\le )$ is a partial order containing a constant 0, $x\in X$. x is said to be a quasi-minimal element, if $\forall a\in X$, $a\le x$ implies $x=a$ or $a=0$.

Definition 16.

A $BCI$- algebra $(X,\le ,\ast ,0)$ is called a QM-$BCI$ algebra, if all elements of X are quasi-minimal elements.

Theorem 7.

Assume that $(X,\le ,\ast ,0)$ is a $BCI$-algebra. X is a QM-$BCI$ algebra if it meets: $\forall x,y\in X-\left\{0\right\}$,

$$x\le y\mathit{\text{ implies }}x=y$$

Proof. 

$(\Rightarrow )$$\forall x,y\in X-\left\{0\right\}$, assume that $x\le y$, according to Definition 15, $y=x$ or $x=0$. However, $x\ne 0$. So $x=y$.

$(\Leftarrow )$ Assume that $x,y\in X$, $x\le y$. If $y=0$, then $x\le y=0$, we can get $x=y=0$. If $x\ne 0$, $y\ne 0$, there is $x=y$ by condition. Therefore, y is a quasi-minimal element of X. Thus, X is a QM-$BCI$ algebra. □

Theorem 8.

Assume that $(X,\le ,\ast ,0)$ is a $BCI$-algebra, $B\left(X\right)$ is a $BCK$-part of X, AG(X) is a generalized associative part of X. Then, the below conditions are equivalent:

(1) 

X is a QM-$BCI$ algebra;

(2) 

$B\left(X\right)$ is quasi-alter $BCK$-algebra and $AG\left(X\right)=(X-B(X\left)\right)\cup \left\{0\right\}$;

(3) 

X is a generalized quasi-left alter $BCI$-algebra.

Proof. 

(1)⇒(2) Assume that X is a QM-$BCI$ algebra. Then, for any $x,y\in B\left(X\right)$, if $x=y$, $x\ast y=0$. If $x\ne y$, it could be divided into the following three cases:

Case 1: $x=0$, $y\ne 0$, $x\ast (x\ast y)=0\ast (0\ast y)=0\ast 0=0$, that is, $x\le x\ast y$; $(x\ast y)\ast x=(0\ast y)\ast 0=0\ast 0=0$, that is, $x\ast y\le x$. According to Definition 4, $x=x\ast y$.

Case 2: $x\ne 0$, $y=0$, $x\ast (x\ast y)=x\ast (x\ast 0)=x\ast x=0$, that is, $x\le x\ast y$; $(x\ast y)\ast x=(x\ast 0)\ast x=x\ast x=0$, that is, $x\ast y\le x$. According to Definition 4, $x=x\ast y$.

Case 3: $x,y\ne 0$, $(x\ast y)\ast x=(x\ast x)\ast y=0\ast y=0$, that is, $x\le x\ast y$. Because $x\ast y\ne 0$ and $x\ne 0$, according to Theorem 7, $x\ast y=x$.

According to Definition 6, $B\left(X\right)$ is a quasi-alter $BCK$-algebra. If $x\in X-B\left(X\right)$, then $0\ast x\ne 0$ and $0\ast (0\ast x)\ne 0$. As $(0\ast (0\ast x\left)\right)\ast x=(0\ast x)\ast (0\ast x)=0$, that is, $0\ast (0\ast x)\le x$. According to Theorem 7, $0\ast (0\ast x)=x$. Thus, $(X-B(X\left)\right)\cup \left\{0\right\}\subseteq AG\left(X\right)$. On the other hand, $AG\left(X\right)\subseteq (X-B(X\left)\right)\cup \left\{0\right\}$; then, $AG\left(X\right)=(X-B(X\left)\right)\cup \left\{0\right\}$.

(2)⇒(3) $\forall x,y\in X$, $x\ne y$,

Case 1: $\forall x,y\in B\left(X\right)$, $x\ast (x\ast y)=x\ast x=0$, $0\ast (0\ast y)=0\ast 0=0$. Then, $x\ast (x\ast y)=0\ast (0\ast y)$.

Case 2: $\forall x,y\in AG\left(X\right)$, assume that $x\le y$, that is, $x\ast y=0$. Then,

$$0\ast (y\ast x)=(x\ast x)\ast (y\ast x)\le x\ast y=0.$$

Therefore, $0\ast (y\ast x)=0$. Then, $y\ast x=0\ast (0\ast (y\ast x\left)\right)=0\ast 0=0$, and $y\ast x=0$. That is, $y\le x$. So $x=y$. According to Definition 4, we can see that $x\ast (x\ast y)\le y$; that is, $x\ast (x\ast y)=y$. Additionally, $0\ast (0\ast y)=y$. So $x\ast (x\ast y)=0\ast (0\ast y)$.

Case 3: $\forall x\in B\left(X\right),\forall y\in AG\left(X\right)$, according to Theorem 4, $x\ast y=0\ast y$; then, $x\ast (x\ast y)=0\ast (0\ast y)$.

Case 4: $\forall x\in AG\left(X\right),\forall y\in B\left(X\right)$, according to Definition 6, $x\ast y=x$; then, $x\ast (x\ast y)=x\ast x=0$, and $0\ast (0\ast y)=0\ast 0=0$. So $x\ast (x\ast y)=0\ast (0\ast y)$.

Above all, X is a generalized quasi-left alter $BCI$-algebra.

(3)⇒(1) If X is a generalized quasi-left alter $BCI$-algebra, assume that $x\le y$ and $x\ne y$. Then,

$$x=x\ast 0=x\ast (x\ast y)=0\ast (0\ast y).$$

$\forall y\in X$,

Case 1: If $y\in B\left(X\right)$, then $x=0\ast (0\ast y)=0\ast 0=0$;

Case 2: If $y\in AG\left(X\right)$, then $x=0\ast (0\ast y)=y$. As $x\ne y$, then $x=0$. Thus, for any $y\in X$, y is a quasi-minimal element of X. □

In the following, we propose the adjoint semigroup of hyper $BCI$-algebra and replace the singleton set $\left\{x\right\}$ with x. Additionally, the concepts of generalized quasi-left alter-hyper $BCI$-algebra and QM-hyper $BCI$ are shown.

Let $(H,\circ )$ be a hyper $BCI$-algebra. $\forall a,x\in H$, denote a map:

$${\rho }_a:H\to P^{\ast }\left(H\right);x\mapsto x\circ a.$$

where $P^{\ast }\left(H\right)$ represents the non-empty subset of H.

$\forall a,b\in H$, denote ${\rho }_a⊛{\rho }_b$: $H\to P^{\ast }\left(H\right)$ as follows: $\forall x\in H$,

$$x\mapsto ({\rho }_a⊛{\rho }_b)\left(x\right)=\bigcup _{\forall y\in {\rho }_b\left(x\right)}{\rho }_a\left(y\right).$$

where ⊛ represents the composition operation.

Theorem 9.

Assume that $(H,\circ )$ is a hyper $BCI$-algebra. Denote $M\left(H\right)$ as a set of finite products ${\rho }_{a_1}⊛...⊛{\rho }_{a_s}$ ($a_1,...,a_s\in H$), where ⊛ represents the composition operation of mappings. Then, $M\left(H\right)$ is a commutative semigroup.

Proof. 

$\forall x\in H$, $a,b,c\in H$, for any $s\in (({\rho }_a⊛{\rho }_b)⊛{\rho }_c)\left(x\right)$, $\exists y\in {\rho }_c\left(x\right)$ s.t. $s\in ({\rho }_a⊛{\rho }_b)\left(y\right)$. Then $\exists u\in {\rho }_b\left(y\right)$ s.t. $u\in {\rho }_b\left({\rho }_c\left(x\right)\right)={\rho }_b⊛{\rho }_c\left(x\right)$ and $s\in {\rho }_a\left(u\right)$. Then $s\in ({\rho }_a⊛({\rho }_b⊛{\rho }_c))\left(x\right)$ and $(({\rho }_a⊛{\rho }_b)⊛{\rho }_c)\left(x\right)\subseteq ({\rho }_a⊛({\rho }_b⊛{\rho }_c))\left(x\right)$.

For any $t\in ({\rho }_a⊛({\rho }_b⊛{\rho }_c))\left(x\right)$, there exists $m\in {\rho }_b⊛{\rho }_c\left(x\right)$ such that $t\in {\rho }_a\left(m\right)$. Then $\exists n\in {\rho }_c\left(x\right)$ s.t. $m\in {\rho }_b\left(n\right)$ and $t\in {\rho }_a\left({\rho }_b\left(n\right)\right)={\rho }_a⊛{\rho }_b\left(n\right)$. Then $t\in (({\rho }_a⊛{\rho }_b)⊛{\rho }_c)\left(x\right)$ and $({\rho }_a⊛({\rho }_b⊛{\rho }_c))\left(x\right)\subseteq (({\rho }_a⊛{\rho }_b)⊛{\rho }_c)\left(x\right)$.

Therefore, $({\rho }_a⊛({\rho }_b⊛{\rho }_c))\left(x\right)=(({\rho }_a⊛{\rho }_b)⊛{\rho }_c)\left(x\right)$. Then, $M\left(H\right)$ satisfies the associative law.

For any $m,n\in H$, $x\in H$. ${\rho }_m⊛{\rho }_n\left(x\right)=(x\circ n)\circ m=(x\circ m)\circ n={\rho }_n⊛{\rho }_m\left(x\right)$. Therefore, $M\left(H\right)$ satisfies the commutative law. Thus, $M\left(H\right)$ is a commutative semigroup. □

Example 3.

Assume that $H=\{0,1,2,3,4\}$. Define the operation on H in

Table 5. Hyper $BCI$-algebra.

∘	0	1	2	3	4
0	0	0	0	4	3
1	$\{1,2\}$	$\{0,1,2\}$	$\{0,1,2\}$	4	3
2	2	$\{1,2\}$	$\{0,1,2\}$	4	3
3	3	3	3	0	4
4	4	4	4	3	0

,

Clearly, $(H,\circ )$ is a hyper $BCI$-algebra.

That is,

${\rho }_0:H\to H:x\mapsto x\circ 0.$ So ${\rho }_0=\{0,\{1,2\},2,3,4\}$;

${\rho }_1:H\to H:x\mapsto x\circ 1.$ So ${\rho }_1=\{0,\{0,1,2\},\{1,2\},3,4\}$;

${\rho }_2:H\to H:x\mapsto x\circ 2.$ So ${\rho }_2=\{0,\{0,1,2\},\{0,1,2\},3,4\}$;

${\rho }_3:H\to H:x\mapsto x\circ 3.$ So ${\rho }_3=\{4,4,4,0,3\}$;

${\rho }_4:H\to H:x\mapsto x\circ 4.$ So ${\rho }_4=\{3,3,3,4,0\}$.

Denote ${\rho }_{34}={\rho }_3⊛{\rho }_4=\{0,0,0,3,4\}$.

We can verify the following:

${\rho }_0⊛{\rho }_0={\rho }_0,{\rho }_0⊛{\rho }_1={\rho }_1,{\rho }_0⊛{\rho }_2={\rho }_2,{\rho }_0⊛{\rho }_3={\rho }_3,{\rho }_0⊛{\rho }_4={\rho }_4,{\rho }_0⊛{\rho }_{34}={\rho }_{34}$;

${\rho }_1⊛{\rho }_0={\rho }_1,{\rho }_1⊛{\rho }_1={\rho }_2,{\rho }_1⊛{\rho }_2={\rho }_2,{\rho }_1⊛{\rho }_3={\rho }_3,{\rho }_1⊛{\rho }_4={\rho }_4,{\rho }_1⊛{\rho }_{34}={\rho }_{34}$;

${\rho }_2⊛{\rho }_0={\rho }_2,{\rho }_2⊛{\rho }_1={\rho }_2,{\rho }_2⊛{\rho }_2={\rho }_2,{\rho }_2⊛{\rho }_3={\rho }_3,{\rho }_2⊛{\rho }_4={\rho }_4,{\rho }_2⊛{\rho }_{34}={\rho }_{34}$;

${\rho }_3⊛{\rho }_0={\rho }_3,{\rho }_3⊛{\rho }_1={\rho }_3,{\rho }_3⊛{\rho }_2={\rho }_3,{\rho }_3⊛{\rho }_3={\rho }_4,{\rho }_3⊛{\rho }_4={\rho }_{34},{\rho }_3⊛{\rho }_{34}={\rho }_3$;

${\rho }_4⊛{\rho }_0={\rho }_4,{\rho }_4⊛{\rho }_1={\rho }_4,{\rho }_4⊛{\rho }_2={\rho }_4,{\rho }_4⊛{\rho }_3={\rho }_{34},{\rho }_4⊛{\rho }_4={\rho }_3,{\rho }_4⊛{\rho }_{34}={\rho }_4$;

${\rho }_{34}⊛{\rho }_0={\rho }_{34},{\rho }_{34}⊛{\rho }_1={\rho }_{34},{\rho }_{34}⊛{\rho }_2={\rho }_{34},{\rho }_{34}⊛{\rho }_3={\rho }_3,{\rho }_{34}⊛{\rho }_4={\rho }_4,{\rho }_{34}⊛{\rho }_{34}={\rho }_{34}$.

Then $M\left(H\right)=\{{\rho }_0,{\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4,{\rho }_{34}\}$, $\left(M\right(H),⊛)$ is a commutative semigroup, and the operation on it is shown in

Table 6. The adjoint semigroup of hyper $BCI$-algebra.

⊛	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$	${\mathit{\rho }}_4$	${\mathit{\rho }}_{34}$
${\rho }_0$	${\rho }_0$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_{34}$
${\rho }_1$	${\rho }_1$	${\rho }_2$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_{34}$
${\rho }_2$	${\rho }_2$	${\rho }_2$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_{34}$
${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_4$	${\rho }_{34}$	${\rho }_3$
${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_{34}$	${\rho }_3$	${\rho }_4$
${\rho }_{34}$	${\rho }_{34}$	${\rho }_{34}$	${\rho }_{34}$	${\rho }_3$	${\rho }_4$	${\rho }_{34}$

.

Definition 17.

In hyper $BCI$-algebra $(H,\circ )$, if $\forall x,y\in H$, and $x\ne y$,

$$x\circ (x\circ y)=0\circ (0\circ y).$$

Then, it is a generalized quasi-left alter-hyper $BCI$-algebra.

Theorem 10.

Assume that $(H,\circ )$ is a generalized quasi-left alter-hyper $BCI$-algebra. Hence, $(H,\circ )$ is $BCI$-algebra.

Proof. 

Assume that $(H,\circ )$ is a generalized quasi-left alter-hyper $BCI$-algebra. $\forall x\in H$ and $x\ne 0$, $0=0\circ 0\in (x\circ x)\circ 0=(x\circ 0)\circ x$. Then, $\exists p\in x\circ 0$, such that $0\in p\circ x$; that is, $p\ll x$. According to Definition 14, $x\circ p\subseteq x\circ (x\circ 0)=0\circ (0\circ 0)=0$. That is to say, $x\ll p$. According to Definition 14, $x=p$. Therefore, $x\in x\circ 0$. Then, $x\circ x\subseteq x\circ (x\circ 0)=0\circ (0\circ 0)=0$.

$\forall x,y\in H$, assume that $|x\circ y|>1$. Let $m,n\in x\circ y$ and $m\ne n$. Then, $m\circ n\subseteq (x\circ y)\circ (x\circ y)\ll x\circ x=0,n\circ m\subseteq (x\circ y)\circ (x\circ y)\ll x\circ x=0$. According to Definition 13, $m\circ n=0,n\circ m=0$, that is, $m\ll n,n\ll m$. Additionally, $m=n$. Therefore, $|x\circ y|=1$.

As $x\in x\circ 0$, $x=x\circ 0$. Additionally, H is $BCI$-algebra. □

Definition 18.

Let $(H,\ll )$ be a partial order containing 0 in hyper structure. x is said to be a quasi-minimal element in H. If for any element a in H, $a\ll x\Rightarrow x=a$ or $a=0$.

Definition 19.

A hyper $BCI$- algebra $(H,\ll ,\circ ,0)$ is called a QM-hyper $BCI$ algebra if all elements of H are quasi-minimal elements.

Theorem 11.

A hyper $BCI$-algebra $(H,\ll ,\circ ,0)$ is a QM-hyper $BCI$ algebra if it meets: $\forall x,y\in H-\left\{0\right\}$,

$$x\ll y\mathit{\text{ implies }}x=y.$$

Proof. 

$(\Rightarrow )$$\forall x,y\in H-\left\{0\right\}$, assume that $x\ll y$, according to Definition 18, $y=x$ or $x=0$. However, $x\ne 0$. Therefore, $x=y$.

$(\Leftarrow )$ Assume that $x,y\in H$, $x\ll y$. If $y=0$, then $x\ll y=0$, we can obtain $x=y=0$. If $x\ne 0$, $y\ne 0$, there is $x=y$ by condition. Therefore, y is a quasi-minimal element of H. Thus, H is a QM-hyper $BCI$ algebra. □

Theorem 12.

If $(H,\circ )$ is a generalized quasi-left alter-hyper $BCI$-algebra, then it is QM-hyper $BCI$-algebra.

Proof. 

Assume that $(H,\circ )$ is a generalized quasi-left alter-hyper $BCI$-algebra. According to Theorem 10, H is a generalized quasi-left alter $BCI$-algebra. Let $B\left(H\right)$ be $BCK$-part of H, and G(H) be $BCK$-remainder of H. Then, for any $x,y\in H$, assume that $x\ll y$ and $x\ne y$. That is,

$$x=x\circ 0=x\circ (x\circ y)=0\circ (0\circ y).$$

(1) When $y\in B\left(H\right)$, $x=0\circ (0\circ y)=0\circ 0=0$.

(2) When $y\in G\left(H\right)$, $x=0\circ (0\circ y)=y$, but $x\ne y$. Therefore, y is a quasi-minimal element of H. As y is arbitrary, H is QM-hyper $BCI$ algebra. □

However, not every QM-hyper $BCI$ algebra is a generalized quasi-left alter-hyper $BCI$-algebra, see Example 4.

Example 4.

Let $H=\{0,1,2,3,4\}$. The operation ∘ on H is shown in Table 5. Clearly, H is QM-hyper $BCI$-algebra.

However, H is not a generalized quasi-left alter-hyper $BCI$-algebra, since $1\circ (1\circ 0)=\{0,1\},0\circ (0\circ 0)=0,1\ne 0$.

Above all, we prove that generalized quasi-left alter $BCI$-algebra, QM-$BCI$ algebra and generaized quasi-left alter hyper $BCI$-algebra are equivalent to one another. Additionally, they are QM-hyper $BCI$-algebra.

4. Quasi-Hyper $\mathbf{BCI}$-Algebra

At the beginning of this part, we introduce the definition of quasi-hyper $BCI$-algebras.

Definition 20.

Let $(H,\circ )$ be a hyper groupoid containing 0. If it meets the following conditions:

• $\left(QHCI1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y,\forall x,y,z\in H$;
• $\left(QHCI2\right)$$(x\circ y)\circ z=(x\circ z)\circ y,\forall x,y,z\in H$;
• $\left(QHCI3\right)$$x\ll x,\forall x\in H$;
• $\left(QHCI4\right)$$x\ll y$ and $y\ll x\Rightarrow x=y,\forall x,y\in H$
• $\left(QHCI5\right)$$x\ll x\circ 0,\forall x\in H$
• $\left(QHCI6\right)$$x\ll 0\Rightarrow x=0,\forall x\in H$.
• At that time, it is a quasi-hyper $BCI$-algebra.

Remark 1.

(1)

Every weak hyper $BCI$-algebra is a quasi-hyper $BCI$-algebra;

(2)

Every hyper $BCI$-algebra is a quasi-hyper $BCI$-algebra.

In thia sectin, we give some examples of quasi-hyper $BCI$-algebra, and they show that not every quasi-hyper $BCI$-algebra is a (weak) hyper $BCI$-algebra.

Example 5.

(1) Assume that $H=\{0,1,2,3,4\}.$ The operation on H is defined in

Table 7. Quasi-hyper $BCI$ algebra.

∘	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{3,4\}$	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	$\{3,4\}$	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$

:

Clearly, $(H,\circ ,0)$ is a quasi-hyper $BCI$-algebra. However, it is not a weak hyper $BCI$-algebra, since $0\circ (0\circ 1)=\{0,1,2,3,4\}$ and $2\ll 1$ is not true.

(2) Assume that $H=\{0,1,2,3,4\}.$ The operation on H is defined in

Table 8. Quasi-hyper $BCI$-algebra.

∘	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$

:

Clearly, $(H,\circ ,0)$ is a quasi-hyper $BCI$-algebra. However, it is not a hyper $BCI$-algebra, since $0\circ (0\circ 0)=\{0,1,2,3,4\}$ and $0\notin 1\circ 0$.

Proposition 3.

In quasi-hyper BCI-algebra $(H,\circ )$, the following holds:$\forall x,y,z\in H$ and, for all non-empty subsets, A and B of H,

(1)

$0\circ (x\circ y)\ll y\circ x$,

(2)

$A\ll A$,

(3)

$A\subseteq B\Rightarrow A\ll B$,

(4)

$A\ll 0\Rightarrow A=0$,

(5)

$x\circ y=0\Rightarrow (x\circ z)\circ (y\circ z)=0$ and $x\circ z\ll y\circ z$,

(6)

$A\circ 0=0\Rightarrow A=0$,

(7)

$x\ll y\Rightarrow 0\ll y\circ x$,

(8)

$x\circ x=0\Rightarrow |y\circ z|=1$.

(9)

$x\circ 0\ll \left\{y\right\}\Rightarrow x\ll y$,

(10)

$y\ll z\Rightarrow x\circ z\ll x\circ y$.

Proof. 

(1) By (QHCI1) and (QHCI3), $0\circ (x\circ y)\subseteq (y\circ y)\circ (x\circ y)\ll y\circ x$. That is, $0\circ (x\circ y)\ll y\circ x$.

(2) By (QHCI3), for any $x\in A,x\ll x$, that is $0\in x\circ x$. Then, $A\ll A$.

(3) Let $a\in A$. Then, $a\in B$. By (QHCI3), $x\ll x$ and $0\in x\circ x$. Then, $A\ll B$.

(4) Let $a\in A$. Then, $a\ll 0$ and so $a=0$. Then, $A=\left\{0\right\}$.

(5) By (QHCI1), $(x\circ z)\circ (y\circ z)\ll x\circ y=\left\{0\right\}$. By (4), $(x\circ z)\circ (y\circ z)=\left\{0\right\}$. Therefore, $x\circ z\ll y\circ z$.

(6) Assume that $A\circ 0=0$; then, $A\ll 0$. Therefore, $A=0$.

(7) Assume that $x\ll y$. Then, $0\in x\circ y$, and so $0\in 0\circ 0\subseteq (y\circ y)\circ (x\circ y)\ll y\circ x$. Hence, $0\ll y\circ x$.

(8) $\forall x\in H$, let $x\circ x=\left\{0\right\}$. Assume that $|y\circ z|>1$; let $m,n\in y\circ z$, and $m\ne n$. Then,

$$m\circ n\subseteq (y\circ z)\circ (y\circ z)\ll y\circ y=0\mathrm{and}n\circ m\subseteq (y\circ z)\circ (y\circ z)\ll y\circ y=0$$

thus, $m\circ n\ll 0$, $n\circ m\ll 0$ and $m\ll n,n\ll m$. Hence, $m=n$, and so $|y\circ z|=1$.

(9) According to Definition 20, $0\in (x\circ 0)\circ y=(x\circ y)\circ 0$, there exists $p\in x\circ y$ s.t. $0\in p\circ 0$, that is, $p\ll 0$. By (QHCI6), $p=0$. So $0\in x\circ y$ and $x\ll y$.

(10) As $y\ll z$, then $(x\circ z)\circ 0\subseteq (x\circ z)\circ (y\circ z)\ll x\circ y$. Therefore, $x\circ z\ll x\circ y$. □

Proposition 4.

In any quasi-hyper $BCI$-algebra $(H,\circ )$ satisfying $0\circ 0=0$, the following holds:

(1)

$(0\circ x)\circ (0\circ x)=0$, $\forall x\in H$;

(2)

$0\circ x$ is a singleton set, $\forall x\in H$;

(3)

$(0\circ x)\circ 0=0\circ x$, $\forall x\in H$.

Proof. 

$\left(1\right)$$\forall x\in H$, $(0\circ x)\circ (0\circ x)\ll 0\circ 0=0$. According to Proposition 3(4), $(0\circ x)\circ (0\circ x)=0$.

$\left(2\right)$$\forall m,n\in 0\circ x$, and $m\ne n$. $m\circ n\subseteq (0\circ x)\circ (0\circ x)=0$, $n\circ m\subseteq (0\circ x)\circ (0\circ x)=0$, by (QHCI4), $m\ll n,n\ll m$, then $m=n$. Therefore, $0\circ x$ is a singleton set.

$\left(3\right)$ By (2), let $0\circ x=m$. By (QHCI5), $m\ll m\circ 0$ and $m\circ m=0$. Assume that $|m\circ 0|>1$, let $a,b\in m\circ 0$. $a\circ b\subseteq (m\circ 0)\circ (m\circ 0)\ll m\circ m=0$, $b\circ a\subseteq (m\circ 0)\circ (m\circ 0)\ll m\circ m=0$, so $a\ll b$, $b\ll a$. Additionally, $a=b$. Therefore, $|m\circ 0|=1$. As $m\circ 0=(0\circ x)\circ 0\subseteq (0\circ x)\circ (x\circ x)\ll 0\circ x=m$, and $|m\circ 0|=1$, $m=m\circ 0$. That is, $(0\circ x)\circ 0=0\circ x$. □

Proposition 5.

In any quasi-hyper $BCI$-algebra $(H,\circ ,0)$, if $\forall x\in H(x\ne 0)$ satisfying $x\circ 0=x$. At that time, it is a weak hyper $BCI$-algebra. $\forall x\in H$ satisfying $x\circ 0=x$; then, it is a hyper $BCI$-algebra.

Proof. 

Firstly, for any $m\in H$ and $m\ne 0$, $m\circ 0=m$. Then, $0\circ (0\circ m)\subseteq (m\circ m)\circ (0\circ m)\ll (m\circ 0)=m.$ Therefore, $(H,\circ ,0)$ is a weak hyper $BCI$-algebra. For any $m\in H$, $0\circ (0\circ m)\subseteq (m\circ m)\circ (0\circ m)\ll (m\circ 0)=x.$ So $(H,\circ ,0)$ is a hyper $BCI$-algebra. □

Definition 21.

A quasi-hyper $BCI$-algebra $(H,\circ )$ is called standard if $\forall x\in H$, we have $x\circ 0=x$.

Proposition 6.

Every standard quasi-hyper $BCI$-algebra is a hyper $BCI$-algebra.

Proof. 

This follows from Proposition 5. □

The concept of $H_v$-group was introduced by T. Vougiouklis [35]: Let $(H,·)$ be a hyperstructure. If it satisfies: (i) $(x·y)·z\cap x·(y·z)\ne ⌀$,$\forall x,y,z\in H$, (ii) $a·H=H·a=H$, $\forall a\in H$, then it is a $H_v$-group.

Firstly, define $x·y$ through $x·y=x\circ (0\circ y)$, $\forall x,y\in H$ in a quasi-hyper $BCI$-algebra.

Theorem 13.

Assume that $(H,\circ )$ is a quasi-hyper $BCI$-algebra and meets these conditions:

(1)

$x\in m\circ (m\circ x)$, $\forall x,m\in H$;

(2)

$x·y\cap y·x\ne ⌀$, $\forall x,y\in H$;

(3)

$x·(y·z)=y·(x·z)$, $\forall x,y,z\in H$. We find that $(H,·)$ is a $H_v$-group.

Proof. 

Obviously, $(x·y)·z=(x·z)·y$ for all $x,y,z\in H$. As

$$(x·y)·z=(x\circ (0\circ y\left)\right)\circ (0\circ z)=(x\circ (0\circ z\left)\right)\circ (0\circ y)=(x·z)·y.$$

According to (1), $x\in m\circ (m\circ x)\subseteq m\circ (0\circ (0\circ (m\circ x)\left)\right)=m·0\circ (m\circ x)\subseteq m·H$. Hence, $H\subseteq m·H$ and $H=a·H$. Moreover, $x\in m\circ (m\circ x)\subseteq (0\circ (0\circ m\left)\right)\circ (m\circ x)=(0\circ (m\circ x\left)\right)\circ (0\circ m)=(0\circ (m\circ x\left)\right)·m\subseteq H·m$. Thus, $x·H=H=H·x$.

According to (2), $x·z\cap z·x\ne ⌀$, and thus $\exists p\in x·z\cap z·x$. According to (2), there is $q\in p·y\cap y·p$ and thus $q\in p·y\subseteq (x·z)·y=(x·y)·z$. On the other hand, $q\in y·p\subseteq y·(x·z)=x·(y·z)$ by (3). So $q\in (x·y)·z\cap x·(y·z)$.

According to the definition of $H_v$-group, we can see that $(H,·)$ is a $H_v$-group. □

Example 6.

Assume that $H=\{0,1,2,3,4\}$ be a quasi-hyper $BCI$-algebra satisfying those conditions in Theorem 13. Additionally, the operation ∘ on H is shown in

Table 9. Quasi-hyper $BCI$ algebra.

∘	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{3,4\}$	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	4	4	4	$\{0,1,2,3,4\}$

:

Then, we obtain a $H_v$-group $(H,·)$ and the operation "·" is shown in

Table 10. $H_v$-group.

·	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$

:

Theorem 14.

Assume that $(H,\circ )$ is a quasi-hyper $BCI$-algebra and meets those conditions: $\forall x,a,y\in H$,

(1)

$x\in a\circ (a\circ x)$,

(2)

$x\circ (0\circ y)=y\circ (0\circ x)$.

We can see that $(H,·)$ is a hypergroup.

Proof. 

According to Theorem 13, $x·H=H=H·x$. Morever, $x·y=y·x$ and $(x·y)·z=(x·z)·y$. Then $(x·y)·z=z·(x·y)=z·(y·x)=(y·x)·z=(y·z)·x=x·(y·z).$ So $(H,·)$ is a hypergroup, and is commutative. □

Example 7.

Let $H=\{0,1,2,3,4\}$ be a quasi-hyper $BCI$-algebra satisfying the conditions in Theorem 14. The operation ∘ on H is shown in

Table 11. Quasi-hyper $BCI$-algebra.

∘	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{2,3,4\}$	$\{1,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{3,4\}$	$\{2,4\}$	$\{1,2,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	$\{3,4\}$	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$

:

Then, we obtain a commutative hypergroup $(H,·)$ and the operation "·" is shown in

Table 12. Hyper group.

·	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$

:

Theorem 15.

Assume that $(H,\circ )$ is a quasi-hyper $BCI$-algebra. Denote $M\left(H\right)$ as a set of finite products ${\rho }_{a_1}⊛...⊛{\rho }_{a_s}$ ($\forall a_1,...,a_s\in H$), where ⊛ represents the composition operation of mappings. Then, $M\left(H\right)$ is a commutative semigroup.

Proof. 

According to Theorem 9. □

Example 8.

Let $H=\{0,1,2,3,4\}$. Define the operation ∘ on H in

Table 13. Quasi-hyper $BCI$-algebra.

∘	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{2,3,4\}$	$\{2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{3,4\}$	$\{3,4\}$	$\{3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	4	4	$\{3,4\}$	$\{0,1,2,3,4\}$

,

Clearly, $(H,\circ )$ is a quasi-hyper $BCI$-algebra.

That is,

${\rho }_0:H\to H:x\mapsto x\circ 0.$ So ${\rho }_0=\{\{0,1,2,3,4\},\{1,2,3,4\},\{2,3,4\},\{3,4\},4\}$;

${\rho }_1:H\to H:x\mapsto x\circ 1.$ So ${\rho }_1=\{\{0,1,2,3,4\},\{0,1,2,3,4\},\{2,3,4\},\{3,4\},4\}$;

${\rho }_2:H\to H:x\mapsto x\circ 2.$ So ${\rho }_2=\{\{0,1,2,3,4\},\{0,1,2,3,4\},\{0,1,2,3,4\},\{3,4\},4\}$;

${\rho }_3:H\to H:x\mapsto x\circ 3.$ So ${\rho }_3=\{\{0,1,2,3,4\},\{0,1,2,3,4\},\{0,1,2,3,4\},\{0,1,2,3,4\},4\}$;

${\rho }_4:H\to H:x\mapsto x\circ 4.$

Therefore, ${\rho }_4=\{\{0,1,2,3,4\},\{0,1,2,3,4\},\{0,1,2,3,4\},\{0,1,2,3,4\},\{0,1,2,3,4\}\}$.

We can verify the following:

${\rho }_0⊛{\rho }_0={\rho }_0,{\rho }_0⊛{\rho }_1={\rho }_1,{\rho }_0⊛{\rho }_2={\rho }_2,{\rho }_0⊛{\rho }_3={\rho }_3,{\rho }_0⊛{\rho }_4={\rho }_4$;

${\rho }_1⊛{\rho }_0={\rho }_1,{\rho }_1⊛{\rho }_1={\rho }_1,{\rho }_1⊛{\rho }_2={\rho }_2,{\rho }_1⊛{\rho }_3={\rho }_3,{\rho }_1⊛{\rho }_4={\rho }_4$;

${\rho }_2⊛{\rho }_0={\rho }_2,{\rho }_2⊛{\rho }_1={\rho }_2,{\rho }_2⊛{\rho }_2={\rho }_2,{\rho }_2⊛{\rho }_3={\rho }_3,{\rho }_2⊛{\rho }_4={\rho }_4$;

${\rho }_3⊛{\rho }_0={\rho }_3,{\rho }_3⊛{\rho }_1={\rho }_3,{\rho }_3⊛{\rho }_2={\rho }_3,{\rho }_3⊛{\rho }_3={\rho }_3,{\rho }_3⊛{\rho }_4={\rho }_4$;

${\rho }_4⊛{\rho }_0={\rho }_4,{\rho }_4⊛{\rho }_1={\rho }_4,{\rho }_4⊛{\rho }_2={\rho }_4,{\rho }_4⊛{\rho }_3={\rho }_4,{\rho }_4⊛{\rho }_4={\rho }_4$.

Then, $M\left(H\right)=\{{\rho }_0,{\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4\}$, and $\left(M\right(H),⊛)$ is a commutative semigroup. The operation is shown in

Table 14. The adjoint semigroup of quasi-hyper $BCI$ algebra.

⊛	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$	${\mathit{\rho }}_4$
${\rho }_0$	${\rho }_0$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$
${\rho }_1$	${\rho }_1$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$
${\rho }_2$	${\rho }_2$	${\rho }_2$	${\rho }_2$	${\rho }_3$	${\rho }_4$
${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_4$
${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$

.

Definition 22.

In quasi-hyper $BCI$-algebra $(H,\circ )$, if $\forall x,y\in H$, and $x\ne y$,

$$x\circ (x\circ y)=0\circ (0\circ y).$$

Then, this is a generalized quasi-left alter quasi-hyper $BCI$ algebra.

Proposition 7.

Assume that $(H,\circ )$ is a generalized quasi-left alter quasi-hyper $BCI$ algebra satisfying $0\circ 0=0$. Hence, H is a $BCI$-algebra.

Proof. 

According to Theorem 10. □

Example 9.

Assume that $H=\{0,1,2,3,4\}$. The operation on H is defined in

Table 15. Generalized quasi-left alter quasi-hyper $BCI$-algebra.

∘	0	1	2	3	4
0	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
1	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	$\{2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$

,

Then, $(H,\circ )$ is a generalized quasi-left alter quasi-hyper $BCI$-algebra.

Definition 23.

A quasi-hyper $BCI$- algebra $(H,\ll ,\circ ,0)$ is called a QM-quasi hyper $BCI$-algebra, if all elements of H are quasi-minimal.

Theorem 16.

Assume that $(H,\ll ,\circ ,0)$ be a quasi-hyper $BCI$-algebra. This is a QM-quasi hyper $BCI$-algebra if $\forall x,y\in H-\left\{0\right\}$,

$$x\ll y\Rightarrow x=y.$$

Proof. 

According to Theorem 11. □

Example 10.

Let $H=\{0,1,2,3,4\}$. The operation on H is defined in

Table 16. Quasi-hyper $BCI$ algebra.

∘	0	1	2	3	4
0	0	0	0	4	3
1	1	$\{0,1\}$	1	4	3
2	2	2	$\{0,2\}$	4	3
3	3	3	3	0	4
4	4	4	4	3	0

,

Clearly, $(H,\circ )$ is a QM-quasi hyper $BCI$-algebra. However, it is not a generalized quasi-left alter quasi-hyper $BCI$-algebra, since $1\circ (1\circ 0)=\{0,1\}$, $0\circ (0\circ 0)=0$, $1\ne 0$.

According to Definition, we know that both QM-$BCI$ algebra and QM-hyper $BCI$-algebra are QM-quasi hyper $BCI$-algebra, but not every QM-quasi hyper $BCI$-algebra is QM-$BCI$ algebra and QM-hyper $BCI$-algebra; see Example 11.

Example 11.

Let $H=\{0,1,2\}$. The operation on H is defined in

Table 17. Quasi-hyper $BCI$ algebra.

∘	0	1	2
0	$\{0,1\}$	$\{0,1\}$	2
1	1	$\{0,1\}$	2
2	2	2	$\{0,1\}$

,

Clearly, $(H,\circ )$ is a QM-quasi hyper $BCI$-algebra. However, it is not QM-hyper $BCI$ algebra, since $0\circ (0\circ 0)=\{0,1\}$, $1\ll 0$ is not true.

5. Discussion

Firstly, we discuss the adjoint semigroup of a generalized quasi-left alter $BCI$-algebra, which is a commutative Clifford semigroup. Then, we introduced QM-$BCI$ algebra and proved that the generalized quasi-left alter $BCI$-algebra is equivalent to QM-$BCI$ algebra. Furthermore, we proved that the generalized quasi-left alter-hyper $BCI$-algebra is a generalized quasi-left alter $BCI$-algebra. In the last part, we proposed the notion of quasi-hyper $BCI$ algebra and discuss its properties. We explored the subalgebras of quasi-hyper $BCI$ algebra and the relationships between quasi-hyper $BCI$ algebra and the hypergroup, $H_v$-group. In general, this paper discusses the relationship between (hyper) logical algebra and classical abstract algebra, and the description of the structure of (hyper) logical algebra is more clear. As a further research topic, we can consider exploring the internal connections between (hyper) $BCI$-algebras, $BI$-algebras and $CA$-semihypergroups (see [36,37,38]).