QM-BZ-Algebras and Quasi-Hyper BZ-Algebras

Abstract

$BZ$-algebra, as the common generalization of $BCI$-algebra and $BCC$-algebra, is a kind of important logic algebra. Herein, the new concepts of QM-$BZ$-algebra and quasi-hyper $BZ$-algebra are proposed and their structures and constructions are studied. First, the definition of QM-$BZ$-algebra is presented, and the structure of QM-$BZ$-algebra is obtained: Each QM-$BZ$-algebra is KG-union of quasi-alter $BCK$-algebra and anti-grouped $BZ$-algebra. Second, the new concepts of generalized quasi-left alter (hyper) $BZ$-algebras and QM-hyper $BZ$-algebra are introduced, and some characterizations of them are investigated. Third, the definition of quasi-hyper $BZ$-algebra is proposed, and the relationships among $BZ$-algebra, hyper $BZ$-algebra, quasi-hyper $BCI$-algebra, and quasi-hyper $BZ$-algebra are discussed. Finally, several special classes of quasi-hyper $BZ$-algebras are studied in depth and the following important results are proved: (1) an anti-grouped quasi-hyper $BZ$-algebra is an anti-grouped $BZ$-algebra; (2) every generalized anti-grouped quasi-hyper $BZ$-algebra corresponds to a semihypergroup.

1. Introduction

BCI-algebra as well as $BCK$-algebra are two kinds of algebraic structures closely related to combinatorial logic and fuzzy logic, which have been studied extensively and deeply (see [1,2,3,4,5,6,7,8,9,10]). As a extension of $BCI$-algebra, pseudo-$BCI$-algebra was proposed in 2008 by Wieslaw A. Dudek and Young Bae Jun (see [11]). Then, X.Y. Wu and X.H. Zhang introduced a quasi-maximal element into pseudo-$BCI$ algebra and studied the structure of QM-pseudo-$BCI$ algebra (see [12]). Dually, X.H. Zhang and Y.D. Du proposed introducing a quasi-minimal element into $BCI$-algebra and proved the adjoint semigroup of QM-$BCI$-algebra is a commutative Clifford semigroup in [13]. As another generalization of $BCI$-algebra, $BZ$-algebra was first proposed by Ye (see [14]). Naturally, we introduce a quasi-minimal element into $BZ$-algebra and study the structure theorem of QM-$BZ$-algebra.

In 1934, the definition of hyperstructure (also called multialgebra) was proposed by F. Marty (see [15]), and hyperstructures have been used widely in pure and applied sciences (see [16,17,18,19]). Naturally, the idea of hyperstructures is also applied to the study of non-classical logic algebras. In 2000, Young Bae Jun et al. proposed the concept of hyper BCK-algebra, and investigated hyper BCK-ideals and some related hyper algebras, such as hyper K-algebra and hyper MV-algebra (see [20,21,22,23,24,25]). In 2006, Jun and Borzooei et al. independently proposed the new concept of hyper BCC-algebra; also in 2006, Xin introduced hyper BCI-algebra, and since then, many research papers on hyper logical algebras have emerged (see [26,27,28,29,30,31,32,33]). In 2021, Y.D. Du and X.H. Zhang introduced the hyper structure into $BZ$-algebra and discussed the relationships between hyper $BZ$-algebra and semihypergroups (see [34]). In this paper, they gave the concepts of anti-grouped hyper $BZ$-algberas as well as generalized anti-grouped hyper $BZ$-algebras and discussed the connection between them and $BZ$-algebras. Additionally, in [13], X.H. Zhang and Y.D. Du proposed quasi-hyper $BCI$-algebra and discussed relationships among quasi-hyper $BCI$-algebra, $H_v$-groups and hyper groups. Similarly, we present the definition of quasi-hyper $BZ$-algebra, which is an extension of hyper $BZ$-algebra and quasi-hyper $BCI$-algebra.

The arrangement of the whole paper is as below. In Section 2, we give the definitions and properties of logical algebras and some related hyper structures. In Section 3, we firstly introduce a method to construct a $BZ$-algebra by using $BCC$-algebra and anti-grouped $BZ$-algebra. Then, QM-$BZ$-algebra is defined and the relationships among QM-$BZ$-algebra, QM-hyper $BZ$-algebra, and generalized quasi-left alter (hyper) $BZ$-algebra are discussed. In Section 4, we introduce quasi-hyper $BZ$-algebra and study its properties, discussing the relationships among $BZ$-algebra, hyper $BZ$-algebra, quasi-hyper $BCI$-algebra, and quasi-hyper $BZ$-algebra. Moreover, we investigate some kinds of quasi-hyper $BZ$-subalgebra and propose the definition of anti-grouped quasi-hyper $BZ$-algebra, generalized anti-grouped quasi-hyper $BZ$-algebra, and associative quasi-hyper $BZ$-algebra, and discuss their relationships. Finally, we introduce QM-quasi-hyper $BZ$-algebra and generalized quasi-left alter quasi-hyper $BZ$-algebra.

2. Preliminaries

Firstly, we give some concepts of some logical algebras and connections between $BZ$-algebra and $BCI$-algebra that play an important role in discussions of special $BZ$-algebra. In $BCK/BCI/BCC/BZ$-algebra, define ≤: $x\le y$ iff $x\ast y=0$.

Definition 1

([1,2]). Assume that <$X;\ast ,0$> is an algebraic structure; then, it is a BCI-algebra if it meets: $\forall x,y,z\in X,$

$\left(1\right)$$\left(\right(x\ast z)\ast (y\ast z\left)\right)\ast (x\ast y)=0$;

$\left(2\right)$$(x\ast (x\ast y\left)\right)\ast y=0$;

$\left(3\right)$$x\ast 0=x$;

$\left(4\right)$$x\ast y=0$ and $y\ast x=0\Rightarrow x=y$.

If a BCI-algebra meets the below condition: $\forall x\in X$,

$\left(5\right)$$0\ast x=0$,

we call it a BCK-algebra.

Definition 2

([10]). Assume that <$X;\ast ,0$> is an algebraic structure; then, it is a BCC-algebra if it meets: $\forall x,y,z\in X,$

$\left(1\right)$$\left(\right(x\ast z)\ast (y\ast z\left)\right)\ast (x\ast y)=0$;

$\left(2\right)$$x\ast 0=x$;

$\left(3\right)$$x\ast x=0$;

$\left(4\right)$$0\ast x=0$;

$\left(5\right)$$x\ast y=0$ and $y\ast x=0$ imply $x=y$.

Definition 3

([14]). Assume that <$X;\ast ,0$> is an algebraic structure; then, it is a BZ-algebra if it meets: $\forall x,y,z\in X,$

$\left(1\right)$$\left(\right(x\ast z)\ast (y\ast z\left)\right)\ast (x\ast y)=0$;

$\left(2\right)$$x\ast 0=x$;

$\left(3\right)$$x\ast y=0$, $y\ast x=0$ imply $x=y$.

An algebra of type (2,0) means that this algebra is composed of a non-empty set and an operation as well as a constant. Clearly, all the above algebras are algebras of type (2,0).

Definition 4

([9]). An algebraic structure <$X;\ast ,0$> is a quasi-alter BCK-algebra iff it meets: $\forall x,y\in X$, if $x=y$, $x\ast y=0$, otherwise, $x\ast y=x$.

Theorem 1

([35]). Assume that <$X;\ast ,0$> is a $BZ$-algebra; it is a $BCI$-algebra iff $\forall x,y\in X$,

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

In paper [36], Xiaohong Zhang and Ruifen Ye introduced a kind of $BZ$-algebra and revealed the connection between $BZ$-algebra and a general group.

Definition 5

([36]). A BZ-algebra <$X;\ast ,0$> is called an anti-grouped BZ-algebra if, for any $x\in X$, it satisfies $0\ast (0\ast x)=x$.

Theorem 2

([36]). A BZ-algebra <$X;\ast ,0$> is anti-grouped iff it meets:

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

Theorem 3

([36]). Let <$X;\ast ,0$> be an anti-grouped BZ-algebra. Define “⊛”:

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

Then, <$X;⊛,0$> is a group.

Theorem 4

([36]). Assume that <$G;\circ ,e$> is a group. Define "·":

$$x·y=x\circ y^{-1},\forall x,y\in G.$$

Then, <$G;·,e$> is an anti-grouped BZ-algebra.

In the following, we give the definitions of semihypergroup and some hyper logical algebras.

Definition 6

([16]). Assume that $(H,\circ )$ is a hypergroupoid. If $\forall x,y,z\in H$, we have $(x\circ y)\circ z=x\circ (y\circ z)$, so $(H,\circ )$ is a semihypergroup. Then, there is

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

Note that, if $(H,\circ )$ is a semihypergroup, there is $(A\circ B)\circ C=A\circ (B\circ C)$ for all $A,B,C\in P^{\ast }\left(H\right)$, where $P^{\ast }\left(H\right)$ represents nonempty subset of H.

Definition 7

([16]). Assume that $(H,\circ )$ is a semihypergroup. $(H,\circ )$ is called a hypergroup if ($\forall a\in H$) $a\circ H=H\circ a=H$.

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

Definition 8

([20]). Assume that $(H,\circ )$ is a hypergroupoid 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 $y\ll x$ imply $x=y,$

then it is a hyper $BCK$-algebra.

Definition 9

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

$\left(HBC1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y;$

$\left(HBC2\right)$$x\ll x;$

$\left(HBC3\right)$$x\circ y\ll x;$

$\left(HBC4\right)$$x\ll y$ and $y\ll x$ imply $x=y$,

then it is a hyper BCC-algebra (Jun’s definition).

Definition 10

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

$\left(HC1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y;$

$\left(HBC2\right)$$0\circ x=\left\{0\right\};$

$\left(HBC3\right)$$x\circ 0=\left\{x\right\};$

$\left(HBC4\right)$$x\ll y$ and $y\ll x$ imply $x=y$,

then it is a hyper BCC-algebra (Borzooei’s definition).

Jun and Borzooei, in [26,27], gave different definitions of hyper $BCC$-algebra, respectively. However, in this paper, we mainly use Definition 10 for hyper $BCC$-algebra.

Definition 11

([28]). Assume that $(H,\circ )$ is a hypergroupoid 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 $y\ll x$ imply $x=y;$

$\left(HBI5\right)$$0\circ (0\circ x)\ll x$,

then it is a hyper $BCI$-algebra.

Definition 12

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

$\left(QHCI1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y;$

$\left(QHCI2\right)$$(x\circ y)\circ z=(x\circ z)\circ y;$

$\left(QHCI3\right)$$x\ll x;$

$\left(QHCI4\right)$$x\ll y$ and $y\ll x$ imply $x=y;$

$\left(QHCI5\right)$$x\ll x\circ 0;$

$\left(QHCI6\right)$$x\ll 0$ implies $x=0$,

then it is a quasi-hyper $BCI$-algebra.

Definition 13

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

$\left(HZ1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y;$

$\left(HZ2\right)$$x\ll x;$

$\left(HZ3\right)$$x\ll y$ and $y\ll x$ imply $x=y;$

$\left(HZ4\right)$$0\circ (0\circ x)\ll x;$

$\left(HZ5\right)$$x\ll x\circ 0,$

then it is a hyper $BZ$-algebra.

Proposition 1

([34]). In any hyper BZ-algebra $(H,\circ )$, the following holds: For all $x,y,z\in H$ and for all non-empty subsets A and B of H,

$\left(1\right)$$x\ll 0\Rightarrow x=0$;

$\left(2\right)$$0\circ (x\circ y)\ll y\circ x$;

$\left(3\right)$$A\ll A$;

$\left(4\right)$$A\subseteq B\Rightarrow A\ll B$;

$\left(5\right)$$A\ll 0\Rightarrow A=0$;

$\left(6\right)$$0\circ 0=0$;

$\left(7\right)$$(0\circ x)\circ (0\circ x)=0$;

$\left(8\right)$$0\circ x$ is a singleton set;

$\left(9\right)$$x\circ y=0\Rightarrow (x\circ z)\circ (y\circ z)=0$ and $x\circ z\ll y\circ z$;

$\left(10\right)$$A\circ 0=0\Rightarrow A=0$;

$\left(11\right)$$x\ll y\Rightarrow 0\ll y\circ x$;

$\left(12\right)$$0\circ (0\circ (0\circ x\left)\right)\ll 0\circ x$;

$\left(13\right)$$x\circ x=0\Rightarrow |y\circ z|=1$.

Definition 14

([34]). A hyper $BZ$-algebra $(H,\circ )$ is called a generalized anti-grouped hyper $BZ$-algebra if $\forall x,y,z\in H$, it satisfies $(x\circ (0\circ y\left)\right)\circ (0\circ z)=x\circ (0\circ (y\circ (0\circ z)\left)\right)$.

3. QM-BZ-Algebra

Firstly, we introduce a method to construct $BZ$-algebra by using $BCC$-algebra and anti-grouped $BZ$-algbera.

Proposition 2.

Assume that $(K,\ast ,0)$ is a $BCC$-algebra and $(G,·,0)$ is an anti-grouped $BZ$-algebra. Denote $A=K\cup G$, and $K\cap G=\left\{0\right\}$. An operation on A is as below:

$$x\circ y=\left\{\begin{array}{ccc}\hfill x\ast y,& & x,y\in K\hfill \\ \hfill x·y,& & x,y\in G\hfill \\ \hfill 0·y,& & x\in K,y\in G-\left\{0\right\}\hfill \\ \hfill x,& & x\in G-\left\{0\right\},y\in K\hfill \end{array}\right.$$

Therefore, $(A,\circ ,0)$ is $BZ$-algebra.

Proof. 

(1) $\forall x,y,z\in A$,

Case 1: $\forall x,y,z\in K$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=\left(\right(x\ast z)\ast (y\ast z\left)\right)\ast (x\ast y)=0$;

Case 2: $\forall x,y,z\in G$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=\left(\right(x·z)·(y·z\left)\right)·(x·y)=0$;

Case 3: $\forall x,y\in K$, $\forall z\in G-\left\{0\right\}$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=\left(\right(0·z)\circ (0·z\left)\right)\circ (x\circ y)=\left(\right(0·z)·(0·z\left)\right)\circ (x\circ y)=0\circ (x\ast y)=0$;

Case 4: $\forall x\in K$, $\forall y\in G-\left\{0\right\}$, $\forall z\in K$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=\left(\right(x\ast z)\circ y)\circ (0·y)=(0·y)\circ (0·y)=(0·y)·(0·y)=0$;

Case 5: $\forall x\in G-\left\{0\right\}$, $\forall y\in K$, $\forall z\in K$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=(x\circ (y\ast z\left)\right)\circ x=x\circ x=x·x=0$;

Case 6: $\forall x,y\in G-\left\{0\right\}$, $\forall z\in K$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=(x\circ y)\circ (x\circ y)=(x·y)·(x·y)=0$;

Case 7: $\forall x\in G-\left\{0\right\}$, $\forall y\in K$, $\forall z\in G-\left\{0\right\}$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=\left(\right(x·z)\circ (0·z\left)\right)\circ x=\left(\right(x·z)·(0·z\left)\right)·x=(x·0)·x=x·x=0$;

Case 8: $\forall x\in K$, $\forall y\in G-\left\{0\right\}$, $\forall z\in G-\left\{0\right\}$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=\left(\right(0·z)\circ (y·z\left)\right)\circ (0·y)=\left(\right(0·z)·(y·z\left)\right)·(0·y)=(0·y)·(0·y)=0$.

Above all, (BZ1) holds.

(2) Obviously, $\forall x\in A$, $x\circ x=0$, that is, (BZ2) holds.

(3) $\forall x,y\in A$, (i) $\forall x,y\in K$, $x\circ y=0$, and $y\circ x=0$ imply $x=y$; (ii) $\forall x,y\in G$, $x\circ y=0$, and $y\circ x=0$ imply $x=y$; (iii) $\forall x\in G-\left\{0\right\}$, $\forall y\in K$, if $x\circ y=0$, there is $x=0$. Because $x\in G-\left\{0\right\}$, there does not exist $x\circ y=0$. Then, (BZ3) holds.

To sum up, $(A,\circ ,0)$ is a $BZ$-algebra. □

Definition 15.

Assume that K is a $BCC$-algebra and G is an anti-grouped $BZ$-algebra, $K\cap G=\left\{0\right\}$. Denote $A=K\cup G$ according to Proposition 2; then, A is $BZ$-algebra. Then, A is called the $KG$-union of K and G, and is written as $A=K{\oplus }_{KG}G$.

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

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

$\forall a,b\in X$, $\forall x\in X$, denote ${\rho }_a\ast {\rho }_b$:

$$({\rho }_a\ast {\rho }_b)\left(x\right)={\rho }_a\left(y\right)\mathrm{and}y={\rho }_b\left(x\right),$$

where ∗ means the composition operation of mappings.

Theorem 5.

Denote $M\left(X\right)$ as a set which is all compositional results of finite mappings which are for all $a\in H$; thus, we have ${\rho }_a$. Then, $M\left(X\right)$ is a monoid.

Proof. 

$\forall a,b,c\in X$, and $\forall x\in X$; therefore:

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

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

Obviously, $({\rho }_a\ast {\rho }_b)\ast {\rho }_c\left(x\right)={\rho }_a\ast ({\rho }_b\ast {\rho }_c)\left(x\right)$. So, $\left(M\right(X),\ast )$ satisfies associative law, that is, $\left(M\right(X),\ast )$ is a semigroup.

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

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

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

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

Example 1.

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

Table 1. $BZ$-algebra.

∗	0	1	2	3	4	5
0	0	0	0	0	5	4
1	1	0	3	3	5	4
2	2	2	0	2	5	4
3	3	0	3	0	5	4
4	4	4	4	4	0	5
5	5	5	5	5	4	0

.

Then, <$X,\ast ,0$> is a $BZ$-algebra and $M\left(X\right)=\{{\rho }_0,{\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4,{\rho }_5,{\rho }_{12}\}$, where ${\rho }_{12}={\rho }_1\ast {\rho }_2$.

We can verify the following:

${\rho }_0\ast {\rho }_0={\rho }_0,{\rho }_0\ast {\rho }_1={\rho }_1,{\rho }_0\ast {\rho }_2={\rho }_2,{\rho }_0\ast {\rho }_3={\rho }_3,{\rho }_0\ast {\rho }_4={\rho }_4,{\rho }_0\ast {\rho }_5={\rho }_5,{\rho }_0\ast {\rho }_{12}={\rho }_{12};$

${\rho }_1\ast {\rho }_0={\rho }_1,{\rho }_1\ast {\rho }_1={\rho }_1,{\rho }_1\ast {\rho }_2={\rho }_{12},{\rho }_1\ast {\rho }_3={\rho }_1,{\rho }_1\ast {\rho }_4={\rho }_4,{\rho }_1\ast {\rho }_5={\rho }_5,{\rho }_1\ast {\rho }_{12}={\rho }_{12}$;

${\rho }_2\ast {\rho }_0={\rho }_2,{\rho }_2\ast {\rho }_1={\rho }_{12},{\rho }_2\ast {\rho }_2={\rho }_2,{\rho }_2\ast {\rho }_3={\rho }_1,{\rho }_2\ast {\rho }_4={\rho }_4,{\rho }_2\ast {\rho }_5={\rho }_5,{\rho }_2\ast {\rho }_{12}={\rho }_{12}$;

${\rho }_3\ast {\rho }_0={\rho }_3,{\rho }_3\ast {\rho }_1={\rho }_1,{\rho }_3\ast {\rho }_2={\rho }_{12},{\rho }_3\ast {\rho }_3={\rho }_2,{\rho }_3\ast {\rho }_4={\rho }_4,{\rho }_3\ast {\rho }_5={\rho }_5,{\rho }_3\ast {\rho }_{12}={\rho }_{12}$;

${\rho }_4\ast {\rho }_0={\rho }_4,{\rho }_4\ast {\rho }_1={\rho }_4,{\rho }_4\ast {\rho }_2={\rho }_4,{\rho }_4\ast {\rho }_3={\rho }_4,{\rho }_4\ast {\rho }_4={\rho }_5,{\rho }_4\ast {\rho }_5={\rho }_{12},{\rho }_4\ast {\rho }_{12}={\rho }_4$;

${\rho }_5\ast {\rho }_0={\rho }_5,{\rho }_5\ast {\rho }_1={\rho }_5,{\rho }_5\ast {\rho }_2={\rho }_5,{\rho }_5\ast {\rho }_3={\rho }_5,{\rho }_5\ast {\rho }_4={\rho }_{12},{\rho }_5\ast {\rho }_5={\rho }_4,{\rho }_5\ast {\rho }_{12}={\rho }_5$;

${\rho }_{12}\ast {\rho }_0={\rho }_{12},{\rho }_{12}\ast {\rho }_1={\rho }_{12},{\rho }_{12}\ast {\rho }_2={\rho }_{12},{\rho }_{12}\ast {\rho }_3={\rho }_{12},{\rho }_{12}\ast {\rho }_4={\rho }_4,{\rho }_{12}\ast {\rho }_5={\rho }_5,{\rho }_{12}\ast {\rho }_{12}={\rho }_{12}$.

Then, $M\left(X\right)$ is a monoid, and the operation ∗ on it is shown in

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

∗	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$	${\mathit{\rho }}_4$	${\mathit{\rho }}_5$	${\mathit{\rho }}_{12}$
${\rho }_0$	${\rho }_0$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_4$	${\rho }_5$	${\rho }_{12}$
${\rho }_1$	${\rho }_1$	${\rho }_1$	${\rho }_{12}$	${\rho }_1$	${\rho }_4$	${\rho }_5$	${\rho }_{12}$
${\rho }_2$	${\rho }_2$	${\rho }_{12}$	${\rho }_2$	${\rho }_1$	${\rho }_4$	${\rho }_5$	${\rho }_{12}$
${\rho }_3$	${\rho }_3$	${\rho }_1$	${\rho }_{12}$	${\rho }_2$	${\rho }_4$	${\rho }_5$	${\rho }_{12}$
${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_4$	${\rho }_5$	${\rho }_{12}$	${\rho }_4$
${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_5$	${\rho }_{12}$	${\rho }_4$	${\rho }_5$
${\rho }_{12}$	${\rho }_{12}$	${\rho }_{12}$	${\rho }_{12}$	${\rho }_{12}$	${\rho }_4$	${\rho }_5$	${\rho }_{12}$

. However, it is not commutative, since ${\rho }_3\ast {\rho }_2={\rho }_{12}\ne {\rho }_1={\rho }_2\ast {\rho }_3$.

Definition 16.

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

Definition 17.

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

Theorem 6.

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

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

Proof. 

$(\Rightarrow )$$\forall x,y\in X-\left\{0\right\}$, assume that $x\le y$, according to Definition 16, $y=x$ or $x=0$. However, $x\ne 0$. So, $x=y$.

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

Theorem 7.

Assume that $(X,\le ,\ast ,0)$ is a $BZ$-algebra, $K\left(X\right)$ is a $BCC$-part of X, and AG(X) is an anti-grouped part of X. Then, the below conditions are equivalent:

(1) X is QM-$BZ$-algebra;

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

Proof. 

(1)⇒(2) Assume that X is a QM-$BZ$-algebra. Then, $\forall x,y\in K\left(X\right)$, if $x=y$, $x\ast y=0$. If $x\ne y$, it can be divided into the below 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 3, $x=x\ast y$;

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 3, $x=x\ast y$;

Case 3: $x\ne 0$, $y\ne 0$, $(x\ast y)\ast x=\left(\right(x\ast y)\ast 0)\ast x=\left(\right(x\ast y)\ast (0\ast y\left)\right)\ast (x\ast 0)=0$, that is, $x\ast y\le x$. Because $x\ast y\ne 0$ and $x\ne 0$, according to Theorem 6, $x\ast y=x$.

According to Definition 4, $K\left(X\right)$ is a quasi-alter $BCK$-algebra. If $x\in X-K\left(X\right)$, then $0\ast x\ne 0$ and $0\ast (0\ast x)\ne 0$. Because $(0\ast (0\ast x\left)\right)\ast x=(0\ast x)\ast (0\ast x)=0$, that is, $0\ast (0\ast x)\le x$. According to Theorem 6, $0\ast (0\ast x)=x$. Thus, $(X-K(X\left)\right)\cup \left\{0\right\}\subseteq AG\left(X\right)$. Additionally, $AG\left(X\right)\subseteq (X-K(X\left)\right)\cup \left\{0\right\}$; then, $AG\left(X\right)=(X-K(X\left)\right)\cup \left\{0\right\}$.

(2)⇒(1) Let (2) hold; then, $\forall x,y\in X$, $x\ne y$:

Case 1: $x,y\in K\left(X\right)$, assume that $x\le y$, $x=x\ast y=0$, and it does not hold. So, $x=y$;

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

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

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

Case 3: $x\in K\left(X\right),y\in AG\left(X\right)$, assume that $x\le y$, that is, $x\ast y=0$. Then,

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

Clearly, $0\ast y=0$ does not hold. Then, $x=y$.

Case 4: $x\in AG\left(X\right),y\in K\left(X\right)$, assume that $x\le y$, that is, $x\ast y=0$. Then,

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

Clearly, $x=0$ does not hold. Then, $x=y$.

According to Theorem 6, X is a QM-$BZ$-algebra. □

According to Theorem 7, the $KG$-union of quasi-alter $BCK$-algebra and anti-grouped $BZ$-algebra is QM-$BZ$-algebra.

Theorem 8.

Assume that $(X,\le ,\ast ,0)$ is QM-$BZ$-algebra, that $K\left(X\right)$ is the $BCC$-part, and that $G\left(X\right)$ is the $BCC$-remainder. Then:

(1) $x\ast y=x$, $x\in G\left(X\right),y\in K\left(X\right)$;

(2) $x\ast y=0\ast y$, $x\in K\left(X\right),y\in G\left(X\right)$.

Proof. 

(1) $\forall x\in G\left(X\right),y\in K\left(X\right)$, assume that $x\ast y\in K\left(X\right)$. Then, $0=0\ast 0=0\ast (0\ast (x\ast y\left)\right)=(0\ast (0\ast x\left)\right)\ast (0\ast (0\ast y\left)\right)=x\ast 0=x$. Clearly, $x=0$ does not hold. So, $x\ast y\in G\left(X\right)$, that is, $x\ast y=0\ast (0\ast (x\ast y\left)\right)=(0\ast (0\ast x\left)\right)\ast (0\ast (0\ast y\left)\right)=x\ast 0=x$.

(2) $\forall x\in K\left(X\right),y\in G\left(X\right)$, assume that $x\ast y\in K\left(X\right)$. Then, $0=0\ast 0=0\ast (0\ast (x\ast y\left)\right)=(0\ast (0\ast x\left)\right)\ast (0\ast (0\ast y\left)\right)=0\ast y$. Clearly, $0\ast y=0$ does not hold. So, $x\ast y\in G\left(X\right)$, that is, $x\ast y=0\ast (0\ast (x\ast y\left)\right)=(0\ast (0\ast x\left)\right)\ast (0\ast (0\ast y\left)\right)=0\ast y$. □

Definition 18.

$BZ$-algebra <$X;\ast ,0$> is called generalized quasi-left alter $BZ$-algebra if it meets: $\forall x,y\in X$,

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

Theorem 9.

Let <$X,\ast ,0$> be a generalized quasi-left alter $BZ$-algebra. $\forall x\in X$, either $0\ast x\ne 0$, or $0\ast (0\ast x)=x$.

Proof. 

Let $0\ast x\ne 0$. Assume that $0\ast (0\ast x)=x$. Because $0\ast (0\ast (0\ast x\left)\right)=0\ast x$, then:

$$x\ast (x\ast (0\ast (0\ast x)\left)\right)=0\ast (0\ast (0\ast (0\ast x)\left)\right)=0\ast (0\ast x).$$

There is $x\ast (0\ast (0\ast x\left)\right)\ne x$, because if $x\ast (0\ast (0\ast x\left)\right)=x$, $0\ast (0\ast x)=x\ast (x\ast (0\ast (0\ast x)\left)\right)=x\ast x=0$. Then, $0\ast x=0\ast (0\ast (0\ast x\left)\right)=0\ast 0=0$, and this conflicts with assumption $0\ast x\ne 0$. So, $x\ast (0\ast (0\ast x\left)\right)\ne x$.

According to Definition 18:

$$\begin{array}{ccc}\hfill x\ast (x\ast (x\ast (0\ast (0\ast x\left)\right)\left)\right)& =& 0\ast (0\ast (x\ast (0\ast (0\ast x\left)\right)\left)\right)\hfill \\ & =& (0\ast (0\ast x\left)\right)\ast (0\ast (0\ast (0\ast (0\ast x\left)\right)\left)\right)\hfill \\ & =& (0\ast (0\ast x\left)\right)\ast (0\ast (0\ast x\left)\right)=0.\hfill \end{array}$$

Therefore, $x\ast (x\ast (x\ast (0\ast (0\ast x\left)\right)\left)\right)=0$. In addition, $x\ast (x\ast (x\ast (0\ast (0\ast x\left)\right)\left)\right)=x\ast (0\ast (0\ast x\left)\right)=0$, $0\ast (0\ast x)=x\ast (x\ast (0\ast (0\ast x)\left)\right)=x\ast 0=x$. This conflicts with assumption $0\ast (0\ast x)\ne x$, so $0\ast (0\ast x)=x$. □

According to Theorem 9, let $K\left(X\right)$ be the $BCC$-part of a generalized quasi-left alter $BZ$-algebra, $G\left(X\right)$ be $BCC$-remainder of a generalized quasi-left alter $BZ$-algebra. Then, $G\left(X\right)\cup \left\{0\right\}$ is an anti-grouped $BZ$-subalgebra.

Theorem 10.

Assume that <$X,\ast ,0$> is a generalized quasi-left alter $BZ$-algebra. Therefore, X is $BCI$-algebra.

Proof. 

Let $K\left(X\right)$ be the $BCC$-part of X and $G\left(X\right)$ be the $BCC$-remainder of X. $\forall x,y\in X$:

If $x=y$, $(x\ast (x\ast y\left)\right)\ast y=(x\ast (x\ast x\left)\right)\ast x=(x\ast 0)\ast x=x\ast x=0$, that is, $x\ast (x\ast y)\le y$.

If $x\ne y$, there are two cases according to Theorem 9:

Case 1: $y\in K\left(X\right)$, $(x\ast (x\ast y\left)\right)\ast y=(0\ast (0\ast y\left)\right)\ast y=0\ast y=0$, that is, $x\ast (x\ast y)\le y$;

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

To sum up, $\forall x,y\in X$, $x\ast (x\ast y)\le y$. According to Theorem 1, X is $BCI$-algebra. □

Theorem 11.

Assume that <$X,\ast ,0$> is a generalized quasi-left alter $BZ$-algebra. Then, X is a QM-$BZ$-algebra.

Proof. 

Let $K\left(X\right)$ be the $BCC$-part of X, and $G\left(X\right)$ be the $BCC$-remainder of X. Then, $\forall x,y\in X$, assume that $x\le y$ and $x\ne y$. Therefore,

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

(1) When $y\in K\left(X\right)$, $x=0\ast (0\ast y)=0\ast 0=0$.

(2) When $y\in G\left(X\right)$, $x=0\ast (0\ast y)=y$, but $x\ne y$. So, y is a quasi-minimal element of X. Because y is arbitrary, X is a QM-$BZ$-algebra. □

In the following, the concepts of QM-hyper $BZ$-algebra and generalized quasi-left alter hyper $BZ$-algebra are shown.

Definition 19.

Let $(H,\ll )$ be a partial order that contains a constant 0 in a hyper structure. x is said to be a quasi-minimal element in H, if for any element a in H, $a\ll x$ implies $x=a$ or $a=0$.

Definition 20.

A hyper $BZ$-algebra $(H,\ll ,\circ ,0)$ is said to be QM-hyper $BZ$-algebra if all elements of H are quasi-minimal elements.

Theorem 12.

Assume that $(H,\ll ,\circ ,0)$ is a hyper $BZ$-algebra. Then, H is a QM-hyper $BZ$-algebra iff it meets: for all $x,y\in H-\left\{0\right\}$,

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

Proof. 

$(\Rightarrow )$ For all $x,y\in H-\left\{0\right\}$, let $x\ll y$, according to Definition 19, $y=x$ or $x=0$. However, $x\ne 0$. So, $x=y$.

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

Definition 21.

Hyper $BZ$-algebra $(H,\circ )$ is called a generalized quasi-left alter hyper $BZ$-algebra if, $\forall x,y\in H$,

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

Theorem 13.

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

Proof. 

Assume that $(H,\circ )$ is a generalized quasi-left alter hyper $BZ$-algebra. Let $B\left(H\right)$ be hyper $BCC$-part of H. $\forall x\in B\left(H\right)$ and $x\ne 0$, $0\circ (x\circ 0)=(0\circ 0)\circ (x\circ 0)\ll 0\circ x=0$, according to Proposition 1, $0\circ (x\circ 0)=0$. According to Definition 21, $x\circ (x\circ 0)=0\circ (0\circ 0)=0$. Assume that $x\ne x\circ 0$, $x\circ 0=x\circ (x\circ (x\circ 0\left)\right)=0\circ (0\circ (x\circ 0\left)\right)=0\circ 0=0$; clearly, $x\ne x\circ 0$ is not true. So, $x=x\circ 0$.

If $x\notin B\left(H\right)$, and $0\circ (x\circ 0)\ne 0$, there is $x\circ (x\circ 0)=0\circ (0\circ 0)=0$. Assume that $x\ne x\circ 0$, $x\circ 0=x\circ (x\circ (x\circ 0\left)\right)=0\circ (0\circ (x\circ 0\left)\right)$. So, $(x\circ 0)\circ (x\circ 0)=(0\circ (0\circ (x\circ 0)\left)\right)\circ (0\circ (0\circ (x\circ 0)\left)\right)\ll 0\circ 0=0$, according to Proposition 1, $(x\circ 0)\circ (x\circ 0)=0$. Therefore, $0\circ \left(\right(x\circ 0)\circ x)\subset (x\circ x)\circ \left(\right(x\circ 0)\circ x)\ll x\circ (x\circ 0)=0$. That is, $0\circ \left(\right(x\circ 0)\circ x)=0$. Beasuse $0\circ x\ne 0$, $0\circ (x\circ 0)\ne 0$, so $0\circ \left(\right(x\circ 0)\circ x)=0$ is not true. That is, $x\ne x\circ 0$. So, $x=x\circ 0$.

Above all, $\forall x\in H$, $x=x\circ 0$. Moreover, $x\circ x=x\circ (x\circ 0)=0\circ (0\circ 0)=0$, according to Proposition 1, $(H,\circ )$ is $BZ$-algebra. According to Theorem 10, $(H,\circ )$ is $BCI$-algebra. □

Theorem 14.

Let $(H,\circ )$ be generalized quasi-left alter hyper $BZ$-algebra. Then, H is a QM-hyper $BZ$ algebra.

Proof. 

Assume that $(H,\circ )$ be a generalized quasi-left alter hyper $BZ$-algebra. According to Theorem 13, H is $BCI$-algebra. Let $K\left(H\right)$ be the $BCC$-part of H, and G(H) be the $BCC$-remainder of H. Then, $\forall x,y\in H$, let $x\ll y$ and $x\ne y$. Therefore,

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

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

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

However, not every QM-hyper $BZ$-algebra is generalized quasi-left alter hyper $BZ$-algebra; see Example 2.

Example 2.

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

Table 3. QM-hyper $BZ$-algebra.

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

.

Clearly, $(H,\circ )$ is QM-hyper $BZ$-algebra, but it is not a generalized quasi-left alter hyper $BZ$-algebra, since $2\circ (2\circ 0)=\{0,2\}$, $0\circ (0\circ 0)=0$, $2\ne 0$.

Additionally, Example 2 shows that not every QM-hyper $BZ$-algebra is QM-$BZ$-algebra.

4. Quasi-Hyper BZ-Algebras

In this part, we propose the definition of quasi-hyper BZ-algebras. In the following, we replace the singleton set $\left\{x\right\}$ with x.

Definition 22.

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

$\left(QHZ1\right)$$(x\circ z)\circ (y\circ z)\ll x\circ y;$

$\left(QHZ2\right)$$x\ll x;$

$\left(QHZ3\right)$$x\ll y$ and $y\ll x\Rightarrow x=y;$

$\left(QHZ4\right)$$x\ll x\circ 0;$

$\left(QHZ5\right)$$x\ll 0\Rightarrow x=0$,

then it is a quasi-hyper $BZ$-algebra.

Remark 1.

$\left(1\right)$ Every BZ-algebra is a quasi-hyper BZ-algebra;

$\left(2\right)$ Every hyper $BZ$-algebra is a quasi-hyper BZ-algebra;

$\left(3\right)$ Every quasi-hyper $BCI$-algebra is a quasi-hyper $BZ$-algebra.

Now, we give some examples about quasi-hyper $BZ$-algebras and some examples which explain that not every quasi-hyper $BZ$-algebra is a hyper $BZ$-algebra.

Example 3.

$\left(1\right)$ Assume that $H=\{0,1,2,3,4\}$. Define an operation ∗ on H in

Table 4. $BZ$-algebra.

∗	0	1	2	3	4
0	0	0	2	2	0
1	1	0	3	2	0
2	2	2	0	0	2
3	3	2	1	0	2
4	4	4	3	3	0

,

Then, $(H,\ast ,0)$ is a $BZ$-algebra, and it is a quasi-hyper $BZ$-algebra.

$\left(2\right)$ Assume that $H=\{0,1,2,3,4\}$. Define an operation ∘ on H in

Table 5. Quasi-hyper $BZ$-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	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$

,

Then, $(H,\circ ,0)$ is a quasi-hyper $BZ$-algebra, but it is not a hyper $BZ$-algebra since $0\circ (0\circ 0)=\{0,1\}$ and $0\notin 1\circ 0(i.e.,1\ll 0$ is not true); this means that the condition (HZ4) in Definition 13 does not hold.

$\left(3\right)$ Assume that $H=\{0,1,2,3,4\}$. Define an operation ∘ on H in

Table 6. Quasi-hyper $BZ$-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	$\{2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$

,

Then $(H,\circ ,0)$ is a quasi-hyper $BZ$-algebra, but it is not a quasi-hyper $BCI$-algebra, since $(1\circ 0)\circ 1=\{2,3,4\}$, $(1\circ 1)\circ 0=\{0,1,2,3,4\}$ and $0\ne 1$.

Proposition 3.

In any quasi-hyper BZ-algebra $(H,\circ )$, the followings hold: for all $x,y,z\in H$ and for all non-empty subsets A and B of H:

$\left(1\right)$$0\circ (x\circ y)\ll y\circ x$;

$\left(2\right)$$A\ll A$;

$\left(3\right)$$A\subseteq B\Rightarrow A\ll B$;

$\left(4\right)$$A\ll 0\Rightarrow A=0$;

$\left(5\right)$$x\circ y=0\Rightarrow (x\circ z)\circ (y\circ z)=0$ and $x\circ z\ll y\circ z$;

$\left(6\right)$$A\circ 0=0\Rightarrow A=0$;

$\left(7\right)$$x\ll y\Rightarrow 0\ll y\circ x$;

$\left(8\right)$$x\circ x=0\Rightarrow |x\circ y|=1$.

Proof. 

(1) By (QHZ1) and (QHZ2), $0\circ (x\circ y)\subseteq (y\circ y)\circ (x\circ y)\ll y\circ x$. Then, $0\circ (x\circ y)\ll y\circ x$.

(2) By (QHZ2), for any $x\in A$, there is $x\ll x$, that is $0\in x\circ x$. Then, $A\ll A$.

(3) Let $a\in A$. Then, $a\in B$. By (QHZ2), $x\ll x$ 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 (QHZ1), $(x\circ z)\circ (y\circ z)\ll x\circ y=\left\{0\right\}$. By (4), $(x\circ z)\circ (y\circ z)=\left\{0\right\}$. So, $x\circ z\ll y\circ z$.

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

(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) For any $x\in H$, let $x\circ x=\left\{0\right\}$. $\forall y\in H$, assume that $|x\circ y|>1$, let $a,b\in x\circ y$, and $a\ne b$. Then,

$$a\circ b\subseteq (x\circ y)\circ (x\circ y)\ll x\circ x=0\mathrm{and}b\circ a\subseteq (x\circ y)\circ (x\circ y)\ll x\circ x=0;$$

thus, $a\circ b\ll 0$, $b\circ a\ll 0$, and $a\ll b,b\ll a$. Thus, $a=b$, and so $|x\circ y|=1$. □

Proposition 4.

In any quasi-hyper $BZ$-algebra $(H,\circ )$ satisfying $0\circ 0=0$, the followings hold: $\forall x\in H$,

$\left(1\right)$$(0\circ x)\circ (0\circ x)=0$;

$\left(2\right)$$0\circ x$ is a singleton set;

$\left(3\right)$$(0\circ x)\circ 0=0\circ x$.

Proof. 

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

$\left(2\right)$ For any $a,b\in 0\circ x$, and $a\ne b$. $a\circ b\subseteq (0\circ x)\circ (0\circ x)=0$, $b\circ a\subseteq (0\circ x)\circ (0\circ x)=0$, by (QHZ3), $a\ll b,b\ll a$, so $a=b$. Thus, $0\circ x$ is a singleton set.

$\left(3\right)$ By (2), let $0\circ x=m$. By(QHZ4), $m\ll m\circ 0$ 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 $a\ll b$, $b\ll a$. In addition, $a=b$. So, $|m\circ 0|=1$. Because $m\circ 0=(0\circ x)\circ 0\subseteq (0\circ x)\circ (x\circ x)\ll 0\circ x=m$, and $|m\circ 0|=1$, $m=m\circ 0$. That is, $(0\circ x)\circ 0=0\circ x$. □

In the following, we give the concepts of standard quasi-hyper $BZ$-algebra and transitive quasi-hyper $BZ$-algebra.

Definition 23.

A quasi-hyper BZ algebra $(H,\circ )$ is called a standard quasi-hyper BZ algebra if, $\forall x\in H$, it satisfies $x\circ 0=x$.

Proposition 5.

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

Proof. 

Assume that $(H,\circ )$ is a standard quasi-hyper $BZ$ algebra. For all $x\in H$, $0\circ (0\circ x)\subseteq (x\circ x)\circ (0\circ x)\ll x\circ 0=x$. That is, $0\circ (0\circ x)\ll x$. So, $(H,\circ )$ is a hyper $BZ$-algebra. □

Definition 24.

A quasi-hyper BZ-algebra $(H,\circ )$ is called a transitive quasi-hyper BZ algebra if, $\forall x,y,z\in H$, it satisfies $x\ll y$ and $y\ll z\Rightarrow x\ll z$.

Proposition 6.

In any transitive quasi-hyper BZ-algebra $(H,\circ )$, the following conditions hold: for all $x,y,z,u\in H$, and for all non-empty subsets $A,B$, and C of H:

$\left(TQHZ1\right)$$A\ll B$ and $B\ll C$ imply $A\ll C$;

$\left(TQHZ2\right)$$x\circ y\ll z$ implies $(x\circ u)\circ (y\circ u)\ll z$.

Proof. 

(TQHZ1) Let $a\in A$. $\exists b\in B$ s.t. $a\ll b$. Additionally, for any $b\in B$, $\exists c\in C$ such that $b\ll c$. So, $a\ll c$. Then, for any $a\in A$, $\exists c\in C$ s.t. $a\ll c$, that is $A\ll C$.

(TQHZ2) By (QHZ1), $(x\circ u)\circ (y\circ u)\ll x\circ y$, and $x\circ y\ll z$. So, $(x\circ u)\circ (y\circ u)\ll z$. □

Example 4.

$\left(1\right)$ Assume that $H=\{0,1,2,3,4\}$. Define an operation ∘ on H in

Table 7. Transitive quasi-hyper $BZ$-algebra.

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

,

Then, $(H,\circ )$ is a transitive quasi-hyper $BZ$-algebra and a transitive hyper $BZ$-algebra.

$\left(2\right)$ Assume that $H=\{0,1,2,3,4\}$. Define an operation ∘ on H in

Table 8. Transitive quasi-hyper $BZ$-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	4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	4	4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	4	4	4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	4	4	4	4	$\{0,1,2,3,4\}$

,

Then, $(H,\circ )$ is a transitive quasi-hyper $BZ$-algebra, but it is not standard.

Proposition 7.

Every transitive standard quasi-hyper $BZ$-algebra is a hyper $BZ$-algebra.

Proof. 

According to Proposition 5. □

In the following, we investigate some kinds of quasi-hyper $BZ$-subalgebra.

Definition 25.

Assume that $(H,\circ )$ is a quasi-hyper $BZ$-algebra, that S is a subset of H, and that S contains 0. If S is a quasi-hyper $BZ$-algebra under the hyper operation "∘" on H, then S is a hyper subalgebra of H.

Proposition 8.

Assume that S is a nonempty subset of a quasi-hyper $BZ$-algebra $(H,\circ )$. If $\forall x,y\in S$ $x\circ y\subseteq S$, there is $0\in S$.

Proof. 

Let $\forall x,y\in S$, $x\circ y\subseteq S$ and $a\in S$. Because $a\ll a$, there is $0\in a\circ a\subseteq S$. □

Theorem 15.

Assume that S is a nonempty subset of a quasi-hyper $BZ$-algebra $(H,\circ )$. S is a hyper subalgebra of H iff $\forall x,y\in S$, $x\circ y\subseteq S$.

Proof. 

(⇒) This is clear.

(⇐) $\forall x,y\in S$, assume that $x\circ y\subseteq S$. By Proposition 8, $0\in S$. $\forall x,y,z\in S$, there is $x\circ z\subseteq S$, $y\circ z\subseteq S$, and $x\circ y\subseteq S$. Thus,

$$(x\circ z)\circ (y\circ z)=\bigcup _{a\in x\circ z,b\in y\circ z}a\circ b\subseteq S.$$

Because $S\subseteq H$, $(x\circ z)\circ (y\circ z)\ll x\circ y\subseteq S.$ Therefore, in S, (QHZ1) holds. In a similar way, (QHZ2), (QHZ3), (QHZ4), and (QHZ5) hold in S. So, S is a hyper subalgebra of H. □

Theorem 16.

Assume that $(H,\circ )$ is a quasi-hyper $BZ$-algebra. Then,

$$S_I:=\{m\in H|m\circ m=0\}$$

is a hyper subalgebra of H and $\forall x,y\in S_I$, $x\circ y$ is a singleton set whenever $S_I\ne ⌀$.

Proof. 

Assume that $x,y\in S_I$ and $a\in x\circ y$. Then, there is $(x\circ y)\circ (x\circ y)\ll x\circ x=0$ and $(x\circ y)\circ (x\circ y)=0$ and $a\circ a\subseteq (x\circ y)\circ (x\circ y)=0$. Therefore, $x\circ y\subseteq S_I$. By Theorem 15, $S_I$ is a hyper subalgebra of H.

According to Proposition 3 (8), $\forall x,y\in S_I$, $x\circ y$ is a singleton set.

However, $S_I$ is not necessarily a $BZ$-algebra; see Example 5.

Example 5.

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

Table 9. Quasi-hyper $BZ$-algebra.

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

,

Then, $(H,\circ )$ is a quasi-hyper $BZ$-algebra, and $S_I=\{0,2,3\}$ is a quasi-hyper $BZ$-subalgebra. But it is not a $BZ$-algebra, since $3\circ 0=2\ne 3$.

Theorem 17.

Assume that $(H,\circ )$ is a standard quasi-hyper $BZ$-algebra. Then,

$$BCC\left(H\right):=\{m\in H|0\circ m=0\},$$

is a hyper $BCC$-subalgebra of H.

Theorem 18.

Assume that $(H,\circ )$ is a transitive standard quasi-hyper $BZ$-algebra. Then,

$$AG\left(H\right):=\{m\in H|0\circ (0\circ m)=m\},$$

is a quasi-hyper $BZ$-subalgebra of H and an anti-grouped $BZ$-algebra.

Proof. 

Obviously, $0\in AG\left(H\right)$. Let $x,y\in AG\left(H\right)$. According to Proposition 7 and (TSHZ4), $0\circ (0\circ (x\circ y\left)\right)=(0\circ (0\circ x\left)\right)\circ (0\circ (0\circ y\left)\right)$, then $0\circ (0\circ (x\circ y\left)\right)=x\circ y$. So, $x\circ y\subseteq AG\left(H\right)$. By Theorem 15, $AG\left(H\right)$ is a hyper subalgebra of H. $\forall x\in AG\left(H\right)$, by Proposition 4 (2), $0\circ x$ is a singleton set, □$x\circ x=(0\circ (0\circ x\left)\right)\circ (0\circ (0\circ x\left)\right)\ll 0\circ 0=0$. So, for any $x,y\in AG\left(H\right)$, $|x\circ y|=1$. Because $AG\left(H\right)$ is a standard quasi-hyper $BZ$-algebra, it is a $BZ$-algebra. According to Definition 5, we know $\left(AG\right(H),\circ )$ is an anti-grouped $BZ$-algebra. □

In the following, we study the connection between quasi-hyper $BZ$-algebra and semigroups.

Assume that $(H,\circ )$ is a quasi-hyper $BZ$-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 a nonempty subset of H.

$\forall a,b\in H$, $\forall x\in H$, denote ${\rho }_a\circ {\rho }_b$:

$$({\rho }_a\circ {\rho }_b)\left(x\right)=\bigcup _{\forall y\in {\rho }_b\left(x\right)}{\rho }_a\left(y\right),$$

where ∘ means the composition operation of mappings.

Theorem 19.

Denote $M\left(H\right)$ as a set which is all compositional results of finite mappings which are for all $a\in H$; therefore, ${\rho }_a$. Then, $M\left(H\right)$ is a semigroup.

Proof. 

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

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

So, $({\rho }_a\circ ({\rho }_b\circ {\rho }_c))\left(x\right)=(({\rho }_a\circ {\rho }_b)\circ {\rho }_c)\left(x\right)$. Then, $M\left(H\right)$ satisfies associative law. □

Example 6.

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

Table 10. Quasi-hyper $BZ$-algebra.

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

,

Then, $(H,\circ )$ is a quasi-hyper $BZ$-algebra, and $M\left(H\right)=\{{\rho }_0,{\rho }_1,{\rho }_2,{\rho }_3,{\rho }_2^2\}$, where ${\rho }_2^2={\rho }_2\circ {\rho }_2$.

We can verify the following:

${\rho }_0\circ {\rho }_0={\rho }_0,{\rho }_0\circ {\rho }_1={\rho }_1,{\rho }_0\circ {\rho }_2={\rho }_2,{\rho }_0\circ {\rho }_3={\rho }_3,{\rho }_0\circ {\rho }_2^2={\rho }_2^2$;

${\rho }_1\circ {\rho }_0={\rho }_1,{\rho }_1\circ {\rho }_1={\rho }_1,{\rho }_1\circ {\rho }_2={\rho }_3,{\rho }_1\circ {\rho }_3={\rho }_3,{\rho }_1\circ {\rho }_2^2={\rho }_2^2$;

${\rho }_2\circ {\rho }_0={\rho }_2,{\rho }_2\circ {\rho }_1={\rho }_2,{\rho }_2\circ {\rho }_2={\rho }_2^2,{\rho }_2\circ {\rho }_3={\rho }_2^2,{\rho }_2\circ {\rho }_2^2={\rho }_3$;

${\rho }_3\circ {\rho }_0={\rho }_3,{\rho }_3\circ {\rho }_1={\rho }_3,{\rho }_3\circ {\rho }_2={\rho }_2^2,{\rho }_3\circ {\rho }_3={\rho }_2^2,{\rho }_3\circ {\rho }_2^2={\rho }_3$;

${\rho }_2^2\circ {\rho }_0={\rho }_2^2,{\rho }_2^2\circ {\rho }_1={\rho }_2^2,{\rho }_2^2\circ {\rho }_2={\rho }_3,{\rho }_2^2\circ {\rho }_3={\rho }_3,{\rho }_2^2\circ {\rho }_2^2={\rho }_2^2$.

Then $\left(M\right(H),\circ )$ is a semigroup, but it is not commutative, since ${\rho }_1\circ {\rho }_2={\rho }_3\ne {\rho }_2={\rho }_2\circ {\rho }_1$. The operation on $M\left(H\right)$ is shown in

Table 11. The adjoint semigroup of quasi-hyper $BZ$-algebra.

∘	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$	${\mathit{\rho }}_2^2$
${\rho }_0$	${\rho }_0$	${\rho }_1$	${\rho }_2$	${\rho }_3$	${\rho }_2^2$
${\rho }_1$	${\rho }_1$	${\rho }_1$	${\rho }_3$	${\rho }_3$	${\rho }_2^2$
${\rho }_2$	${\rho }_2$	${\rho }_2$	${\rho }_2^2$	${\rho }_2^2$	${\rho }_3$
${\rho }_3$	${\rho }_3$	${\rho }_3$	${\rho }_2^2$	${\rho }_2^2$	${\rho }_3$
${\rho }_2^2$	${\rho }_2^2$	${\rho }_2^2$	${\rho }_3$	${\rho }_3$	${\rho }_2^2$

.

In the following, we give the concepts of anti-grouped quasi-hyper $BZ$-algebra and generalized anti-grouped quasi-hyper $BZ$-algebra. Moreover, we study the relations between quasi-hyper $BZ$-algebra and semihypergroups by generalized anti-grouped quasi-hyper $BZ$-algebra and associative quasi-hyper $BZ$-algebra.

Definition 26.

A quasi-hyper $BZ$-algebra $(H,\circ )$ is called an anti-grouped quasi-hyper $BZ$-algebra if, $\forall x,y,z\in H$, it satisfies $(x\circ z)\circ (y\circ z)=x\circ y$.

Proposition 9.

Assume that $(H,\circ )$ is an anti-grouped quasi-hyper $BZ$-algebra. Therefore, it is an anti-grouped $BZ$-algebra.

Proof. 

Assume that $|0\circ 0|>1$, let $0\circ 0=\{0,m\}$, and $m\ne 0$. According to Definition 26, $(0\circ 0)\circ (0\circ 0)=0\circ 0$, and $(0\circ 0)\circ (0\circ 0)=\{0\circ 0,0\circ m,m\circ 0,m\circ m\}$, $0\circ 0=\{0,m\}$. Because $m\circ 0\ne 0$, $m\circ 0=m$. Then, $(m\circ 0)\circ (0\circ 0)=m\circ (0\circ 0)=\{m\circ 0,m\circ m\}$, $m\circ 0=m$, according to Definition 26, $\{m\circ 0,m\circ m\}=m$. So, $0\in m\circ m=m$, but $m\ne 0$. So, $|0\circ 0|=1$ and $0\circ 0=0$. According to Proposition 4, $\forall x\in H$, $0\circ x$ is singleton set.

$\forall x,y\in H$, assume that $x\ll y$, that is, $0\in x\circ y$. Then, $0=0\circ 0=(y\circ y)\circ (x\circ y)=y\circ x$, and $y\ll x$. According to Definition 22, $x=y$. Then, $x\ll y$ implies $x=y$. By Definition 22, $x\ll x\circ 0$, then $x\circ 0=x$. Then, $0\circ (0\circ x)\subset (x\circ x)\circ (0\circ x)=x\circ 0=x$. So, $0\circ (0\circ x)=x$, because $0\circ x$ is singleton set. Therefore, $x\circ x=(0\circ (0\circ x\left)\right)\circ (0\circ (0\circ x\left)\right)=0$. According to Proposition 3, $\forall x,y,z\in H$, $\left(\right(x\circ z)\circ (y\circ z\left)\right)\circ (x\circ y)=0$, and $x=x\circ 0$. So, H is anti-grouped $BZ$-algebra. □

Definition 27.

A quasi-hyper $BZ$-algebra $(H,\circ )$ is called a generalized anti-grouped quasi-hyper $BZ$-algebra if, $\forall x,y,z\in H$, it satisfies $(x\circ (0\circ y\left)\right)\circ (0\circ z)=x\circ (0\circ (y\circ (0\circ z)\left)\right)$.

Remark 2.

Every generalized anti-grouped hyper $BZ$-algebra is a generalized anti-grouped quasi-hyper $BZ$-algebra.

The following example shows that not every generalized anti-grouped quasi-hyper $BZ$-algebra is a generalized anti-grouped hyper $BZ$-algebra.

Example 7.

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

Table 12. Generalized anti-grouped quasi-hyper $BZ$-algebra.

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

,

Then, $(H,\circ )$ is a generalized anti-grouped quasi-hyper $BZ$-algebra. However, it is not a hyper $BZ$-algebra, since $0\circ (0\circ 0)=\{0,1\}$ and $1\ll 0$ is not true.

Proposition 10.

In any generalized anti-grouped quasi-hyper $BZ$-algebra $(H,\circ )$, $\forall x,y\in H$, define “⊕”:

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

Then, $(H,\oplus ,0)$ is a semihypergroup.

Proof. 

$\forall x,y,z\in H$, $(x\oplus y)\oplus z=(x\circ (0\circ y\left)\right)\circ (0\circ z)=x\circ (0\circ (y\circ (0\circ z)\left)\right)=x\oplus (y\oplus z)$. Then, $(H,\oplus ,0)$ is a semihypergroup. □

Example 8.

$\left(1\right)$ Assume that $H=\{0,1,2,3,4\}$. An operation ∘ on H is shown in

Table 13. Generalized anti-grouped quasi-hyper $BZ$-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	4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
2	4	4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	4	4	4	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	$\{1,2,3,4\}$	$\{2,3,4\}$	$\{3,4\}$	4	$\{0,1,2,3,4\}$

,

Then, $(H,\circ )$ is a generalized anti-grouped quasi-hyper $BZ$-algebra. According to Proposition 10, we get a semihypergroup $(H,\oplus )$ and the operation ⊕ on it is shown in

Table 14. Semihypergroup derived from Table 13.

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

.

$\left(2\right)$ Assume that $H=\{0,1,2,3\}$. An operation ∘ on H in shown in

Table 15. Generalized anti-grouped quasi-hyper $BZ$-algebra.

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

.

Then $(H,\circ )$ is a generalized anti-grouped quasi-hyper $BZ$-algebra. According to Proposition 10, we get a semihypergroup $(H,\oplus )$ and the operation ⊕ on it is shown in

Table 16. Semihypergroup derived from Table 15.

⊕	0	1	2	3
0	0	0	0	0
1	$\{1,2\}$	$\{1,2\}$	$\{1,2\}$	$\{1,2\}$
2	2	2	2	2
3	3	3	3	3

.

Definition 28.

A quasi-hyper $BZ$-algebra $(H,\circ )$ is called associative if, $\forall x,y,z\in H$, it satisfies $x\circ (y\circ z)=(x\circ y)\circ z$.

Remark 3.

Every associative quasi-hyper $BZ$ algebra is a generalized anti-grouped quasi-hyper $BZ$ algebra.

Example 7 shows that not every generalized anti-grouped quasi-hyper $BZ$ algebra is an associative quasi-hyper $BZ$ algebra, since $1\circ (0\circ 0)=\{0,1\}\ne 1=(1\circ 0)\circ 0$.

Proposition 11.

Assume that $(H,\circ )$ is an associative quasi-hyper $BZ$-algebra. Then, the below conditions hold: $\forall x\in H$,

$\left(1\right)$$0\circ 0=0$;

$\left(2\right)$$(0\circ x)\circ (0\circ x)=0$;

$\left(3\right)$$0\circ x$ is a singleton set;

$\left(4\right)$$(0\circ x)\circ 0=0\circ x$;

$\left(5\right)$$0\circ x=x$;

$\left(6\right)$$x\circ x=0$;

$\left(7\right)$$x\circ 0=x$.

Proof. 

(1) Assume that $0\circ 0\ne 0$, and let $0\circ 0=\{0,m\}$ and $m\ne 0$. Therefore, $0\in m\circ m\subseteq m\circ (0\circ 0)=(m\circ 0)\circ 0$. So, $\exists p\in m\circ 0$ s.t. $0\in p\circ 0$. That is, $p\ll 0$, and by (QHZ5), $p=0$. Therefore, $0\in m\circ 0$. That is, $m\ll 0$ and $m=0$. So, $m\ne 0$ is not true, and $0\circ 0=0$.

(2) This follows from Proposition 4.

(3) This follows from Proposition 4.

(4) This follows from Proposition 4.

(5) By (2) and (4), for any $x\in H$, $0\in 0\circ 0\subseteq 0\circ (x\circ x)=(0\circ x)\circ x=\left(\right(0\circ x)\circ 0)\circ x=(0\circ x)\circ (0\circ x)=0$. Because $0\circ x$ is a singleton set, $0\circ x\ll x$. For any $x\in H$, $0\in (x\circ 0)\circ (x\circ 0)=x\circ \left(\right(0\circ x)\circ 0)=x\circ (0\circ x)$, so $x\ll 0\circ x$. By (QHZ3), $x=0\circ x$.

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

(7) According to (6) and Proposition 3 (8), $\forall x,y\in H$, $|x\circ y|=1$. Then, $x\circ 0=(0\circ x)\circ (x\circ x)\ll 0\circ x=x$. By (QHZ4), $x\ll x\circ 0$, and by (QHZ3), $x=x\circ 0$. □

Theorem 20.

Assume that $(H,\circ )$ is an associative quasi-hyper $BZ$-algebra. Then, it is an associative $BCI$-algebra.

Proof. 

By Propositions 11 (6) and 3 (8), $\forall x,y,z\in H$, $\left(\right(x\circ y)\circ (x\circ z\left)\right)\circ (z\circ y)=(x\circ y)\circ (x\circ (z\circ z)\circ y)=(x\circ y)\circ \left(\right(x\circ 0)\circ y)=(x\circ y)\circ (x\circ y)=0$, and Definition 1 (1) holds. $(x\circ (x\circ y\left)\right)\circ y=(x\circ x)\circ (y\circ y)=0\circ 0=0,$ and Definition 1 (2) holds. Obviously, $x\circ x=0$ and Definition 1 (3) holds. Clearly, Definition 1 (4) holds. So, $(H,\circ )$ is an associative $BCI$-algebra. □

In the following, we give the concepts of generalized quasi-left alter quasi-hyper $BZ$-algebra and QM-quasi-hyper $BZ$-algebra.

Definition 29.

A quasi-hyper $BZ$-algebra $(H,\ll ,\circ ,0)$ is called QM-quasi-hyper $BZ$-algebra if every element of H is a quasi-minimal element.

Theorem 21.

Assume that $(H,\ll ,\circ ,0)$ is a quasi-hyper $BZ$-algebra. Then, H is a QM-quasi-hyper $BZ$-algebra iff it meets: $\forall x,y\in H-\left\{0\right\}$,

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

Proof. 

The proof is similar to Theorem 12. □

According to Remark 1, we know that both $BZ$-algebra and hyper $BZ$-algbera are quasi-hyper $BZ$-algebra. So, both QM-$BZ$-algebra and QM-hyper $BZ$-algebra are QM-quasi-hyper $BZ$-algebra, but not every QM-quasi-hyper $BZ$-algebra is QM-$BZ$-algebra and QM-hyper $BZ$-algebra (see Example 9).

Example 9.

Assume that $H=\{0,1,2\}$. An operation on H is shown in

Table 17. QM-quasi-hyper $BZ$-algebra.

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

.

Then, $(H,\circ )$ is a QM-quasi-hyper $BZ$-algebra, but it is not a QM-hyper $BZ$-algebra since $0\circ (0\circ 0)=\{0,2\}$,$2\ll 0$ is not true. Moreover, it is not a QM-$BZ$-algebra.

Definition 30.

Quasi-hyper $BZ$-algebra $(H,\circ )$ is called a generalized quasi-left alter quasi-hyper $BZ$ algebra if, $\forall x,y\in H$,

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

Example 10.

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

Table 18. Generalized quasi-center alter quasi-hyper $BZ$-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	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
3	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{0,1,2,3,4\}$	$\{0,1,2,3,4\}$
4	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{1,2,3,4\}$	$\{2,3,4\}$	$\{0,1,2,3,4\}$

,

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

Proposition 12.

Let $(H,\circ )$ be a generalized quasi-left alter quasi-hyper $BZ$-algebra satisfying $0\circ 0=0$. Therefore, H is a $BCI$-algebra.

Proof. 

The proof is similar to Theorem 13. □

According to Theorems 11 and 14, we know generalized quasi-left alter $BZ$-algebra is QM-$BZ$-algebra and generalized quasi-left alter hyper $BZ$-algebra is QM-hyper $BZ$-algebra. However, not every generalized quasi-left alter quasi-hyper $BZ$-algebra is QM-quasi-hyper $BZ$-algebra; see Example 10.

In addition, not every QM-quasi-hyper $BZ$ algebra is a generalized quasi-left alter quasi-hyper $BZ$-algbera; see Example 11.

Example 11.

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

Table 19. Generalized quasi-center alter quasi-hyper $BZ$-algebra.

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

.

Then, $(H,\circ )$ is QM-quasi-hyper $BZ$-algebra, but it is not a generalized quasi-left alter quasi-hyper $BZ$-algebra since $1\circ (1\circ 0)=\{0,1,2,3,4\}$, $0\circ (0\circ 0)=0$ and $1\ne 0$.

5. Discussion

In this paper, we firstly proposed a construction theorem of $BZ$-algebra through $BCC$-algebra and anti-grouped $BZ$-algebra. Secondly, the definition of QM-$BZ$-algebra was proposed and their properties were studied. Moreover, the structure of QM-$BZ$-algebra was obtained: each QM-$BZ$-algebra is a KG-union of quasi-alter $BCK$-algebra and anti-grouped $BZ$-algebra. Thirdly, we introduced generalized quasi-left alter $BZ$-algebra and proved that every generalized quasi-left alter $BZ$-algebra is QM-$BZ$-algebra. Forthly, we introduced the notion of quasi-minimal elements into hyper $BZ$-algebra, and QM-hyper $BZ$-algebra was obtained. Additionally, the relationship between QM-hyper $BZ$-algebra and generalized quasi-left alter hyper $BZ$-algebra was discussed. Next, quasi-hyper $BZ$-algebra, which is an extension of hyper $BZ$-algebra and quasi-hyper $BCI$-algebra, was proposed. We also gave the concepts of anti-grouped quasi-hyper $BZ$-algebra, generalized anti-grouped quasi-hyper $BZ$-algebra, and associative quasi-hyper $BZ$-algebra, and proved that: (1) an anti-grouped quasi-hyper $BZ$-algebra is an anti-grouped $BZ$-algebra; (2) every generalized anti-grouped quasi-hyper $BZ$-algebra corresponds to a semihypergroup; (3) every associative quasi-hyper $BZ$-algebra is an associative $BCI$-algebra.

The above research results are helpful for revealing the connections among the relevant logic algebras (and their hyper structures), and can be used for reference for other non-classical logic algebras. As a further research topic, we will consider the relationship among (quasi-hyper) $BZ$-algebras, semihypergroups (see [37]) and $BI$-algebras (basic implication algebras, see [38,39]) as well as their ideal (filter) theories.