Some New Results on N(2,2,0)-Algebras

1. Introduction

Fuzzy algebras and their axiomatizations have become important topics in theoretical research and in the applications of fuzzy logic[1]. Wu introduced a class of fuzzy implication algebras, FI-algebras for short, in 1990 [2]. Recently, researchers such as [3,4] have studied fuzzy implications from different perspectives. Naturally, it is meaningful to investigate the common properties of some important fuzzy implications used in fuzzy logic. Consequently, various interesting properties of FI-algebras, regular FI-algebras, commutative FI-algebras, and some concepts of filter, ideal and fuzzy filter of FI-algebras were proposed, some relations between these FI-algebras and several famous fuzzy algebras were systematically discussed [5].

In the last decades, the development algebraic models for non-commutative multiple-valued logics has become a central topic in the study of fuzzy systems. Various logical algebras have been introduced as the semantical systems of nonclassical logic systems. The notion of NA-algebras was introduced by Deng and Xu in 1996 [7]. We investigate the basic properties of NA-algebra $(S,\ast ,△,1)$, and showed that if the class of operation ∗ is idempotent, then $(S,\ast ,△,1)$ is a rewriting systems. Analogously. If the operation △ is nilpotent, then $(S,△,1)$ is an associated BCI-algebra. Furthermove, some relationships between nilpotent NA-algebra and other algebras with the type of(2,0)are obtained [8].

Quantum B-algebras were introduced by Rump and Yang(2014)[19]. A quantum B-algebra is a partially ordered set with two binary operations → and ⇝ satisfying four conditions (see Sect.4). Logically, → and ⇝ represent the left and right implications, while ≤ corresponds to the entailment relation. In commutative logic, the two implications → and ⇝ coincide.

The present paper is organized as follows: In Section 2, we review some basic concepts, present the main properties of NA-algebras, and show relationship between the NA-algebras and the related fuzzy logic algebras. In Section 3, we consider subalgebras, ideal, stabilizer and congruence kernels and NA-morphism theorem in NA-algebras. In Section 4, we investigate the relationships among NA-algebra and other related logical algebraic structures, such as Q-algebras[14], CI-algebras[15] , Quantum B-algebras[19,20,21], pseudo-BCH-algebras[22], and RM-algebras[23].

2. Basis Concepts and Properties of NA-Algebras

The numerical verification in this paper can be referred to the algorithm program in [30]. We recall some basic definitions and results that are necessary for this paper.

Let S be a semigroup. An element a of S is idempotent if $a^2=a$. The set of idempotents of a subset A of S is denoted by $E\left(A\right)$.

An element z of S is a left (respectively, right)zero of S if $zs=z$ (respectively, $sz=z$) for all $s\in S$; z is a zero of S if it is both a left and right zero of S. A semigroup all of whose elements are left (respectively, right) zeros is a left (respectively, right) zero semigroup. Elements a and b of S commute if $ab=ba$; S is commutative if any two of its elements commute[6].

Definition 2.1. ([2]) A fuzzy implication algebra X with a constant 0 and a binary operation $“\to ”$ satisfying the following axioms: for any $x,y,z\in X$,

$\left(I_1\right)x\to (y\to z)=y\to (x\to z)$;

$\left(I_2\right)(x\to y)\to ((y\to z)\to (x\to z))=1$;

$\left(I_3\right)x\to x=1$;

$\left(I_4\right)x\to y=y\to x=1\Rightarrow x=y$;

$\left(I_5\right)0\to x=1$, where $1=0\to 0$.

In a fuzzy implication algebra $(X,\to ,0)$:

(I) The order relation ≤ satisfying $x\le y$ iff $x\to y=1$ is a partial order[2];

(II) A multiplication ⚬ induced by the adjoint relation

$$u\le a\to b\iff a⚬u\le b.$$

(1)

$a,b,u\in X$, where $(⚬,\to )$ is an adjoint pair on X[1].

In [7], we proved that in a fuzzy implication algebra $(X,\to ,0)$, for all $a,b,u\in X$ if, the following equalities hold:

$$u\to (a⚬b)=b\to (u\to a)$$

(2)

$$(a⚬u)\to b=u\to (a\to b),$$

(3)

then $(X,⚬)$ is a semigroup [7,8].

By generalizing the expressions (2) and (3), we replace the symbol → by the symbol ∗, and the symbol ⚬ by the symbol △, we obtain the basic equations of NA-algebra.

Definition 2.2.([7]) An algebra $(S,\ast ,△,1)$ of type (2,2,0) is called an NA-algebra if for all $a,b,c\in S$ the following axioms hold:

$\left(N_1\right)a\ast (b△c)=c\ast (a\ast b)$;

$\left(N_2\right)(a△b)\ast c=b\ast (a\ast c)$;

$\left(N_3\right)1\ast a=a$.

In order to distinguish NA-algebras and Fuzzy implication algebras, we have introduced constant 1 in NA-algebra, and axioms $\left(N_1\right),\left(N_2\right)$ are conditions of the multiplication ∗ forming a semigroup, which induced by the adjointness relation expressions (1) in fuzzy implication algebra $(X,\to ,1)$.

Example 1. Let $S=\{1,a,b,c\}$. Define the operations ∗ and △ on S by cayley Tables 1 and 2 below:

Table 1. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	1	a	b	c
b	c	b	b	c
c	c	b	b	c

Table 1. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	1	a	b	c
b	c	b	b	c
c	c	b	b	c

Table 2. △ table.

△	1	a	b	c
1	1	1	c	c
a	a	a	b	b
b	b	b	b	b
c	c	c	c	c

Table 2. △ table.

△	1	a	b	c
1	1	1	c	c
a	a	a	b	b
b	b	b	b	b
c	c	c	c	c

Based on the Cayley Tables 1 and 2, $\forall a,b,c\in S$, we have

$a\ast (b△c)=a\ast b=b,c\ast (a\ast b)=c\ast b=b$, so $\left(N_1\right)$ holds.

$(a△b)\ast c=b\ast c=c,b\ast (a\ast c)=b\ast c=c$, hence $\left(N_2\right)$ holds.

Similarly, other cases can be verified and established.

The following propositions gives some basic properties of NA-algebra.

Theorem 2.1. Let $(S,\ast ,△,1)$ be an NA-algebra. Then for all $a,b,c\in S$, the following hold:

$\left(N_4\right)a\ast b=b△a$;

$\left(N_5\right)(a\ast b)\ast c=a\ast (b\ast c)$ (∗ associative law), $a\ast (b\ast c)=b\ast (a\ast c)$ (∗ left commutative law);

$\left(N_6\right)(a△b)△c=a△(b△c)$ (△ associative law); $a△b)△c=(a△c)△b$ (△ right commutative law).

Proof. $\left(N_4\right)$: $a△b\stackrel{\left(N_3\right)}{=}1\ast (a△b)\stackrel{\left(N_1\right)}{=}b\ast (1\ast a)\stackrel{\left(N_3\right)}{=}b\ast a$;

$\left(N_5\right)$: $(a\ast b)\ast c\stackrel{\left(N_4\right)}{=}(b▵a)\ast c\stackrel{\left(N_2\right)}{=}a\ast (b\ast c)$; $a\ast (b\ast c)\stackrel{\left(N_4\right)}{=}a\ast (c△b)\stackrel{\left(N_1\right)}{=}b\ast (a\ast c)$;

$\left(N_6\right)$: $(a▵b)▵c\stackrel{\left(N_4\right)}{=}c\ast (a△b)\stackrel{\left(N_1\right)}{=}b\ast (c\ast a)\stackrel{\left(N_5\right)}{=}c\ast (b\ast a)\stackrel{\left(N_5\right)}{=}(c\ast b)\ast a\stackrel{\left(N_4\right)}{=}(b△c)\ast a\stackrel{\left(N_4\right)}{=}a△(b△c)$;

$(a△b)△c\stackrel{\left(N_4\right)}{=}c\ast (a△b)\stackrel{\left(N_1\right)}{=}b\ast (c\ast a)\stackrel{\left(N_4\right)}{=}(c\ast a)△b\stackrel{\left(N_4\right)}{=}(a△c)△b$.

It is easy to verity that $(S,\ast ,△,1)$ is an NA-algebra. This completes the proof.

Corollary 2.2.([7,8]) If $(S,\ast ,△,0)$ is an NA-algebra, then $(S,\ast ,1)$ and $(S,△,1)$ are semigroups.

Let $(S,⚬)$ be an arbitrary semigroup. A semigroup $(S,\ast )$ is called a dual semigroup to $(S,⚬)$ if $x\ast y=y⚬x$ for all $x,y\in S$.

A semigroup $(S,⚬)$ is called left commutative (respectively, right commutative) if it satisfies the identity $x⚬(y⚬a)=y⚬(x⚬a)$ (respectively, $(a⚬x)⚬y=(a⚬y)⚬x$)[6].

Obviously, the NA-algebra $(S,\ast ,△,0)$ is an algebra system with a pair of dual semigroups, and $(S,\ast )$ is a left commutative semigroup, $(S,△)$ is a right commutative semigroup.

Several interesting properties of NA-algebra have been discussed earlier [7,8].

Theorem 2.3. Let $(S,\ast ,▵,1)$ be an NA-algebra and for every $x,y\in S$, then the following hold:

(1)If $x\ast x=1$, then $x\ast y=x△y$. In this case, two binary operations ∗ and △ coincide, and $(S,\ast {,}^{-1},1)$ is an abelian group, where “1” is an unit element, $x^{-1}=x$, for all x.

(2)If $x\ast 1=1$, then $x\in E\left(S\right)$ and semigroup $(S,\ast ,1)$ is a right zero semigroups.

Proof. (1)By theorem 2.1 $\left(N_5\right)$, we have operation∗ is associative, and $x\ast y=(x\ast y)\ast 1=x\ast (y\ast 1)=y\ast (x\ast 1)=y\ast x=x△y$. In this case, ∗ and △ coincide, $(S,\ast ,1)$ is a commutative semigroups. Since $x\ast x=1$, then $x\ast 1=x\ast (x\ast x)=(x\ast x)\ast x=1\ast x=x$. So, $x\ast 1=1\ast x=x$. Hence, 1 is unique unit element. Since $1^{\prime }=1\ast 1^{\prime }=1^{\prime }\ast 1=1$, we get 1 is unique unit in S.

From that $x\ast x=1$, we have x is an inverse x, if there exists an inverse element y of x such that $x\ast y=1$, then $x=1\ast x=(x\ast y)\ast x=x\ast (y\ast x)=y\ast (x\ast x)=y\ast 1=y$. Hence, for any $x\in S$, we have x is itself inverse and unique. So, for an NA-algebras, if $x\ast x=1$, then $(S,\ast ,1)$ is an Abelian group.

(2) For every $x\in S$, if it satisfies $x\ast 1=1$, then

$$x^2=x\ast x=x\ast (1\ast x)=(x\ast 1)\ast x=1\ast x=x.$$

We get $x\ast 1=1\Rightarrow x^2=x$, i.e., $x\in E\left(S\right)$.

Moreover, for any $x,y\in S$, we have $x\ast y=x\ast (1\ast y)=(x\ast 1)\ast y=1\ast y=y$. Hence, $(S,\ast ,1)$ is a right zero semigroup.

Theorem 2.4. Let $(S,\ast ,△,1)$ be an NA-algebra. For any $x,y\in S$, then $x\ast y=y\ast x$ if and only if $x\ast 1=x$.

Proof. $(⟸)$ $x\ast y=1\ast (x\ast y)=(x\ast y)\ast 1=x\ast (y\ast 1)=y\ast (x\ast 1)=y\ast x$.

$(⟹)$ Let us suppose that $x\ast 1\ne x$, but $1\ast x=x$, by $x\ast y=y\ast x$, a contradiction.

It is easy to check that the following results are true.

Remark 1. Let $(S,\ast ,△,1)$ be an NA-algebra. Then, for any $x\in S$, satisfy simultaneously $x\ast 1=x$ and $x\ast 1=1$ if and only if $x=1$. In this case, NA-algebra is trivial, and also it shows the condition $x\ast x=1$ are very strong.

Remark 2. Let $(S,\ast ,△,1)$ be an NA-algebra. For all $x\in S$, such that $x\ast x=1$ is a sufficient but unnecessary condition semigroup for $(S,\ast ,1)$ as commutative semigroup.

We illustrate this conclusion with the following example:

Example 2.([7],[8]) Let $S=\{1,a,b,c\}$. Define the operations ∗ and △ on S by Tables 3 and 4 below:

Table 3. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Table 3. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Table 4. △ table.

△	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Table 4. △ table.

△	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Where $\forall x\in S$, such that $x\ast x=1$, we can verify that $(S,\ast ,△,1)$ is an NA-algebra, and $(S,\ast ,1)$ is a commutative semigroup.

Example 3. Let $X=\{1,a,b\}$. Define the operations ∗ and △ on X by Tables 5 and 6 below:

Table 5. ∗ table.

∗	1	a	b
1	1	a	b
a	a	1	b
b	b	b	b

Table 5. ∗ table.

∗	1	a	b
1	1	a	b
a	a	1	b
b	b	b	b

Table 6. △ table.

△	1	a	b
1	1	a	b
a	a	1	b
b	b	b	b

Table 6. △ table.

△	1	a	b
1	1	a	b
a	a	1	b
b	b	b	b

It is easy to verity that $(X,\ast ,△,1)$ is an NA-algebra, where $b=b\ast b\ne 1$, but $\forall x\in X,x\ast 1=x$ , by theorem 2.4, then $(X,\ast ,1)$ is a commutative semigroup.

So, for all $x\in S$ satisfies $x\ast x=1$ is a sufficient but unnecessary condition semigroup for $(S,\ast ,1)$ as commutative semigroup.

Remark 3. If a fuzzy implication algebra $(X,\to ,1)$ with a partial order $“\le ”$, such that any $a,b,u\in X,a\le b\iff a\to b=1$, $u\le a\to b\iff a\ast u\le b$, and for all $a,b,u\in X$, the following conditions hold:

$$u\to (a\ast b)=b\to (u\to a),$$

$$(a\ast u)\to b=u\to (a\to b),$$

then $(X,\to ,\ast ,1)$ is an NA-algebra.

Remark 4. Let $(S,\ast ,▵,1)$ be an NA-algebra, then semigroups $(S,\ast ,1)$ and $(S,△,1)$ are a pair of dual semigroup. A pair of dual operations $(\ast ,△)$ form an adjoint pair $(\to ,\ast )$, ie., $u\le a\to b\iff a\ast u\le b$, for every $a,b,u\in S$.

Theorem 2.5. Let $(S,\ast ,▵,1)$ be an NA-algebra and for every $x,y,z$ in S, then the following hold:

$$x▵(y\ast z)=y\ast (x△z),x\ast (y△z)=y△(x\ast z).$$

Proof. By Theorem 2.1, we have $x△y=y\ast x$. Hence, $x△(y\ast z)=(y\ast z)\ast x=y\ast (z\ast x)=y\ast (x△z)$. Similarly, $x\ast (y△z)=y△(x\ast z)$.

Note that the associativity of ∗ and △ of NA-algebra $(S,\ast ,▵,1)$ is equivalent to the equation:

$$x\ast (y△z)=y△(x\ast z).$$

Theorem 2.6. ([8]) Let $(S,\ast ,▵,1)$ be an NA-algebra. For every $x\in S$ , if the binary operation ∗ satisfies $x\ast x=1$, then the following hold:

(1) $x\ast y\ne 1\iff x\ne y,\forall x,y\in S$;

(2) $x\ast y=x\iff y=1,\forall x,y\in S$;

(3) $x\ast (x\ast y)=y,y\ast (x\ast y)=x,\forall x,y\in S$.

Theorem 2.7. Let S be a set with a consistent 1 and two binary operations $\ast ,△$, for all $x,y,z\in S$, satisfying

(I)$x\ast (y△z)=y△(x\ast z),$

(II)$x△y=y\ast x,$

(III)$x\ast (y\ast z)=y\ast (x\ast y),$ (left commutative)

(IV) $1\ast x=x.$

Then $(S,\ast ,△,1)$ is an NA-algebra.

Proof. By (I) and (II), for all $x,y,z\in S$, we have

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

Thus, for any $x,y,z\in S$, we have $x\ast (z\ast y)=(x\ast z)\ast y$.

Similarly, we can prove $x△(z△y)=(x△z)△y.$

So, “∗” and “△” with associative.

Using the hypothesis, we have

$x\ast (y△z)=x\ast (z\ast y)=z\ast (x\ast y)$, hence, axiom $\left(N_1\right)$ holds.

$(x△y)\ast z=(y\ast x)\ast z=y\ast (x\ast z)$, so axiom $\left(N_2\right)$ holds.

by hypothesis (IV), we get axiom $\left(N_3\right)$ holds. Thus, $(S,\ast ,△,1)$ is an NA-algebra.

Let $(S,\ast ,△,1)$ be a NA-algebra. If operations ∗ and △ of $(S,\ast ,△,1)$ coincide, then the NA-algebra becomes a semigroup. Thus, every NA-algebras are a generalization of semigroups.

Let $(L,\odot ,1)$ is a monoid, the operations ⊙,→ and ⇝ satisfy the so-called right adjointness property:

$$x\odot y\le z\iff x\le y\to z,$$

and the left adjointness property:

$$x\odot y\le z\iff y\le x⇝z.$$

Remark 5. Let $(S,\ast ,▵,1)$ be an NA-algebra, for any $x,y,z\in S$, by using $x\ast y=y△x$, and

$$x\le y\to z\iff y\ast x\le z,$$

$$x\le y\to z\iff x△y\le z$$

we obtain ∗ satisfy the left adjoint property, and △ satisfy the right adjoint property.

3. Subalgebras, Ideal and Stabilizer of NA-Algebras

3.1. Subalgebras, Ideal of NA-Algebras

In this section, we will introduce ideals and congruences of NA-algebra $(S,\ast ,▵,1)$. We define two quasi-ordering relations and denote them by ${\le }_r$ and ${\le }_l$ on $E\left(S\right)$ as follows, respectively.

$$e{\le }_{r}f\iff f\ast e=e\iff e△f=e$$

(4)

$$e{\le }_{l}f\iff e\ast f=e\iff f△e=e$$

(5)

Definition 3.1. Let A be an nonempty subset of NA-algebra $(S,\ast ,▵,1)$ is called a subalgebra of S if $x\ast y,x△y\in A$ hold for any $x,y\in A$.

Definition 3.2. Let A be an nonempty subset of NA-algebra $(S,\ast ,▵,1)$ is called an NA-ideal of S, if

(1) $1\in A$;

(2) for any $x\in A$, and $x\ast y,x△y\in A\Rightarrow y\in A.$

It is easy to verify that $\left\{1\right\}$ and S are two ordinary ideals of NA-algebra $(S,\ast ,▵,1)$.

Example 4. Let $S=\{1,a,b\}$. Define the operations ∗ and △ on X by Tables 7 and 8 below:

Table 7. ∗ table.

∗	1	a	b
1	1	a	b
a	a	a	a
b	b	a	a

Table 7. ∗ table.

∗	1	a	b
1	1	a	b
a	a	a	a
b	b	a	a

Table 8. △ table.

△	1	a	b
1	1	a	b
a	a	a	a
b	b	a	a

Table 8. △ table.

△	1	a	b
1	1	a	b
a	a	a	a
b	b	a	a

Then, it is easy to verity that $(S,\ast ,△,1)$ is an NA-algebra. One can easily check that $A_1=\left\{1\right\},A_2=S,A_3=\{1,a\},A_4=\{1,b\}$ are ideals of NA-algebra $(S,\ast ,△,1)$.

Proposition 3.1. Let $(S,\ast ,▵,1)$ be an NA-algebra. Then we have

$$A\left(a\right)=\left\{x\right|x,a\in E\left(S\right),a{\le }_{r}x\}$$

and

$$B\left(1\right)=\left\{x\right|x\ast (x\ast 1)=1\}$$

are ideals of S.

Proof. (1) By equation (4) and $1\ast a=a$, we have $a{\le }_{r}1$. Hence, $1\in A\left(a\right)$. Now, if $x\in A\left(a\right)$, then $a{\le }_{r}x$, i.e., $x\ast a=a$, and $x\ast y\in A\left(a\right)$, so $a{\le }_{r}x\ast y$, i.e, $(x\ast y)\ast a=a$.

By $(x\ast y)\ast a=x\ast (y\ast a)=y\ast (x\ast a)=y\ast a$, so $y\ast a=a$, thus $a{\le }_{r}y\Rightarrow y\in A\left(a\right)$. Therefore $A\left(a\right)$ is an ideal of S.

(2) Using $1\ast (1\ast 1)=1$, so $1\in B\left(1\right)$. If $x\in B\left(1\right)$, and $x\ast y\in B\left(1\right)$, then we have

$(x\ast y)\ast \left(\right(x\ast y)\ast 1)=1\Rightarrow (x\ast y)\ast (x\ast (y\ast 1\left)\right)=(x\ast y)\ast (y\ast (x\ast 1\left)\right)=1\Rightarrow (y\ast y)\ast (x\ast (x\ast 1\left)\right)=1\Rightarrow y\ast (y\ast 1)=1\Rightarrow y\in B\left(1\right)$. Hence, $B\left(1\right)$ is an ideal of S.

Example 5. Let $S=\{1,a,b,c,d,e\}$. Define the operations ∗ and △ on S by Tables 9 and 10 below:

Table 9. ∗ table.

∗	1	a	b	c	d	e
1	1	a	b	c	d	e
a	1	a	b	c	d	e
b	b	b	b	c	b	b
c	c	c	c	c	c	c
d	d	d	b	c	d	d
e	1	a	b	c	d	e

Table 9. ∗ table.

∗	1	a	b	c	d	e
1	1	a	b	c	d	e
a	1	a	b	c	d	e
b	b	b	b	c	b	b
c	c	c	c	c	c	c
d	d	d	b	c	d	d
e	1	a	b	c	d	e

Table 10. △ table.

△	1	a	b	c	d	e
1	1	1	b	c	d	1
a	a	a	b	c	d	a
b	b	b	b	c	b	b
c	c	c	c	c	c	c
d	d	d	b	c	d	d
e	e	e	b	c	d	e

Table 10. △ table.

△	1	a	b	c	d	e
1	1	1	b	c	d	1
a	a	a	b	c	d	a
b	b	b	b	c	b	b
c	c	c	c	c	c	c
d	d	d	b	c	d	d
e	e	e	b	c	d	e

Then, it is easy to check that $(S,\ast ,△,1)$ is an NA-algebra, and

$$A\left(1\right)=A\left(a\right)=A\left(e\right)=\{1,a,e\},A\left(c\right)=S,A\left(b\right)=\{1,a,b,d,e\},A\left(d\right)=\{1,a,d,e\},B\left(1\right)=\{1,a,e\}$$

are NA-ideals.

Proposition 3.2. Let $(S,\ast ,▵,1)$ be an NA-algebra. If Q is an ideal of S, and for all $x,y\in Q$, $\exists z\in S$, such that $x\ast (y\ast z)=1$, then $x\ast z,y\ast z\in Q$.

Proof. Since Q is an ideal of S, we have $1\in Q$. If $x\in Q,\exists z\in S$, such that $x\ast (y\ast z)=1\in Q$, then by Definition 3.1, we have $y\ast z\in Q$. Similarly, $y\in Q,x\ast (y\ast z)=y\ast (x\ast z)=1\in Q$, then we have $y\ast z\in Q$.

Proposition 3.3. Let $(S,\ast ,▵,1)$ be an NA-algebra. If for any $x\in S$, such that $x\ast 1=x$, then we have

$$S_g=\left\{x\right|g\in E\left(S\right),x\ast g\ast x=g\}$$

(6)

is an ideal of S.

Proof. (1) By equation(6) and $1\ast (g\ast 1)=g\ast (1\ast 1)=g\ast 1=g$, we have $1\in S_g$ holds.

(2) If $x\in S_g$, then, $x\ast g\ast x=g$. Let $x\ast y\in S_g$, then we get $(x\ast y)\ast g\ast (x\ast y)=g$, it follows that $y\ast (x\ast g\ast x)\ast y=g\Rightarrow y\ast g\ast y=g$. Therefore, $S_g$ is an ideal of S.

Example 6. Let $S=\{1,a,b,c\}$. Define the operations ∗ and △ on S by Tables 11 and 12 below:

Table 11. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	1	a	b	c
b	c	b	b	c
c	c	b	b	c

Table 11. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	1	a	b	c
b	c	b	b	c
c	c	b	b	c

Table 12. △ table.

△	1	a	b	c
1	1	1	c	c
a	a	a	b	b
b	b	b	b	b
c	c	c	c	c

Table 12. △ table.

△	1	a	b	c
1	1	1	c	c
a	a	a	b	b
b	b	b	b	b
c	c	c	c	c

Then, it is easy to verity that $(S,\ast ,△,1)$ is an NA-algebra. One can easily check that

$$S_1=\left\{1\right\},S_a=\left\{a\right\},S_b=\{a,b\},S_c=\{1,c\}$$

are ideals of NA-algebra $(S,\ast ,△,1)$.

An element $a\in S$ is called nilpotent provided that $a^2=a\ast a=1$.

Proposition 3.4. Let $(S,\ast ,▵,1)$ be an NA-algebra, N be set of all nilpotent element S. Then N is an ideal of S.

Proof. $1\in N$ is obvious. For any $x\in N,x\ast y\in N$, we have $1={(x\ast y)}^2=x^2\ast y^2=1\ast y^2=y^2$, hence $y\in N$. Therefore, N is an ideal of S.

From Proposition 3.4 it follows that if N be a set of all nilpotent element on NA-algebra $(S,\ast ,▵,1)$, then N is an ideal of S. For any $s,t\in N$, we define

$$s\sim t\iff s\ast t,s△t\in N,$$

then, the relation ∼ in N is an equivalence relation.

In fact, if $x\ast x=1\in N$, we get $x\sim x$, that is ∼ satisfying reflexive.

Secondly, from the define of relation ∼, for any $x,y\in N,x\sim y\Rightarrow y\sim x$, it follows that we get ∼ satisfying symmetric.

Finally, suppose that $x\sim y,y\sim z$, that is , $x\ast y,y\ast z\in N,y\ast x,z\ast y\in N$, we have

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

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

That is, $x\ast z\in N$. This case of $z\ast x\in N$ can be proved similarly. Hence, ∼ is transitive. Therefore, the relation ∼ in N is an equivalence relation.

In what follows, let us check the equivalence classes partition of the set S. Call $N_e$ the equivalence class of an element e belong to the S. So $N_e$ consists of the elements t such that $e\sim t$:

$$N_e=\{t\in S\mid e\in N,\exists t\in S,s.t.,e\ast t,e△t\in N\}.$$

We denote by $S/N$ the set of equivalence class. Two operations $⊛,\copyright$ in $S/N$ be defined as follows:

$$N_e⊛N_f=N_{e\ast f},N_e\copyright N_f=N_{e△f}.$$

Then, it is easy to verify that $(S/N,⊛,\copyright ,N_1)$ is an NA-algebras.

3.2. Multiplicative Stabilizers in NA-Algebras

From a logic point of view, stabilizer can be used in studying the consequence connectives in the correspondence logic system, which has a very close relationship with fuzzy set. Since stabilizer was successful in some distinct tasks in various branches of mathematics (Roudabri and Torkzadeh)[11], it has been extended to various logical algebras; for example, Haveshki and Mohamadhasani first introduced the notion of stabilizers in BL-algebras and then, they studied several properties of them[12]. Michiro Kondo considered two types of stabilizers, implicative and multiplicative stabilizers in residuated lattices and proved their fundamental properties[13].

In this section, we will introduce some stabilizers and study related properties of them in NA-algebras.

Let S be an NA-algebras. For an non-empty subset $\varnothing \ne X\subseteq S$, $a\in S$, the multiplicative stabilizers $X_a$ are defined as follows:

$$X_a=\{x\in S|x\ast a=a=a△x\}$$

(7)

for all $a\in X.$

Proposition 3.5. Let S be an NA-algebras. Then, we have the following basic results about multiplicative stabilizers:

(1) $X_1=E\left(S\right)$;

(2) $X_a$ is a subalgebra and an ideal of S;

(3) If $x,y\in S$, s.t., $x\ast y=y△x=1$, then x and y is a multiplicative stabilizer of the same element.

Proof. (1) By Definition 2.2 $\left(N_3\right)$, we have $1\ast x=x$, hence $1\ast 1=1$, so we get $x\ast 1=1$. It follows that $1\in X_1.$

From $X_1=\{x\in S\mid x\ast 1=1\}$, we have $\forall x\in S,x^2=x\ast x=x\ast (1\ast x)=(x\ast 1)\ast x=1\ast x=x\Rightarrow x^2=x$. Therefore, $X_1=E\left(S\right)$.

(2) Let $\forall x,y\in X_a$. Then, $x\ast a=a=a△x,y\ast a=a=a△y$. We have $(x\ast y)\ast a=x\ast (y\ast a)=x\ast a=a\Rightarrow x\ast y,y△x\in X_a$. In a similar way, we have $y\ast x,x△y\in X_a$. Since $x\ast y=y△x$, we get $(X_a,\ast ,△,0)$ is a subalgebra of $(S,\ast ,△,0)$.

It is obvious that $1\in X_a$, because of $1\ast a=a$.

Next, if $x,x\ast y,y△x\in X_a$, then $x\ast a=a,(x\ast y)\ast a=a,a△x=a,a△(y△x)=a$, we get

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

that is $y\in X_a$. Hence, $X_a$ is an ideal of S.

(3) Suppose that $x\in X_a,y\in X_b$. Since $x\ast y=1$, it follows that $x\ast a=a\Rightarrow y\ast (x\ast a)=y\ast a\Rightarrow x\ast (y\ast a)=y\ast a\Rightarrow (x\ast y)\ast a=y\ast a\Rightarrow a=1\ast a=y\ast a$, we get $y\ast a=a$, we show $y\in X_a$. $y\in X_b\Rightarrow x\in X_b$ can be proved similarly.

Proposition 3.6. Let S be an NA-algebras. We defined two types products of S and $S_a$ as follows:

$$S\ast S_a=\{s\ast x\mid s\in S,x\in S_a\},$$

$$S\ast S_a\ast S=\{s_1\ast x\ast s_2\mid s_1,s_2\in S,x\in S_a\}.$$

Then, the following statements hold.

(1) $(S\ast S_a,\ast ,△,1)$ is a subalgebra of NA-algebras S. In particular, for any positive integer n, we get a NA-algebras sequence as follows:

$$\dots \le (S^{n+1},\ast ,△,1)\le (S^n,\ast ,△,1)\le (S^{n-1},\ast ,△,1)\le \dots \le (S,\ast ,△,1).$$

(2) $(S\ast S_a\ast S,\ast ,△,1)$ is a subalgebra of NA-algebras S. In addition, $(S_a^n,\ast ,△,0)=(S_a,\ast ,△,1).$

(3) For any $a,b\in S$, we have

(A) $(S_1,\ast ,△,1)\subseteq (S_a\bigcap S_b,\ast ,△,1)\subseteq (S_a\ast S_b\bigcap S_b\ast S_b,\ast ,△,1)\subseteq (S_{a\ast b}\bigcap S_{b\ast a},\ast ,△,1);$

(B) $(S_a,\ast ,△,1)\subseteq (S_a\ast S_b\bigcap S_b\ast S_b,\ast ,△,1);$

(C) If $a\ast b=1$, then $(S_a,\ast ,△,1)=(S_b,\ast ,△,1)=(S_1,\ast ,△,1).$

Proof. We only show the cases (1) and (C) of (3). Other cases can be proved easily.

(1) It follows from Proposition 3.5 (2) that $X_a$ is a subalgebra of S, we have $1=1\ast 1\in S\ast X_a$.

For any $s_1\ast x_1,s_2\ast x_2\in S\ast X_a$, where $s_1,s_2\in S,x_1,x_2\in X_a$, we have

$(s_1\ast x_1)\ast (s_2\ast x_2)=s_1\ast (x_1\ast (s_2\ast x_2))=s_1\ast (s_2\ast (x_1\ast x_2))=(s_1\ast s_2)\ast (x_1\ast x_2)\in S\ast X_a.$

By using the operational duality between ∗ and △, we get $(s_1\ast x_1)△(s_2\ast x_2)\in S△X_a$. Therefore, $(S\ast S_a,\ast ,△,0)$ is a subalgebra of NA-algebras S.

Obviously, for any positive integer n, the subalgebras sequence of NA-algebras $(S,\ast ,△,1)$

$$\dots \le (S^{n+1},\ast ,△,1)\le (S^n,\ast ,△,1)\le (S^{n-1},\ast ,△,1)\le \dots \le (S,\ast ,△,1)$$

holds.

(3) We prove that case (C) of (3). Let $a,b\in S$, if $a,b\in S$, satisfies $a\ast b=1$, then, for any $x\in S_a$, we have $x\ast b=x\ast (1\ast b)=x\ast \left(\right(a\ast b)\ast b)=(x\ast a)\ast (b\ast b)=a\ast (b\ast b)=(a\ast b)\ast b=1\ast b=b$, i.e., $x\in S_b$. Therefore, $S_a\subseteq S_b$. In a similar way, we can show that $x\in S_b$ implies $x\in S_a$, i.e., $S_b\subseteq S_a$. Hence, $S_a=S_b$. Moreover, for any $x\in S_b$, $x\ast 1=x\ast (a\ast b)=a\ast (x\ast b)=a\ast b=1$ implies $x\in S_1$. It follows from (A) of (3), we get $S_1\subseteq S_a$. Therefore, we have $(S_a,\ast ,△,1)=(S_b,\ast ,△,1)=(S_1,\ast ,△,1).$

3.3. Congruence Decomposition of NA-Algebras

We define congruence decomposition of NA-algebras, which plays a central role in the results that follow.

Definition 3.3. A congruence on an NA-algebras $(S,\ast ,△,1)$ is an equivalence relation $\rho$ on S such that $x\rho x^{\prime }$ and $y\rho y^{\prime }$, then $(x\ast y)\rho (x^{\prime }\ast y^{\prime })$ and $(x△y)\rho (x^{\prime }△y^{\prime })$.

The condition for an equivalence relation $\rho$ on S to be a congruence ensures that the set $S/\rho$ of equivalence classes under $\rho$ has a well-defined monoid structure inherited from S.

Definition 3.4. Let $(S,\ast ,△,1)$ be an NA-algebra. For all $x,y\in S$, if there exists some $u\in S$ with $x\ast u\ast y=x\ast y$, then u is called middle unit of S.

The all middle unit set of S will be denoted by $M\left(S\right)$. Obviously, $1\in M\left(S\right)$.

Proposition 3.7. Let $(S,\ast ,△,0)$ be an NA-algebras. Then, the following statements hold.

(1) $\left(M\right(S),\ast ,△,1)$ is a subalgebra of $(S,\ast ,△,1)$.

(2) If $\forall x,y\in S,\exists u\in S$ satisfies $x^2\ast u\ast y^2=x\ast y$, then $u\in M\left(S\right)$ if and only if $x\ast y\in E\left(S\right)$.

Proof. (1) By $1\in M\left(S\right)$, we have $\varnothing \ne M\left(S\right)\subseteq S$. $\forall u,v\in M\left(S\right),x,y\in S$, using Definition 3.4, this implies that

$$x\ast u\ast y=x\ast y,x\ast v\ast y=x\ast y,$$

$x\ast (u\ast v)\ast y=(x\ast u)\ast (v\ast y)=v\ast \left(\right(x\ast u)\ast y)=v\ast (x\ast u\ast y)=v\ast (x\ast y)=x\ast v\ast y=x\ast y$, hence, $u\ast v\in M\left(S\right)$. In a similar way, we can show that $v\ast u\in M\left(S\right)$. By using the operational duality between ∗ and △, we get $u△v,v△u\in M\left(S\right).$ Therefore, $\left(M\right(S),\ast ,△,1)$ is a subalgebra of $(S,\ast ,△,1)$.

(2)Suppose that $u\in M\left(S\right)$. Then $\forall x,y\in S$, we have $x\ast u\ast y=x\ast y$. On the other hand, from $x^2\ast u\ast y^2=x\ast y$, we get $x\ast y=x^2\ast u\ast y^2=x\ast x\ast u\ast y\ast y=x\ast (x\ast u\ast y)\ast y=x\ast x\ast y\ast y=x\ast y\ast x\ast y={(x\ast y)}^2$, and hence that $x\ast y\in E\left(S\right)$. Conversely, we assume that $x^2\ast u\ast y^2=x\ast y$, for all $x,y\in S$, i.e., $u\ast (x^2\ast y^2)=u\ast {(x\ast y)}^2=x\ast y$ hold. If $x\ast y\in E\left(S\right)$, then, for any $x,y\in S$, $x\ast u\ast y=x\ast (u\ast y)=u\ast (x\ast y)=u\ast {(x\ast y)}^2=x\ast y$. This implies that $u\in M\left(S\right)$.

Proposition 3.8. Let $(S,\ast ,△,1)$ be an NA-algebras and $M\left(S\right)$ is all middle unit set of S. Define the relation $\rho$ as follows: $\forall x,y\in S$, $(x,y)\in \rho \iff$ there exists $u_1,u_2\in M\left(S\right)$ such that $u_1\ast x=u_2\ast y$. then $\rho$ is a congruence on S.

Proof. It is easy to verify that reflexive and symmetric are satisfied. We only show the proof of transitivity.

From the relation $\rho$ on S, it is defined by the following rule:

$(x,y)\in \rho$ if and only if there exists $u_1,u_2\in M\left(S\right)$ such that $u_1\ast x=u_2\ast y$, for any $x,y\in S$.

In fact, if $\forall x,y,z\in S,(x,y)\in \rho ,(y,z)\in \rho$, then there exists $u_1,u_2,u_3,u_4\in S$, such that

$$u_{1}x=u_{2}y,u_{3}y=u_{4}z.$$

It follows that $(u_1\ast u_3)\ast x=u_3\ast (u_1\ast x)=u_3\ast (u_2\ast y)=u_2\ast (u_3\ast y)=u_2\ast (u_4\ast z)$.

From proposition 3.7(1), $\left(M\right(S),\ast ,▵,1)$ is a subalgebra of $(S,\ast ,▵,0)$, we get $u_1\ast u_3,u_2\ast u_4\in M\left(S\right)$, Thus $(x,z)\in \rho$. Therefore, $\rho$ is an equivalence relation on S.

Next, we show that $\rho$ is a congruence on S. Suppose that $x_1,x_2,y_1,y_2\in S,(x_1,y_1)\in \rho ,(x_2,y_2)\in \rho$, i.e., $a,b,c,d\in M\left(S\right)$, such that

$$a\ast x_1=b\ast y_1,c\ast x_2=d\ast y_2,$$

we have $(a\ast c)\ast (x_1\ast x_2)=(a\ast x_1)\ast (c\ast x_2)=(b\ast y_1)\ast (d\ast y_2)=(b\ast d)\ast (y_1\ast y_2)$. Hence, $\rho$ is a congruence on S.

We denote equivalence classes belonging to x by $\left[x\right]$ . $S/\rho =\left\{\right[x\left]\right|x\in S\}$. For any $\left[x\right],\left[y\right]\in S/\rho$, we define $\left[x\right]\ast \left[y\right]=[x\ast y],\left[x\right]△\left[y\right]=[x△y],$ then, $(S/\rho ,\ast ,△,[1\left]\right)$ form an NA-algebras, which is a quotient algebra of $(S,\ast ,△,1)$.

Suppose that a is a fixed element in S. Now, we define another congruence relation on S by using $S_a$ as follow.

Proposition 3.9. Let $(S,\ast ,△,1)$ be an NA-algebras. $\forall x,y\in S,(x,y)\in \theta \iff \exists h_1,h_2\in S_a$, such that $x\ast h_1=y\ast h_2$. Then, $\theta$ is a congruence relation on S as follow.

Proof. It is easy to verify that reflexive and symmetric are satisfied. We only show the proof of transitivity.

In fact, let $\forall x,y,z\in S,(x,y)\in \theta ,(y,z)\in \theta$, then there exists $h_1,h_2,h_3,h_4\in S_a$, such that

$$x\ast y_1=y\ast h_2,y\ast h_3=z\ast h_4.$$

It follows that:

$x\ast (h_1\ast h_3)=(x\ast h_1)\ast h_3=(y\ast h_2)\ast h_3=y\ast (h_2\ast h_3)=h_2\ast (y\ast h_3=h_2\ast (z\ast h_4)=z\ast (h_2\ast h_4)$.

From Proposition 3.5(2), we get $h_1\ast h_3,h_2\ast h_4\in S_a$. Thus $(x,z)\in \theta$. Therefore, $\theta$ is a equivalence relation on S.

Let now $x_1,x_2,y_1,y_2\in S,x_1\theta y_1,x_2\theta y_2$, then there exists $h_1,h_2,h_3,h_4\in S_a$, such that $x_1\ast h_1=y_1\ast h_2,x_2\ast h_3=y_2\ast h_4$. Thus, we have $(x_1\ast x_2)\ast (h_1\ast h_3)=x_1\ast (x_2\ast (h_1\ast h_3))=x_1\ast (h_1\ast (x_2\ast h_3))=(x_1\ast h_1)\ast (x_2\ast h_3)=(y_1\ast h_2)\ast (y_2\ast h_4)=y_1\ast (h_2\ast (y_2\ast h_4))=y_1\ast (y_2\ast (h_2\ast h_4))=(y_1\ast \left(y_2\right)\ast (h_2\ast h_4)$. Hence, $\theta$ is a congruence relation on S.

Similar to Proposition 3.8, we denote equivalence classes belong to x by $\left[x\right]$ . $S/\theta =\left\{\right[x\left]\right|x\in S\}$. For any $\left[x\right],\left[y\right]\in S/\theta$, we define

$$\left[x\right]\ast \left[y\right]=[x\ast y],\left[x\right]△\left[y\right]=[x△y],$$

Then, $(S/\theta ,\ast ,△,[1\left]\right)$ forms an NA-algebra, which is a quotient algebra of $(S/\theta ,\ast ,△,[1\left]\right)$.

3.4. Congruence Kernels of NA-Algebras

In this section, we give a characterization of congruence kernels in an NA-algebra. Let $\theta$ be a binary relation on an NA-algebra $(S,\ast ,△,1)$. We denote $\{x\in S\mid (x,1)\in \theta \}$, by ${\left[1\right]}_{\theta }$. If $\theta$ is a congruence relation on S, then ${\left[0\right]}_{\theta }$ is called a congruence kernel.

Proposition 3.10. Let $(S,\ast ,△,1)$ be an NA-algebras. We denote natural homomorphism $f:S\to S/\theta ,x⟼\left[x\right]$. Then, we have

$$ker f=\{x\mid x\in S,f(x)=[1\left]\right\}.$$

is a subalgebra and an ideal of $(S,\ast ,△,1)$.

Proof. In fact, $ker f=\{x\mid x\in S,\left[x\right]=\left[1\right]\}=\{x\mid x\in S,x\theta 1\}=\{x\mid x\in S,\exists h_1,h_2\in S_a,x\ast h_1=1\ast h_2\}$.

Obviously, $1\in ker f$. For all $x,y\in ker f$, there exists $h_1,h_2,h_3,h_4\in S_a$, such that

$$x\ast h_1=1\ast h_2=h_2,y\ast h_3=1\ast h_4=h_4.$$

This means that $(x\ast y)\ast (h_1\ast h_3)=x\ast (y\ast (h_1\ast h_3))=x\ast (h_1\ast (y\ast h_3))=(x\ast h_1)\ast (y\ast h_3)=h_2\ast h_4$. Hence $x\ast y\in ker f$.

By using the operational duality between ∗ and △, we can show that $x\Delta y\in ker f$. Therefore, $ker f$ is a subalgebra of $(S,\ast ,△,1)$.

Next, suppose $x,x\ast y\in ker f$, then there exists $h_1,h_2,h_3,h_4\in S_a$, such that

$$x\ast h_1=1\ast h_2=h_2,x\ast y\ast h_3=1\ast h_4=h_4.$$

Hence $y\ast (h_2\ast h_3)=y\ast ((x\ast h_1)\ast h_3)=(x\ast h_1)\ast (y\ast h_3)=x\ast (h_1\ast (y\ast h_3))=x\ast (y\ast (h_1\ast h_3))=(x\ast y)\ast (h_1\ast h_3))=h_1\ast ((x\ast y)\ast h_3))=h_1\ast h_4.$ This means that $y\in ker f$.

Therefore $ker f$ is an ideal of $(S,\ast ,△,1).$

3.5. NA-Morphisms

Let $X:=(X,\ast ,△,1)$ and $Y:=(Y,\overline{\ast },\overline{△},\overline{1})$ be two NA-algebras unless otherwise specified.

Definition 3.5. A mapping $\varrho :X\to Y$ is called an NA-morphism if it satisfies: for all $x,y\in X$,

$$\varrho (x\ast y)=\varrho \left(x\right)\overline{\ast }\varrho \left(y\right),\varrho (x△y)=\varrho \left(x\right)\overline{△}\varrho \left(y\right).$$

Example 7. Let $X=\{1,a,b,c\}$, define the operations ∗ and △ with Cayley Tables 13 and 14:

Table 13. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	1	a	b	c
b	c	b	b	c
c	c	b	b	c

Table 13. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	1	a	b	c
b	c	b	b	c
c	c	b	b	c

Table 14. △ table.

△	1	a	b	c
1	1	1	c	c
a	a	a	b	b
b	b	b	b	b
c	c	c	c	c

Table 14. △ table.

△	1	a	b	c
1	1	1	c	c
a	a	a	b	b
b	b	b	b	b
c	c	c	c	c

Let $Y=\{\overline{1},w,s,t\}$, define the operations $\overline{\ast }$ and $\overline{△}$ with Cayley Tables 15 and 16:

Table 15. ∗ table.

$\overline{\ast }$	$\overline{1}$	w	s	t
$\overline{1}$	$\overline{1}$	w	s	t
w	$\overline{1}$	w	s	t
s	t	s	s	t
t	t	s	s	t

Table 15. ∗ table.

$\overline{\ast }$	$\overline{1}$	w	s	t
$\overline{1}$	$\overline{1}$	w	s	t
w	$\overline{1}$	w	s	t
s	t	s	s	t
t	t	s	s	t

Table 16. △ table.

$\overline{△}$	$\overline{1}$	w	s	t
$\overline{1}$	$\overline{1}$	$\overline{1}$	t	t
w	w	w	s	s
s	s	s	s	s
t	t	t	t	t

Table 16. △ table.

$\overline{△}$	$\overline{1}$	w	s	t
$\overline{1}$	$\overline{1}$	$\overline{1}$	t	t
w	w	w	s	s
s	s	s	s	s
t	t	t	t	t

It is routine to verify that $(X;\ast ,△,1)$ and $(Y;\overline{\ast },\overline{△},\overline{1})$ both are NA-algebras.

Now, let $\varrho :X\to Y$ be a mapping defined by

$$\varrho \left(x\right)=\left\{\begin{array}{c}\hfill \overline{1},x\in \{1,a,b\}\\ \hfill w,x\in \left\{c\right\}\end{array}\right.$$

(8)

It is routine to verify that $\varrho$ is an NA-morphism[30].

Proposition 3.11. If $\delta :X\to Y$ is an NA-morphism, then $\delta \left(1\right)=1$ and $\delta \left(x\right)\le \delta \left(y\right)$ for all $x,y\in X$ with $x\le y$.

Proof. We have $\delta \left(1\right)=\delta (1\ast 1)=\delta \left(1\right)\ast \delta \left(1\right)=1$. Let $x,y\in X$ be such that $x\le y$. Then $x\ast y=1$, and so $\delta \left(x\right)\ast \delta \left(y\right)=\delta (x\ast y)=\delta \left(1\right)=1$. This shows that $\delta \left(x\right)\le \delta \left(y\right)$.

Let $f:X\to Y$ be an NA-morphism. For any subset B of Y , the set

$$f^{-1}\left(B\right):=\{x\in X\mid f\left(x\right)\in B\}$$

is called the preimage of B under f . The preimage of $f^{-1}\left(1\right):=\{x\mid f\left(x\right)=1\}$ under f is called the kernel of f and is denoted by $f^{-1}\left(1\right)$ or $ker\left(f\right)$, and so

$$f^{-1}\left(1\right)=ker\left(f\right)=\{x\in X|f\left(x\right)=1\}.$$

Proposition 3.12. The preimage of an NA-ideal under an NA-morphism is an NA-ideal.

Proof. Let $\varphi :X\to Y$ be an NA-morphism and let G be an NA-ideal of Y. It is clear that $0\in {\varphi }^{-1}\left(G\right)$. Let $x,y\in X$ be such that $x\ast y\in {\varphi }^{-1}\left(G\right)$ and $x\in {\varphi }^{-1}\left(G\right)$. Then $\varphi \left(x\right)\in G$ and $\varphi \left(x\right)\ast \varphi \left(y\right)=\varphi (x\ast y)\in G$. It follows from Definition 3.2 that $\varphi \left(y\right)\in B$. Hence $y\in {\varphi }^{-1}\left(G\right)$, and therefore ${\varphi }^{-1}\left(G\right)$ is an NA-ideal of X. Now, let G be an NA-ideal of Y.

Given a constant element c and an non-empty subset H of an NA-algebra S, we consider a special set:

$$H_c:=\{x\in S\mid c\ast x\in H\}.$$

Theorem 3.13. If H is an NA-ldeal of S and $S/H$ is the quotient NA-algebra induced by H, then the map $\phi :S\to S/H,x\mapsto H_x$ is an NA-morphism.

Proof. Assume that H is an NA-ldeal of S. We define a map $\phi :S\to S/H,x\mapsto H_x$. This implies that

$\forall x,y\in S$, we have

$$x\mapsto H_x=\{a\in S\mid x\ast a\in H\},$$

$$y\mapsto H_y=\{b\in S\mid x\ast b\in H\},$$

$$x\ast y\mapsto H_{x\ast y}=\{c\in S\mid (x\ast y)\ast c\in H\}.$$

Obvious that

$$H_{x\ast y}=H_x\ast H_y.$$

We say that $\phi$ is the natural NA-morphism.

In a NA-algebra $(S,\ast ,△,1)$, a binary relation “≤” is defined by $(\forall x,y\in S)(x\le y\iff x\ast y=1$ or $(x\le y\iff y▵x=1)$.

Definition 3.6. (1) An NA-algebra is antisymmetric if the binary relation “≤” is antisymmetric.

(2)An NA-algebra S is said to be transitive if it satisfies: $\forall x,y,z\in S$

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

In fact, by the definition of binary relation “≤”, we have $x\ast x=1$, and $x\ast y=y\ast x=1\Rightarrow x=1\ast x=(y\ast x)\ast x=y\ast (x\ast x)=y\ast 1=y\ast (x\ast y)=(y\ast x)\ast y=1\ast y=y$.

Therefore, the binary relation “≤” is antisymmetric.

Theorem 3.14. Let S be an NA-algebra. For every $x\in S$ if it satisfies $x\ast x=1$, then S is a transitive NA-algebra.

Proof. Let S be an NA-algebra and $x,y,z\in S$. Then

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

Hence $x\ast y\le (z\ast x)\ast (z\ast y)$, and therefore S is a transitive NA-algebra.

Theorem 3.15. Let $\mu :X\to Y$ and $\nu :X\to U$ be NA-morphisms where U is an antisymmetric NA-algebra. If $\mu$ is onto and $ker\left(\mu \right)\subseteq ker\left(\nu \right)$, then there exists a unique NA-morphism $\omega :Y\to U$ such that the following diagram commutes.

Proof. For every $y\in Y$, we can take $x\in X$ such that $\mu \left(x\right)=y$ since $\mu$ is onto. Define $\omega :Y\to U$ by $\omega \left(y\right)=\nu \left(x\right)$. We first show that $\omega$ is well-defined. Let $x_1,x_2\in X$ be such that $\mu \left(x_1\right)=\mu \left(x_2\right)$. Then $\mu (x_1\ast x_2)=\mu \left(x_1\right)\ast \mu \left(x_2\right)=1$, and thus $x_1\ast x_2\in ker\left(\mu \right)\subseteq ker\left(\nu \right).$

It follows that $\nu \left(x_1\right)\ast \nu \left(x_2\right)=\mu (x_1\ast x_2)=1$. Similarly, we get $\nu \left(x_2\right)\ast \nu \left(x_1\right)=1$. Hence $\nu \left(x_1\right)=\nu \left(x_2\right)$. Since U is an antisymmetric NA-algebra, this shows that $\omega$ is well-defined, and $\nu \left(x\right)=\omega \left(y\right)=\omega \left(\mu \right(x\left)\right)=(\omega ⚬\mu )\left(x\right)$. Let $y_1,y_2\in Y$. For every $x_1,x_2\in X$ with $\mu \left(x_1\right)=y_1$ and $\mu \left(x_2\right)=y_2$, we have $\omega (y_1\ast y_2)=\omega (\mu \left(x_1\right)\ast \mu \left(x_2\right))=\omega \left(\mu (x_1\ast x_2)\right)=\nu (x_1\ast x_2)=\nu \left(x_1\right)\ast \nu \left(x_2\right)=\omega \left(\mu \left(x_1\right)\right)\ast \omega \left(\mu \left(x_2\right)\right)=\omega \left(y_1\right)\ast \omega \left(y_2\right)$. Hence $\omega$ is an NA-morphism. The uniqueness of $\omega$ is directly verified since $\nu$ is an onto NA-morphism. This completes the proof.

In the next theorem, we consider the NA-morphism theorem.

Theorem 3.16.(NA-morphism theorem) Let $(S;\ast ,△,1)$ and $(T;\overline{\ast },\overline{△},\overline{1})$ be arbitrary NA-algebras. $\mu :S\to T$ is an onto NA-morphism where S is transitive and T is antisymmetric, then there exists an one-one and onto NA-morphism $\nu :S/ker\left(\mu \right)\to T$ , and so $S/ker\left(\mu \right)$ is isomorphic to T such that the following diagram commutes.

Proof. Recall that $ker\left(\mu \right)$ is an NA-ideal of S. Thus $S/ker\left(\mu \right)$ is an NA-algebra where $ker\left(\mu \right)=F$. Define a mapping $\nu :S/ker\left(\mu \right)\to T$ by $\nu \left(F_x\right)=\nu \left(x\right)$ for all $F_x\in S/ker\left(\mu \right)$. For every $y\in T$ , there exists $x\in S$ such that $\mu \left(x\right)=y$ since $\mu$ is onto. Thus $\nu \left(F_x\right)=\mu \left(x\right)=y$ which shows that $\nu$ is onto. Assume that $F_x=F_y\in S/ker\left(\mu \right)$. Then $x{\sim }_{ker\left(\mu \right)}y$, and so $x\ast y\in ker\left(\mu \right)$ and $y\ast x\in ker\left(\mu \right)$. It follows that $\mu \left(x\right)\ast \mu \left(y\right)=\mu (x\ast y)=1$ and $\mu \left(y\right)\ast \mu \left(x\right)=\mu (y\ast x)=1$. Since T is antisymmetric, we get $\nu \left(F_x\right)=\mu \left(x\right)=\mu \left(y\right)=\nu \left(F_y\right)$. Thus $\nu$ is a well-defined mapping. Suppose that $F_x\ne F_y\in S/ker\left(\mu \right)$. Then x and y cannot belong to the same equivalence class. Hence $x\ast y\notin ker\left(\mu \right)$ or $y\ast x\notin ker\left(\mu \right)$. If $x\ast y\notin ker\left(\mu \right)$, then $\mu \left(x\right)\ast \mu \left(y\right)=1$ and so $\mu \left(x\right)=\mu \left(y\right)$. If $y\ast x\notin ker\left(\mu \right)$, then $\mu \left(y\right)\ast \mu \left(x\right)=1$ and so $\mu \left(y\right)=\mu \left(x\right)$. Hence $\nu$ is an one-one mapping. Since $\nu (F_x\ast F_y)=\nu \left(F_{x\ast y}\right)=\mu (x\ast y)=\mu \left(x\right)\ast \mu \left(y\right)=\nu \left(F_x\right)\ast \nu \left(F_y\right)$ and $\nu (F_x△F_y)=\nu \left(F_{x△y}\right)=\mu (x△y)=\mu \left(x\right)△\mu \left(y\right)=\nu \left(F_x\right)△\nu \left(F_y\right)$ for all $F_x,F_y\in S/ker\left(\mu \right)$.

This completes the proof.

4. Relation Between NA-Algebras and Other Related Logical Algebras

This section aims to systematically position NA-algebras in the spectrum of non-classical logical algebras.

An algebra is called a bicommutative algebra [9], if it satisfies the following identities

$$x\left(yz\right)=y\left(xz\right),\left(xy\right)z=\left(xz\right)y.$$

One-sided commutative algebras first appeared in the paperby Cayley in 1857[10] . The variety of right commutative algebras is defined by the following identity:

$$\left(xy\right)z=\left(xz\right)y.$$

Similarly, the variety of left commutative algebras is defined by the following identity:

$$x\left(yz\right)=y\left(xz\right).$$

Thus, for a given NA-algebra $(S,\ast ,▵,1)$, $(S,\ast ,1)$ is a left commutative algebras, and $(S,▵,1)$ is a right commutative algebra. If two binary operations ∗ and △ coincide, then NA-algebra $(S,\ast ,▵,1)$ is a bicommutative algebra.

Meanwhile, NA-algebra $(S,\ast ,▵,1)$ has close connections with several important non-classical logical algebraic structures:

Relations with CI-algebras and Q-algebras: When operations in a NA-algebra satisfy the condition of power-zero (i.e. $x\ast x=1,\forall x\in S)$, its algebraic structure $(S,\ast ,1$ forms a CI-algebra, and its dual structure $(S,△,1)$ forms a Q-algebra. This shows that a NA-algebra which satisfies certain conditions be embedded into the category of CI/Q-algebras.

Relation with quantum B-algebras: This is a central conclusion of the paper. Any power-zero NA-algebra which satisfies $x\ast x=1$ forms a quantum B-algebra. This makes NA-algebras a subclass of the wide quantum B-algebraic framework (which contains-effect algebras, residuated lattices, etc.).

Relation with pseudo-BCH-algebras: In a NA-algebra, let $x\ast x=1$ and use its properties, one can derive a pseudo-BCH-algebra. The converse is not true, and there are examples which are pseudo-BCH-algebras but not-algebras, showing that pseudo-BCH-algebras are a broader class than NA-algebras.

Relation with RM-algebras: Also, a NA-algebra $(S,\ast ,1)$ which satisfies $x\ast x=1$ forms an RM-algebra. However, a general RM-algebra is not necessarily a NA-algebra. Therefore a NA-algebra which satisfies the power-zero condition forms a proper subclass of RM-algebras.

Below we discuss the relationship between NA-algebras and CI/Q-algebras, quantum B-algebras, pseudo BCH-algebras, RM-algebras, etc.

4.1. Relations with CI-Algebras and Q-Algebras

J.Neggers, S.S.Ahn and H.S. Kim introduced a new notion, called an Q-algebra, which is a generalization of the idea of BCH/BCI/BCK-algebras and they generalized some theorems discussed in BCI-algebras[14]. CI-algebras as a generalization of BE-algebras and BCK/BCI/BCH-algebras was initiated by B.L.Meng in 2009, and studied its important properties and relations with BE-algebras[15]. Arsham Borumand Saeid proved that CI-algebra is equivalent to dual Q-algebra[16].

In this section, we investigate the relation between NA-algebras and CI-algebras and Q-algebras.

Definition 4.1.([15]) An algebra $(X;\ast ,1)$ of type (2, 0) is called a CI-algebra if it satisfies the following axioms: for all $x,y,z\in X$

$\left(C{I}_1\right)x\ast x=1$ for all $x\in X$;

$\left(C{I}_2\right)1\ast x=x$ for all $x\in X$;

$\left(C{I}_3\right)x\ast (y\ast z)=y\ast (x\ast z)$, for all $x,y,z\in X$.

A CI–algebra X satisfying the condition $x\ast 1=1$ is called a BE-algebra. In any CI-algebra X one can define a binary relation $“\le ”$ by $x\le y$ if and only if $x\ast y=1$.

A CI-algebra X has the following properties:

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

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

$\left(iii\right)$ if $1\le x$, then $x=1$, for all $x,y\in X$.

Definition 4.2.([14]) An algebra $(X;\ast ,1)$ of type (2, 0) is called a Q-algebra if it satisfies the following axioms: for all $x,y,z\in X$

$\left(Q_1\right)x▵x=1$ for all $x\in X$;

$\left(Q_2\right)x▵1=x$ for all $x\in X$;

$\left(Q_3\right)(x▵y)▵z=(x▵z)▵y$ for all $x,y,z\in X$.

Definition 4.3.([14,18]) Let $(X,△,1)$ be a Q-algebra and binary operation $“\ast ”$ on X is defined as follows:

$$x△y=y\ast x.$$

Then $(X,\ast ,1)$ is called dual Q-algebra. In fact, the axioms of that are as follows:

$\left(D{Q}_1\right)x\ast x=1$;

$\left(D{Q}_2\right)1\ast x=x$;

$\left(D{Q}_3\right)x\ast (y\ast z)=y\ast (x\ast z)$, for all $x,y,z\in X$.

Example 8. Let $X=\{1,a,b,c\}$. We define binary operations △ on X with the following Tables 17 and 18:

Table 17. △ table.

△	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Table 17. △ table.

△	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

It is not complicated to check that $(X;△,1)$ is an Q-algebra, this is also a Klein four-group. By Definition 4.3, we can immediately write down the dual algebra of the Q-algebra $(X;\ast ,1)$ with the following Table 18:

Table 18. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Table 18. ∗ table.

∗	1	a	b	c
1	1	a	b	c
a	a	1	c	b
b	b	c	1	a
c	c	b	a	1

Obviously, here $\forall a,b\in X,a△b=a\ast b$. Therefore $(X;\ast ,1)$ is a self-dual Q-algebras.

Theorem 4.1.([16,17]) Any CI-algebra is equivalent to the dual Q-algebra.

Applying Theorem 2.1, based on Definition 4.1-4.3, for all $x,y,z\in S$, the following hold,

Theorem 4.2. Let $(S,\ast ,▵,1)$ be an NA-algebra and for all $x\in S$, it holds that $x\ast x=1$, then

(1) $(S,\ast ,1)$ is a CI-algebra;

(2) $(S,△,1)$ is a Q-algebra.

Proof. The proof is straightforward.

Definition 4.4.([29]) An algebra$(X;\ast ,△,1)$ of type (2,2,0) is called a pseudo-CI algebra if, for all $x,y,z\in X$, it satisfies the following axioms:

$\left(psC{I}_1\right)x\ast x=x△x=1$,

$\left(psC{I}_2\right)1\ast x=1△x=x$,

$\left(psC{I}_3\right)x\ast (y△z)=y△(x\ast z)$,

$\left(psC{I}_4\right)x\ast y=1\iff x△y=1$.

It follows from Definition 2.2 and Theorem 2.1 that the following conclusion is obtained easily:

Theorem 4.3. An NA-algebra $(S,\ast ,▵,1)$ if, for all $x\in S$, it satisfies $x\ast x=1$, then $(S,\ast ,▵,1)$ is a pseudo-CI algebra, i.e, every pseudo-CI algebra is contained in the class of NA-algebras.

4.2. Relation with Quantum B-Algebras

Rump and Yang introduced the concept of quantum B-algebras[19], and proved that the quantum B-algebras can provide a unified semantic for non-commutative algebraic logic. Almost all implicational algebras studied before - pseudo-effect algebras, residuated lattices, pseudo MV/BL/MTL-algebras, bounded non-commutative R-monoids, pseudo-hoops, pseudo BCK/BCI-algebras are quantum B-algebras. Now, let us proceed by stating the definitions, some of them being well known.

Definition 4.5.([19]) A quantum B-algebra is a partially ordered set X with two binary operations → and ⇝ such that the following are satisfied for $x,y,z\in X:$

$$y\le z\Rightarrow x\to y\le x\to z$$

(9)

$$x\le y\to z\iff y\le x⇝z$$

(10)

$$x\to (y⇝z)=y⇝(x\to z).$$

(11)

Both → and ⇝ stand for logical implication, a left one and a right one, which have to be distinguished in an non-commutative framework. The partial order relation ≤ stands for entailment.

Proposition 4.4. ([19,20]) Let $(X,\le ,\to ,⇝)$ be a quantum B-algebra. The following hold, for all $x,y,z\in X$,

$\left(1^{⚬}\right)$$x\le (x\to y)⇝y,x\le (x⇝y)\to y$;

$\left(2^{⚬}\right)$$y\le z$ implies $x⇝y\le x⇝z$;

$\left(3^{⚬}\right)$$x\le y$ implies $y\to z\le x\to z,y⇝z\le x⇝z$;

$\left(4^{⚬}\right)$$x\to y=\left(\right(x\to y)⇝y)\to y,x⇝y=\left(\right(x⇝y)\to y)⇝y$;

$\left(5^{⚬}\right)$$x\to y\le (y\to z)⇝(x\to z),x⇝y\le (y⇝z)\to (x⇝z)$.

Now, we study the relationship between NA-algebra and quantum B-algebras as follows.

Theorem 4.5. Let $(X,\ast ,△,1)$ be an NA-algebra which satisfies $x\ast x=1$ for any $x\in X$, we have

$\left(I_1\right)$ the order is given by $x\le y\iff x\ast y=1$(or $y△x=1)$ is a partial order.

$\left(I_2\right)$$(X,\ast ,△,1)$ is a trivial quantum B-algebra, the element x is group-like element.

Proof. $\left(I_1\right)$ 1)By $x\ast x=1$, we have $x\le x$;

2) Assume that $x\ast y=1=y\ast x$ holds for given $x,y\in X$. Then it implies that $x\ast (y\ast x)=(x\ast y)\ast x=1\ast x=x\Rightarrow x\ast 1=x\Rightarrow y=y\ast 1=y\ast (x\ast x)=x\ast (y\ast x)=x\ast 1=x$. Hence $x\le y,y\le x\Rightarrow x=y$.

3) Assume that $x,y\in X$ satisfies $x\ast y=1,y\ast z=1\Rightarrow 1=1\ast 1=(x\ast y)\ast (y\ast z)=x\ast \left(\right(y\ast y)\ast z)=x\ast (1\ast z)=x\ast z$.

Hence $x\le y,y\le z\Rightarrow x\le z.$ So the order relation $“\le ”$ is a partial order in X.

$\left(I_2\right)$ In the semigroup $(X,\ast ,1)$ of NA-algebra $(X,\ast ,△,1)$, for all $x,y\in X$, we have $x\ast y=y△x$ holds, hence for all $x,y,z\in X$,

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

(12)

Using equality (12), by ∗ instead of →, and use △ instead of ⇝, we obtain equality(11)is verified.

From that $y\le z\iff y\ast z=1\Rightarrow (x\to y)\to (x\to z)=(x\ast y)\ast (x\ast z)=x\ast \left(\right(x\ast y)\ast z)=(x\ast x)\ast (y\ast z)=1\ast 1=1.$ It follows that (9) $y\le z\Rightarrow x\to y\le x\to z.$

By $x\ast x=1\Rightarrow x\ast 1=x\to (x\ast x)=(x\ast x)\ast x=1\ast x=x\Rightarrow x\ast y=x\ast (y\ast 1)=y\ast (x\ast 1)=y\ast x=x△y$, i.e., $x\ast y=y\ast x=x△y)$. In this case, $x\to y=x⇝y$. and if $x\to (y\to z)=1$, then $y\to (x⇝z)=1$, the converse is also true. Thus, we obtain $x\le y\to z\iff y\le x⇝z.$ So, the inequality (10) is verified.

This proves $(X,\ast ,△,1)$ is a quantum B-algebra.

Furthermore, we can be obtain:

$$(x\ast y)△y=y\ast (x\ast y)=x\ast (y\ast y)=x\ast 1=x,$$

$$(x△y)\ast y=(y\ast x)\ast y=y\ast (x\ast y)=x\ast (y\ast y)=x\ast 1=x.$$

Therefore, $(X,\ast ,△,1)$ is a commutative quantum B-algebra, and the element x is group-like element.

Remark 6. In an NA-algebra $(S,\ast ,△,1)$, a order relation is defined by $“\le ”$ is defined by $x\le y\iff x\ast y=1$. We can prove that only if condition $x\ast x=1$ holds, this order relation is a partial order. Therefore, to construct a quantum B-algebra using NA-algebra, an additional condition $“x\ast x=1”$ must be imposed to ensure reflexivity.

The following example shows that a nilpotent NA-algenra is a quantum B-algebra.

Let $S=\{a,1\}$. Define the operations ∗ and △ on S by cayley Tables 19 and 20 below:

Table 19. ∗ table.

∗	a	1
a	1	a
1	a	1

Table 19. ∗ table.

∗	a	1
a	1	a
1	a	1

Table 20. ∗ table.

△	a	1
a	1	a
1	a	1

Table 20. ∗ table.

△	a	1
a	1	a
1	a	1

Here $(S,\ast ,△,1)$ is a NA-algebra. We putting $\ast =\to$,$△=⇝$, and $a=1$, then $(S,\to ,⇝,1)$ is also a quantum B-algebra.

In [20], Shengwei Han and Xiaoting Xu use the identity (11) to propose the concept of C-algebras and show that a C-algebra is a group if and only if each of its elements is dualizing and by dualizing elements of a C-algebra X define different binary operations on X such that X is a moniod. In this section, we will to define C-algebras and investigate the relation between C-algebras and NA-algebras.

Definition 4.6.([20]) A set X with two binary operations → and ⇝ is called a C-algebra if → and ⇝ satisfy the condition (11).

Theorem 4.6. Every NA-algebra $(S,\ast ,△,1)$ is a C-algebra.

Proof. By Theorem 2.1, we putting $\ast =\to$, $△=⇝$, for all $x,y,z\in S$, we have

$$x⇝(y\to z)=x△(y\ast z)=x△(z△y),$$

$$y\to (x⇝z)=y\ast (x△z)=(x△z)△y)=x△(z△y),$$

Hence, $x⇝(y\to z)=y\to (x⇝z)$. So, axiom (11) hold. The proof is complete.

4.3. Relation with Pseudo-BCH-Algebras

In order to characterize fuzziness, non-commutability or uncertainty in quantum logic, it is necessary to break through the rigid structures of traditional Boolean algebra, such as excluded neutrality and commutative law. Pseudo-prefix algebra, such as pseudo-Boolean algebra and pseudo-BCI algebra, has become an effective mathematical tool for describing complex logical relationships by weakening axiomatic conditions such as non-commutativity and non-associativity. 2015, Andrzej Walendziak introduced the concept of pseudo-BCH-algebras, which is a theoretical exploration in the evolution of non-commutative logic algebra. Its core value lies in extending the BC/BCI framework to adapt to non-commutative logic scenarios and filling the theoretical gap between strict Boolean algebra and completely free algebra[22]. Definition 4.7.([22]) A pseudo-BCH-algebra is an algebra $X=(X;\ast ,\diamond ,1)$ of type (2,2,0) satisfying the axioms:

(pBCH-1) $x\ast x=x\diamond x=1$;

(pBCH-2) $(x\ast y)\diamond z=(x\diamond z)\ast y$;

(pBCH-3) $x\ast y=y\diamond x=1\Rightarrow x=y$;

(pBCH-4) $x\ast y=1\iff x\diamond y=1$.

Replacing ⋄ by △ in Definition 4.7 we obtain

Remark 7. Observe that if $(S;\ast ,△,1)$ is a NA-algebra, letting $x\ast x:=1$. By Theorem 2.5 and Theorem 2.6, then we have $x\ast y=y\ast x=x△y$, and produces a pseudo-BCH-algebra $(S;\ast ,△,1)$. Its inverse is not generally true as shown in the following example.

Example 9. Let $X=\{1,a,b,c\}$. Define the operations ∗ and ⋄ on X by Tables 21 and 22 below:

Table 21. ∗ table.

∗	1	a	b	c
1	1	1	1	1
a	a	1	a	1
b	b	b	1	1
c	c	b	c	1

Table 21. ∗ table.

∗	1	a	b	c
1	1	1	1	1
a	a	1	a	1
b	b	b	1	1
c	c	b	c	1

Table 22. ⋄ table.

⋄	1	a	b	c
1	1	1	1	1
a	a	1	a	1
b	b	b	1	1
c	c	c	a	1

Table 22. ⋄ table.

⋄	1	a	b	c
1	1	1	1	1
a	a	1	a	1
b	b	b	1	1
c	c	c	a	1

Then, it is easy to verity that $(X;\ast ,\diamond ,1)$ is a pseudo-BCH-algebra, which is not a NA-algebras, since $b\ast c=1\ne a=c\diamond b$.

4.4. Relation with RM-Algebras

In 2019, Walendziak, A. introduced the notion of RM algebras and investigated its elementary properties[23]. The class of all RM algebras contains BCK, BCI, BCH, BZ, RME, pre-BZ algebras and many others.

In this section, for all $x,y,z\in X$, we apply the identity

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

to NA-algebras $(S,\ast ,△,1)$ and investigate some relations between this condition with other axioms in some algebras of logic and some examples are given to illustrate them. The relations between NA-algebras and RM-algebras are given.

Definition 4.8.([23])An RM-algebra is an algebra $(X,\to ,1)$ of type (2,0) satisfying the following axioms: for all $x,y\in X$

(R) $x\to x=1$,

(M) $1\to x=x$.

Example 10. Let $X=\{1,a,b\}$. Define the operations → on X with Cayley Table 23:

Table 23. → table.

→	1	a	b
1	1	a	b
a	a	1	b
b	b	b	1

Table 23. → table.

→	1	a	b
1	1	a	b
a	a	1	b
b	b	b	1

then $(X,\to ,1)$ is an RM-algebra and satisfies condition(I)[24], but $b\ast (a\ast b)=b\ast b=b\ast b=1\ne a\ast (b\ast b)=a\ast 1=a$, hence it does not satisfy

(E) $x\to (y\to z)=y\to (x\to z)$.

So, RM-algebra is not NA-algebra.

Theorem 4.7. If $(S,\ast ,△,1)$ is a NA-algebra with $x\ast x=1$, then $(S,\ast ,1)$ is an RM-algebra.

Proof. By $\left(N_3\right)$, we have (M) holds. Based on the assumption $x\ast x=1$, we obtain $(S,\ast ,1)$ is an RM-algebra.

Theorem 4.8. Let $(S,\ast ,△,1)$ be an NA-algebra. If it satisfies condition(I), for all $x,y,z\in S$, we have:

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

Proof. Let $x,y,z\in S$. By using $x\ast (y\ast z)=y\ast (x\ast z)$, if S satisfies condition(I), we have

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

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

Proposition 4.8. If NA-algebra $(S,\ast ,△,0)$ satisfies $x\ast x=1$ and $x\to 1=1$, then $S=\left\{1\right\}$.

Proof. Let $x\in S$. suppose $x\to x=1$, then $x=1\to x=(x\to x)\to x=x\to (x\to x)=x\to 1$. If S satisfies $x\to 1=1$, we have $x=1$, and $S=\left\{1\right\}.$

Proposition 4.9. Let $(S,\ast ,△,0)$ be an NA-algebra, we have

(1) if it satisfies condition(I), then $x\to 1=x$, for any $x\in S$.

(2) if for all $x,y,z\in S$, then $(z\ast x)\ast y=(z\ast y)\ast x$ iff $x\ast y=y\ast x$.

Proof. (1) For any $x\in S$, we apply identity (I) to NA-algebra $(S,\ast ,△,0)$, we obtain

$$x\ast 1=1\ast (x\ast 1)=(1\ast 1)\ast =1\ast x=x.$$

(2) Put $z=1$ with identity $(z\ast x)\ast y=(z\ast y)\ast x$, we get $(1\ast x)\ast y=(1\ast y)\ast x\iff x=y$. Conversely, if all $x,y\in S$, have $x\ast y=y\ast x$, then $(z\ast x)\ast y=y\ast (z\ast x)=z\ast (y\ast x)=(z\ast y)\ast x$. The proof is complete.

Proposition 4.10. ([24]) If $(X,\to ,1)$ is an RM-algebra satisfying (I), then (E) holds.

From theorem 4.7, Proposition 4.9 and Proposition 4.10, we have: The RM-algebras are a special subclass of the NA-algebras with $x\ast x=1$ for all $x\in S$.

According to the above analysis, the following diagram of inclusion relation can be outlined (where “Nilpotent NA” denotes a nilpotent NA-algebra satisfying $x\ast x=1$):

Figure 1. The relationships between NA-algebras and other algebras

Figure 1. The relationships between NA-algebras and other algebras

5. Conclusions and Future Work

In this paper, we have introduced the concept of NA-algebras and investigated some of their useful properties. It is well known that the ideals with special properties play an important role in the logic system. The aim of this article is to investigate ideals and stabilizers in NA-algebra, we establish the NA-morphism theorem. We characterize congruence kernels in a NA-algebra. Finally, the relationships between NA-algebras and other algebras, like Q-algebras, CI-algebras, GE-algebras and quantum B-algebra are also investigated. As a direction of research, one can investigate and extend these results to the other algebraic structures.

We believe that these results are very useful in developing algebraic structures also these definitions and main results can be similarly extended to some other algebraic systems such as lattices and Lie algebras etc. We hope this work would serve as a foundation for further studies on the structure of NA-algebras like fuzzy NA-algebras, soft NA-algebras and hyper NA-algebras.

NA-algebras have applications in the dialgebra theory. It is also closely connected with dimonoids and Digroups introduced by Loday[25,26,28]. Digroups which form a subclass of dimonoids, play an important role in the theory of Leibniz algebras[28]. Next, we will embed the NA-algebra into a vector space, deeply further study its relationship with other algebraic systems, and expand the field of NA-algebra.