Some Operators on Quantum B-Algebras

Abstract

The aim of this paper is to investigate several operators on quantum B-algebras. At first, we introduce closure and interior operators on quantum B-algebras and consider their relations on bounded quantum B-algebras. Furthermore, we discuss very true operators on quantum B-algebras by three cases via the unit element, and present some similar conclusions and different results. Finally, by constructing a very true operator on a quotient very true perfect quantum B-algebra, we establish a homomorphism theorem on very true perfect quantum B-algebras.

1. Introduction

Quantum B-algebras, the partially ordered implicational algebras arising as subreducts of quantales, were introduced by Rump and Yang [1]. They provided a uniform semantics for nearly all of non-commutative logical algebras, including pseduo-BCK-algebras, pseudo-MV-algebras, po-groups, quantales, BL-algebras, residuated lattices and GPE-algebras [1]. After that, many researchers studied this perfect algebraic structures from various aspects. Botur and Paseka [2] studied filters on quantum B-algebras and considered filters in pseudo-hoops, then they established an embedding of a cartesian product of polars of a pseudo-hoop into itself. Zhang-Borzooei-Jun [3] constructed the quotient structures by using normal q-filters in perfect quantum B-algebras and investigated the relation between basic implication algebras and quantum B-algebras. Han-Xu-Qin [4] constructed untial quantum B-algebra from non-unital quantum B-algebras. Motivated by Rump and Yang’s work [1], Han-Wang-Xu gave the injective hulls of quantum B-algebras in terms of the injective hulls of posemigroups in [5]. Moreover, through the framework of quantum B-algebras, Rump and Yang [6] transformed multiplication of ideals into composition of functions, then obtained a perfect fundamental theorem of arithmetic under non-commutative situation. Moreover, quantum B-algebras have two kinds of binary operations, they have symmetry, so when studying some properties of quantum B-algebras, in some cases, it is enough to study only one operation, which makes the study simple and effective. However, both operations are necessary, because quantum B-algebras are generally noncommutative.

As a tool to reduce the number of possible logical values in multi-valued fuzzy logic, the definition of very true operators was put forward by Hájek [7], it is actually the same idea of hedge that was proposed by Zadeh [8]. It presented a positive answer to the question of “whether natural axiomatization is possible and to what extent the standard methods of mathematical logic can capture this fuzzy logic”. Further, it has been successful in several different tasks in various logical algebras, such as MV-algebras [9], Rl-monoids [10], pseudo-BCK algebras [11], effect algebras [12] and equality algebras [13].

Based on above knowledge, very true operators were applied to many logical algebras and their non-commutative versions, so it could be significant to study very true operators on a unified algebraic structures. In fact, quantum B-algebras were the more general structure exactly. In [1], Rump and Yang classified three prototypes of non-commutative algebraic logic together with all their descendants were special cases of quantum B-algebras via the unit element. Motivated by the above, in this paper, we consider very true operators on quantum B-algebras in three cases and analyse their different properties, whose results will unify and enrich previous works on very true operators of quantum structure. In addition, since closure operators and interior operators play a important role in topology on universal algebras, the properties in this paper about closure operators may provide a basis for the future study of topology on quantum B-algebras.

This paper is organized as follows. After Preliminaries in Section 2 and Section 3 studies closure (interior) operators on quantum B-algebras, then the relationship between closure operators and interior operators on bounded quantum B-algebras (Theorem 1) is considered. Section 4 investigates very true operators on quantum B-algebras by three cases via the unit element and present some similar conclusions as well as different results. In Section 5, by giving a very true normal q-filter of a very true perfect quantum B-algebra, we construct quotient structures on very true perfect quantum B-algebras. Finally, we establish a homomorphism theorem on very true perfect quantum B-algebras (Theorem 3).

2. Preliminaries

We first review some basic knowledge about quantum B-algebras.

Definition 1.

Refs. [1,14] A quantum B-algebra is a quadruple $(X,\le ,\to ,⇝)$ such that the ordered pair $(X,\le )$ is a poset and both → and ⇝ are binary operations in X, satisfying the following conditions:

$$\begin{array}{ccc}& & \hfill y\to z\le (x\to y)\to (x\to z),\hfill \end{array}$$

(1)

$$\begin{array}{ccc}& & \hfill y⇝z\le (x⇝y)⇝(x⇝z),\hfill \end{array}$$

(2)

$$\begin{array}{ccc}& & \hfill y\le z\Rightarrow x\to y\le x\to z,\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& & \hfill x\le y\to z\iff y\le x⇝z\hfill \end{array}$$

(4)

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

Let X be a quantum B-algebra, if X has the unit element u, which satisfies $u\to x=u⇝x=x$ for all $x\in X$, then it is called unital. Notice that the unit element u is always unique. In fact, if $u,u^{{}^{\prime }}$ are unit elements, then $u\le u^{{}^{\prime }}\to u$, it implies that $u^{{}^{\prime }}\le u⇝u=u$. Similarly, we have $u\le u^{{}^{\prime }}⇝u^{{}^{\prime }}=u^{{}^{\prime }}$. So $u=u^{{}^{\prime }}$.

If u exists, then Definition 1 can be rewritten in another form:

$$\begin{array}{ccc}& & \hfill x⇝(y\to z)=y\to (x⇝z),\hfill \end{array}$$

(5)

$$\begin{array}{ccc}& & \hfill y\to z\le (x\to y)\to (x\to z),\hfill \end{array}$$

(6)

$$\begin{array}{ccc}& & \hfill x\le y\iff u\le x\to y\iff u\le x⇝y.\hfill \end{array}$$

(7)

It is well-known that the unit element u always represents the “true” proposition, that is, the true proposition has a fixed value. However, that does not happen for a unital quantum B-algebra. In fact, for a quantum B-algebra X, the element u is not necessarily the greatest element, but (7) can make us transform $x\le y$ into $u\le x\to y$.

Naturally, the quantum-B logic $\left(\mathrm{qB}\right)$ [14] can be shown by the following axioms:

$$\begin{array}{ccc}\hfill \left(B1\right)& & \hfill (\phi \to \omega )⇝\left(\right(\varphi \to \phi )\to (\varphi \to \omega \left)\right),\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill \left(C1\right)& & \hfill (\varphi \to (\phi ⇝\omega \left)\right)\to (\phi ⇝(\varphi \to \omega \left)\right),\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill \left(C2\right)& & \hfill (\varphi ⇝(\phi \to \omega \left)\right)⇝(\phi \to (\varphi ⇝\omega \left)\right),\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill \left(I\right)& & \hfill \varphi \to \varphi ,\hfill \end{array}$$

(11)

together with the following inference rules:

$$\begin{array}{ccc}\hfill \left(MP\right)& & \hfill \varphi ,\varphi \to \phi ⊢\phi ,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill \left(M{P}_1\right)& & \hfill \varphi \to \phi ⊢\varphi ⇝\phi ,\hfill \end{array}$$

(13)

$$\begin{array}{ccc}\hfill \left(M{P}_2\right)& & \hfill \varphi ⇝\phi ⊢\varphi \to \phi .\hfill \end{array}$$

(14)

Proposition 1.

Ref. [1] Assume that X is a quantum B-algebra. Then for all $x,y,z\in X$, we have:

$$\begin{array}{ccc}& & \hfill y\le z\Rightarrow x⇝y\le x⇝z,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}& & \hfill y\le z\Rightarrow z⇝x\le y⇝x,\hfill \end{array}$$

(16)

$$\begin{array}{ccc}& & \hfill y\le z\Rightarrow z\to x\le y\to x,\hfill \end{array}$$

(17)

$$\begin{array}{ccc}& & \hfill x\le (x⇝y)\to y,x\le (x\to y)⇝y,\hfill \end{array}$$

(18)

$$\begin{array}{ccc}& & \hfill x\to y=\left(\right(x\to y)⇝y)\to y;x⇝y=\left(\right(x⇝y)\to y)⇝y,\hfill \end{array}$$

(19)

$$\begin{array}{ccc}& & \hfill x\to (y⇝z)=y⇝(x\to z).\hfill \end{array}$$

(20)

Recall that for a quantum B-algebra X, if $x\to y=x⇝y$ for all $x,y\in X$, then X is called commutative. A subset Y of a quantum B-algebra X is called a subalgebra if Y is closed with respected to → and ⇝. A quantum B-algebra X is called bounded if X admits a smallest 0 and that implies it has a greatest element 1 [1].

For a bounded quantum B-algebra X, we define two negations: $x^{-}:=x\to 0$, $x^{\sim }:=x⇝0$ for all $x\in X$; denote by $\mathrm{Reg}\left(X\right)=\{x\in X|x^{-\sim }=x^{\sim -}=x\}$ the set of all regular elements of X and $\mathrm{Den}\left(X\right)=\{x\in X|x^{-\sim }=x^{\sim -}=1\}$ the set of all dense elements of X; suppose that X satisfies $x^{-\sim }=x^{\sim -}$ for all $x\in X$, then X is called a good quantum B-algebra; assume that X satisfies ${(x\to y)}^{-\sim }=x\to y^{-\sim }$ and ${(x⇝y)}^{-\sim }=x⇝y^{-\sim }$, then we say that X has the Glivenko property.

Definition 2.

Ref. [3] For a quantum B-algebra X, a nonempty subset F is called a normal q-filter if the following conditions hold for all $x,y\in X$:

$$\begin{array}{ccc}& & \hfill \forall x\in X,y\in F,y\le x\Rightarrow x\in F,\hfill \end{array}$$

(21)

$$\begin{array}{ccc}& & \hfill \forall x\in X,y\in F,y\to x\in F\Rightarrow x\in F,\hfill \end{array}$$

(22)

$$\begin{array}{ccc}& & \hfill \forall x\in F,x\to x\in F,x⇝x\in F,\hfill \end{array}$$

(23)

$$\begin{array}{ccc}& & \hfill x\to y\in F\iff x⇝y\in F.\hfill \end{array}$$

(24)

Remark 1.

Obviously, F is an upper set. If it only satisfies (21) and (22), then F is called a filter [2]. For all $x,y\in X$, if there exists a binary relation ${\theta }_F$ satisfies $x{\theta }_{F}y\iff x\to y\in F,y\to x\in F$, then ${\theta }_F$ is called a congruence relation on X [3]. For details about normal q-filters on X, we refer the reader to [3]. For a bounded quantum B-algebra X, denote by $NF\left(X\right)$ the set of normal q-filters, then we obtain $\mathrm{Den}\left(X\right)\subseteq NF\left(X\right)$.

Definition 3.

Ref. [1] Assume that $X_1$ and $X_2$ are two quantum B-algebras. We call $\gamma :X_1\to X_2$ a morphism if γ is a monotonic map which satisfies the following inequalities:

$$\begin{array}{cc}\hfill & \hfill \gamma (x\to y)\le \gamma \left(x\right)\to \gamma \left(y\right)\hfill \end{array}$$

(25)

for all $x,y\in X_1$.

Obviously, $\gamma$ satisfies $\gamma (x⇝y)\le \gamma \left(x\right)⇝\gamma \left(y\right)$. If the inequalities are equations, then $\gamma$ is called exact.

3. Closure (Interior) Operators on Quantum B-Algebras

In what follows, closure (interior) operators on quantum B-algebras are studied, and the relationship between closure operators and interior operators on bounded quantum B-algebras is considered. First, we give the definition of closure operators.

Definition 4.

Assume that X is a quantum B-algebra. A mapping $f:X\to X$ is called a closure operator on X if it satisfies the following conditions for all $x,y\in X$:

$$\begin{array}{ccc}\hfill \left(C{O}_1\right)& & \hfill x\le f\left(x\right),\left(increasing\right)\hfill \\ \hfill \left(C{O}_2\right)& & \hfill ff\left(x\right)=f\left(x\right),\left(idempotent\right)\hfill \\ \hfill \left(C{O}_3\right)& & \hfill x\le y\Rightarrow f\left(x\right)\le f\left(y\right).\left(monotone\right)\hfill \end{array}$$

Example 1.

Consider $X=\{0,p,q,r,1\}$, together with the order $0<p<r<1$, $0<p<q<1$ as the following Hasse diagram (Figure 1) shows. The binary relations “→” and “⇝” are given by

Table 1. → on X of Example 1.

→	0	p	q	r	1
0	1	1	1	1	1
p	p	1	q	1	1
q	0	p	1	1	1
r	0	p	q	1	1
1	0	p	q	r	1

and

Table 2. ⇝ on X of Example 1.

⇝	0	p	q	r	1
0	1	1	1	1	1
p	q	1	q	1	1
q	0	p	1	1	1
r	0	p	q	1	1
1	0	p	q	r	1

.

One can check that X is a bounded quantum B-algebra. Closure operators $f_i:X\to X,i=1,\dots ,4$ on X are given in

Table 3. Closure operators of Example 1.

f	0	p	q	r	1
$f_1\left(x\right)$	0	p	q	r	1
$f_2\left(x\right)$	p	p	q	r	1
$f_3\left(x\right)$	0	p	r	r	1
$f_4\left(x\right)$	0	p	q	1	1

:

Remark 2.

If we replace $\left(C{O}_1\right)$ by $f\left(x\right)\le x$ (decreasing), then f is called an interior operator.

In what follows we present some properties of closure operators on bounded quantum B-algebras.

Proposition 2.

Assume that X is a quantum B-algebra. For any closure operator f, g on X, we have $fg=g\iff f\le g$.

Proof. 

Let $x\in X$, if $f\le g$, then $fg\left(x\right)\le gg\left(x\right)=g\left(x\right)$, and by $\left(C{O}_1\right)$, we have $g\left(x\right)\le fg\left(x\right)$. Thus $fg=g$. Conversely, assume that $fg=g$, then $f\left(x\right)\le fg\left(x\right)=g\left(x\right)$, which implies that $f\le g$. □

Proposition 3.

Assume that X is a quantum B-algebra. Consider any closure operator f, g on X, then the following are equivalent:

(i) 

$fg=gf$,

(ii) 

fg and gf are closure operators,

(iii) 

$fgfg=fg$, $gfgf=gf$.

Proof. 

$\left(i\right)\Rightarrow \left(ii\right)$. Suppose that $fg=gf$, then for all $x,y\in X$ one could obtain:

(1)

$x\le g\left(x\right)\le fg\left(x\right)$,

(2)

$fgfg\left(x\right)=ffgg\left(x\right)=ffg\left(x\right)=fg\left(x\right)$,

(3)

$x\le y\Rightarrow g\left(x\right)\le g\left(y\right)\Rightarrow fg\left(x\right)\le fg\left(y\right)$.

It implies that $fg$ satisfies $\left(C{O}_1\right),\left(C{O}_2\right),\left(C{O}_3\right)$. Similarly, we can verify that $gf$ is a closure operator.

$\left(ii\right)\Rightarrow \left(iii\right)$. Since $fg$ and $gf$ are closure operators, by $\left(C{O}_2\right)$, we have $fgfg=fg$ and $gfgf=gf$.

$\left(iii\right)\Rightarrow \left(i\right)$. Assume that $fgfg=fg$ and $gfgf=gf$, then for all $x\in X$, we obtain

$$\begin{array}{ccc}& & \hfill gf\left(x\right)\le gfg\left(x\right)\le fgfg\left(x\right)=fg\left(x\right),\hfill \\ & & \hfill fg\left(x\right)\le fgf\left(x\right)\le gfgf\left(x\right)=gf\left(x\right).\hfill \end{array}$$

Therefore, $fg=gf$. □

Proposition 4.

Assume that X is a quantum B-algebra. Then for any closure operator f on X and for all $x,y\in X$, we have:

(i) 

$f\left(x\right)\to y\le x\to f\left(y\right),f\left(x\right)⇝y\le x⇝f\left(y\right)$,

(ii) 

$f\left(x\right)\to y\le f(x\to y),f\left(x\right)⇝y\le f(x⇝y)$.

If X is bounded, then:

(iii) 

$f\left(1\right)=1$,

(iv) 

$f{\left(x\right)}^{-}\le f\left(x^{-}\right),f{\left(x\right)}^{\sim }\le f\left(x^{\sim }\right)$,

(v) 

$x\le {\left(f\left(x^{-}\right)\right)}^{\sim },x\le {\left(f\left(x^{\sim }\right)\right)}^{-}$.

Proof. 

(i).

Since $x\le f\left(x\right)$, by (17), we can verify that $f\left(x\right)\to f\left(y\right)\le x\to f\left(y\right)$. Note that $y\le f\left(y\right)$, applying (3), we obtain that $f\left(x\right)\to y\le f\left(x\right)\to f\left(y\right)$. Hence, $f\left(x\right)\to y\le x\to f\left(y\right)$. Similarly, we have $f\left(x\right)⇝y\le x⇝f\left(y\right)$.

(ii).

By (17), we obtain that $f\left(x\right)\to y\le x\to y$, notice that $x\to y\le f(x\to y)$, so $f\left(x\right)\to y\le f(x\to y)$. Similarly, we can verify that $f\left(x\right)⇝y\le f(x⇝y)$.

(iii).

Since X is bounded, that is, $0\le x\le 1$, together with $x\le f\left(x\right)$, we have $1\le f\left(1\right)$. Hence $f\left(1\right)=1$.

(iv).

From (2), take $y=0$, we have $f\left(x\right)\to 0\le f(x\to 0)$, that is, $f{\left(x\right)}^{-}\le f\left(x^{-}\right)$. Similarly, we have $f{\left(x\right)}^{\sim }\le f\left(x^{\sim }\right)$.

(v).

Since $x\le (x\to y)⇝y$, take $y=0$, that is, $x\le x^{-\sim }$. Notice that $x^{-}\le f\left(x^{-}\right)$ and $x^{\sim }\le f\left(x^{\sim }\right)$, by (3), we have $x^{-}⇝0\le f\left(x^{-}\right)⇝0$, which implies that $x^{-\sim }\le {\left(f\left(x^{-}\right)\right)}^{\sim }$. Hence $x\le {\left(f\left(x^{-}\right)\right)}^{\sim }$. Similarly, we have $x\le {\left(f\left(x^{\sim }\right)\right)}^{-}$.

□

Proposition 5.

Assume that X is a good quantum B-algebra which satisfies Glivenko property, and for any closure operator f:

(i) 

Define $f_1:\mathrm{Reg}\left(X\right)\to \mathrm{Reg}\left(X\right)$ by $f_1\left(x\right)=f{\left(x\right)}^{-\sim }$, then $f_1$ is a closure operator;

(ii) 

Define $f_2:X/\mathrm{Den}\left(X\right)\to X/\mathrm{Den}\left(X\right)$ by $f_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)={\left[f\left(x\right)\right]}_{\mathrm{Den}\left(X\right)}$, then $f_2$ is a closure operator.

Proof. 

(i).

Assume that $x\in \mathrm{Reg}\left(X\right)$, note that $x\le f\left(x\right)$ and X is a good quantum B-algebra, then we obtain that $x=x^{-\sim }\le f{\left(x\right)}^{-\sim }=f_1\left(x\right)$, so $f_1$ satisfies $\left(C{O}_1\right)$. Suppose that $x,y\in \mathrm{Reg}\left(X\right)$ satisfy $f\left(x\right)\le f\left(y\right)$ and $f_1{\left(x\right)}^{-\sim }\le f_1{\left(y\right)}^{-\sim }$, that is, $f_1\left(x\right)\le f_1\left(y\right)$, which implies that $\left(C{O}_3\right)$ holds. At last, $f_1{f}_1\left(x\right)=f_1\left(f{\left(x\right)}^{-\sim }\right)={\left(f{\left(x\right)}^{-\sim }\right)}^{-\sim }=f{\left(x\right)}^{-\sim }=f_1\left(x\right)$. That is, $\left(C{O}_2\right)$ holds. Therefore, $f_1$ is a closure operator.

(ii).

We first check that $f_2$ is well defined. Since X satisfies Glivenko property, ${\left[x\right]}_{Den\left(X\right)}={\left[x^{-\sim }\right]}_{Den\left(X\right)}$ and ${\left[x\right]}_{Den\left(X\right)}={\left[y\right]}_{Den\left(X\right)}$ iff $x^{-\sim }=y^{-\sim }$. In fact, ${\left[x\right]}_{Den\left(X\right)}={\left[y\right]}_{Den\left(X\right)}$ iff $x\to y,y\to x\in Den\left(X\right)$ iff ${(x\to y)}^{-\sim }={(y\to x)}^{-\sim }=1$ iff $x^{-\sim }\to y^{-\sim }=y^{-\sim }\to x^{-\sim }=1$ iff $x^{-\sim }\le y^{-\sim }$ and $y^{-\sim }\le x^{-\sim }$ iff $x^{-\sim }=y^{-\sim }$.

Assume that ${\left[x\right]}_{Den\left(X\right)}={\left[y\right]}_{Den\left(X\right)}$, then we have $f_2\left({\left[x\right]}_{Den\left(X\right)}\right)=f_2\left({\left[x^{-\sim }\right]}_{Den\left(X\right)}\right)={\left[f\left(x^{-\sim }\right)\right]}_{Den\left(X\right)}={\left[f\left(y^{-\sim }\right)\right]}_{Den\left(X\right)}=f_2\left({\left[y^{-\sim }\right]}_{Den\left(X\right)}\right)=f_2\left({\left[y\right]}_{DenX}\right)$. So $f_2$ is well defined.

Since $x\le f\left(x\right)$, we have ${\left[x\right]}_{\mathrm{Den}\left(X\right)}\le {\left[f\left(x\right)\right]}_{\mathrm{Den}\left(X\right)}$, so ${\left[x\right]}_{\mathrm{Den}\left(X\right)}\le f_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)$, $\left(C{O}_1\right)$ holds. For any $x,y\in X$ such that $x\le y$, we have $f\left(x\right)\le f\left(y\right)$, so ${\left[f\left(x\right)\right]}_{\mathrm{Den}\left(X\right)}\le {\left[f\left(y\right)\right]}_{\mathrm{Den}\left(X\right)}$, that is, $f_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)\le f_2\left({\left[y\right]}_{\mathrm{Den}\left(X\right)}\right)$, $\left(C{O}_3\right)$ holds. Also, $ff\left(x\right)=f\left(x\right)$ and ${\left[ff\left(x\right)\right]}_{\mathrm{Den}\left(X\right)}={\left[f\left(x\right)\right]}_{\mathrm{Den}\left(X\right)}$. So $f_2{f}_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)=f_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)$, $\left(C{O}_2\right)$ holds. Hence, $f_2$ is a closure operator. □

We now consider the relationship between closure operators and interior operators on bounded quantum B-algebras.

Theorem 1.

Assume that X is a bounded quantum B-algebra, if f is an interior operator, then mappings $f^p\left(x\right)={\left(f\left(x^{-}\right)\right)}^{\sim }$ and $f^q\left(x\right)={\left(f\left(x^{\sim }\right)\right)}^{-}$ are closure operators on X.

Proof. 

Note that $x\le x^{-\sim }\le {\left(f\left(x^{-}\right)\right)}^{\sim }$, then $\left(C{O}_1\right)$ holds. $f^p\left(f^p\left(x\right)\right)=f^p\left({\left(f\left(x^{-}\right)\right)}^{\sim }\right)={\left(f\left({\left(f\left(x^{-}\right)\right)}^{\sim -}\right)\right)}^{\sim }\le {\left(f\left(f\left(x^{-}\right)\right)\right)}^{\sim }={\left(f\left(x^{-}\right)\right)}^{\sim }=f^p\left(x\right)$, so $\left(C{O}_2\right)$ holds. Let $x,y$ satisfy $x\le y$, then $y^{-}\le x^{-}$ and $f\left(y^{-}\right)\le f\left(x^{-}\right)$, hence $f^p\left(x\right)={\left(f\left(x^{-}\right)\right)}^{\sim }\le {\left(f\left(y^{-}\right)\right)}^{\sim }=f^p\left(y\right)$. That is, $\left(C{O}_3\right)$ holds. So $f^p$ is a closure operator. Similarly, we have $f^q$ is a closure operator on X. □

Remark 3.

However, assume that X is a bounded quantum B-algebra; if f is a closure operator, then mappings ${\left(f\left(x^{-}\right)\right)}^{\sim }$ and ${\left(f\left(x^{\sim }\right)\right)}^{-}$ need not be interior operators on X. In fact, if we consider Example 1, note that $f_2$ is a closure operator on X, however, ${\left(f_2\left(p^{-}\right)\right)}^{\sim }={\left(f_2\left(p\right)\right)}^{\sim }=p⇝0=q\nleqq p$, that is, ${\left(f_2\left(p^{-}\right)\right)}^{\sim }$ is not an interior operator on X, and ${\left(f_3\left(r^{\sim }\right)\right)}^{-}={\left(f_3\left(0\right)\right)}^{-}=0\to 0=1\nleqq r$ is not an interior operator on X.

4. Very True Quantum B-Algebras

In what follows, we investigate very true operators on quantum B-algebras and study their properties. Motivated by [1], we classify very true quantum B-algebras into three cases via the unit element. Notice that a quantum B-algebra is not always unital. See Example 2.

Example 2.

Ref. [4] Consider the commutative quantum B-algebra $(X_1,\le ,\to ,⇝)$, where $X_1=\{0,p,q,1\}$ is a poset determined by Figure 2 and operation $\to =⇝$ by

Table 4. → on $X_1$ of Example 2.

→	0	p	q	1
p	0	p	0	1
q	0	0	q	1
1	0	0	0	1

:

Obviously, $X_1$ is not unital.

Next, we study very true quantum B-algebras in three cases via the unit element.

Case 1: If the unit element is the greatest one, then the so-called pseudo-BCK algebras, MV-algebras, the algebras of Łukasiewicz’ infinite-valued logic, and their non-commutative extensions are included in this case. At first, we give the concept of very true operators of quantum B-algebras in this case.

Definition 5.

Assume that X is a bounded quantum B-algebra whose unit element is the greatest one. We call $\alpha :X\to X$ a very true operator on X if the following conditions hold for all $x,y\in X$:

$$\begin{array}{ccc}\hfill \left(V{T}_1\right)& & \hfill \alpha \left(1\right)=1,\hfill \\ \hfill \left(V{T}_2\right)& & \hfill \alpha \left(x\right)\le x,\hfill \\ \hfill \left(V{T}_3\right)& & \hfill \alpha \left(x\right)\le \alpha \alpha \left(x\right),\hfill \\ \hfill \left(V{T}_4\right)& & \hfill \alpha (x\to y)\le \alpha \left(x\right)\to \alpha \left(y\right),\alpha (x⇝y)\le \alpha \left(x\right)⇝\alpha \left(y\right).\hfill \end{array}$$

We call $(X,\alpha )$ a very true quantum B-algebra, and if $\alpha$ is a very true operator of a bounded quantum B-algebra X, then $\mathrm{Ker}\left(\alpha \right)=\{x\in X|\alpha \left(x\right)=1\}$ is called the kernel of $\alpha$.

Remark 4.

We illustrate above definition from a logical point of view. $\left(V{T}_1\right)$ means that absolutely true is very true, which is the standard axiom for obtaining classical logic. $\left(V{T}_2\right)$ stands for ‘suppose x is very true then it is true.’ $\left(V{T}_3\right)$ represents that very true in very true is very true, and this is an inevitability about very true connections. $\left(V{T}_4\right)$ shows that if both x, $x\to y$ and $x⇝y$ are very true then y is very true too.

Example 3.

Ref. [5] Assume that $X=\{0,p,q,r,1\}$ satisfies the following table (

Table 5. → on X of Example 3.

→	0	p	q	r	1
0	1	1	1	1	1
p	0	1	q	r	1
q	0	p	1	r	1
r	r	1	1	1	1
1	0	p	q	r	1

) and Hasse diagram (Figure 3):

One can check that X is a bounded and commutative quantum B-algebra.

Through calculation, we have very true operators: ${\alpha }_i:X\to X,i=1,\dots ,4$, which given by

Table 6. Very true operators of Example 3.

$\mathit{\alpha }$	0	p	q	r	1
${\alpha }_1\left(x\right)$	0	p	q	r	1
${\alpha }_2\left(x\right)$	0	1	q	r	1
${\alpha }_3\left(x\right)$	0	p	1	r	1
${\alpha }_4\left(x\right)$	0	1	1	r	1

:

Proposition 6.

Assume that $(X,\alpha )$ is a very true quantum B-algebra in this case, then we have:

(1)

$\alpha \left(0\right)=0$,

(2)

$\alpha \left(x\right)=1\iff x=1$,

(3)

$x\le y\Rightarrow \alpha \left(x\right)\le \alpha \left(y\right)$,

(4)

$\alpha \alpha \left(x\right)=\alpha \left(x\right)$,

(5)

$\alpha \left(x\right)\le y\iff \alpha \left(x\right)\le \alpha \left(y\right)$.

Proof. 

(1)

By $\left(V{T}_2\right)$, we have $\alpha \left(0\right)\le 0$, so $\alpha \left(0\right)=0$.

(2)

If $\alpha \left(x\right)=1$, then by $\left(V{T}_2\right)$, we have $1=\alpha \left(x\right)\le x$, so $x=1$. Conversely, using $\left(V{T}_1\right)$, we have $\alpha \left(x\right)=1$.

(3)

Since each unital quantum B-algebra admits a pseudo-BCK subalgebra, and there is a natural partial order: $x\le y\iff x\to y=x⇝y=1.$ Using $\left(V{T}_4\right)$, we have $1=\alpha \left(1\right)=\alpha (x\to y)\le \alpha \left(x\right)\to \alpha \left(y\right)$, it implies that $\alpha \left(x\right)\to \alpha \left(y\right)=1$, hence $\alpha \left(x\right)\le \alpha \left(y\right)$.

(4)

By $\left(V{T}_2\right)$, we have $\alpha \alpha \left(x\right)\le \alpha \left(x\right)$, and by $\left(V{T}_3\right)$, we obtain $\alpha \alpha \left(x\right)=\alpha \left(x\right)$.

(5)

If $\alpha \left(x\right)\le \alpha \left(y\right)$ and by $\left(V{T}_2\right)$, then we have $\alpha \left(x\right)\le \alpha \left(y\right)\le y$. If $\alpha \left(x\right)\le y$, then $\alpha \left(x\right)=\alpha \alpha \left(x\right)\le \alpha \left(y\right)$.

□

Theorem 2.

Assume that X is a bounded quantum B-algebra and $\alpha ,\beta$ are two very true operators on X. Then $\alpha \beta ,\beta \alpha$ are very true operators if and only if $\alpha \beta =\beta \alpha$.

Proof. 

$(\Rightarrow )$. Let $\alpha ,\beta$ be two very true operators on X such that $\alpha \beta$ and $\beta \alpha$ are very true operators. By Proposition 6 (4), we have $\alpha \beta \alpha \beta =\alpha \beta$, $\beta \alpha \beta \alpha =\beta \alpha$. Using $\left(V{T}_2\right)$, we obtain

$$\alpha \beta \left(x\right)=\alpha \beta \alpha \beta \left(x\right)\le \beta \alpha \beta \left(x\right)\le \beta \alpha \left(x\right)$$

$$\beta \alpha \left(x\right)=\beta \alpha \beta \alpha \left(x\right)\le \alpha \beta \alpha \left(x\right)\le \alpha \beta \left(x\right).$$

So $\alpha \beta =\beta \alpha$.

$(\Leftarrow )$. Conversely, suppose that $\alpha$ and $\beta$ are two very true operators such that $\alpha \beta =\beta \alpha$. Then we have:

$\alpha \beta \left(1\right)=\alpha \left(1\right)=1$ i.e. $\left(V{T}_1\right)$ holds

$\alpha \beta \left(x\right)\le \beta \left(x\right)\le x$ i.e. $\left(V{T}_2\right)$ holds.

As for $\left(V{T}_3\right)$, in fact, we obtain that $\alpha \beta \left(x\right)\le \alpha \beta \alpha \beta \left(x\right)$. This is because $\alpha$ and $\beta$ are very true operators on X, then $\alpha \left(x\right)=\alpha \alpha \left(x\right)$, so $\alpha \beta \left(x\right)=\alpha \alpha \beta \left(x\right)=\alpha \alpha \beta \beta \left(x\right)=\alpha \beta \alpha \beta \left(x\right)$, obviously, $\alpha \beta \left(x\right)\le \alpha \beta \alpha \beta \left(x\right)$, that is, $\left(V{T}_3\right)$ holds.

Since $\alpha$ and $\beta$ are very true operators on X, so $\alpha \beta (x\to y)=\alpha \left(\beta \right(x\to y\left)\right)\le \alpha \left(\beta \right(x)\to \beta (y\left)\right)\le \alpha \left(\beta \right(x\left)\right)\to \alpha \left(\beta \right(y\left)\right)=\alpha \beta \left(x\right)\to \alpha \beta \left(y\right)$. Analogously, we have $\alpha \beta (x⇝y)\le \alpha \beta \left(x\right)⇝\alpha \beta \left(y\right)$, thus $\left(V{T}_4\right)$ holds.

Hence, $\alpha \beta =\beta \alpha$ is a very true operator on X. □

Similar to the result given in Section 3 that about Glivenko property related to closure operators, the following proposition shows that it also has a close connection to very true operators.

Proposition 7.

Assume that X is a good bounded quantum B-algebra whose unit is the greatest one and satisfies Glivenko property, then for any very operator α:

(1) 

Define ${\alpha }_1:\mathrm{Reg}\left(X\right)\to \mathrm{Reg}\left(X\right)$ by ${\alpha }_1\left(x\right)=\alpha {\left(x\right)}^{-\sim }$, then ${\alpha }_1$ is a very true operator on $\mathrm{Reg}\left(X\right)$.

(2) 

Define ${\alpha }_2:X/\mathrm{Den}\left(X\right)\to X/\mathrm{Den}\left(X\right)$ by ${\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)={\left[\alpha \left(x\right)\right]}_{\mathrm{Den}\left(X\right)}$, then ${\alpha }_2$ is a very true operator on $X/\mathrm{Den}\left(X\right)$.

Proof. 

(1)

Obviously, $\mathrm{Reg}\left(X\right)$ is a bounded quantum B-algebra whose unit is the greatest one. Suppose that $x\in \mathrm{Reg}\left(X\right)$, since $\alpha \left(x\right)\le x$, ${\alpha }_1\left(x\right)={\left(\alpha \left(x\right)\right)}^{-\sim }\le x^{-\sim }=x$, so $\left(V{T}_2\right)$ holds. ${\alpha }_1\left(1\right)=(1\to 0)⇝0=1$, that is, $\left(V{T}_1\right)$ holds. As for $\left(V{T}_3\right)$, we can verify that ${\alpha }_1\left(x\right)={\alpha }_1{\alpha }_1\left(x\right)$. This is because ${\alpha }_1{\alpha }_1\left(x\right)={\alpha }_1\left(\alpha {\left(x\right)}^{-\sim }\right)={\left(\alpha {\left(x\right)}^{-\sim }\right)}^{-\sim }=\alpha {\left(x\right)}^{-\sim }={\alpha }_1\left(x\right)$, obviously, ${\alpha }_1\left(x\right)\le {\alpha }_1{\alpha }_1\left(x\right)$, that is, $\left(V{T}_3\right)$ holds. Note that X has Glivenko property and $\alpha$ is a very true operator on X. Thus we have ${\alpha }_1(x\to y)=\alpha {(x\to y)}^{-\sim }\le {(\alpha \left(x\right)\to \alpha \left(y\right))}^{-\sim }=\alpha {\left(x\right)}^{-\sim }\to \alpha {\left(y\right)}^{-\sim }$. Similarly, we have ${\alpha }_1(x⇝y)\le {\alpha }_1\left(x\right)⇝{\alpha }_1\left(y\right)$, so $\left(V{T}_4\right)$ holds. Therefore, ${\alpha }_1$ is a very true operator on $\mathrm{Reg}\left(X\right)$.

(2)

Similar to Proposition 5 (2), we can check that ${\alpha }_2$ is well defined. Here, we omit the proof. We just prove ${\alpha }_2$ is a very true operator.

Obviously, ${\alpha }_2\left({\left[1\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)={\left[1\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}$, $\left(V{T}_1\right)$ holds. Let $x\in X$, since ${\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)\le {\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}$, so ${\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)\le {\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}$, that is, $\left(V{T}_2\right)$ holds. For $x\in X$, $\alpha \alpha \left(x\right)=\alpha \left(x\right)$ and ${\left[\alpha \alpha \left(x\right)\right]}_{\mathrm{Den}\left(X\right)}={\left[\alpha \left(x\right)\right]}_{\mathrm{Den}\left(X\right)}$. So ${\alpha }_2{\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)={\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(X\right)}\right)$, hence $\left(V{T}_3\right)$ holds. Further, ${\alpha }_2({\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\to {\left[y\right]}_{\mathrm{Den}\left(\mathrm{X}\right)})={\alpha }_2\left({[x\to y]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)={\left[\alpha (x\to y)\right]}_{\mathrm{Den}\left(\mathrm{X}\right)})\le {[\alpha \left(x\right)\to \alpha \left(y\right)]}_{\mathrm{Den}\left(\mathrm{X}\right)})={\left[\alpha \left(x\right)\right]}_{\mathrm{Den}\left(\mathrm{X}\right)})\to {\left[\alpha \left(y\right)\right]}_{\mathrm{Den}\left(\mathrm{X}\right)})={\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)\to {\alpha }_2\left({\left[y\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)$.

Similarly, we have ${\alpha }_2({\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}⇝{\left[y\right]}_{\mathrm{Den}\left(\mathrm{X}\right)})\le {\alpha }_2\left({\left[x\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)⇝{\alpha }_2\left({\left[y\right]}_{\mathrm{Den}\left(\mathrm{X}\right)}\right)$, so $\left(V{T}_4\right)$ holds.

Hence, ${\alpha }_2$ is a very true operator on $X/\mathrm{Den}\left(X\right)$. □

Case 2: If the unit element is placed in an intermediate position, then this class quantum B-algebras consist of the monoids $(X,·)$ with two residuals and

$$\begin{array}{c}\hfill x·y\le z\iff x\le y\to z\iff y\le x⇝z,\end{array}$$

(26)

holds, the so called residuated poset [15]. Moreover, if X is a lattice, then X is called a residuated lattice [16].

Notice that (26) can be defined by the equivalent:

$$\begin{array}{c}\hfill a\to b=\bigvee \{c\in Q:c·a\le b\},a⇝b=\bigvee \{c\in Q:a·c\le b\},\end{array}$$

(27)

for Q is a quantale [17] where $a,b\in Q$. That is, every quantale can be viewed as a residuated lattice, and hence it is a quantum B-algebra.

Consider the concept of very true operators of quantum B-algebras in this case, $\left(V{T}_1\right)$ in Definition 5 can be replaced by $\left(V{T}_1^{\prime }\right):\alpha \left(u\right)=u$. So, we can analyse this case using the same method as Case 1.

Case 3: Assume that the unit element is the smallest one.

In this case, it consists of effect algebras [18] and their non-commutative versions. The unit element u of $\tilde{E}$ is the smallest element of E, where E denotes a generalized pseudo-effect algebra [19], not of $\tilde{E}$ where 0 is the smallest element. Note that u is only the smallest possible unit element, but $u=0$ is not possible, in general, u is bigger than 0, that is $0<u$.

Consider the concept of very true operator of quantum B-algebras in this case, we also apply Definition 5, that is, if $\beta$ is a very true operator of a bounded quantum B-algebra whose unit element is the smallest one, then it has to be satisfied $\left(V{T}_1\right)-\left(V{T}_4\right)$.

Although the concept of closure operators in this case is the same with Case 1, in the following we observe that there exist some differences in the concrete nature.

Remark 5.

Obviously, one can check that $\left(1\right),\left(2\right),\left(4\right),\left(5\right)$ and $\left(6\right)$ in Proposition 6 hold for all $x,y\in X$ in this case. Moreover, we have

$$\beta \left(x^{-}\right)\le \beta {\left(x\right)}^{-},\beta \left(x^{\sim }\right)\le \beta {\left(x\right)}^{\sim }.$$

(28)

In fact, since $x^{-}=x\to 0$, by $\left(V{T}_4\right)$ and $\left(1\right)$, we have $\beta \left(x^{-}\right)=\beta (x\to 0)\le \beta \left(x\right)\to \beta \left(0\right)=\beta \left(x\right)\to 0=\beta {\left(x\right)}^{-}$. Similarly, we have $\beta \left(x^{\sim }\right)\le \beta {\left(x\right)}^{\sim }$.

However, (3) in Proposition 6 does not holds in this case. See Example 4.

Example 4.

Consider Example 1 in [12]. An effect algebra E is determined by $E=\{0,a,b,c,d,e,f,g,1\}$ together with operation · and Hasse diagram are given by

Table 7. · on E of Example 4.

·	0	a	b	c	d	e	f	g	1
0	0	a	b	c	d	e	f	g	1
a	a	−	d	e	−	−	1	−	−
b	b	d	g	f	−	1	−	e	−
c	c	e	f	−	1	−	−	−	−
d	d	−	−	1	−	−	−	−	−
e	e	−	1	−	−	−	−	−	−
f	f	1	−	−	−	−	−	−	−
g	g	−	e	−	−	−	−	1	−
1	1	−	−	−	−	−	−	−	−

and Figure 4:

According to [1], if $a·b$ is undefined, then we stipulate $a·b=1$. And then $a\le b\iff \exists c\in E:c·a=b$, so we can check that it is a partial order, also, if we put $a\to b:=c$, then by Proposition 20 in [1], which implies that E together with $\{0,1\}$ is a commutative quantum B-algebra.

Consider the very true operator $\beta :E\to E$ by the following

Table 8. A very true operator of Example 4.

x	0	a	b	c	d	e	f	g	1
$\beta \left(x\right)$	0	a	b	c	a	b	c	b	1

:

We can check that $c\le e$, but $\beta \left(c\right)\nleqq \beta \left(e\right)$.

5. Very True Normal Q-Filters and Very True Perfect Quantum B-Algebras Homomorphism

In what follows, we study very true normal q-filters and very true homomorphisms of a very true perfect quantum B-algebra $(X,\alpha )$. By giving a very true normal q-filter F of a very true perfect quantum B-algebra $(X,\alpha )$, we construct a very true operator on a quotient quantum B-algebra $X/F$. First, we recall the definition of perfect quantum B-algebras.

Definition 6.

Ref. [3] A quantum B-algebra X is called perfect if the following conditions hold:

(1) 

for $x,y\in X$, $\exists a\in X$, ${[x\to y]}_F={[a\to a]}_F$⇔$\exists b\in X$, ${[x⇝y]}_F={[b⇝b]}_F$, where F is a normal q-filter,

(2) 

$(X/\sim ,{\to }_F,{⇝}_F,\le )$ is also a quantum B-algebra, where ${\to }_F$ and ${⇝}_F$ are defined: ${\left[x\right]}_F{\to }_F{\left[y\right]}_F:={[x\to y]}_F$ and ${\left[x\right]}_F{⇝}_F{\left[y\right]}_F:={[x⇝y]}_F$; ≤ is defined: ${\left[x\right]}_F\le {\left[y\right]}_F\iff$$\exists a\in X$, ${[x\to y]}_F={[a\to a]}_F$⇔$\exists b\in X$, ${[x⇝y]}_F={[b⇝b]}_F$.

Analogously to Section 4, in what follows, very true normal q-filters of perfect quantum B-algebras are considered by three cases.

Case 1: In the case that the unit element in the quantum B-algebras is the greatest one.

Definition 7.

Assume that $(X,\alpha )$ is a very true perfect quantum B-algebra, where α is given by Definition 5. We call F a very true normal q-filter of a perfect quantum B-algebra X if $\alpha \left(F\right)\subseteq F$. Denote by $N{F}^{\alpha }\left(X\right)$ the set of all $\alpha -$normal q-filters of X.

Example 5.

(1) 

Trivially, X and $Ker\left(\alpha \right)$ are very true normal q-filters, where α is a very true operator of a perfect quantum B-algebras X.

(2) 

(Ref. [3], Example 2). Consider a perfect quantum B-algebra $X_1$ with: $q\le p\le 1$, $t\le s\le r$, and operations → and ⇝ are defined as

Table 9. → on $X_1$ of Example 5 (2).

→	p	q	r	s	t	1
p	1	p	r	r	r	1
q	1	1	r	r	r	1
r	r	r	1	p	q	r
s	r	r	1	r	1	r
t	r	r	1	1	1	r
1	p	q	r	s	t	1

and

Table 10. ⇝ on $X_1$ of Example 5 (2).

⇝	p	q	r	s	t	1
p	1	p	r	r	s	1
q	1	1	r	r	r	1
r	r	r	1	p	p	r
s	r	r	1	1	p	r
t	r	r	1	1	1	r
1	p	q	r	s	t	1

.

Then we have very true operators ${\alpha }_i:X_1\to X_1,i=1,\dots ,9$, see

Table 11. Very true operators of Example 5 (2).

x	p	q	r	s	t	1
${\alpha }_1\left(x\right)$	p	q	r	s	t	1
${\alpha }_2\left(x\right)$	p	q	r	r	s	1
${\alpha }_3\left(x\right)$	p	p	r	r	s	1
${\alpha }_4\left(x\right)$	p	q	r	r	r	1
${\alpha }_5\left(x\right)$	1	q	r	r	r	1
${\alpha }_6\left(x\right)$	p	p	r	r	r	1
${\alpha }_7\left(x\right)$	1	p	r	r	r	1
${\alpha }_8\left(x\right)$	p	1	r	r	r	1
${\alpha }_9\left(x\right)$	1	1	r	r	r	1

:

Notice that $F_1=\left\{1\right\},F_2=\{a,b,1\},F_3=\left\{X_1\right\}$ are all normal q-filters of $X_1$, so we obtain $N{F}^{{\alpha }_1}\left(X_1\right)=N{F}^{{\alpha }_2}\left(X_1\right)=N{F}^{{\alpha }_4}\left(X_1\right)=NF\left(X_1\right)$, $N{F}^{{\alpha }_3}\left(X_1\right)=N{F}^{{\alpha }_5}\left(X_1\right)=N{F}^{{\alpha }_6}\left(X_1\right)=N{F}^{{\alpha }_7}\left(X\right)=N{F}^{{\alpha }_8}\left(X_1\right)=N{F}^{{\alpha }_9}\left(X_1\right)=\left\{\left\{1\right\}\right\}$.

Proposition 8.

Assume that $(X_1,\alpha )$ is a very true perfect quantum B-algebra and $F\in N{F}^{\alpha }\left(X\right)$. Define $\overline{\alpha }:X/F\to X/F$ by $\overline{\alpha }\left({\left[x\right]}_F\right)={\left[\alpha \left(x\right)\right]}_F$. Then $\overline{\alpha }$ is a very true operator on $X/F$.

Proof. 

Firstly, we verify that $\overline{\alpha }$ is well-defined. In fact, if ${\left[x\right]}_F={\left[y\right]}_F$, then by the definition of normal q-filter, we have $x\to y,y\to x\in F$. Note that $F\in N{F}^{\alpha }\left(X\right)$, by means of Definition 7, it implies that $\alpha (x\to y),\alpha (y\to x)\in F$. Since F is an upper set, by $\left(V{T}_4\right)$, we have $\alpha \left(x\right)\to \alpha \left(y\right),\alpha \left(y\right)\to \alpha \left(x\right)\in F$, it follows that ${\left[\alpha \left(x\right)\right]}_F={\left[\alpha \left(y\right)\right]}_F$, hence $\overline{\alpha }$ is well-defined on $X/F$.

Obviously, $\left(V{T}_1\right)$ holds. Applying $\left(V{T}_1\right)$, we obtain $\overline{\alpha }\left({\left[x\right]}_F\right)={\left[\alpha \left(x\right)\right]}_F\le {\left[x\right]}_F$, so $\left(V{T}_2\right)$ is satisfied. For all $x\in X$, we have $\overline{\alpha }\left({\left[x\right]}_F\right)={\left[\alpha \left(x\right)\right]}_F\le {\left[\alpha \alpha \left(x\right)\right]}_F=\overline{\alpha }\left({\left[\alpha \left(x\right)\right]}_F\right)=\overline{\alpha }\overline{\alpha }\left({\left[x\right]}_F\right)$, so $\left(V{T}_3\right)$ holds. $\overline{\alpha }({\left[x\right]}_F\to {\left[y\right]}_F)=\overline{\alpha }\left({[x\to y]}_F\right)={\left[\alpha (x\to y)\right]}_F\le {[\alpha \left(x\right)\to \alpha \left(y\right)]}_F={\left[\alpha \left(x\right)\right]}_F\to {\left[\alpha \left(y\right)\right]}_F=\overline{\alpha }\left({\left[x\right]}_F\right)\to \overline{\alpha }\left({\left[y\right]}_F\right)$. Similarly, we can verify that $\overline{\alpha }({\left[x\right]}_F⇝{\left[y\right]}_F)\le \overline{\alpha }\left({\left[x\right]}_F\right)⇝\overline{(}{\left[y\right]}_F)$. That is, $\left(V{T}_4\right)$ holds. So we complete the proof. □

Definition 8.

Assume that $(X_1,\alpha )$ and $(X_2,\beta )$ are two very true perfect quantum B-algebras, $\gamma :X_1\to X_2$ is an exact morphism. Then we call γ a very true perfect quantum B-algebra homomorphism if $\gamma \left(\alpha \right(x\left)\right)=\beta \left(\gamma \right(x\left)\right)$.

Denote by $VHOM((X_1,\alpha ),(X_2,\beta ))$ the set of all very true perfect quantum B-algebra homomorphisms. For $\gamma \in VHOM((X_1,\alpha ),(X_2,\beta ))$, $Ker\left(\gamma \right)=\{x\in X|\gamma \left(x\right)=1\}$ is called the kernel of $\gamma$.

Proposition 9.

Assume that $(X_1,\alpha )$ and $(X_2,\beta )$ are two very true perfect quantum B-algebras together with $\gamma \in VHOM((X_1,\alpha ),(X_2,\beta ))$. Then we have:

(i) 

if $F\in N{F}^{\beta }\left(X_2\right)$, then ${\gamma }^{-1}\left(F\right)\in N{F}^{\alpha }\left(X_1\right)$,

(ii) 

if γ is surjective and $F\in N{F}^{\alpha }\left(X_1\right)$, then $\gamma \left(F\right)\in N{F}^{\beta }\left(X_2\right)$.

Proof. 

(i)

Let $F\in N{F}^{\beta }\left(X_2\right)$, obviously, ${\gamma }^{-1}\left(F\right)\in N{F}^{\alpha }\left(X_1\right)$. If $x\in {\gamma }^{-1}\left(F\right)$, then $\gamma \left(x\right)\in F$, which implies that $\beta \left(\gamma \right(x\left)\right)\in F$, so $\gamma \left(\alpha \right(x\left)\right)\in F$, then $\alpha \left(x\right)\in {\gamma }^{-1}\left(F\right)$, that is, ${\gamma }^{-1}\left(F\right)\in N{F}^{\alpha }\left(X_1\right)$.

(ii)

Observing that $\gamma$ is surjective, then for $F\in N{F}^{\alpha }\left(X_1\right)$, we have $\gamma \left(F\right)\in NF\left(X_2\right)$. For $x\in F$, $y\in \gamma \left(F\right)$, then $\gamma \left(x\right)=y$. Notice that $F\in N{F}^{\alpha }\left(X_1\right)$ i.e. $\alpha \left(x\right)\in F$, so $\gamma \left(\alpha \right(x\left)\right)=\gamma \left(F\right)$, which implies that $\beta \left(y\right)=\beta \left(\gamma \right(x\left)\right)=\gamma \left(\alpha \right(x\left)\right)\in \gamma \left(F\right)$. Hence $\gamma \left(F\right)\in N{F}^{\beta }\left(X_2\right)$.

□

Proposition 10.

Suppose that $(X,\alpha )$ is a very true perfect quantum B-algebra together with $F\in N{F}^{\alpha }\left(X\right)$. Define $\pi :X\to X/F$ by $\pi \left(x\right)={\left[x\right]}_F$ and $\overline{\alpha }:X/F\to X/F$ by $\overline{\alpha }\left({\left[x\right]}_F\right)={\left[\alpha \left(x\right)\right]}_F$. Then we have $\pi \in VHOM((X,\alpha ),(X/F,\overline{\alpha }))$.

Proof. 

From Proposition 8, we have $\overline{\alpha }$ is a very true operator of $X/F$. Clearly, $\pi \in HOM(X,X/F)$ and so $\pi \left(\alpha \left(x\right)\right)={\left[\alpha \left(x\right)\right]}_F=\overline{\alpha }\left({\left[x\right]}_F\right)=\overline{\alpha }\left(\pi \left(x\right)\right)$. This completes the proof. □

The following theorem presents the same result about very true perfect quantum B-algebras which is similar to the conclusion of the homomorphism theorem in universal algebras.

Theorem 3.

Suppose that $(X_1,\alpha )$ and $(X_2,\beta )$ are two very true perfect quantum B-algebras and $\gamma \in VHOM((X_1,\alpha ),(X_2,\beta ))$. Consider $F\in N{F}^{\alpha }\left(X_1\right)$ such that $F\subseteq Ker\left(\gamma \right)$, and let $\pi \in VHOM((X_1,\alpha ),(X_1/F,\overline{\alpha }))$, where $\overline{\alpha }$ is given in Proposition 10. Then we can find a unique $\overline{\gamma }\in VHOM((X_1/F,\overline{\alpha }),(X_2,\beta ))$ satisfying

Proof. 

Define $\overline{\gamma }:X_1/F\to X_2$ by $\overline{\gamma }\left({\left[x\right]}_F\right)=\gamma \left(x\right)$. We first prove $\overline{\gamma }$ is well-defined. In fact, for $x,y\in X_1$ such that ${\left[x\right]}_F={\left[y\right]}_F$, which follows that $x\to y,y\to x\in F\subseteq Ker\left(\gamma \right)$, that is,

$$1=\gamma (x\to y)=\gamma \left(x\right)\to \gamma \left(y\right),$$

$$1=\gamma (y\to x)=\gamma \left(y\right)\to \gamma \left(x\right).$$

So $\gamma \left(x\right)\le \gamma \left(y\right)$ and $\gamma \left(y\right)\le \gamma \left(x\right)$ which implies that $\gamma \left(x\right)=\gamma \left(y\right)$. So $\overline{\gamma }$ is well-defined.

Since $\gamma \in VHOM((X_1,\alpha ),(X_2,\beta ))$, we have $\overline{\gamma }\left(\overline{\alpha }\left({\left[x\right]}_F\right)\right)=\overline{\gamma }\left({\left[\alpha \left(x\right)\right]}_F\right)=\gamma \left(\alpha \left(x\right)\right)=\beta \left(\gamma \left(x\right)\right)=\beta \left(\overline{\gamma }\left({\left[x\right]}_F\right)\right)$. Thus $\overline{\gamma }\in VHOM((X_1/F,\overline{\alpha }),(X_2,\beta ))$, and clearly, $\overline{\alpha }\pi =\gamma$.

Assume that there exists $\beta \in VHOM((X_1,\overline{\alpha }),(X_2,\beta ))$ such that $\overline{\alpha }\pi =\beta \pi$, next we prove $\beta =\overline{\gamma }$. In fact, note that $\pi$ is surjective, then for $y\in X_1/F$, we can find $x\in X_1$ such that $y=\pi \left(x\right)$, observing that $\overline{\gamma }\left(\pi \left(x\right)\right)=\beta \left(\pi \left(x\right)\right)$, so $\beta =\overline{\gamma }$. □

Case 2: In the case that the unit element in quantum B-algebras is placed in an intermediate position. Note that a quantum B-algebra X is a partially ordered set, if it is a lattice-ordered, then we call X a lattice-ordered quantum B-algebra. Since any residuated poset is a quantum B-algebra, if it is a residuated lattice, then the join and meet operations given by $x\vee y=sup\{x,y\},x\wedge y=inf\{x,y\}$.

Assume that X is a lattice-ordered perfect quantum B-algebra, we observe that many conclusions of very true operators in Case 1 also hold in this case. However, we have a more profound result.

Proposition 11.

Assume that X is a lattice-ordered perfect quantum B-algebra and 1 is the greatest element on X. Then a very true perfect quantum B-algebra $(X,\alpha )$ is a subdirect product of a family ${\{X_i,{\alpha }_i\}}_{i\in I}$ of perfect very true quantum B-algebras if and only if there is a family ${\left\{F_i\right\}}_{i\in I}$ of normal q-filters of X such that $\left(i\right):X_i\cong X/F_i$ and $\left(ii\right):{\bigcap }_{i\in I}{F}_i=\left\{1\right\}$.

Proof. 

Assume that $(X,\alpha )$ is a subdirect product of a family ${\{X_i,{\alpha }_i\}}_{i\in I}$ of perfect very true quantum B-algebras. Let $h:X\to {\prod }_{i\in I}{X}_i$ (homomorphism respectively) and ${\pi }_i:{\prod }_{i\in I}{X}_i\to X_i$ ($ith$ projection function), for each $i\in I$, let $F_i=Ker\left({\pi }_{i}h\right)$, then we have $h{\left(x\right)}_i\cong X/F_i$. If $x\in {\bigcap }_{i\in I}{F}_i$, then ${\pi }_i\left(h\left(x\right)\right)=1$ for each $i\in I$. So $h\left(x\right)=1$, and since h is injective, we have $x=1$, that is, ${\bigcap }_{i\in I}{F}_i=\left\{1\right\}$.

Conversely, let $F_i$ be a family of normal q-filters of $X_i$ such that $\left(i\right)$ and $\left(ii\right)$. Let $h:X\to {\prod }_{i\in I}{X}_i$ be defined by ${\left(h\left(x\right)\right)}_i=x/F_i$, which follows that $Ker\left(h\right)=\left\{1\right\}$, that is, h is injective. Since for each $i\in I$ and $\xi \in X/F_i$, there exists $x\in X$ such that $\xi =x/F_i$, which follows that ${\pi }_{i}h:X\to X/F_i$ is a surjective homomorphism. So X is a subdirect product of a family ${\{X_i,{\alpha }_i\}}_{i\in I}$ of very true perfect quantum B-algebras. □

Case 3: In the case that the unit element in quantum B-algebras is the smallest one.

In this case, the definition of a very true operator of a quantum B-algebra also uses Definition 5. Observe that it is not order preserving from Example 4, but we can also analyse the very true perfect quantum B-algebra similar to Case 1, and we have many similar results, so we omit this process.

6. Conclusions

In this paper, we study closure (interior) operators and very true operators on quantum B-algebras. We first study the relationship between closure operators and interior operators on bounded quantum B-algebras. Then we investigate very true operators on quantum B-algebras by three cases via the unit element and present some similar conclusions as well as different results. At last, by giving a very true normal q-filter of a very true perfect quantum B-algebra, we construct quotient structures on very true perfect quantum B-algebras. Finally, we establish a homomorphism theorem on very true perfect quantum B-algebras.

As a future work, it makes sense to study topological structure on quantum B-algebras, since closure operators have a close relation to topology.