# On Soft Quantum B-Algebras and Fuzzy Soft Quantum B-Algebras

Authors: Xiongwei Zhang; Sultan Aljahdali; Ahmed Mostafa Khalil
Journal: Journal of Mathematics
 (2021)

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## Abstract

This paper aims to make a combination between the quantum B-algebras (briefly,X-As) and two interesting theories (e.g., soft set theory and fuzzy soft set theory). Firstly, we propose the novel notions of soft quantum B-algebras (briefly, SℚB-As), a soft deductive system of ℚB-As, and deducible soft quantum B-algebras (briefly, DSℚB-As). Then, we discuss the relationship between SℚB-As and DSℚB-As. Furthermore, we investigate the union and intersection operations of DSℚB-As. Secondly, we introduce the notions of a fuzzy soft quantum B-algebras (briefly, FSℚB-As), a fuzzy soft deductive system of ℚB-As, and present some characterizations of FSℚB-As, along with several examples. Finally, we explain the basic properties of homomorphism image of FSℚB-As.


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## Body

## 1. Introduction

In 1999, Molodtsov [1] introduced the notion called soft sets (briefly, SS) (i.e., which reduce the uncertainty and vagueness of knowledge). Maji et al. [2] presented the fuzzy soft sets (briefly, FSS). Since then, many researchers studied further on SS and FSS as in the following published articles (e.g., [3–9]).In 2014, Rump and Yang [10] proposed the notion of ℚB-As (i.e., a partial ordered implication algebras). Rump [11, 12] investigated many implication algebras (for example, pseudo-BCK-algebras, po-groups, BL-algebras, MV-algebras, GPE-algebras, and resituated lattices). Botur and Paseka [13] studied filters on integral ℚB-As, and Zhang et al. [14] established the quotient structures by using q-filters in ℚB-As and investigated the relation between basic implication algebras and ℚB-As. Han et al. [15] constructed the unitality of ℚB-As and explained the injective hulls of ℚB-As in [16]. By the framework of ℚB-As, there are many published papers on ℚB-As (e.g., [17–23]).Regarding these developments, as the motivation of this paper, we will combineℚB-As with SS and FSS (i.e., enrich the previous work on hybrid soft set and fuzzy soft set theories algebras with quantum structures). We introduce the notions of SℚB-As and the soft deductive system of ℚB-As and consider the relation between SℚB-As and DSℚB-As. Furthermore, some conditions are given to ensure the operations union and intersection holds of soft deductive of ℚB-As. Then, we investigate the homomorphism image of deductive SℚB-As. Lastly, we define FSℚB-As and fuzzy soft deductive system of ℚB-As and give an example to illustrate its derive properties.In the following, we have arranged the sections as follows. In Section2, we briefly recall many notions related to ℚB-As, SS, and FSS as indicated in Definitions 1–7, which are used in the sequel. In Section 3, we propose the notions of SℚB-As, soft deductive system of ℚB-As, and DSℚB-As. In Section 4, we present the notions of FSℚB-As and a fuzzy soft deductive system of ℚB-As and discuss the homomorphism image of FSℚB-As. The conclusions are explained in Section 5.


## 2. Preliminaries

We give some basic notions ofℚB-As, SS, and FSS before defining SℚB-As in Section 3.Definition 1 (cf. [10]).
(1)
ℚB-As is a partially ordered set X,≤ with two binary operations ⟶ and ⇝ which satisfy (∀x,y,z∈X):(1)y⟶z≤x⟶y⟶x⟶z,y⇝z≤x⇝y⇝x⇝z,y≤z⟹x⟶y≤x⟶z,x≤y⟶z⟺y≤x⇝z.(2)
ℚB-A is a commutative (briefly, ℂℚB-A) if x⟶y=x⇝y∀x,y∈X.(3)
A subsetY of a ℚB-AX is a subalgebra if x⟶y,x⇝y∈Y∀x,y∈X.
In what follows, denote byX a ℚB-A unless otherwise specified.Definition 2 (cf. [10]).
LetX1 and X2 be two ℚB-As. Then, ψ:X1⟶X2 is a morphism of ℚB-As if it satisfies (∀x,y∈X1):(2)ψx⟶y≤ψx⟶ψy,ψx⇝y≤ψx⇝ψy.
We say morphismψ is exact if the inequalities become equations.Definition 3 (cf. [1]).
Assume thatX be a set and K be a set of parameters. SK (called SS) is a mapping given by S:K⟶2X (i.e., 2X is the power set of X).Definition 4 (cf. [3]).
Assume thatSK1 and SK2 are two SS over X. SK1 is a subset of SK2 (denoted by SK1⊂˜SK2) if(1)
K1⊂K2(2)
For everyk∈K1,SK1k and SK2k are identical approximationsDefinition 5 (cf. [3]).
Assume thatSK1,SK2, and SK3 are three SS over X. SK3 is the intersection of SK1 and SK2 (denoted by SK3=SK1∩˜SK2) if(1)
K3=K1∩K2(2)
∀k∈K3,SK3k=SK1k or SK2k (as both are same sets)Definition 6 (cf. [3]).
Assume thatSK1,SK2, and SK3 are three SS over X. SK3 is called the union of SK1 and SK2 (denoted by SK3=SK1∪˜SK2) if(1)
K3=K1∪K2.(2)
k∈K3,(3)SK3k=SK1k,k∈K1\K2SK2k,k∈K2\K1,SK1k∪SK2k,k∈K1∩K2.Definition 7 (cf. [2]).
FSS (called FSS) S^K is a mapping given by S^:K⟶IX (i.e., IX is the set of all fuzzy sets [24] of X).


## 3.SℚB-As

We define theSℚB-As and give several examples based on SℚB-As. Also, we will study the union and intersection operations between two SℚB-As as follows .Table 1
The binary operation⟶.
⟶k1k2k31k11k1k11k211k11k311k111k1k2k31Table 2
The binary operation⟶.
⟶k1k2k31k11111k2k1k211k3k1k1111k1k1k31Definition 8.
SK is a SℚB-As over X if SKx∀x∈K are subalgebras of X (i.e., in case K=X).Example 1.
(1)
SupposeX (i.e., X=k1,k2,k3,1) with the order k2,k3<k1<1. Now, we show, by Table 1, the binary operation ⟶.Clearly,X is a ℂℚB-A. We define SK∀x∈K (i.e., K=X) by(4)SKx=y∈X|x⟶y⟶y∈k1,1.From Table1, we can get on SKk1=X,SKk2=SK,k3=k1,k3,1,andSK1=X, and then, SKxx∈K are all subalgebras of X. Consequently, SK is a SℚB-As over X.(2)
SupposeX (i.e., X=k1,k2,k3,1) with the order k1<k2<k3<1. Now, we show, by Table 2, the binary operation ⟶.
Clearly,X is a ℂℚB-A. We define SK∀x∈K (i.e., K=X) by(5)SKx=y∈X|xℛy⟺x⟶x⟶y∈k3,1.
From Table2, we can get on HTML translation failed, and then, SKxx∈K are all subalgebras of X. Consequently, SK is a SℚB-As over X.
We ensure the operations (i.e., union and intersection) are holding onSℚB-As by the following suggested theorem.Theorem 1.
Assume thatSK1 and SK2 are SℚB-As over X. Then,(1)
IfK3=K1∩K2, then SK3=SK1∩˜SK2 is called a SℚB-A over X(2)
IfK1∩K2=∅, then SK1∪˜SK2 is called a SℚB-A over XProof.
(1)
IfK3=K1∩K2 and by Definition 5, we obtain SK3x=SK1x or SK3x=SK2x, for all x∈K3. Since SK1 and SK2 are SℚB-As over X, which implies that SK3 is a SℚB-As over X, that is, SK3x=SK1x or SK3x=SK2x are both subalgebras of X∈K3, therefore, SK3=SK1∩˜SK2 is a SℚB-A over X.(2)
IfK3=K1∪K2 and by Definition 6, we obtain(6)SK3x=SK1x,x∈K1\K2,SK2x,x∈K2\K1,SK1x∪SK2x,x∈K1∩K2.
Forx∈K1\K2 and since SK1 is a SℚB-A, then we have SK3x=SK1x is a subalgebra of X. Similarly, for x∈K2\K1, then SK3x=SK2x is a subalgebra of X due to SK2 is a SℚB-A. Again, for K1∩K2=∅, so x∈K1∩K2 or x∈K2∩K1, for all x∈K3. Thus, SK3=SK1∪˜SK2 is a SℚB-A over X.Remark 1.
IfK1∩K2≠∅, then Theorem 1 (2) does not hold by the following example.Example 2.
SupposeX (i.e., X=0,k1,k2,k3,k4,1). Now, we show, by Tables 3 and 4, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. Then,(i)
We defineSK1∀x∈K1 (i.e., K1=X) by(7)SK1x=y∈X|xℛy⟺x⟶x⟶yx⇝x⇝y∈k3,k4,1.From Table3, we can get SK10=XandSK1k1=SK1k2=SK1k3=SK1k4=SK11=k3,k4,1, and then, SK1xx∈K1 are all subalgebras of X. Consequently, SK1 is a SℚB-As over X.(ii)
We defineSK2∀x∈K2 (i.e., K2=k2) by(8)SK2x=y∈K2|xℛy⟺x⟶y=k2,x⇝y=k2.From Table4, we can get SK2k2=k2 is the subalgebra of X. Consequently, SK2 is a SℚB-As over X.
From (i) and (ii) andK1∩K2=k2≠∅, then we have SK3k2=SK1k2∪SK2k2=k3,k4,1∪k2=k2,k3,k4,1 is not a subalgebra over X. Thus, SK3 is not a SℚB-A.Table 3
The binary operation⟶.
⟶0k1HTML translation failedk3k410111111k10k20k411k200k2k4k41k3000111k4000k4111000k4k41Table 4
The binary operation⇝.
⇝0k1k2k3k410111111k1000111k20k1k2k3k41k3000111k40001111000k3k41


### 3.1. Soft Deductive Systems ofSℚB-As

Based on Definition8, we will propose the notion of soft deductive systems of SℚB-As as indicated below.Definition 9.
Assume thatX=X,⟶,⇝,≤ be a SℚB-A. A nonempty subset D⊆X is a deductive system of X if it satisfies(1)
∀x∈D,x⟶x∈D,x⇝x∈D(2)
∀x,y∈X,x∈D,x⟶y∈D⟹y∈DDefinition 10.
LetX be a SℚB-A and Y a subalgebra of X. A subset D of X is a deductive system of X related to Y (i.e., Y-deductive system of X), denoted by D⋈Y, and satisfies the following two conditions:(1)
∀x∈D,x⟶x∈D,x⇝x∈D(2)
∀y∈Y,x∈D,x⟶y∈D⟹y∈DRemark 2.
According to Definitions9 and 10, we obtain that any deductive system of X is Y-deductive system if Y is a subalgebra of X.
The converse of Remark2 does not hold by Example 3 (i.e., Y is a subalgebra of X and Y-deductive system is not a deductive system).Example 3.
SupposeX (i.e., X=0,k1,k2,k3,1) with partial order 0<k1<k3<1 and 0<k1<k2<1. Now, we show, by Tables 5 and 6, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. Consider a subalgebra Y=k1,1 and a subset D=k1,k2,1; we can see that D⋈Y. However, D is not a deductive system of X since k3⟶1=1∈D and k3∉D.Table 5
The binary operation⟶.
⟶0k1k2k31011111k101k211k2k1k1111k30k1k21110k1k2k31Table 6
The binary operation⇝.
⇝0k1k2k31011111k1k21k211k20k1111k30k1k21110k1k2k31Definition 11.
Assume thatSK is a SℚB-A over X. SD (i.e., SS) over X is a soft deductive system of SK, denoted by SD⊳⊲˜SK, and satisfies the following two conditions:(1)
D⊆K(2)
∀x∈D,SDx⊳⊲SKx
Now, we will give an example to illustrate Definition11 as follows.Example 4.
SupposeX (i.e., X=k1,k2,k3,k4,1) with partial order k1<k2<k3<k4<1. Now, we show, by Tables 7 and 8, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. We define SK∀x∈K (i.e., K=X) by(9)SKx=y∈X|xℛy⟺x⟶y⇝y=1.
From Tables7 and 8, we can get on SKk1=SKk2=1,SKk3=k2,1,SKk4=k2,k3,1,andSK1=X, and then, SKxx∈K are all subalgebras of X. Consequently, SK is a SℚB-As over X.
Next, for a subsetD=k2,k4, we define SD∀x∈D by(10)SDx=1∪y∈X|y≤x.
Then, we obtainSDk2=k1,k2,1⊳⊲1=SKk2 and SDk4=X⊳⊲k2,k3,1=SKk4. Consequently, SD is a soft deductive system of SK.Table 7
The binary operation⟶.
⟶k1k2k3k41k111111k2k31111k3k2k2111k4k2k2kc111k1k2k3k41Table 8
The binary operation⇝.
⇝k1k2k3k41k111111k2k41111k3k2k2111k4k1k2k3111k1k2k3k41Theorem 2.
Assume thatSK is a SℚB-A over X and SD1 and SD2 are two SS. Then,(1)
IfD1∩D2≠∅, then SD1⊳⊲˜SK,SD2⊳⊲˜SK⟹SD1∩˜SD2⊳⊲˜SK(2)
IfD1∩D2=∅, then SD1⊳⊲˜SK,SD2⊳⊲˜SK⟹SD1∪˜SD2⊳⊲˜SKProof.
(1)
Follow from Definition5.(2)
IfSD1⊳⊲˜SK,SD2⊳⊲˜SK, then, by Definition 6, we have D3=D1∩D2 (i.e., x∈D3), SD1∪˜SD2=SD3, and(11)SD3x=SD1x,x∈D1\D2SD2x,x∈D2\D1,SD1x∪SD2x,x∈D1∩D2.
SinceD1∩D2=∅, we obtain either x∈D1\D2 or x∈D2\D1. Then, we have the following:Case 1: ifx∈D1\D2, since SD1⊳⊲˜SK, then SD3x=SD1x⊳⊲SKxCase 2: ifx∈D2\D1 and SD2⊳⊲˜SK, then SD3x=SD2x⊳⊲SKx
Consequently, for allx∈D3, we have SD3x⊳⊲SKx, which implies that SD1∪˜SD2=SD3⊳⊲˜SK.Remark 3.
IfK1∩K2≠∅, then Theorem 2 (2) does not hold by the following example.Example 5.
SupposeX (i.e., X=0,k1,k2,k3,k4,1). Now, we show, by Table 9, the binary operations ⟶.
Clearly,X is a ℂℚB-A. Then,(i)
We defineSK∀x∈K (i.e., K=X) by(12)SKx=y∈X|xℛy⟺x⟶y⟶y∈k1,k2,1.From Table9, we can get SK0=k1,k2,1,SKk1=X,SKk2=SKk3=k1,1,andSKk4=k1,k2,k3,1,SK1=X, and then, SKx∀x∈K are all subalgebras of X. Consequently, SK is a SℚB-As over X.(ii)
We defineSK1∀x∈K1 (i.e., K1=k1,k2,k3) by(13)SK1x=y∈X|xℛy⟺x⟶y=1.Then, we can getSK1k1=k1,1⊳⊲X=SKk1,SK1k2=k2,1⊳⊲a,1=SKk2,andSK1k3=k2,k3,1⊳⊲k1,1. Therefore, SK1 is a soft deductive system over SK.(iii)
We defineSK2∀x∈K2 (i.e., K2=k1) by(14)SK2x=y∈X|xℛy⟺y⟶x=k1.Then, we can getSK2k1=k2,k3,1⊳⊲X=SK2k1. Therefore, SK2 is a soft deductive system over SK.
From (i)–(iii), we haveSK3=SK1∪˜SK2 which is not a soft deductive system of SK, where SK3k1=SK1k1∪SK2k1=k1,k2,k3,1 is not a SKa-deductive system because k2⟶k4=k1∈k1,k2,k3,1 and k4∉k1,k2,k3,1.Table 9
The binary operation⟶.
⟶0k1k2k3k410111111k1k31k2k3k21k2k4k11k2k11k3k1k111k11k4k211k21110k1k2k3k41


### 3.2.DSℚB-As

We will give the notion ofDSℚB-As and investigate homomorphism image of DSℚB-As as indicated below.Definition 12.
Assume thatSK is a SℚB-A over X. If SKx∀x∈K is a deductive system of X, then SK is called a DSℚB-A over X.Example 6 (continued from Example1 (2)).
Clearly,SK is DSℚB-A over X.Definition 13.
(1)
SupposeX be a ℚB-A with the greatest element 1 (i.e., X just only a poset); for any x∈X, the order of element x is defined as(15)Ox=minp,q∈N|x⟶px=1,x⇝qx=1,i,whereN is a natural number and x⟶px=x⟶x⟶⋯⟶x,x⇝qx=x⇝x⇝⋯⇝x.(2)
Ifp,q∈N does not exist to satisfy the above condition (i), then x∀x∈X is called infinite order.Remark 4.
Assume thatSK and SK1 be two SℚB-As over X such that K1⊆K⊆X. If SK is a DSℚB-A over X, then SK1 is a DSℚB-A.
The converse of Remark4 does not hold by the following Example 7.Example 7 (continued from Example2).
We defineSK∀x∈K (i.e., K=X) by(16)SKx=y∈X|Ox=Oy.
Then, we get onSK0=SKk3=SKk4=SKk1=0,k3,k4,1,SKk1=k1,andSKk2=k2. However, k3⟶k1=0∈0,k3,k4,1 and k1∉0,k3,k4,1 imply that SK is not DSℚB-A. If we take K1=k3,k4,1⊆K and we define SK1=y∈X|Ox=Oy∀x∈K1, then SK1 is DSℚB-A.Definition 14.
Assume thatSK is SℚB-A over X with the greatest element 1. If SKx=X∀x∈K, then SK is called whole DSℚB-A.Example 8.
SupposeX (i.e., X=0,k1,k2,1) with partial order 0<k1<k2<1. Now, we show, by Tables 10 and 11, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. We define SK∀x∈K (i.e., K=X) by(17)SKx=y∈X|Ox=Oy.
From Tables10 and 11, we can get on SKx=X∀x∈K. Thus, SK is a whole DSℚB-A over X.
Now, we will study homomorphism image ofDSℚB-As by the following two theorems.Table 10
The binary operation⟶.
⟶0k1k2101111k1k1111k2k1k11110k1k21Table 11
The binary operation⇝.
⇝0k1k2101111k1k2111k20k11110k1k21Theorem 3.
Assume thatψ:X⟶Y be a surjective exact morphism of ℚB-A and X is a ℚB-As. If SK is a DSℚB-A over X, then ψSK is alsoDSℚB-A over Y.Proof.
SinceSKxx∈K is a deductive system of X and ψ is surjective, then ψSKx=ψSKx is a deductive system of Y which implies that ψSK is a DSℚB-A over X.Theorem 4.
Assume thatψ:X⟶Y be a surjective exact morphism of ℚB-A and SK a DSℚB-A over X. Then,(1)
IfSKx=kerψ, for all x∈K, then ψSK is the whole DSℚB-A over Y(2)
If is wholeDSℚB-A over X, then ψSK is the whole DSℚB-A over YProof.
(1)
Assume thatSKx=kerψ, where kerψ=x∈X|ψx=x⟶x,ψx=x⇝x. Since ψ is surjective, then, from Theorem 3, we have ψSKx=ψSKx=ψX=Yx∈K. Thus, ψSK is the whole DSℚB-A over Y.(2)
Clearly,SKx=X since SK is whole DSℚB-A over Xx∈K. Thus, ψSKx=ψSKx=ψX=Yx∈K. By Theorem 3, we have ψSK is the whole DSℚB-A over Y


## 3.1. Soft Deductive Systems ofSℚB-As

Based on Definition8, we will propose the notion of soft deductive systems of SℚB-As as indicated below.Definition 9.
Assume thatX=X,⟶,⇝,≤ be a SℚB-A. A nonempty subset D⊆X is a deductive system of X if it satisfies(1)
∀x∈D,x⟶x∈D,x⇝x∈D(2)
∀x,y∈X,x∈D,x⟶y∈D⟹y∈DDefinition 10.
LetX be a SℚB-A and Y a subalgebra of X. A subset D of X is a deductive system of X related to Y (i.e., Y-deductive system of X), denoted by D⋈Y, and satisfies the following two conditions:(1)
∀x∈D,x⟶x∈D,x⇝x∈D(2)
∀y∈Y,x∈D,x⟶y∈D⟹y∈DRemark 2.
According to Definitions9 and 10, we obtain that any deductive system of X is Y-deductive system if Y is a subalgebra of X.
The converse of Remark2 does not hold by Example 3 (i.e., Y is a subalgebra of X and Y-deductive system is not a deductive system).Example 3.
SupposeX (i.e., X=0,k1,k2,k3,1) with partial order 0<k1<k3<1 and 0<k1<k2<1. Now, we show, by Tables 5 and 6, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. Consider a subalgebra Y=k1,1 and a subset D=k1,k2,1; we can see that D⋈Y. However, D is not a deductive system of X since k3⟶1=1∈D and k3∉D.Table 5
The binary operation⟶.
⟶0k1k2k31011111k101k211k2k1k1111k30k1k21110k1k2k31Table 6
The binary operation⇝.
⇝0k1k2k31011111k1k21k211k20k1111k30k1k21110k1k2k31Definition 11.
Assume thatSK is a SℚB-A over X. SD (i.e., SS) over X is a soft deductive system of SK, denoted by SD⊳⊲˜SK, and satisfies the following two conditions:(1)
D⊆K(2)
∀x∈D,SDx⊳⊲SKx
Now, we will give an example to illustrate Definition11 as follows.Example 4.
SupposeX (i.e., X=k1,k2,k3,k4,1) with partial order k1<k2<k3<k4<1. Now, we show, by Tables 7 and 8, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. We define SK∀x∈K (i.e., K=X) by(9)SKx=y∈X|xℛy⟺x⟶y⇝y=1.
From Tables7 and 8, we can get on SKk1=SKk2=1,SKk3=k2,1,SKk4=k2,k3,1,andSK1=X, and then, SKxx∈K are all subalgebras of X. Consequently, SK is a SℚB-As over X.
Next, for a subsetD=k2,k4, we define SD∀x∈D by(10)SDx=1∪y∈X|y≤x.
Then, we obtainSDk2=k1,k2,1⊳⊲1=SKk2 and SDk4=X⊳⊲k2,k3,1=SKk4. Consequently, SD is a soft deductive system of SK.Table 7
The binary operation⟶.
⟶k1k2k3k41k111111k2k31111k3k2k2111k4k2k2kc111k1k2k3k41Table 8
The binary operation⇝.
⇝k1k2k3k41k111111k2k41111k3k2k2111k4k1k2k3111k1k2k3k41Theorem 2.
Assume thatSK is a SℚB-A over X and SD1 and SD2 are two SS. Then,(1)
IfD1∩D2≠∅, then SD1⊳⊲˜SK,SD2⊳⊲˜SK⟹SD1∩˜SD2⊳⊲˜SK(2)
IfD1∩D2=∅, then SD1⊳⊲˜SK,SD2⊳⊲˜SK⟹SD1∪˜SD2⊳⊲˜SKProof.
(1)
Follow from Definition5.(2)
IfSD1⊳⊲˜SK,SD2⊳⊲˜SK, then, by Definition 6, we have D3=D1∩D2 (i.e., x∈D3), SD1∪˜SD2=SD3, and(11)SD3x=SD1x,x∈D1\D2SD2x,x∈D2\D1,SD1x∪SD2x,x∈D1∩D2.
SinceD1∩D2=∅, we obtain either x∈D1\D2 or x∈D2\D1. Then, we have the following:Case 1: ifx∈D1\D2, since SD1⊳⊲˜SK, then SD3x=SD1x⊳⊲SKxCase 2: ifx∈D2\D1 and SD2⊳⊲˜SK, then SD3x=SD2x⊳⊲SKx
Consequently, for allx∈D3, we have SD3x⊳⊲SKx, which implies that SD1∪˜SD2=SD3⊳⊲˜SK.Remark 3.
IfK1∩K2≠∅, then Theorem 2 (2) does not hold by the following example.Example 5.
SupposeX (i.e., X=0,k1,k2,k3,k4,1). Now, we show, by Table 9, the binary operations ⟶.
Clearly,X is a ℂℚB-A. Then,(i)
We defineSK∀x∈K (i.e., K=X) by(12)SKx=y∈X|xℛy⟺x⟶y⟶y∈k1,k2,1.From Table9, we can get SK0=k1,k2,1,SKk1=X,SKk2=SKk3=k1,1,andSKk4=k1,k2,k3,1,SK1=X, and then, SKx∀x∈K are all subalgebras of X. Consequently, SK is a SℚB-As over X.(ii)
We defineSK1∀x∈K1 (i.e., K1=k1,k2,k3) by(13)SK1x=y∈X|xℛy⟺x⟶y=1.Then, we can getSK1k1=k1,1⊳⊲X=SKk1,SK1k2=k2,1⊳⊲a,1=SKk2,andSK1k3=k2,k3,1⊳⊲k1,1. Therefore, SK1 is a soft deductive system over SK.(iii)
We defineSK2∀x∈K2 (i.e., K2=k1) by(14)SK2x=y∈X|xℛy⟺y⟶x=k1.Then, we can getSK2k1=k2,k3,1⊳⊲X=SK2k1. Therefore, SK2 is a soft deductive system over SK.
From (i)–(iii), we haveSK3=SK1∪˜SK2 which is not a soft deductive system of SK, where SK3k1=SK1k1∪SK2k1=k1,k2,k3,1 is not a SKa-deductive system because k2⟶k4=k1∈k1,k2,k3,1 and k4∉k1,k2,k3,1.Table 9
The binary operation⟶.
⟶0k1k2k3k410111111k1k31k2k3k21k2k4k11k2k11k3k1k111k11k4k211k21110k1k2k3k41


## 3.2.DSℚB-As

We will give the notion ofDSℚB-As and investigate homomorphism image of DSℚB-As as indicated below.Definition 12.
Assume thatSK is a SℚB-A over X. If SKx∀x∈K is a deductive system of X, then SK is called a DSℚB-A over X.Example 6 (continued from Example1 (2)).
Clearly,SK is DSℚB-A over X.Definition 13.
(1)
SupposeX be a ℚB-A with the greatest element 1 (i.e., X just only a poset); for any x∈X, the order of element x is defined as(15)Ox=minp,q∈N|x⟶px=1,x⇝qx=1,i,whereN is a natural number and x⟶px=x⟶x⟶⋯⟶x,x⇝qx=x⇝x⇝⋯⇝x.(2)
Ifp,q∈N does not exist to satisfy the above condition (i), then x∀x∈X is called infinite order.Remark 4.
Assume thatSK and SK1 be two SℚB-As over X such that K1⊆K⊆X. If SK is a DSℚB-A over X, then SK1 is a DSℚB-A.
The converse of Remark4 does not hold by the following Example 7.Example 7 (continued from Example2).
We defineSK∀x∈K (i.e., K=X) by(16)SKx=y∈X|Ox=Oy.
Then, we get onSK0=SKk3=SKk4=SKk1=0,k3,k4,1,SKk1=k1,andSKk2=k2. However, k3⟶k1=0∈0,k3,k4,1 and k1∉0,k3,k4,1 imply that SK is not DSℚB-A. If we take K1=k3,k4,1⊆K and we define SK1=y∈X|Ox=Oy∀x∈K1, then SK1 is DSℚB-A.Definition 14.
Assume thatSK is SℚB-A over X with the greatest element 1. If SKx=X∀x∈K, then SK is called whole DSℚB-A.Example 8.
SupposeX (i.e., X=0,k1,k2,1) with partial order 0<k1<k2<1. Now, we show, by Tables 10 and 11, the binary operations ⟶ and ⇝, respectively.
Clearly,X is a ℂℚB-A. We define SK∀x∈K (i.e., K=X) by(17)SKx=y∈X|Ox=Oy.
From Tables10 and 11, we can get on SKx=X∀x∈K. Thus, SK is a whole DSℚB-A over X.
Now, we will study homomorphism image ofDSℚB-As by the following two theorems.Table 10
The binary operation⟶.
⟶0k1k2101111k1k1111k2k1k11110k1k21Table 11
The binary operation⇝.
⇝0k1k2101111k1k2111k20k11110k1k21Theorem 3.
Assume thatψ:X⟶Y be a surjective exact morphism of ℚB-A and X is a ℚB-As. If SK is a DSℚB-A over X, then ψSK is alsoDSℚB-A over Y.Proof.
SinceSKxx∈K is a deductive system of X and ψ is surjective, then ψSKx=ψSKx is a deductive system of Y which implies that ψSK is a DSℚB-A over X.Theorem 4.
Assume thatψ:X⟶Y be a surjective exact morphism of ℚB-A and SK a DSℚB-A over X. Then,(1)
IfSKx=kerψ, for all x∈K, then ψSK is the whole DSℚB-A over Y(2)
If is wholeDSℚB-A over X, then ψSK is the whole DSℚB-A over YProof.
(1)
Assume thatSKx=kerψ, where kerψ=x∈X|ψx=x⟶x,ψx=x⇝x. Since ψ is surjective, then, from Theorem 3, we have ψSKx=ψSKx=ψX=Yx∈K. Thus, ψSK is the whole DSℚB-A over Y.(2)
Clearly,SKx=X since SK is whole DSℚB-A over Xx∈K. Thus, ψSKx=ψSKx=ψX=Yx∈K. By Theorem 3, we have ψSK is the whole DSℚB-A over Y


## 4.FSℚB-As

We give the definition ofFSℚB-As; a concrete example is given to illustrate its derive properties. Furthermore, we study the homomorphism image and preimage of FSℚB-As. Now, we first propose the definition of fuzzy quantum B-algebra (briefly, FℚB-A) as indicated below.Definition 15.
We callFℚB-A (or a fuzzy set μ^ in ℚB-A) if it satisfies (∀x,y∈X,X is ℚB-A):(18)μ^x⟶y≥minμ^x,μ^y,μ^x⇝y≥minμ^x,μ^y.Definition 16.
We callμ^ is a fuzzy deductive system of X if it satisfies ∀x,y∈X:(19)μ^x⟶x≥μ^x,μ^x⇝x≥μ^x,μ^y≥minμ^x⟶y,μ^x.Definition 17.
Assume thatS^K be a FSS over X. Then,(1)
If there existsμ^∈K such that S^Kμ is a FℚB-A (i.e., fuzzy deductive system) in a ℚB-A over X, then S^K is called a S^K-A (i.e., fuzzy soft deductive system FSDS) which depends on a parameter set μ^ over X(2)
IfS^Kμ is a FℚB-A (i.e., fuzzy deductive system) of X based on all parameters, then we say that S^K is a FSℚB-A (i.e., FSDS) of X
In the following, a concrete example is given to illustrate Definition17.Example 9.
Suppose that there are five-class cars:(20)X=BMW,Audi,Toyota,Jeep,Cadilac.
Let⊕ and ⊗ be two soft machines to characterize two cars, defined by the following manner.
BMW⊕x=Cadilac, forallx∈X,(21)Audi⊕y=Jeep,y=BMW,Cadilac,y∈Audi,Toyouta,Jeep,Cadilac,Toyota⊕z=Toyota,z=BMW,Jeep,z=Audi,Cadilac,z∈Toyota,Jeep,Cadilac,Jeep⊕s=Toyota,s=BMW,Jeep,s∈Audi,Toyoya,Cadilac,s∈Jeep,Cadilac,Cadilac⊕t=BMW,t=BMW,Audi,t=Audi,Toyota,t=Toyota,Jeep,t=Jeep,Cadilac,t=Cadilac,(22)BMW⊗x=Cadilac forallx∈X,Audi⊗y=Jeep,y=BMW,Cadilac,y∈Audi,Toyouta,Jeep,Cadilac,Toyota⊗z=Jeep,z∈BMW,Audi,Cadilac,z∈Toyouta,Jeep,Cadilac,Jeep⊗s=Audi,s=BMW,Jeep,s∈Audi,Toyouta,Cadilac,s∈Jeep,Cadilac,Cadilac⊗t=BMW,t=BMW,Audi,t=Audi,Toyota,t=Toyota,Jeep,t=Jeep,Cadilac,t=Cadilac.
Then,X,⊕,⊗,≤ is a ℚB-A. Now, we consider a set of parameters: μ^=(Excellent, Good, Moderate)∈K. Then, we have the following:(1)
We defineS^Kμ^ over X (i.e., S^K[Excellent], S^K[Good], and S^K[Moderate] are fuzzy sets) by Table 12.Therefore, we can see thatS^K[Excellent], S^K[Good], and S^K[Moderate] are all FSℚB-As based on parameters “Excellent,” “Good,” and “Moderate” over X. Thus, S^K is a FSℚB-A over X.(2)
We defineS^K1μ^ over X (i.e., S^K1[Excellent], S^K1[Good], and S^K1[Moderate] are fuzzy sets) by Table 13.However,S^K1μ^ is not a FSℚB-A based on a parameter “Excellent” over HTML translation failed, where S^K1ExcellentToyota⊕BMW=S^K1ExcellentToyota=0.1≱0.2=min0.2,0.4=minS^K1ExcellentAudi,S^K1ExcellentBMW. Also, we obtain that S^K1μ^ is a FSℚB-A based on both the parameter “Good” and “Moderate” over X.(3)
We defineS^K2μ^ over X (i.e., S^K2Excellent and S^K2 [Good] are fuzzy sets) by Table 14.Then,S^K2μ^ is a FSDS on parameters “Excellent.” However, S^K2μ is not a fuzzy deductive system of X based on parameter “Good,” where S^K2[Good](Toyota)=0.3<0.5=minS^K2GoodJeep⊕Toyota,S^K2GoodJeep.(4)
We defineS^K3μ^ over X (i.e., S^K3Excellent and S^K2Moderate are fuzzy sets) by Table 15.Then,S^K3μ^ is a FSDS of X.
Now, we will present several characterizations ofFSℚB-As.
By Definition17, if S^K is a FSℚB-A of ℚB-A over X based on all parameters, then we say that S^K is a FSℚB-A of X, that is,(23)S^Kμ^x⟶y≥minS^Kμ^x,S^Kμ^y,S^Kμ^x⇝y≥minS^Kμ^x,S^Kμ^y.Table 12
Fuzzy setsS^Kμ^ over X.
S^KBMWAudiToyotaJeepCadilacExcellent0.20.20.50.60.8Good0.10.20.30.50.7Moderate0.10.10.40.40.6Table 13
Fuzzy setsS^K1μ^ over X.
S^K1BMWAudiToyotaJeepCadilacExcellent0.40.20.10.60.8Good0.20.20.30.50.7Moderate0.10.10.40.50.9Table 14
Fuzzy setsS^K2μ^ over X.
S^K2BMWAudiToyotaJeepCadilacExcellent0.20.20.20.20.6Good0.20.20.30.50.7Table 15
Fuzzy setsS^K3μ^.
S^K3BMWAudiToyotaJeepCadilacExcellent0.30.30.30.30.3Moderate0.10.10.10.10.7Proposition 1.
AssumeX be a ℚB-A. If S^K is FSℚB-A over X, then, for all t∈0,1,S^Kt≠∅ is the subalgebra of X, in which(24)S^Kt=S^Kμ^t|μ^∈K.Proof.
LetS^Kμ^t≠∅. Then, ∀x,y∈S^Kμ^t; since S^K is a FSℚB-A, then S^Kμ^x≥t,S^Kμ^y≥t. So,(25)S^Kμ^x⟶y≥minS^Kμ^x⟶y,S^Kμ^x⇝y≥minS^Kμ^x,S^Kμ^y≥t.
Similarly, we haveS^Kμ^x⇝y≥t. Therefore, x⟶y,x⇝y∈S^Kμ^t. This implies that S^Kμ^t is the subalgebra of X.
Analogously, we can get Proposition2 as follows.Proposition 2.
Assume thatSK1 and SK2 are two FSℚB-A over X. Then, SK1∩˜SK2 and SK1∪˜SK2 are FSℚB-As over X.Definition 18.
Letα,β be a fuzzy soft map from ℚB-A over X to ℚB-A over Y. Then,(1)
Ifα is an exact morphism from X to Y, then α,β is called a FSℚB-A exact morphism from X to Y(2)
Ifα is an isomorphism from X to Y and β is a bijective from K1 to K2, then α,β is a called an isomorphism between FSℚB-AsProposition 3.
LetX and Y be two ℚB-As. SK is a FSℚB-A over Y and α,β a FSℚB-A exact morphism from X to Y; then, α,β−1SK is FSℚB-A over X.Proof.
Forμ^∈β−1K,(26)α−1SKμ^x⟶y=SKβμ^αx⟶y=SKβμ^αx⟶αy≥minSKβμ^αx,SKβμ^αy=minα−1SKμ^x,α−1SKμ^x.
Consequently,α,β−1SK is a FSℚB-A over X.
Similarly, we can get Proposition4 as follows.Proposition 4.
LetX and Y be two ℚB-As. SK is a FSℚB-As over X and α,β a FSℚB-As isomorphism from X to Y; then, α,βSK is the FSℚB-As over Y.


## 5. Conclusions

In this paper, we introduce the concept ofSℚB-As, and some examples are given to illustrate this definition. Also, we investigate the union and intersection operations between two SℚB-As and give some conditions for the operation holds. With the help of the definition of SℚB-As, we define soft deductive systems of SℚB-As and then investigate the relation between them. As a further step, we define DSℚB-As and investigate the homomorphism image of DSℚB-As. Moreover, we define FSℚB-As. Finally, a concrete example is given to illustrate its derive properties; besides, homomorphism image and preimage of FSℚB-As are discussed.As a future work, it makes sense to applySℚB-As to medical diagnosis (for example, [25, 26]) in practice. Furthermore, it would be interesting if we study hybrid soft lattice-ordered quantum B-algebras.



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