Some properties of generalized soft rough model
In this section, we will generalize the soft rough model on a complete atomic Boolean lattice B by defining modified soft rough sets on B. Also, we will present some properties of modified soft rough approximation operators on B, and introduce a new soft rough model on B, which is an improvement of previous models [14, 20].
Definition 7
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let φ:A(B)→℘(A) be another map defined as φ(b)={a∈A:b≤f(a)}. Then the pair (A(B),φ) is called MSRapproximation space on B and for any element x∈B, lower MSRapproximation on B is defined as

\(x^{\vee }_{\varphi }=\bigvee {\{a\in {A(B)}: a\leq {x} \hspace {0.1cm},\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm}b\in A(B)\hspace {0.1cm}s.t\hspace {0.1cm}b\not \leq x\}}\),

and its upper MSRapproximation over B is defined as

\(x^{\wedge }_{\varphi }=\bigvee {\{a\in {A(B)}:\varphi (a)= \varphi (b) \hspace {0.1cm} for \hspace {0.1cm}some \hspace {0.1cm} b\in A(b)\hspace {0.1cm}s.t\hspace {0.1cm} b\leq x\}}.\)

If \(x^{\vee }_{\varphi }\neq x^{\wedge }_{\varphi }\), Then x is said to be MSRelement on B.
Remark 1
Lower MSRapproximation of x over B can be defined as \(x^{\vee }_{\varphi }=\bigvee {\{a\in {A(B)}}: {\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm}b\in A(B)\hspace {0.1cm}s.t\hspace {0.1cm}b\not \leq x\}}\) because if we let φ(a)≠φ(b)∀b∈A(B) s.t b≦̸x and a≦̸x, then by hypothesis φ(a)≠φ(a) which is impossible.
Lemma 4
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then for all c∈A(B) and x∈B

(i)
\({c\leq {x^{\vee }_{\varphi }}}\Longleftrightarrow {c\leq x\hspace {0.1cm}}\) and φ(c)≠φ(b) ∀ b∈A(B) s.t b≦̸x;

ii)
\({c\leq {x^{\wedge }_{\varphi }}}\Longleftrightarrow { \varphi (c)=\varphi (b)\hspace {0.1cm}}\) for some b∈A(B) s.t b≤x.
Proof
(i) (⇒) Suppose that \(c\leq {x^{\vee }_{\varphi }}=\bigvee \left \{a\in {A(B)}: a\leq {x} \hspace {0.1cm},\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm}b\in A(B)\hspace {0.1cm} s.t\hspace {0.1cm}b\not \leq x\right \}\). So c≤x. If φ(c)≠φ(b)∀ b∈A(B) s.t b≦̸x, then \(c\wedge {{x^{\vee }_{\varphi }}}=c\wedge {\bigvee {\{a\in {A(B)}: a\leq {x} \hspace {0.1cm},\varphi (a)\neq \varphi (b)}} {{ \hspace {0.1cm}\forall \hspace {0.1cm}b\in A(B)\hspace {0.1cm} s.t\hspace {0.1cm}b\not \leq x\}}}\)= \(\bigvee {\{a\wedge c: a\in {A(B)}, c\leq {x} \hspace {0.1cm},\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm}b\in A(B)\hspace {0.1cm} s.t\hspace {0.1cm}b\not \leq x\}}\). Since φ(c)≠φ(b), then c≠a, i.e., c∧a=0. Hence \(c\leq ({x^{\vee }_{\varphi }})^{\prime }\), a contradiction.
(⇐) Suppose that c≤x and φ(c)≠φ(b) ∀ b∈A(B) s.t b≦̸x, then \(c\leq {\bigvee {\{a\in {A(B)}: a\leq {x} \hspace {0.1cm}}}\), \(\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm}b\not \leq x\}=x^{\vee }_{\varphi }\). Condition (ii) can be proved similarly. □
Proposition 2
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. If φ(a)=φ(b) for some a,b∈A(B), then for any x∈B either \(a,b\leq x^{\wedge }_{\varphi }\) or \(a,b\not \leq x^{\wedge }_{\varphi }\).
Proof
If \(a\leq x^{\wedge }_{\varphi }\), then φ(a)=φ(c) for some c∈A(B) s.t c≤x. Since φ(a)=φ(b), then φ(b)=φ(c), c≤x. This implies that \(b\leq x^{\wedge }_{\varphi }\). □
Proposition 3
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then for all x∈B

(i)
\(x^{\vee }_{\varphi }\leq x\leq x^{\wedge }_{\varphi }\)

(ii)
\(0^{\vee }_{\varphi }=0=0^{\wedge }_{\varphi }\)

(iii)
\(1^{\vee }_{\varphi }=1=1^{\wedge }_{\varphi }\)
Proof
(i) Assume that b∈A(B), s.t \(b\leq x^{\vee }_{\varphi }\), then b≤x. For the other inclusion, let b∈A(B) s.t b≤x. Then φ(b)=φ(b) for some b≤x. Thus \(b\leq x^{\wedge }_{\varphi }\).
(ii) \(0^{\vee }_{\varphi }=\bigvee {\{a\in {A(B)}: a\leq {0} \hspace {0.1cm},\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall b\in A(B)\hspace {0.1cm}s.t\hspace {0.1cm}b\not \leq 0\}}=0\). Also, \(0^{\wedge }_{\varphi }=\bigvee {\{a\in {A(B)}:} {\varphi (a)= \varphi (b)} \hspace {0.1cm} for \hspace {0.1cm}some \hspace {0.1cm} b\in A(B)\hspace {0.1cm}s.t\hspace {0.1cm} b\leq 0\}=0.\) Claim (iii) can be proved similarly. □
Proposition 4
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then for all x,y∈B

(i)
The mappings \(^{\vee }_{\varphi }:B\longrightarrow {B}\) and \(^{\wedge }_{\varphi }:B\longrightarrow {B}\) are order preserving.

(ii)
The mappings \(^{\vee }_{\varphi }:B\longrightarrow {B}\) and \(^{\wedge }_{\varphi }:B\longrightarrow {B}\) are mutually dual.
Proof
(i) Assume that x≤y and \(a\leq x^{\vee }_{\varphi }\). Let b∈A(B) s.t b≦̸y. Since x≤y, then b≦̸x. Since \(a\leq x^{\vee }_{\varphi }\), then φ(a)≠φ(b). So \(a\leq y^{\vee }_{\varphi }\) and we get \(x^{\vee }_{\varphi }\leq y^{\vee }_{\varphi }\). Now let \(b\leq x^{\wedge }_{\varphi }\), that is φ(b)=φ(c) for some c∈A(B) s.t c≤x. Since x≤y, then φ(b)=φ(c) for some c∈A(B) s.t c≤y. Thus \(b\leq y^{\wedge }_{\varphi }\). Consequently, \(x^{\wedge }_{\varphi }\leq y^{\wedge }_{\varphi }\).
(ii) We must show that \(x^{\wedge }_{\varphi }=((x^{{\prime }})^{\vee }_{\varphi })^{{\prime }}\) and \(((x^{{\prime }})^{\wedge }_{\varphi })^{{\prime }}=x^{\vee }_{\varphi }\). Let a∈A(B) s.t \(a\leq ((x^{{\prime }})^{\vee }_{\varphi })^{{\prime }}\), then \(a\not \leq (x^{{\prime }})^{\vee }_{\varphi }\). So, either a≦̸x or φ(a)=φ(b) for some b∈A(B) s.t b≤(x^{′})^{′}=x, that is \(a\leq x^{\wedge }_{\varphi }\). Conversely, Let \(a\leq x^{\wedge }_{\varphi }\), then φ(a)=φ(b) for some b∈A(B) s.t b≤x. Thus \(a\not \leq (x^{{\prime }})^{\vee }_{\varphi }\), that is \(a\leq ((x^{{\prime }})^{\vee }_{\varphi })^{{\prime }}\). So \(x^{\wedge }_{\varphi }=((x^{{\prime }})^{\vee }_{\varphi })^{{\prime }}\).
Let a∈A(B) s.t \(a\leq ((x^{{\prime }})^{\wedge }_{\varphi })^{{\prime }}\), then \(a\not \leq (x^{{\prime }})^{\wedge }_{\varphi }\). Hence, φ(a)≠φ(b) for all b∈A(B) s.t b≤x^{′}. That is a≤x because otherwise, if a≤x^{′}, then φ(a)≠φ(a), a contradiction. Thus, a≤x, φ(a)≠φ(b) for all b∈A(B) s.t b≤x^{′} and consequently, \(a\leq x^{\vee }_{\varphi }\). Conversely, let a∈A(B) s.t \(a\leq x^{\vee }_{\varphi }\), then a≤x and φ(a)≠φ(b) for all b∈A(B) s.t b≤x^{′}. Thus \(a\not \leq (x^{{\prime }})^{\wedge }_{\varphi }\), i.e \(a\leq ((x^{{\prime }})^{\wedge }_{\varphi })^{{\prime }}\). So \(((x^{{\prime }})^{\wedge }_{\varphi })^{{\prime }}=x^{\vee }_{\varphi }\).
For all S⊆B, we denote \(S^{\vee }_{\varphi }={\{x^{\vee }_{\varphi }: x\in {S}\}}\) and \({S^{\wedge }_{\varphi }}={\{x^{\wedge }_{\varphi }: x\in {S}\}}\). □
Proposition 5
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B, then

i)
For all S⊆B, \(\vee {S^{\wedge }_{\varphi }}=(\vee {S})^{\wedge }_{\varphi }\) and \(\wedge {S^{\wedge }_{\varphi }}\geq (\wedge {S})^{\wedge }_{\varphi }\).

ii)
For all S⊆B, \(\wedge {S^{\vee }_{\varphi }}=(\wedge {S})^{\vee }_{\varphi }\) and \(\vee {S^{\vee }_{\varphi }}\leq (\vee {S})^{\vee }_{\varphi }\).

(iii)
\((B^{\wedge }_{\varphi },\leq)\) is a complete lattice; 0 is the least element and 1 is the greatest element of \((B^{\wedge }_{\varphi },\leq)\).

(iv)
\((B^{\vee }_{\varphi },\leq)\) is a complete lattice; 0 is the least element and 1 is the greatest element of\((B^{\vee }_{\varphi },\leq)\).

(v)
The kernal \(\Theta ^{\vee }_{\varphi }=\{(x,y):x^{\vee }_{\varphi }=y^{\vee }_{\varphi }\}\) of the map \(^{\vee }_{\varphi }:B\longrightarrow {B}\) is a congruence on the semi lattice (B,∧) such that the \(\Theta ^{\vee }_{\varphi }\)class of any x has a least element.

(vi)
The kernal \(\Theta ^{\wedge }_{\varphi }=\{(x,y):x^{\wedge }_{\varphi }=y^{\wedge }_{\varphi }\}\) of the map \(^{\wedge }_{\varphi }:B\longrightarrow {B}\) is a congruence on the semi lattice (B,∨) such that the \(\Theta ^{\wedge }_{\varphi }\)class of any x has a greatest element.
Proof
(i) Let S⊆B. The mapping \(^{\wedge }_{\varphi }\): B→B is order preserving, which implies that \(\vee {S^{\wedge }_{\varphi }}\leq {(\vee {S})^{\wedge }_{\varphi }}\) and \(\wedge {S^{\wedge }_{\varphi }}\geq (\wedge {S})^{\wedge }_{\varphi }\). Let b∈A(B) and assume that \(a\leq {(\vee {S})^{\wedge }_{\varphi }}\). So, φ(a)=φ(b) for some b∈A(B) s.t b≤∨S. So, φ(a)=φ(b) for some b∈A(B) and x∈S s.t b≤x. So {a∈A(B):φ(a)=φ(b) forsome b∈A(B) s.t b≤∨S}
\(\subseteq {\cup _{_{x\in {S}}}\{a\in {A(B)}:\varphi (a)= \varphi (b)\hspace {0.1cm} \\for some \hspace {0.1cm}b\in A(B)\hspace {0.1cm} s.t \hspace {0.1cm}b\leq x\}}\). Then
\((\vee {S})^{\wedge }_{\varphi }=\bigvee {\{a\in {A(B)}:\varphi (a)= \varphi (b) \hspace {0.1cm} for \hspace {0.1cm}some \hspace {0.1cm} b\in A(b)\hspace {0.1cm}s.t\hspace {0.1cm} b\leq \vee {S}\}}\)
\(\hspace {1.3 cm}\leq {\bigvee (\cup _{_{x\in {S}}}\{a\in {A(B)}:\varphi (a)= \varphi (b) \hspace {0.1cm} for \hspace {0.1cm}some \hspace {0.1cm} b\in A(B)\hspace {0.1cm}s.t\hspace {0.1cm} b\leq x\}}\))
\(\hspace {1.3 cm}={\bigvee _{_{x\in {S}}}(\bigvee \{a\in {A(B)}:\varphi (a)= \varphi (b) \hspace {0.1cm} for \hspace {0.1cm}some \hspace {0.1cm} b\in A(b)\hspace {0.1cm}s.t\hspace {0.1cm} b\leq x\}}\)) (by Lemma 1)
\(=\bigvee \{x^{\wedge }_{\varphi }:x\in {S}\}=\bigvee {S^{\wedge }_{\varphi }}\)
(ii) Let S⊆B. The mapping \(^{\vee }_{\varphi }\): B→B is order preserving, which implies that \((\wedge {S})^{\vee }_{\varphi }\leq {\wedge {S^{\vee }_{\varphi }}}\) and \(\vee {S^{\vee }_{\varphi }}\leq (\vee {S})^{\vee }_{\varphi }\). Let a∈A(B) s.t \(a\leq {\wedge {S^{\vee }_{\varphi }}}=\wedge {\{x^{\vee }_{\varphi }: x\in {S}\}}\). So, a≤x and φ(a)≠φ(b) for all b∈A(B) s.t b≦̸x for every x∈S. Hence φ(a)≠φ(b) for all b∈A(B) s.t b≦̸∧S. In fact if b≦̸∧S, then ∃x∈S s.t b≦̸x. So, φ(a)≠φ(b). Therefore \(b\leq {(\wedge {S})^{\vee }_{\varphi }}\). Consequently, \(\wedge {S^{\vee }_{\varphi }}\leq {(\wedge {S})^{\vee }_{\varphi }}\). Assertions (iii) and (iv) follow easily from (i), (ii) and Proposition 3(i). The proof of (v) and (vi) follow by (i) and (ii). □
The inequalities in Proposition 5 may be proper. This can be seen in the following example.
Example 2
Let B={0,a,b,c,d,e,f,1} and let the order ≤ be defined as in Fig. 1. Let A={e_{1},e_{2},e_{3},} and f_{A} be a soft set over B defined as follows:
f(e_{1})=0, f(e_{2})=c, and f(e_{3})=d. Then the map φ of MSRapproximation space (A(B),φ) on B will be φ(a)={e_{3}}, φ(b)={e_{3}}, and φ(c)={e_{2}}.
If we take x=e and y=f. Then x∨y=1 and \((x\vee y)^{\vee }_{\varphi }=1\). Also, \(x^{\vee }_{\varphi }=c\), \(y^{\vee }_{\varphi }=c\) and \(x^{\vee }_{\varphi }\vee y^{\vee }_{\varphi }=c\). Thus \((x\vee y)^{\vee }_{\varphi }{>}_{_{_{_{\neq }}}}{x^{\vee }_{\varphi }}\vee y^{\vee }_{\varphi }\).
Now \(x^{\wedge }_{\varphi }=a\vee b\vee c=1\), \(y^{\wedge }_{\varphi }=a\vee b\vee c=1\). So \(x^{\wedge }_{\varphi }\wedge y^{\wedge }_{\varphi }=1\). On the other hand x∧y=e∧f=c and \((x\wedge y)^{\wedge }_{\varphi }=0\). Thus \(x^{\wedge }_{\varphi }\wedge y^{\wedge }_{\varphi }{>}_{_{_{_{\neq }}}}(x\wedge y)^{\wedge }_{\varphi }\).
Proposition 6
Let B=(B,≤) be a complete atomic Boolean lattice and let fn be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then, \((B^{\vee }_{\varphi },\geq)\cong {(B^{\wedge }_{\varphi },\leq)}\)
Proof
We show that \(x^{\wedge }_{\varphi }{\longrightarrow {(x^{{\prime }})^{\vee }_{\varphi }}}\) is the required dual order isomorphism. It is obvious that \(x^{\wedge }_{\varphi }{\longrightarrow {(x^{{\prime }})^{\vee }_{\varphi }}}\) is onto \((B^{\vee }_{\varphi },\geq)\). We show that \(x^{\wedge }_{\varphi }{\longrightarrow {(x^{{\prime }})^{\vee }_{\varphi }}}\) is order embedding. Suppose that \(x^{\wedge }_{\varphi }\leq {y^{\wedge }_{\varphi }}\). Then for all a∈A(B), \(a\leq {x^{\wedge }_{\varphi }}\) implies \(a\leq {y^{\wedge }_{\varphi }}\). So, for all a∈A(B) such that φ(a)=φ(c) for some c∈A(B) s.t c≤x implies φ(a)=φ(b) for some b∈A(B) s.t b≤y. Suppose that \({(y^{{\prime }})^{\vee }_{\varphi }}{\not \leq }{(x^{{\prime }})^{\vee }_{\varphi }}\). So there exists a∈A(B) such that \(a\leq {{(y^{{\prime }})^{\vee }_{\varphi }}}\) and \(a\not \leq {(x^{{\prime }})^{\vee }_{\varphi }}\). Hence a≤y′ and φ(a)≠φ(b) for all b∈A(B) s.t b≤(y^{′})^{′}=y. Also \(a\not \leq (x^{{\prime }})^{\vee }_{\varphi }\) implies either a≦̸x′ or φ(a)=φ(c) for some c∈A(B) s.t c≤(x^{′})^{′}=x, a contradiction. Hence \((y^{{\prime }})^{\vee }_{\varphi }\leq {(x^{{\prime }})^{\vee }_{\varphi }}\). On the other hand, assume that \((y^{{\prime }})^{\vee }_{\varphi }\leq {(x^{{\prime }})^{\vee }_{\varphi }}\) and \(x^{\wedge }_{\varphi }{\not \leq {y^{\wedge }_{\varphi }}}\). So there exists a∈A(B) such that \(a\leq {x^{\wedge }_{\varphi }}\) and \(a\not \leq {y^{\wedge }_{\varphi }}\). So φ(a)=φ(b) for some b∈A(B) s.t b≤x. But φ(a)≠φ(c) for all c∈A(B) s.t c≤y. So \(a\leq (y^{{\prime }})^{\vee }_{\varphi }\) and \(a\not \leq {(x^{{\prime }})^{\vee }_{\varphi }}\), a contradiction. □
Proposition 7
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then for all x∈B

i)
\((x^{\vee }_{\varphi })^{\vee }_{\varphi }=x^{\vee }_{\varphi }\);

ii)
\((x^{\wedge }_{\varphi })^{\wedge }_{\varphi }=x^{\wedge }_{\varphi }\).

iii)
\((x^{\vee }_{\varphi })^{\wedge }_{\varphi }=x^{\vee }_{\varphi }\)

iv)
\((x^{\wedge }_{\varphi })^{\vee }_{\varphi }=x^{\wedge }_{\varphi }\)
Proof
i) \((x^{\vee }_{\varphi })^{\vee }_{\varphi }=\bigvee {\{a\in {A(B)}: a\leq {x^{\vee }_{\varphi }} \hspace {0.1cm},\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm} b\in A(B)\hspace {0.1cm} s.t\hspace {0.1cm}b\not \leq x^{\vee }_{\varphi }\}}\). By Proposition 3\(x^{\vee }_{\varphi }\leq x\) which gives \({x}^{{\prime }}\leq ({x^{\vee }_{\varphi }})^{{\prime }}\). So if b≤x^{′}, then \(b\leq ({x^{\vee }_{\varphi }})^{{\prime }}\). Therefore \((x^{\vee }_{\varphi })^{\vee }_{\varphi }=\bigvee {\{a\in {A(B)}: a\leq {x^{\vee }_{\varphi }} \hspace {0.1cm},\varphi (a)\neq \varphi (b) \hspace {0.1cm}\forall \hspace {0.1cm} b\in A(B)\hspace {0.1cm} s.t\hspace {0.1cm}b\not \leq x\}}=x^{\vee }_{\varphi }\).
ii)By Proposition 3\(x^{\wedge }_{\varphi }\leq (x^{\wedge }_{\varphi })^{\wedge }_{\varphi }\). For the reverse inclusion, let \(a\leq (x^{\wedge }_{\varphi })^{\wedge }_{\varphi }\), then φ(a)=φ(b) for some b∈A(B) s.t \(b\leq x^{\wedge }_{\varphi }\), that is φ(b)=φ(c) for some c∈A(B) s.t c≤x. This implies that φ(a)=φ(c) for some c∈A(B) s.t c≤x. Hence \(a\leq x^{\wedge }_{\varphi }\) and therefore \((x^{\wedge }_{\varphi })^{\wedge }_{\varphi }\leq x^{\wedge }_{\varphi }\). Hence \((x^{\wedge }_{\varphi })^{\wedge }_{\varphi }=x^{\wedge }_{\varphi }\).
iii)\(x^{\vee }_{\varphi }\leq (x^{\vee }_{\varphi })^{\wedge }_{\varphi }\) by Proposition 3. For the converse assume that \(a\leq (x^{\vee }_{\varphi })^{\wedge }_{\varphi }\). So φ(a)=φ(b) for some b∈A(B) s.t \(b\leq x^{\vee }_{\varphi }\). Hence φ(b)≠φ(c) for all c∈A(B) s.t c≤x^{′}. Since φ(a)=φ(b), then φ(a)≠φ(c) for all c∈A(B) s.t c≤x^{′}. Since φ(a)=φ(a), then a≦̸x^{′}, i.e a≤x. So \(a\leq x^{\vee }_{\varphi }\). Hence \((x^{\vee }_{\varphi })^{\wedge }_{\varphi }\leq x^{\vee }_{\varphi }\) and we conclude that \((x^{\vee }_{\varphi })^{\wedge }_{\varphi }=x^{\vee }_{\varphi }\).
iv)By Proposition 3\((x^{\wedge }_{\varphi })^{\vee }_{\varphi }\leq x^{\wedge }_{\varphi }\). Conversely, if \(a\not \leq (x^{\wedge }_{\varphi })^{\vee }_{\varphi }\), then either \(a\not \leq x^{\wedge }_{\varphi }\) or φ(a)=φ(b) for some b∈A(B) s.t \(b\leq (x^{\wedge }_{\varphi })^{{\prime }}\). If \(a\not \leq x^{\wedge }_{\varphi }\), then we get our required result. In later case, φ(a)=φ(b) for some b∈A(B) s.t \(b\leq (x^{\wedge }_{\varphi })^{{\prime }}\). So φ(b)≠φ(z) for all z∈A(B) s.t z≤x. Therefore φ(a)≠φ(z) for all z∈A(B) s.t z≤x. Also, a≤x^{′} because if a≦̸x^{′}, then a≤x. So φ(a)≠φ(a), a contradiction. Hence \(a\leq (x^{{\prime }})^{\vee }_{\varphi }\). That is \(a\not \leq ((x^{{\prime }})^{\vee }_{\varphi })^{{\prime }}=x^{\wedge }_{\varphi }\) by Proposition 4(ii). □
Relations between soft rough approximations on B and current approximations
The following proposition shows a relation between soft lower approximation operators over a complete atomic Boolean lattice B and lower MSRapproximation over B.
Proposition 8
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then for any x∈B, \(x^{\vee }\leq x^{\vee }_{\varphi }\).
Proof
Let b∈A(B) s.t b≤x^{∨}. Then ∃ a∈A s.t b≤f(a)≤x. So a∈φ(b) and b≤x. If \(b\not \leq x^{\vee }_{\varphi }\), then φ(b)=φ(c) for c∈A(B) s.t c≦̸x. Since a∈φ(b) and φ(b)=φ(c), then a∈φ(c) which implies that c≤f(a)≤x. Hence c≤x which is a contradiction. Therefore \(x^{\vee }\leq x^{\vee }_{\varphi }\). □
The following example shows that the relation ≤ between \(x^{\vee }_{\varphi }\) and x^{∨} may be proper.
Example 3
Let B={0,a,b,c,d,e,f,1} and let the order ≤ be defined as in Fig. 1. Let A={e_{1},e_{2},e_{3},} and f_{A} be a soft set over B defined as follows:
f(e_{1})=a, f(e_{2})=0, and f(e_{3})=e. Then the map φ of MSRapproximation space (A(B),φ) on B will be φ(a)={e_{1},e_{3}}, φ(b)=ϕ, and φ(c)={e_{3}}.
Let x=d. Then \(x^{\vee }_{\varphi }=a\vee b=d\) and x^{∨}=a. Thus \(x^{\vee }{<}_{_{_{\neq }}} x^{\vee }_{\varphi }\).
In the following proposition we study a necessary and sufficient condition for \(x^{\vee }_{\varphi }\leq x^{\vee }\) to be hold for any x∈B.
Proposition 9
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then for any x∈B, \(x^{\vee }_{\varphi }\leq x^{\vee }\) iff for every b∈A(B)∃e∈A s.t \(f(e)=\bigvee \{a\in A(B):\varphi (a)=\varphi (b)\}\).
Proof
(⇒) Assume that \(x^{\vee }_{\varphi }\leq x^{\vee }\) for any x∈B. Let b∈A(B) and \(x=\bigvee \{a\in A(B):\varphi (a)=\varphi (b)\}\). So \(x^{\vee }_{\varphi }=\bigvee \{a\in A(B):\varphi (a)=\varphi (c) \hspace {0.1cm}for\hspace {0.1cm} some\hspace {0.1cm}c\in A(B)\hspace {0.1cm}c\leq x \} =\bigvee \{a\in A(B):\varphi (a)=\varphi (b)\}=x\). Since \(b\leq x=x^{\vee }_{\varphi }\) and \(x^{\vee }_{\varphi }\leq x^{\vee }\), then b≤x^{∨}. Therefore ∃e∈A s.t b≤f(e) and f(e)≤x. Also, for all a∈A(B) s.t a≤x, we have φ(a)=φ(b), thus e∈φ(b)=φ(a). So a≤f(e) and therefore x≤f(e). Consequently, \(f(e)=x=\bigvee \{a\in A(B):\varphi (a)=\varphi (b)\}\).
(⇐) Let x∈B and b∈A(B) s.t \(b\leq {x^{\vee }_{\varphi }}\). So for every a∈A(B), if φ(a)=φ(b), then a≤x. Hence \(\bigvee \{a\in A(B):\varphi (a)=\varphi (b)\}\leq x\). By assumption, ∃e∈A s.t \(f(e)=\bigvee \{a\in A(B):\varphi (a)=\varphi (b)\}\). Hence b≤f(e) and f(e)≤x. Therefore b≤x^{∨} and consequently, \(x^{\vee }_{\varphi }\leq x^{\vee }\). □
Remark 2
In general, there is no relation between \(x^{\wedge }_{\varphi }\) and x^{∧}. If x=e in Example 1, then x^{∧}=d and \(x^{\wedge }_{\varphi }=e\not \leq d\).
In Propositions 10 and 11 and Corollary 2 we show that there is a relation between \(x^{\wedge }_{\varphi }\) and x^{∧} if some specific conditions hold.
Proposition 10
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then f_{A} is full iff \(x^{\wedge }_{\varphi }\leq x^{\wedge }\) for every x∈B.
Proof
(⇒) Assume that f_{A} is full and x∈B. Let a∈A(B) s.t \(a\leq {x^{\wedge }_{\varphi }}\), then ∃b∈A(B) s.t b≤x and φ(a)=φ(b). Since \(b\leq {1}={\bigvee _{_{_{_{e\in {A}}}}}}f(e)\), then ∃e∈A s.t b≤f(e). Hence b≤f(e)∧x and thus f(e)∧x≠0. By, b≤f(e), we have e∈φ(b)=φ(a) and therefore a≤f(e). Consequently, \(x^{\wedge }_{\varphi }\leq x^{\wedge }\).
(⇐) Suppose that \(x^{\wedge }_{\varphi }\leq x^{\wedge }\) for every x∈B, we show that \(1\leq {\bigvee _{_{_{_{e\in {A}}}}}}f(e)\). Let a∈A(B), then \(a\leq {a^{\wedge }_{\varphi }}\leq a^{\wedge }\). Therefore ∃e∈A s.t f(e)∧a≠0 and thus a≤f(e) because a∈A(B). Consequently, f_{A} is a full soft set over B. □
Corollary 1
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then f_{A} is full iff x≤x^{∧} for every x∈B.
Proof
(⇒) Assume that f_{A} is full and x∈B. Then \(x\leq {x^{\wedge }_{\varphi }}\leq {x^{\wedge }}\) (by Propositions 3 and 10).
(⇐) Assume that x≤x^{∧} for every x∈B. Let b∈A(B), then \(b\leq {b^{\wedge }}\leq {\bigvee \{f(e): e\in {A}\hspace {0.1cm}} and {f(e)\wedge {b}}\neq {0}\}={\bigvee \{f(e): e\in {A} \hspace {0.1cm}and\hspace {0.1cm} b\leq {f(e)}\}}\). Therefore ∃e∈A s.t b≤f(e) and consequently, f_{A} is full. □
Proposition 11
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. Then \(x^{\wedge }\leq x^{\wedge }_{\varphi }\) for every x∈B iff for every e_{1},e_{2}∈A, f(e_{1})∧f(e_{2})=0 whenever f(e_{1})≠f(e_{2}).
Proof
(⇒) Assume that \(x^{\wedge }\leq x^{\wedge }_{\varphi }\) for every x∈B. Let e_{1},e_{2}∈A, if f(e_{1})∧f(e_{2})≠0, then ∃b∈A(B) s.t b≤f(e_{1})∧f(e_{2}). Since b≤f(e_{1}), then \(f(e_{1})\leq {\bigvee \{f(e): b\leq {f(e)}\}}={\bigvee \{f(e): b\wedge {f(e)}\neq {0}\}}=b^{\wedge }\leq b^{\wedge }_{\varphi }=\bigvee \{a\in A(B): \varphi (a)=\varphi (b)\}\). On the other hand we show that \({\bigvee \{a\in A(B): \varphi (a)=\varphi (b)\}}\leq {f(e_{1})}\). Let c∈A(B) s.t \(c\leq {\bigvee \{a\in A(B): \varphi (a)} {=\varphi (b)\}}\), then φ(c)=φ(b) and thus e_{1}∈φ(b)=φ(c). Therefore c≤f(e_{1}) and consequently, \(f(e_{1})=\bigvee \{a\in A(B): \varphi (a)=\varphi (b)\}\). Similarly, by b≤f(e_{2}), \(f(e_{2})=\bigvee \{a\in A(B): \varphi (a)=\varphi (b)\}\) and hence f(e_{1})=f(e_{2}).
(⇐) Assume that for every e_{1},e_{2}∈A, f(e_{1})∧f(e_{2})=0 whenever f(e_{1})≠f(e_{2}). Let x∈B and a∈A(B) s.t a≤x^{∧}, then ∃e_{1}∈A s.t a≤f(e_{1}) and f(e_{1})∧x≠0. So, ∃b∈A(B) s.t b≤f(e_{1})∧x. We show that φ(a)={e_{2}∈A:f(e_{2})=f(e_{1})}. If f(e_{2})≠f(e_{1}), then f(e_{1})∧f(e_{2})=0 by assumption. Thus a≦̸f(e_{2}) because a≤f(e_{1}) and therefore e_{2}∉φ(a). So φ(a)⊆{e_{2}∈A:f(e_{2})=f(e_{1})}. On the other hand, if f(e_{2})=f(e_{1}), then a≤f(e_{2}) and hence e_{2}∈φ(a). Consequently, φ(a)={e_{2}∈A:f(e_{2})=f(e_{1})}. Similarly we can show that φ(b)={e_{2}∈A:f(e_{2})=f(e_{1})} and thus φ(a)=φ(b). Since b≤x, then \(a\leq x^{\wedge }_{\varphi }\). □
Corollary 2
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Let (A(B),φ) be a MSRapproximation space on B. If f(e)≠0 for every e∈A, then f_{A} is a partition soft set iff \(x^{\wedge }=x^{\wedge }_{\varphi }\) for every x∈B.
Proof
Obvious. □
Deviation of some properties of the previous soft rough approximations and the current approximations
In this section we will introduce the deviations of some properties of soft rough approximations on a complete atomic Boolean lattice B [14] and the current approximations.
On accounting of Proposition 3, 5, 6, and 7[parts (ii) and (iv)], there are deviations between some of the properties of lower and upper approximations on B and MSR lower and upper approximations on B, as follows
Let B=(B,≤) be a complete atomic Boolean lattice and let f_{A} be a soft set over B. Then for all x∈B

(i)
x≦̸x^{∧}

(ii)
\(1^{\vee }_{\varphi }\neq 1\neq 1^{\wedge }\)

(iii)
For all S⊆B, ∧S^{∨}≠(∧S)^{∨}.

(iv)
(x^{∧})^{∧}≦̸x^{∧}.

(v)
(x^{∨})^{∧}≦̸x^{∨}

(vi)
In general (B^{∨},≥)≇(B^{∧},≤)
The following counter example support our claims about the above deviations.
Example 4
Let B={0,a,b,c,d,e,f,1} and let the order ≤ be defined as in Fig. 1. Let A={e_{1},e_{2},e_{3},}.
(1) Let f_{A} be a soft set over B defined as follows:
f(e_{1})=a, f(e_{2})=b, f(e_{3})=d and f(e_{4})=0. So f_{A} is not full. Let x=f, then x^{∧}=d≧̸f. So, x≦̸x^{∧}. Also, \(1^{\vee }_{\varphi }=d=1^{\wedge }\neq 1\).
(2) Let f_{A} be a soft set over B defined as follows: f(e_{1})=a, f(e_{2})=f, f(e_{3})=e and f(e_{4})=0. So f_{A} is not keeping infumum. Let S={f,e}, then ∧S=f∧e=c and (∧S)^{∨}=c^{∨}=0. Also, S^{∨}={f,e} and hence ∧S^{∨}=f∧e=c≦̸0=(∧S)^{∨}. Let x=a, then x^{∧}=e and (x^{∧})^{∧}=e^{∧}=1≦̸e=x^{∧}. Also x^{∨}=a and (x^{∨})^{∧}=e≦̸a=x^{∨}.
Let x=a and y=f; then x^{∧}=e and y^{∧}=1. Therefore x^{∧}≤y^{∧}. On the other hand y^{′∨}=a≦̸f=x^{′∨}. Hence (B^{∨},≥)≇(B^{∧},≤).
Remark 3
(1)It is mentioned that in order to prove that x≤x^{∧} and 1^{∨}=1=1^{∧} in [14] we employed a strong condition on soft set f_{A} over a complete atomic Boolean lattice to be full. However in proving \(x\leq x^{\wedge }_{\varphi }\) and \(1^{\vee }_{\varphi }=1= 1^{\wedge }_{\varphi }\) in Proposition 3 no such condition is required.
(2) Also to prove that for all S⊆B, ∧Sn=(∧S)^{∨} in [14] we employed a strong condition on soft set f_{A} over a complete atomic Boolean lattice to be keeping infimum. However in proving \(\wedge {S^{\vee }_{\varphi }}=(\wedge {S})^{\vee }_{\varphi }\) in Proposition 5 no such condition is required.
(3) Finally to prove that (x^{∨})^{∧}=x^{∨}, (x^{∧})^{∧}=x^{∧} and (B^{∨},≥)≅(B^{∧},≤) we employed a strong condition on soft set f_{A} over a complete atomic Boolean lattice to be a partition. However in proving \((x^{\vee }_{\varphi })^{\wedge }_{\varphi }=x^{\vee }_{\varphi }\), \((x^{\wedge }_{\varphi })^{\wedge }_{\varphi }=x^{\wedge }_{\varphi }\) and \((B^{\vee }_{\varphi },\geq)\cong {(B^{\wedge }_{\varphi },\leq)}\) in Propositions 6 and 7 no such condition is required.
(4) It is clear that MSRelement over a complete atomic Boolean lattice satisfies all the basic properties Järvinen’s approximations [20]. Thus, MSRelement over a complete atomic Boolean lattice provides a good combination of roughness and parametrization.