116 X. Zhang et al. / Knowledge-Based Systems 91 (2016) 114–125
3.1. The first constructive method: ( ∪ , ∩ )-type
Definition 3.1. Let U be a non-empty set, P(U)bethepowersetofU.If
apr
(i)
and apr
(i)
are some mappings from P(U)toP(U), i = 1, 2,...,n.
Then define two new mappings from P(U)toP(U) as follows: for any
X⊆U,
apr
(1..n)
(
∪,∩)
(X) =
n
i=1
apr
(i)
(X); apr
(1..n)
(
∪,∩)
(X) =
n
i=1
apr
(i)
(X).
We call (apr
(1..n)
(
∪,∩)
, apr
(1..n)
(
∪,∩)
) is ( ∪ , ∩ )-type generated pair by operator
pairs
(apr
(i)
, apr
(i)
)(i = 1, 2,...,n), where (1..n)meansthatfrom1
to n.Whenn=2, denote (1..n)by(1,2).
Theorem 3.1. Let U be a non-empty set, apr
(i)
and apr
(i)
be mappings
from P(U) to P
(U), i = 1, 2,...,n.
(1) If for all X, Y ∈ P(U), apr
(i)
(i = 1, 2,...,n) satisfy (Lk),
thenapr
(1..n)
(
∪,∩)
satisfies (Lk),wherek= 1, 2, 3, 5.
(2) If for all X, Y ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (Hk),then
apr
(1..n)
(
∪,∩)
satisfies (Hk),wherek= 1, 2, 3, 5.
(3) If for all X, Y ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (L4), then
apr
(1..n)
(
∪,∩)
satisfies (L4
).
(4) If for all X, Y ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (H4), then
apr
(1..n)
(
∪,∩)
satisfies (H4
).
(5) If for all X, Y ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (L6),
thenapr
(1..n)
(
∪,∩)
satisfies (L6).
(6) If for all X, Y ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (H6), then
apr
(1..n)
(
∪,∩)
satisfies (H6).
(7) If for all X ∈ P(U), apr
(i)
andapr
(i)
(i = 1, 2,...,n) satisfy (L7) or
(H7) , then apr
(i)
and apr
(1..n)
(
∪,∩)
satisfy (L7) or (H7).
(8) If for all X ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (L6) and (L8
),
thenapr
(1..n)
(
∪,∩)
satisfies (L8
).
(9) If for all X ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (H6) and (H8
),
then
apr
(1..n)
(
∪,∩)
satisfies (H8
).
(10) If for all X ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (L1), (L5) or (L6),
and (L8), thenapr
(1..n)
(
∪,∩)
satisfies (L8).
(11) If for all X ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (H1), (H5) or
(H6), and (H8), then
apr
(1..n)
(
∪,∩)
satisfies (H8).
(12) If for all X ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (L1), then
apr
(1..n)
(
∪,∩)
satisfies (L9
).
(13) If for all X ∈ P
(U), apr
(i)
(i = 1, 2,...,n) satisfy (H1), then
apr
(1..n)
(
∪,∩)
satisfies(H9
).
Proof. It is easy to verify that (L1), (L2), (L3), (L5), (H1), (H2), (H3)
and (H5) hold for apr
(1..n)
(
∪,∩)
and apr
(1..n)
(
∪,∩)
) when the corresponding con-
ditions are satisfied. W e prove other properties as follows.
(3): It follows from (1) and Lemma 3.1 (5).
(4): It follows from (2) and Lemma 3.1 (6).
(5): Applying (1) and Lemma 3.1 (9) we can get (5).
(6): Applying (2) and Lemma 3.1 (10) we can get (6).
(7): For all X ∈ P(U), if apr
(i)
and apr
(i)
(i = 1, 2,...,n) satisfy (L7),
from this and Definition 3.1 we have
apr
(1..n)
(
∪,∩)
( ∼ X) =
n
i=1
apr
(i)
( ∼ X) =
n
i=1
( ∼ apr
(i)
(X))
=∼
n
i=1
apr
(i)
(X)
=∼ apr
(1..n)
(
∪,∩)
(X).
This means that apr
(1..n)
(
∪,∩)
and apr
(1..n)
(
∪,∩)
satisfy (L7).
Similarly, we can get that apr
(1..n)
(
∪,∩)
and apr
(1..n)
(
∪,∩)
satisfy (H7).
(8) We only prove the case of n = 2. For all X ∈ P(U), if apr
(i)
(i =
1, 2) satisfy (L6) and (L8
), by Definition 3.1 we have
apr
(1,2)
(
∪,∩)
(apr
(1,2)
(
∪,∩)
(X)) = apr
(1,2)
(
∪,∩)
(apr
(1)
(X) ∪ apr
(2)
(X))
(By Definition 3.1)
⊇ apr
(1,2)
(∪,∩)
(apr
(1)
(X)) ∪ apr
(1,2)
(∪,∩)
(apr
(2)
(X))
(
By (L6) forapr
(1,2)
(
∪,∩)
from (5))
= (apr
(1)
(apr
(1)
(X)) ∪ apr
(2)
(apr
(1)
(X)))
∪ (apr
(1)
(apr
(2)
(X)) ∪ apr
(2)
(apr
(2)
(X)))
(By Definition 3.1)
⊇
(apr
(1)
(apr
(1)
(X))) ∪ (apr
(2)
(apr
(2)
(X)))
⊇ apr
(1)
(X) ∪ apr
(2)
(X)
(
By(L8
) for apr
(1)
and apr
(2)
)
= apr
(1,2)
(
∪,∩)
(X).
This means that apr
(1,2)
(
∪,∩)
satisfies (L8
).
(9) We only prove the case of n = 2. For all X ∈ P(U), if
apr
(i)
(i =
1, 2) satisfy (H6) and (H8
), by Definition 3.1 we have
apr
(1,2)
(
∪,∩)
(apr
(1,2)
(
∪,∩)
(X)) = apr
(1,2)
(
∪,∩)
(apr
(1)
(X) ∩ apr
(2)
(X))
(By Definition 3.1)
⊆
apr
(1,2)
(
∪,∩)
(apr
(1)
(X)) ∩ apr
(1,2)
(
∪,∩)
(apr
(2)
(X))
(
By (H6) forapr
(1,2)
(
∪,∩)
from (6))
= (apr
(1)
(apr
(1)
(X)) ∩ apr
(2)
(apr
(1)
(X)))
∩ (apr
(1)
(apr
(2)
(X)) ∩ apr
(2)
(apr
(2)
(X)))
(By Definition 3.1)
⊆
(apr
(1)
(apr
(1)
(X))) ∩ (apr
(2)
(apr
(2)
(X)))
⊆ apr
(1)
(X) ∩ apr
(2)
(X)
(
By(H8
) for apr
(1)
and apr
(2)
)
= apr
(1,2)
(
∪,∩)
(X).
This means that apr
(1,2)
(
∪,∩)
satisfies (H8
).
(10) By (8) and Lemma 3.1 (7), (9) and (13) we can get (10).
(11) By (9) and Lemma 3.1 (8), (10) and (14) we can get (11).
(12) It follows from (1) and Lemma 3.1 (15).
(13) It follows from (2) and Lemma 3.1 (16).
Corollary 3.1. Let U be a non-empty set,
(apr
(i)
, apr
(i)
) (i = 1, 2,...,n)
be Pawlak’s rough approximation pairs on U. Then apr
(1..n)
(
∪,∩)
and apr
(1..n)
(
∪,∩)
satisfy (L1)–(L3), (L4
), (L5)–(L8), (L9
), (H1)–(H3), (H4
), (H5)–(H8),
and (H9
).
In general case, apr
(1..n)
(
∪,∩)
and apr
(1..n)
(
∪,∩)
do not satisfy (L4), (H4), (L9)
and (H9). Please see the following examples.
Example 3.1. Let
U = {e
1
, e
2
, e
3
, e
4
, e
5
, e
6
, e
7
, e
8
},
U /R
1
= {{e
1
, e
7
}, {e
2
, e
3
, e
4
, e
5
, e
6
}, {e
8
}},
U /R
2
= {{e
1
, e
2
}, {e
3
, e
4
, e
5
}, {e
6
, e
7
, e
8
}}.
If apr
(1)
and apr
(2)
denote Pawlak’s rough approximations by equiva-
lence relations R
1
and R
2
, respectively. Putting
X = {e
1
, e
2
, e
3
, e
4
, e
5
, e
6
}, Y = {e
3
, e
4
, e
5
, e
6
, e
7
, e
8
}.