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In-Depth Information
F
L
ʲ
(
˜
c
1
,
˜
c
2
,...,
˜
c
m
)
=
m
ij
˜
c
,
c
∈
|
ʻ
C
(
u
i
)
=
ʻ
C
(
i
,
j
u
j
)
where,
c
i
is a Boolean variable pertaining to a condition attribute
c
i
∈
˜
C
.
˅
C
(
u
i
)
=
˅
C
(
Function
F
U
ʲ
is true if and only if at least one variable
c
in
m
ij
of
˜
u
j
)
is true. While function
F
L
ʲ
is true if and only if at least one variable
c
in
m
ij
of
˜
ʻ
C
(
u
i
)
=
ʻ
C
(
is true.
Remember that we associate
A
u
j
)
c
A
c
1
,
˜
c
2
,...,
˜
c
m
)
ↆ
C
with a Boolean vector
˜
=
(
˜
as follows:
1
c
k
∈
,
0 otherwise
A
c
k
=
˜
.
Then, we can prove the next theorem from Corollary
1
. Remember that
ˆ
A
is the
term
{˜
a
|
a
∈
A
}
.
Theorem 5
([
20
,
33
])
Let A be the subset of C , and
ʲ
∈[
0
,
0
.
5
)
be an admissible
error rate. We have the following equivalences:
if and only if F
U
c
A
•
A satisfies
(
VPUG
)
as well as
(
VPU
)
with
ʲ
ʲ
(
˜
)
=
1
. Moreover,
ˆ
A
is a prime implicant of F
U
A is a U-reduct with
ʲ
if and only if
,
ʲ
if and only if F
L
c
A
•
A satisfies
(
VPLG
)
as well as
(
VPL
)
with
ʲ
ʲ
(
˜
)
=
1
. Moreover,
ˆ
A
is a prime implicant of F
L
A is an L-reduct with
ʲ
if and only if
.
ʲ
For the preservation of the boundaries, the positive region, and the unpredictable
region, we cannot use discernibility function approach. Because we cannot obtain a
preserving subset
A
C
by determining which pairs of objects should be discerned.
For example, consider a decision table below.
ↆ
c
1
c
2
c
3
X
1
X
2
X
3
P
1
000400
P
2
001020
P
3
010001
P
4
110111
There are 3 condition attributes
C
={
c
1
,
c
2
,
c
3
}
with the value set
V
={
0
,
1
}
,
and 3 decision classes
X
1
,
P
4
are sets of objects, where members
of each set have the same condition attribute values. The distribution of the decision
classes on each set
P
i
is shown in the table, for instance the distribution on
P
1
forms
|
X
2
,
X
3
.
P
1
,
P
2
,
P
3
,
X
1
∩
P
1
|=
4,
|
X
2
∩
P
1
|=
0, and
|
X
3
∩
P
1
|=
0. Consider P-reducts with
ʲ
=
0
.
4.
The positive region of the table is POS
C
(
d
)
=
P
1
∪
P
2
∪
P
3
. When we can make
P
1
and
P
2
be indiscernible and combine
P
1
∪
P
2
, the positive region is still preserved.
Because the distribution on
P
1
∪
P
2
is
(
X
1
,
X
2
,
X
3
)
=
(
4
,
2
,
0
)
, and
μ
X
1
(
P
1
∪
P
2
)
=
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