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In-Depth Information
•
J
1
(μ
B
(
), μ
S
(
))
=
(
−
)
x
y
x
1
y
2
y
2
2
.
•
J
2
(μ
VS
(
), μ
VB
(
))
=
((
−
)
,
)
=[
(
−
,
)
]
x
y
min
1
x
min
1
x
y
Hence,
•
μ
Q
1
(
x
)
=
J
1
(
0
.
4
,
1
−
y
)
=
0
.
4
·
(
1
−
y
)
,
0
.
36
,
if
y
0
.
6
2
y
2
2
•
μ
Q
2
(
x
)
=
J
2
((
1
−
0
.
4
)
,
)
=
(
min
(
0
.
6
,
y
))
=
y
2
,
if
y
<
0
.
6
,
whose graphics are,
µ
Q*
µ
Q2*
µ
Q1*
Finally,
⊧
⊨
0
.
4
(
1
−
y
),
if 0
y
0
.
463
y
2
μ
Q
∗
(
y
)
=
max
(μ
Q
1
(
y
), μ
Q
2
(
y
))
=
,
if 0
.
463
y
0
.
6
⊩
0
.
36
,
if 0
.
6
y
1
.
Example 3.4.13
Let's find the function CRI:
X
ₒ
Y
, in the case with
X
=[
0
,
1
]
,
Y
=[
0
,
1
]
, and
•
=
r1: If
x
is small, then
y
9
•
=
r2: If
x
is big, then
y
2,
1
−
x
,
if
y
=
9
it follows
μ
Q
1
(
y
)
=
(
1
−
x
)μ
{
9
}
(
y
)
=
0
,
if
y
=
9
,
x
,
if
y
=
2
μ
Q
2
(
y
)
=
x
μ
{
2
}
(
y
)
=
0
,
if
y
=
2
,
and
⊧
⊨
x
,
if
y
=
2
μ
Q
∗
(
y
)
=
max
(μ
Q
1
(
y
), μ
Q
2
(
y
))
=
1
−
x
,
if
y
=
9
that gives,
⊩
0
,
otherwise
2
x
+
9
(
1
−
x
)
CRI
(
x
)
=
=
9
−
7
x
,
x
+
1
−
x
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