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⊨
⊨
0
( 1 −
1
1
(μ · ˃ ) ( x ) = 1 − (μ · ˃ )( x ) =
(μ
· ˃ )( x ) =
x
(
x
−
5
)
x
x
−
5
−
,
and
10 )
,
20
2
⊩
⊩
x
10
1 −
x
10
1 −
it results
⊧
⊨
W ∗ (
1
,
0
) =
1
(˃ + (μ
· ˃ ))(
7
−
x
x
−
5
x
x
(
x
−
5
)
W ∗ (
= (μ · ˃ ) (
x
) =
,
(
1
−
10 ) =
1
−
x
)
.
⊩
2
2
20
x
x
10
W ∗ (
0
,
1
−
10 ) =
1
−
=
[
,
]
Example 2.2.53
Predicate F
high fever refers to the interval
37
42
in a clini-
{
,
.
,
,...,
.
,
}
cal thermometer, in which the values
37
37
5
38
41
5
42
are significative.
Asking an expert one obtains the following fuzzy set
μ F =
0
.
3
/
38
.
5
+
0
.
5
/
39
+
0
.
7
/
39
.
5
+
0
.
8
/
40
+
0
.
9
/
40
.
5
+
1
/
41
+
1
/
41
.
5
+
1
/
42
where it is clear that 0
38 is avoided since this values of the body's
temperature are not significative for F . With all that, give the membership function
of P
/
37
+
0
/
37
.
5
+
0
/
=
very high fever , Q
=
more or less high fever , R
=
low fever, S
=
not high
fever .
Solution . With the usual definition
=+ √ μ F , μ lo w F
2
μ v er y F
= μ
F , μ mol F
=
μ F (
37
+
42
−
x
) = μ F (
79
−
x
), μ not F =
1
− μ F ,
it results:
• μ P
=
0
.
09
/
38
.
5
+
0
.
25
/
39
+
0
.
49
/
39
.
5
+
0
.
64
/
40
+
0
.
81
/
40
.
5
+
1
/
41
+
1
/
41
.
5
+
1
/
42.
• μ Q
=
0
.
55
/
38
.
5
+
0
.
7
/
39
+
0
.
84
/
39
.
5
+
0
.
89
/
40
+
0
.
95
/
40
.
5
+
1
/
41
+
1
/
41
.
5
+
1
/
42.
• μ R
=
1
/
37
+
1
/
37
.
5
+
1
/
38
+
0
.
9
/
38
.
5
+
0
.
8
/
39
.
5
+
0
.
7
/
39
.
5
+
0
.
5
/
40
.
5
+
0
.
3
/
40
.
5.
• μ S
=
1
/
37
+
1
/
37
.
5
+
1
/
38
+
0
.
7
/
38
.
5
+
0
.
5
/
39
.
5
+
0
.
3
/
39
.
5
+
0
.
2
/
40
.
5
+
0
.
1
/
40
.
5.
μ S = μ not F μ lo w F
= μ R . An incoherence
Notice the incoherence produced by
μ not F
=
− μ F ,butsome
showing that it cannot be taken the representation
1
μ not F =
N
ⓦ μ F with N
≥
N 0 .
1 − x
For example, if N
(
x
) =
9 x ,itis
1
−
0
.
μ S
=
1
/
37
+
1
/
37
.
5
+
1
/
38
++
0
.
96
/
38
.
5
+
0
.
91
/
39
.
5
+
0
.
81
/
39
.
5
+
0
.
71
/
40
+
0
.
53
/
40
.
5
+
1
/
41
+
1
/
41
.
5
+
1
/
42, showing
μ lo w F μ not F .
Look that
μ F & lo w F =
0
.
3
/
38
.
5
+
0
.
5
/
39
.
5
+
0
.
7
/
39
.
5
+
0
.
5
/
40
+
0
.
3
/
40
.
5
provided
μ F & lo w F =
min
(μ F , μ lo w F )
.
 
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