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verifies:
μ 0 (
μ 0 = μ 1
x
) =
1
μ 0 (
1
x
) =
1:
μ 1 (
μ 1 = μ 0
x
) =
1
μ 1 (
1
x
) =
1
1
=
0:
) ˃ μ
μ ˃
1
˃(
1
x
)
1
μ(
1
x
μ ∗∗ (
μ (
μ ∗∗ = μ
x
) =
1
1
x
) =
1
−[
1
μ(
x
) ]= μ(
x
)
:
.
μ μ can be taken as a “strong negation”
Hence, it could seem that the function
} [ 0 , 1 ] , then
for the fuzzy sets in
[
0
,
1
]
, but it is not the case. Notice that if
μ ∈{
0
,
1
μ ∈{
} [ 0 , 1 ] , that is, if
it should be also
0
,
1
μ
represents a classical subset of
[
0
,
1
]
,
μ should represent not only a classical subset but precisely the complement of
also
1
μ
. But with A
=[
0
,
2 ]ↂ
X ,
1
,
0
x
0
.
5
,
μ A (
x
) =
0
,
0
.
5
<
x
1
,
follows,
1
0
,
0
.
5
<
x
1
,
,
0
.
5
<
x
1
,
μ A (
x
) =
1
μ A (
1
x
) =
1
=
0
,
0
x
0
.
5
,
1
,
0
x
0
.
5
, but not A c
that represents the subset
violates the preservation principle , and hence it cannot be taken into account to
negate fuzzy sets.
[
0
,
0
.
5
]
= (
0
.
5
,
1
]
. The unary operation
2.1.4 Resolution
X ,
Let us denote by
μ r the constant fuzzy sets in
[
0
,
1
]
μ r (
x
) =
r ,for r
∈[
0
,
1
]
X
and all x
X . Notice that in
{
0
,
1
}
there are only the “constants”
μ 0 and
μ 1 , that
correspond to the sets
and X , respectively.
X , let us denote by
μ ( r ) the fuzzy (crisp) set
Given
μ ∈[
0
,
1
]
1
,
if r
μ(
x
),
μ ( r ) (
x
) =
0
,
otherwise,
[ μ ( r ) ]
for all r
∈[
0
,
1
]
, and by
the corresponding classical subset
{
x
X
;
r
μ(
x
) }
.
[ μ ( 0 ) ]=
μ
These sets are called the r-cuts of
X .
For example, in the following figures are shown, respectively, the constant fuzzy
and it is always
set
μ 0 . 5 , and the 0
.
3-cut of two different fuzzy sets.
 
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