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is, obviously, continuous, strictly increasing,
2
˕
˕(
)
=
This function
and verifies
0
(
)
=
, ˕(
)
=
−
(
(
))
=
−
(
)
=
h
0
0
1
1
h
N
1
1
h
0
1. Then
•
If
x
∈[
0
,
n
)
,or
N
(
x
)
∈
(
n
,
1
]
,
˕(
N
(
x
))
=
1
−
h
(
x
)
=
1
−
˕(
x
)
, and
N
(
x
)
=
˕
−
1
(
1
−
˕(
x
))
.
h
−
1
1
2
)
=
˕
−
1
1
)
=
˕
−
1
•
If
x
=
n
,
N
(
n
)
=
n
=
(
(
2
)
,or
N
(
n
(
1
−
˕(
x
)
•
If
x
∈
(
n
,
1
]
,or
N
(
x
)
∈[
0
,
n
)
,
˕(
N
(
x
))
+
˕(
x
)
=
h
(
N
(
x
))
+
1
−
h
(
N
(
x
))
=
1
)
=
˕
−
1
In conclusion,
N
(
x
(
1
−
˕(
x
))
, for all
x
in
[
0
,
1
]
,or
N
=
N
.
˕
Notice that the proof of last theorem shows clearly that the order-automorphism
˕
such that
N
=
N
is not unique. Notice also that with
˕
=
id
it follows
˕
[
0
,
1
]
N
(
x
)
=
1
−
x
, the fundamental strong negation, with which it results
N
=
N
˕
=
˕
−
1
]
)
ⓦ
˕
=
˕
−
1
ⓦ
(
1
−
id
ⓦ
N
ⓦ
˕
, that is, all strong negations belong to the
[
0
,
1
=
˕
−
1
1
family of
N
0
(
x
)
=
1
−
x
. Nevertheless, in all cases it is
n
(
2
)
the fixed point
of
N
˕
.
If
)
=
√
1
x
2
, it results
N
˕(
x
)
=
˕
(
x
−
x
2
, called the circular negation. If
˕(
x
)
=
1
−
x
1
1
−
x
1
2
x
1
˕
−
1
x
)
=
˕
−
1
2
x
1
x
)
=
˕
−
1
x
,or
(
x
)
=
x
,itfollows
N
˕
(
x
(
1
−
(
x
)
=
3
x
,
+
2
−
+
+
+
that is the strong negation
N
3
of the before mentioned Sugeno's negations.
With
1
ʻ
x
ʱ
), ʻ >
−
˕(
x
)
=
ln
(
1
+
ʻ
1
, ʱ >
0, it follows the bi-parametric family
x
ʱ
1
ʱ
, where with
1
−
N
˕
(
x
)
=
(
x
ʱ
)
ʱ
=
1 it is obtained the Sugeno's family of strong
1
+
ʻ
1
−
x
negations
N
ʻ
=
1
+
ʻ
x
(ʻ >
−
1
)
that only depends on one single parameter.
Remark 2.2.31
The only linear strong negation
N
is
N
=
N
0
, since from
N
(
a
)
=
ʱ
a
+
ʲ
, with
N
(
0
)
=
1
=
ʲ
and
N
(
1
)
=
0
=
ʱ
+
1, follows
ʱ
=−
1 and
N
(
a
)
=
1
−
a
.
ax
+
b
Remark 2.2.32
The only “rational” strong negations
N
of the form
N
(
x
)
=
d
,
cx
+
a
,
b
different of 0, are those
N
ʻ
(ʻ >
−
1
)
in the Sugeno's family. It follows from:
b
•
(
)
=
=
=
N
0
1
d
,or
d
b
a
+
b
•
N
(
1
)
=
0
=
d
,or
a
=−
b
c
+
that gives
)
=
−
bx
+
b
b
(
1
−
x
)
1
−
x
N
(
x
=
=
b
x
.
c
cx
+
b
cx
+
b
1
+
c
c
b
To have 0
N
(
x
)
1, it should be 1
−
x
1
+
b
.But
−
1
=
implies
N
(
x
)
=
1,
c
c
that is not an strong negation. Hence it is
−
1
<
b
, and with
ʻ
=
b
,itfollows
1
−
x
N
(
x
)
=
=
N
ʻ
(
x
)
, with
−
1
< ʻ
.
1
+
ʻ
x
2
For
(
x
<
y
)
is evident that
˕(
x
)<˕(
y
)
if either
x
,
y
∈[
0
,
n
]
,or
x
,
y
∈
(
n
,
1
]
.If
x
∈[
0
,
n
]
,
y
∈
(
n
,
1
]
and
x
<
y
,since
h
(
x
)
+
h
(
N
(
x
)) <
1, it is
h
(
x
)<
1
−
h
(
N
(
x
))
,or
˕(
x
)<˕(
y
)
.
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