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=
The family of S
max, only contains this t-conorm.
The family of T pr od
contains all t-conorms of the form
pr od ˕ (
) = ˕ 1
pr od (˕(
))) = ˕ 1
a
,
a
(
a
), ˕(
b
(˕(
a
) + ˕(
b
) ˕(
a
) · ˕(
b
))
The family of W contains all t-conorms of the form
W (
) = ˕ 1
W (˕(
)) = ˕ 1
a
,
a
(
a
), ˕(
b
(
min
(
1
, ˕(
a
) + ˕(
b
)))
Remark 2.2.27
The order-automorphism
˕
plays the role of a functional parameter.
x r , it follows, for example,
By taking,
˕(
x
) =
r max
r min
W ˕ (
a r
b r
a r
b r
W
˕ (
a
,
b
) =
((
0
,
+
1
),
a
,
b
) =
((
1
,
+
)
giving a family of t-norms (t-conorms) depending on the numerical parameter r
>
0.
x r ,
Notice that with
˕(
x
) =
a r
r
b r
Prod ˕ (
a
,
b
) =
·
=
a
·
b
=
Prod
(
a
,
b
),
but
a r
Prod ˕ (
) = ˕ 1
r
b r
a r
b r
a
,
b
(˕(
a
) + ˕(
b
) ˕(
a
) · ˕(
b
)) =
+
·
.
2.2.6 Strong Negations
As it was said before, an strong negation is a function N
:[
0
,
1
]ₒ[
0
,
1
]
such that
N
(
0
) =
1
If a
b , then N
(
b
)
N
(
a
)
,or N 2
N
(
N
(
a
)) =
a , for all a
∈[
0
,
1
]
=
id.
Notice that N 2
N 1 , that shows N is a continuous
=
id is equivalent to N
=
function: It is N
(
1
) =
N
(
N
(
0
)) =
0
,
and if a
<
b it should be N
(
b
)<
N
(
a
)
since
N
(
b
) =
N
(
a
)
would imply N
(
N
(
b
)) =
N
(
N
(
a
))
,or a
=
b . Hence, N is strictly
decreasing.
Since N is continuous, the equation N
(
x
) =
x has solutions, but there is only
one. Suppose N
(
x 1 ) =
x 1 and N
(
x 2 ) =
x 2 . Either x 1
x 2 ,or x 2
<
x 1 .Inthe
first case, it follows N
(
x 2 )
N
(
x 1 )
,or x 2
x 1 , and x 1 =
x 2 . In the second case,
N
(
x 1 )<
N
(
x 2 )
,or x 1 <
x 2 , that is absurd. Then, each strong negation has a single
fixed point N
(
n
) =
n , in the open interval
(
0
,
1
)
, since N
(
0
) =
1
,
N
(
1
) =
0show
that 0 and 1 are not fixed points.
Remark 2.2.28
In the classical case (a Boolean algebra L , or a power set
P (
X
)
), the
transformation
 
 
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