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T is commutative, S is commutative
T is associative, S is associative
N is involutive,
that is:
T
(
a
,
b
) =
T
(
b
,
a
),
S
(
a
,
b
) =
S
(
b
,
a
)
, for all a
,
b in
[
0
,
1
]
T
(
a
,
T
(
b
,
c
)) =
T
(
T
(
a
,
b
),
c
),
S
(
a
,
S
(
b
,
c
)) =
S
(
S
(
a
,
b
),
c
)
, for all a
,
b
,
c in
[
0
,
1
]
N 1 .
N
(
N
(
a
)) =
a , for all a in
[
0
,
1
]
,or N
N
=
id, or N
=
Functions T and S are called t-norms and t-conorms, respectively. Functions N
are strong negations. Hence,
( [
0
,
1
] ,
T
, )
is an ordered semigroup with neutral 1,
( [
,
] ,
, )
and absorbent 0, and
0
1
S
is also an ordered semigroup but with neutral 0
(
) =
and absorbent 1. Since N
0, it seems that this two kind of ordered semigroups
should show some character of duality. This duality goes in the way:
1
If T is a t-norm, T N
=
N
S
(
N
×
N
)
is a t-conorm
is a t-norm
that are easy to prove. Of course, from Sect. 2.1.4 ,
If S is a t-conorm, S N
=
N
S
(
N
×
N
)
If T is a t-norm, T
min, and min is a t-norm
If S is a t-conorm, max
S , and max is a t-conorm
Hence, for all t-norm T and all t-conorm S :
T
min
max
S
,
S . 1
in particular, T
Even more, the function
min
b
,
if a
=
1
(
a
,
b
),
if a
=
1or b
=
1
Z
(
a
,
b
) =
a
,
if b
=
1
=
0
,
otherwise
,
0
,
otherwise
,
is obviously a t-norm such that Z
T for all t-norm T. Consequently,
b
,
if a
=
0
Z (
a
,
b
) =
1
Z
(
1
a
,
1
b
) =
a
,
if b
=
0
1
,
otherwise
,
max
(
a
,
b
),
if a
=
0or b
=
0
=
1
,
otherwise
,
Z
is a t-conorm such that S
for all t-conorm S . Hence, for all t-norm T and all
t-conorm S ,
Z .
Z
T
min
max
S
1
Notice that T
S mean T
(
a
,
b
) S
(
a
,
b
)
,forall
(
a
,
b
) ∈[
0
,
1
]×[
0
,
1
]
.
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