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2.2.8.2 De Morgan Laws
With classical sets it always do hold the De Morgan, or duality, laws
A
c
B
c
c
1.
A
∪
B
=
(
∩
)
A
c
B
c
c
2.
A
∩
B
=
(
∪
)
showing that one of the two operators
can be defined by the other and the
complementation. With fuzzy sets in standard algebras, these laws are
∩
,
∪
μ
·
˃
=
(μ
+
˃
)
,or
(μ
·
˃)
=
μ
+
˃
1.
μ
+
˃
=
(μ
·
˃
)
,or
(μ
+
˃)
=
μ
·
˃
,
2.
that correspond to the functional equations in the unknowns
T
,
S
and
N
•
T
(
a
,
b
)
=
N
(
S
(
N
(
a
),
N
(
b
)))
•
S
(
a
,
b
)
=
N
(
T
(
N
(
a
),
T
(
b
)))
for all
a
.
Obviously, law (1) does hold if and only if
T
,
b
in
[
0
,
1
]
=
N
ⓦ
S
ⓦ
(
N
×
N
)
and law (2)
does hold if and only if
S
, two formulas that are equivalent
since, for example, from the first it follows (with
N
2
=
N
ⓦ
T
ⓦ
(
N
×
N
)
=
id)
N
ⓦ
T
=
S
ⓦ
(
N
×
N
)
or
N
, that is, the second formula.
Hence,
the two
De Morgan
laws hold in an algebra given by the triplet
ⓦ
S
=
T
ⓦ
(
N
×
N
)
(
T
,
S
,
N
)
if and only if T
=
N
ⓦ
S
ⓦ
(
N
×
N
)
, that is, T and S are N -dual.
µ
·
µ
=
µ
0
2.2.8.3 Restricted Non-contradiction Principle
A
c
With classical sets it always holds
A
∩
= ∅
. With fuzzy sets, when is it
μ
·
μ
=
μ
0
?
The equation to be solved is
T
(
a
,
N
(
a
))
=
0
,
for all a
∈[
0
,
1
]
in the unknowns
T
and
N
.
Theorem 2.2.41
If T is a continuous t-norm, and N is a strong negation, it is
T
(
a
,
N
(
a
))
=
0
for all a
∈[
0
,
1
]
, if and only if T
=
W
˕
and N
N
˕
.
Proof
With the fixed point
n
∈
(
0
,
1
)
of
N
,itfollows
T
(
n
,
n
)
=
0, that is,
T
has zero-
))
=
˕
−
1
divisors. Hence,
T
=
W
˕
, and
W
˕
(
a
,
N
(
a
(
max
(
0
, ˕(
a
)
+
˕(
N
(
a
))
−
1
))
=
0
,
or max
(
0
, ˕(
a
)
+
˕(
N
(
a
))
−
1
)
=
0, or
˕(
a
)
+
˕(
N
(
a
))
−
1
0, that implies
)
˕
−
1
˕(
N
(
a
))
1
−
˕(
a
)
,or
N
(
a
(
1
−
˕(
a
))
=
N
˕
(
a
),
for all
a
∈[
0
,
1
]
. Hence
N
N
˕
. The reciprocal is a simple calculation.
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