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Z
∗
)
(
)
(
Notice that min
max
is a continuous t-norm (t-conorm), but
Z
is a discontinuous
[
,
]
t-norm (t-conorm). The operations in
0
1
given by
•
T
pr od
(
a
,
b
)
=
pr od
(
a
,
b
)
=
a
·
b
•
T
W
(
a
,
b
)
=
W
(
a
,
b
)
=
max
(
0
,
a
+
b
−
1
)
=
(
max
(
0
,
Sum
−
1
))(
a
,
b
),
are also continuous t-norms. Then, the dual operations,
T
pr od
(
•
a
,
b
)
=
1
−
T
pr od
(
1
−
a
,
1
−
b
)
=
1
−
(
1
−
a
)
·
(
1
−
b
)
=
a
+
b
−
a
·
b
=
(
Sum
−
pr od
)(
a
,
b
)
W
∗
(
•
a
,
b
)
=
1
−
W
(
1
−
a
,
1
−
b
)
=
min
(
1
,
a
+
b
)
=
min
(
1
,
Sum
(
a
,
b
))
are continuous t-conorms. Since it is easy to prove that
W
T
pr od
min
,
T
pr od
W
∗
, and
it follows max
T
pr od
W
∗
.
Z
W
T
pr od
min
max
Remark 2.2.21
Since
Z
(
0
.
5
,
0
.
5
)
=
0, t-norm
Z
has zero-divisors. Analogously,
from
W
(
a
,
b
)
=
0
⃔
max
(
0
,
a
+
b
−
1
)
=
0
⃔
a
+
b
1
,
it follows that t-norm
W
has zero-divisors, for example,
W
(
0
.
5
,
0
.
4
)
=
0; t-norms
min and
T
pr od
do not have zero-divisors:
•
T
pr od
(
a
,
b
)
=
0
⃔
a
=
0
,
or
b
=
0
•
min
(
a
,
b
)
=
0
⃔
a
=
0
,
or
b
=
0
Proposition 2.2.22
The only idempotent t-norm, i.e. T
(
a
,
a
)
=
a, for all a
∈[
0
,
1
]
,
is T
=
min
.
Proof
If
T
is idempotent, min
(
a
,
b
)
=
T
(
min
(
a
,
b
),
min
(
a
,
b
))
T
(
a
,
b
)
since
(
,
)
,
(
,
)
(
,
)
(
,
)
(
,
)
min
a
b
a
and min
a
b
b
. Hence, min
a
b
T
a
b
min
a
b
=
implies
T
min.
Proposition 2.2.23
The only idempotent t-conorm, i.e. S
(
a
,
a
)
=
a, for all a
∈
[
0
,
1
]
,isS
=
max
.
Proof
If
S
is idempotent, max
(
a
,
b
)
=
S
(
max
(
a
,
b
),
max
(
a
,
b
))
≥
S
(
a
,
b
)
, since
max
(
a
,
b
)
≥
a
,
and max
(
a
,
b
)
≥
b
. Hence, max
(
a
,
b
)
≥
S
(
a
,
b
)
≥
max
(
a
,
b
)
implies
S
=
max.
Remark 2.2.24
t-norms can be continuous, like min,
T
pr od
, and
W
, or discontinuous,
like
Z
. They can have zero-divisors, like
W
and
Z
, or not like min and
T
pr od
.They
can have all elements in
[
0
,
1
]
idempotent (only
T
=
min), only have the idempotents
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