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μ
tall
,aswellas
Notice that it is not needed to fix
S
and
N
, but o
nly a fo
rm for
)
=
√
μ
P
(
2
μ
v
er y P
(
)
=
μ
P
(
)
, μ
mol P
(
),
(
)
=
−
to accept that
x
x
x
x
and
A
x
210
x
.
This last hypotheses is perfectly reasonable since
μ
tall
is non-decreasing, and then
the order
.
Provided we need to know up to which numerical degree
p
and
q
do hold, we can
use the linear function in the figure,
μ
tall
is just the order of
[
0
,
210
]
and fix
N
N
0
. It results
degree up to which
p
is true
=
2
2
=
1
−
(
1
−
μ
tall
(
175
))
=
μ
tall
(
175
)
=
0
.
391
2
degree up to which
q
is true
=
μ
tall
(
180
)
=
0
.
563,
that shows how is
q
more true than
p
.
Example 2.2.50
It is known that the algebra with which fuzzy sets must be combined
should satisfy the laws
μ
+
μ
·
˃
=
μ,
μ
·
(μ
+
˃)
=
μ,
as well as that the negation is linear, Determine the triplet
(
T
,
S
,
N
)
and, with
X
=
[
0
,
10
]
, and
⊧
⊨
0
,
if 0
x
5
x
10
, ˃(
x
−
5
μ(
)
=
)
=
,
x
x
if 5
x
8
6
⊩
x
−
6
,
if 8
x
10
,
4
μ
+
˃
compute
.
Solution
. The first law of absorption
μ
·
˃, μ
+
˃,
and
μ
+
μ
·
˃
=
μ
, implies
S
=
max for
any
T
. The second law of absorption
μ
·
(μ
+
˃)
=
μ
, implies
T
=
min for
any
S
. Hence
(
T
,
S
)
=
(
min
,
max
)
, and the only linear
N
is
N
=
N
0
. Hence,
(
T
,
S
,
N
)
=
(
min
,
max
,
1
−
id
)
. With the graphics of
μ
and
˃
in the figure,
μ
′
μ
˃
7
80:9
:
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