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(μ
∧
˃)(
)
=
(μ
×
˃)(
,
)
(μ
∨
˃)(
)
=
(μ
×
˃)(
,
).
t
Sup
x
y
, and
t
Sup
x
y
t
=
min
(
x
,
y
)
t
=
max
(
x
,
y
)
Without the proof, let's state:
(
R
∗
,
∧
,
∨
)
Theorem 6.3.1
is a distributive lattice, that means:
•
μ
∧
˃, μ
∨
˃
∈ R
∗
•
μ
∧
˃
=
˃
∧
μ, μ
∧
μ
=
μ
•
μ
∨
˃
=
˃
∨
μ, μ
∨
μ
=
μ
•
μ
∧
(μ
∨
˃)
=
μ, μ
∨
(μ
∧
˃)
=
μ
•
μ
∧
(˃
∨
ʻ)
=
(μ
∧
˃)
∨
(μ
∧
ʻ)
•
μ
∨
(˃
∧
ʻ)
=
(μ
∨
˃)
∧
(μ
∨
ʻ)
,
R
∗
.
for all
μ, ˃, ʻ
in
R
∗
can be endowed with the
partial
order given by
Hence,
μ
≤
∗
˃
⃔
μ
∧
˃
=
μ
⃔
μ
∨
˃
=
˃.
6.3.1 Example
With
X
={
1
,
2
,
3
}
, and
μ
=
0
.
8
|
1
+
0
.
7
|
2
+
1
|
3
, ˃
=
0
.
9
|
1
+
1
|
2
+
0
.
6
|
3, compute
μ
∧
˃
,
μ
∨
˃
.
Since,
t
=
min
(
x
,
y
)
and
t
=
max
(
x
,
y
)
belong to
{
1
,
2
,
3
}
, it results:
•
(μ
∧
˃)(
t
)
=
Max
)
[
min
(μ(
x
), ˃(
y
))
]
, and:
t
=
min
(
x
,
y
-
(μ
∧
˃)(
1
)
=
max
(
min
(μ(
1
), ˃(
1
)),
min
(μ(
1
), ˃(
2
)),
min
(μ(
2
), ˃(
1
)),
min
(μ(
1
), ˃(
3
)),
min
(μ(
3
), ˃(
1
))
=
max
(
min
(
0
.
8
,
0
.
9
),
min
(
0
.
8
,
1
),
min
(
0
.
7
,
0
.
9
),
min
(
0
.
8
,
0
.
6
),
min
(
1
,
0
.
9
))
=
max
(
0
.
8
,
0
.
8
,
0
.
7
,
0
.
6
,
0
.
9
)
=
0
.
9
-
(μ
∧
˃)(
2
)
=
max
(
min
(μ(
2
), ˃(
3
)),
min
(μ(
3
), ˃(
2
)))
=
max
(
min
(
0
.
7
,
0
.
6
),
min
(
1
,
1
))
=
max
(
0
.
6
,
1
)
=
1
-
(μ
∧
˃)(
3
)
=
min
(μ(
3
), ˃(
3
))
=
min
(
1
,
0
.
6
)
=
0
.
6
that is
μ
∧
˃
=
0
.
9
|
1
+
1
|
2
+
0
.
6
|
3.
This fuzzy set is different from
μ
·
˃
=
0
.
8
|
1
+
0
.
7
|
2
+
0
.
6
|
3, with the t-norm
min
.
•
(μ
∨
˃)(
t
)
=
Max
)
[
min
(μ(
x
), ˃(
y
))
]
, and:
t
=
max
(
x
,
y
-
(μ
∨
˃)(
1
)
=
min
(μ(
1
), ˃(
1
))
=
min
(
0
.
8
,
0
.
9
)
=
0
.
8
-
(μ
∨
˃)(
2
)
=
max
(
min
(μ(
1
), ˃(
2
)),
min
(μ(
2
), ˃(
1
)))
=
(
(
.
,
),
(
.
,
.
))
=
(
.
,
.
)
=
.
max
min
0
8
1
min
0
7
0
9
max
0
8
0
7
0
8
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