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A collection of the weighed points of all
-unimodal numbers belonging to
-
Λ
Λ
tolerance number is referred to as the weighed set of this
-tolerance number.
Λ
This definition is extended also on
-unimodal numbers.
Λ
~
(
)
The Proposition 6.1.
Weighed set of
-tolerance number
A
≡
a
,
a
,
a
L
a
,
is a
Λ
1
2
R
[
]
segment
A
1
,
A
, where
A
=
a
−
la
,
A
=
a
+
la
. Let us refer this segment as
2
2
R
2
1
1
L
the weighed segment of
-tolerance number
A
.
Λ
~
(
)
The proof
. Let us consider two unimodal fuzzy numbers
,
B
≡
a
,
a
,
0
1
1
L
~
~
(
)
(
)
which belong to
-tolerance number
. Let us
B
≡
a
,
a
Λ
A
≡
a
,
a
,
a
L
a
,
2
2
R
1
2
R
[
]
[
]
~
~
denote sets of
-level of numbers
with
B
α
=
B
1
1
,
a
and
B
=
a
,
B
2
2
,
B
,
B
α
1
2
1
α
1
2
α
2
α
accordingly, and assign weighed points for these numbers in according with (6.4)
1
1
1
(
)
(
)
()
()
∫
1
1
∫
−
1
∫
−
1
A
B
+
a
α
d
α
=
2
a
−
L
α
a
α
d
α
=
a
−
L
α
a
α
d
α
=
a
−
la
;
1
=
α
1
1
L
1
L
1
L
0
0
0
1
1
1
[
]
(
)
()
()
∫
2
2
∫
−
1
∫
−
1
A
a
+
B
α
d
α
=
2
a
+
R
α
a
α
d
α
=
a
+
R
α
a
α
d
α
=
a
+
ra
,
2
=
2
α
2
R
2
R
2
R
0
0
0
1
1
()
()
∫
∫
L
−
1
α
α
d
α
=
l
,
R
−
1
α
α
d
α
=
r
.
Where
0
0
~
(
)
Let us consider arbitrary
-unimodal number
which belongs to
Λ
B
≡
b
,
b
L
b
,
R
~
(
)
[
]
tolerance number
B
, and
the weighed point
~
with
B
. From definition of a membership of one fuzzy
number to another it follows that
A
≡
a
,
a
,
a
L
a
,
. Let us denote
-level set with
α
1
,
2
1
2
R
α
α
1
1
1
2
1
2
2
2
B
≤
B
,
a
≤
B
,
a
≥
B
,
B
≥
B
α
α
1
α
2
α
α
α
B
1
1
+
a
B
1
+
B
2
a
+
B
2
2
B
1
+
B
2
α
1
≤
α
α
,
2
α
≥
α
α
⇒
A
≤
B
,
A
≥
B
.
1
2
2
2
2
2
The proposition 6.1 is proved.
The Proposition 6.2.
Weighed segment of sum of
Λ
-tolerance numbers is equal to
the sum of the weighed segments of these numbers.
~
(
)
A
≡
a
,
a
,
a
1
,
a
The proof
. Let us prove that the sum of
-tolerance numbers
,
Λ
1
2
L
R
1
~
(
)
B
≡
b
,
b
,
b
2
,
b
[
]
with the weighed segments
[
]
,
, accordingly, has
A
1
,
A
B
1
,
B
1
2
L
R
2
2
2
~
~
[
]
the weighed segment
A
+
B
,
A
+
B
. Let us denote the weighed segment
A
+
B
1
1
2
2
[
]
with
. Then in accordance with (6.4)
C
1
,
C
2
1
[
]
(
)
()
()
∫
−
1
−
1
2
C
2
a
+
b
−
L
α
a
−
L
α
b
α
d
α
=
1
=
1
1
1
L
L
1
2
0
