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=
a
+
b
−
l
a
−
l
b
=
A
+
B
;
1
1
1
L
2
L
1
1
1
2
1
[
]
(
)
()
()
∫
−
1
−
1
C
2
a
+
b
+
R
α
a
+
R
α
b
α
d
α
=
2
=
2
2
1
R
2
R
1
2
0
=
a
+
b
+
r
a
+
r
b
=
A
+
B
;
2
2
1
R
2
R
2
2
1
2
1
1
()
()
∫
−
1
∫
−
1
l
=
L
α
α
d
α
,
r
=
R
α
α
d
α
;
1
1
1
1
0
0
1
1
()
()
∫
−
1
∫
−
1
l
=
L
α
α
d
α
,
r
=
R
α
α
d
α
.
2
2
2
2
0
0
[
]
[
]
Thus
. The proposition 6.2 is proved.
C
,
C
=
A
+
B
,
A
+
B
1
2
1
1
2
2
The Proposition 6.3.
Boundaries of the weighed segment of multiplication of
-
tolerance numbers are determined by linear combinations of products of
parameters of these numbers.
Λ
~
(
)
The proof
. Let us consider fuzzy number which is the product of
A
≡
a
,
a
,
a
1
,
a
1
2
L
R
1
~
(
)
~
~
~
and
B
≡
b
,
b
,
b
2
,
b
, and let us denote it with
. Let us write out
-level
D
=
A
×
B
α
1
2
L
R
2
sets
~
and
~
according to (6.3)
[
()
()
]
[
]
A
=
A
1
,
A
2
=
a
−
L
−
1
α
a
,
a
+
R
−
1
α
a
;
α
α
α
1
1
L
2
1
R
1
1
()
()
[
]
[
]
==
According to multiplication operation for fuzzy numbers of
~
and
~
,
B
B
1
,
B
2
b
−
L
−
1
α
b
,
b
+
R
−
1
α
b
.
α
α
α
1
2
L
2
2
R
2
2
-level set
α
~
looks like
(
[
]
)
(
)
1
1
1
2
2
1
2
2
1
1
1
2
2
1
2
2
D
=
min
A
B
,
A
B
,
A
B
,
A
B
,
max
A
B
,
A
B
,
A
B
,
A
B
.
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
a
a
0
Without limiting a generality, let us consider that
−
>
,
b
+
b
<
0
1
L
R
1
2
(
~
— a positive number,
~
— a negative number). Proofs of other cases are
carried out similarly. Let us compute the weighed segment
[
]
D
1
,
D
for
Λ
-
2
-tolerancy of number
~
is proved in the proposition 2.2):
tolerance number
~
(
Λ
1
[
]
()
()
() ()
∫
D
=
2
a
b
−
a
b
L
−
1
2
α
+
a
b
R
−
1
α
−
a
b
L
−
1
α
R
−
1
α
α
d
α
=
1
2
1
2
L
R
1
1
R
L
2
1
2
1
1
2
0
=
a
b
−
l
a
b
+
r
a
b
−
ma
b
;
2
1
2
2
L
1
R
1
R
L
2
1
1
2
1
[
]
()
()
() ()
∫
−
1
−
1
−
1
−
1
D
=
2
a
b
+
a
b
R
α
−
a
b
L
α
−
a
b
R
α
L
α
α
d
α
=
2
1
2
1
R
2
L
2
1
L
R
2
1
2
1
1
2
0
=
a
b
+
r
a
b
−
l
a
b
−
pa
b
;
1
2
2
1
R
1
L
2
L
R
2
1
1
2
1
1
1
1
()
()
()
()
∫
−
1
∫
−
1
∫
−
1
2
∫
−
1
l
=
L
α
α
d
α
,
r
=
R
α
α
d
α
;
l
=
L
α
α
d
α
,
r
=
R
α
α
d
α
;
1
1
1
1
2
2
2
0
0
0
0
