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GIFPWAF w 1 2 ,...,α m )
ρ w j
1
w j
1
ρ
1
1
m
m
μ F n
1
=
,
v F n
κ α j α j ( α j
)
κ α j α j ( α j
)
j
=
1
j
=
1
1
w j
1
1
ρ
m
μ F n
=
,
κ α j α j ( α j
)
j
=
1
ρ
) ) ρ w j
1
1
m
1
(
1
1
v F n
(1.345)
κ α j α j ( α j
j
=
1
Moreover, from Definition 1.28 and the operational laws given in Sect. 1.8 ,we
can easily prove that the aggregated values by using the GIFPWA operators are also
IFVs.
Theorem 1.45 (Xia andXu 2010) If all the IFVs
α j
= α j ,
v
α j )(
j
=
1
,
2
,...,
m
)
are equal, i.e.
, for all j , then
(1) GIFPWAD w 1 2 ,...,α m ) =
α j
= α
D n
κ α α (α)
.
F n
GIFPWA w 1 2 ,...,α m )
(2)
=
κ α α (α)
, where
κ α j
+ λ α j
1,
m .
(3) GIFPWAG w 1 2 ,...,α m ) =
j
=
1
,
2
,...,
G n
κ α α (α)
.
(4) GIFPWAH w 1 2 ,...,α m ) =
H n
κ α α (α)
.
H , n
(5) GIFPWAH , n
w
1 2 ,...,α m ) =
κ α α (α)
.
(6) GIFPWAJ w 1 2 ,...,α m ) =
J n
κ α α (α)
.
J , n
(7) GIFPWAJ , n
w
1 2 ,...,α m ) =
κ α α (α)
.
GIFPWAP w 1 2 ,...,α m )
P n
(8)
=
κ α α (α)
, where
κ α j
+ λ α j
1,
j
=
1
,
2
,...,
m .
GIFPWAQ w 1 2 ,...,α m )
Q n
(9)
=
κ α α (α)
, where
κ α j
+ λ α j
1,
j
=
1
,
2
,...,
m .
Proof We first prove (2), by (2) in Theorem 1.44, we have
GIFPWAF w 1 2 ,...,α m )
w 1 F n
ρ
w 2 F n
ρ
w m F n
ρ
1
ρ
=
κ α 1 α 1 1 )
κ α 2 α 2 2 )
⊕···⊕
κ α m α m m )
= w 1 F n
κ α α (α) ρ
w 2 F n
κ α α (α) ρ ⊕···⊕
w m F n
κ α α (α) ρ
1
ρ
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