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If S α F n <
S
α F n and S α F n >
S
α F n , then by using Xu and Yager
(2006)'s ranking method, we have
α F n
GIFPWAF w 1 2 ,...,α n )<α F n
<
(1.358)
S
α F n , i.e.,
If S α F n =
μ F n
v F n
μ α F n
v
=
max
j
min
j
(1.359)
κ α j α j ( α j
)
κ α j α j ( α j
)
α F n
then by Eqs. ( 1.349 ) and ( 1.355 ), we have
v F n
μ α F n
=
max
j
μ F n
, α F n
=
min
j
(1.360)
κ α j α j ( α j
)
κ α j α j ( α j
)
then
v F n
h
α F n
H α F n = μ α F n +
=
μ F n
+
=
v α F n
max
j
min
j
κ α j α j ( α j
)
κ α j α j ( α j
)
So we have
GIFPWAF w 1 2 ,...,α m ) = α F n
(1.361)
S
α F n , i.e.,
If S α F n =
v F n
μ α F n
v α F n
=
min
j
μ F n
max
j
(1.362)
κ α j α j ( α j
)
κ α j α j ( α j
)
then by Eqs. ( 1.350 ) and ( 1.354 ), we have
μ F n
v F n
μ α F n
=
,
=
min
j
v α F n
max
j
(1.363)
κ α j α j
)
κ α j α j
)
j
j
hence
μ F n
v F n
h
α F n
(1.364)
H α F n = μ α F n +
v
=
min
j
+
max
j
=
α
κ α j α j j )
κ α j α j j )
F n
Thus, it follows that
GIFPWAF w 1 2 ,...,α n ) = α F n
(1.365)
and then from Eqs. ( 1.358 ), ( 1.361 ) and ( 1.365 ), we know that (2) always holds.
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