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Proof
(1) and (2) are obvious, we prove the others:
(
3
)λ(α 1 α 2 )
= λ(
h 1
α 2 )), g 1
(
h
α 1 ) +
h
(g(
v
) + g(
v
)))
α
α
1
2
h 1
h h 1
, g 1
g 1
=
λ
(
h
α 1 ) +
h
α 2 ))
λg
(g(
v
) + g(
v
))
α
α
1
2
)
λα 1 λα 2 = h 1 λ h α 1 ) , g 1 λg( v α 1 ) h 1 λ h α 2 ) , g 1 λg( v α 2 )
= h 1 h h 1 λ h α 1 ) + h h 1 λ h α 2 ) ,
g 1 g g 1 λg(
h 1 λ h
α 2 ) , g 1 λ g(
=
α 1 ) +
h
v
) + g(
v
α
α
1
2
v α 1 ) + g g 1 λg(
v α 2 )
= h 1 λ
α 2 ) , g 1 λg(
v α 2 )
h
α 1 ) + λ
h
v α 1 ) + λg(
= λ(α 1 α 2 ).
h 1
h 1
α )) , g 1
α )) , g 1
(
5
1 α λ 2 α =
1 h
1 g(
v α ))
2 h
2 g(
v α ))
h 1 h h 1
h h 1
=
1 h
α ))
+
2 h
α ))
,
g 1
g 1
g 1
g
1 g(
v α ))
+ g
2 g(
v α ))
h 1
α )) , g 1
=
1 h
α ) + λ 2 h
1 g(
v α ) + λ 2 g(
v α ))
= 1 + λ 2 )α.
Similarly, (4) and (6) can be proven which completes the proof of the theorem.
Theorem 1.28 (Xia et al. 2012c)
c
) λ = (λα)
c ,
λ>
(1)
0.
c
) = λ )
c ,
λ(α
λ>
(2)
0.
c
c
c .
(3)
α
1 α
2 = 1 α 2 )
c
c
c ,
(4)
α
1 α
2 = 1 α 2 )
c
where
α
= (
v
α α )
denotes the complement of the IFV
α
.
Proof Based on the operations defined in Definition 1.21, we have
) λ = g 1
α )) = (λα)
c
h 1
c .
(1)
(λg(
v
α )) ,
h
) = h 1
α )) = λ )
c
α )) , g 1
c .
(2)
λ(α
h
(
v
(λg(
v
2 = h 1 h
) , g 1 g(μ α 1 ) + g(μ α 2 ) = 1 α 2 )
c
c
c .
(3)
α
1 α
(
v
) +
h
(
v
α
α
1
2
2 = g 1 g(
α 2 ) ,
h 1 h
α 1 ) + g(μ α 2 ) = 1 α 2 )
c
c
c ,
(4)
α
1 α
v
α 1 ) + g(
v
which completes the proof.
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