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Proof
(1) and (2) are obvious, we prove the others:
(
3
)λ(α
1
⊕
α
2
)
=
λ(
h
−
1
(μ
α
2
)), g
−
1
(
h
(μ
α
1
)
+
h
(g(
v
)
+
g(
v
)))
α
α
1
2
h
−
1
h
h
−
1
, g
−
1
g
−
1
=
λ
(
h
(μ
α
1
)
+
h
(μ
α
2
))
λg
(g(
v
)
+
g(
v
))
α
α
1
2
)
λα
1
⊕
λα
2
=
h
−
1
λ
h
(μ
α
1
)
, g
−
1
λg(
v
α
1
)
⊕
h
−
1
λ
h
(μ
α
2
)
, g
−
1
λg(
v
α
2
)
=
h
−
1
h
h
−
1
λ
h
(μ
α
1
)
+
h
h
−
1
λ
h
(μ
α
2
)
,
g
−
1
g
g
−
1
λg(
h
−
1
λ
h
(μ
α
2
)
, g
−
1
λ
g(
=
(μ
α
1
)
+
h
v
)
+
g(
v
α
α
1
2
v
α
1
)
+
g
g
−
1
λg(
v
α
2
)
=
h
−
1
λ
(μ
α
2
)
, g
−
1
λg(
v
α
2
)
h
(μ
α
1
)
+
λ
h
v
α
1
)
+
λg(
=
λ(α
1
⊕
α
2
).
h
−
1
h
−
1
(μ
α
)) , g
−
1
(μ
α
)) , g
−
1
(
5
)λ
1
α
⊕
λ
2
α
=
(λ
1
h
(λ
1
g(
v
α
))
⊕
(λ
2
h
(λ
2
g(
v
α
))
h
−
1
h
h
−
1
h
h
−
1
=
(λ
1
h
(μ
α
))
+
(λ
2
h
(μ
α
))
,
g
−
1
g
−
1
g
−
1
g
(λ
1
g(
v
α
))
+
g
(λ
2
g(
v
α
))
h
−
1
(μ
α
)) , g
−
1
=
(λ
1
h
(μ
α
)
+
λ
2
h
(λ
1
g(
v
α
)
+
λ
2
g(
v
α
))
=
(λ
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
,
(α
λ>
(1)
0.
c
)
=
(α
λ
)
c
,
λ(α
λ>
(2)
0.
c
c
c
.
(3)
α
1
⊕
α
2
=
(α
1
⊗
α
2
)
c
c
c
,
(4)
α
1
⊗
α
2
=
(α
1
⊕
α
2
)
c
where
α
=
(
v
α
,μ
α
)
denotes the complement of the IFV
α
.
Proof
Based on the operations defined in Definition 1.21, we have
)
λ
=
g
−
1
(μ
α
))
=
(λα)
c
h
−
1
c
.
(1)
(α
(λg(
v
α
)) ,
(λ
h
)
=
h
−
1
α
))
=
(α
λ
)
c
α
)) , g
−
1
c
.
(2)
λ(α
(λ
h
(
v
(λg(
v
2
=
h
−
1
h
)
, g
−
1
g(μ
α
1
)
+
g(μ
α
2
)
=
(α
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
(μ
α
1
)
+
g(μ
α
2
)
=
(α
1
⊕
α
2
)
c
c
c
,
(4)
α
1
⊗
α
v
α
1
)
+
g(
v
which completes the proof.
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