Information Technology Reference
In-Depth Information
μ
k
⊕
1
β
i
=
i
=
1
β
i
,
v
k
i
=
1
β
i
k
i
=
i
=
1
(
)μ
β
i
)
−
i
=
1
(
1
+
(γ
−
1
1
−
μ
β
i
)
=
−
μ
β
i
)
,
i
=
1
(
)
i
=
1
(
1
+
(γ
−
1
)μ
β
i
)
+
(γ
−
1
1
γ
i
=
1
v
β
i
(1.250)
i
=
1
(
)
i
=
1
v
1
+
(γ
−
1
)(
1
−
v
β
i
))
+
(γ
−
1
β
i
Then we prove that Eq. (
1.249
) holds for
n
=
k
+
1, that is,
k
i
=
1
β
i
⊕
β
k
+
1
⎛
⎝
1
+
(γ
−
1
)μ
1
−
μ
1
+
(γ
−
1
)μ
β
k
+
1
−
1
−
μ
β
k
+
1
k
i
=
1
β
i
k
i
=
1
β
i
1
+
(γ
−
1
)μ
1
+
(γ
−
1
)μ
β
k
+
1
+
(γ
−
1
)
1
−
μ
1
−
μ
β
k
+
1
,
=
k
i
=
1
β
i
k
i
=
1
β
i
⎞
⎠
γ
v
k
i
=
1
β
i
v
β
k
+
1
1
1
1
)
1
v
β
k
+
1
+
(γ
−
+
(γ
−
1
)
−
v
k
i
=
1
β
i
+
(γ
−
1
−
1
)
v
k
i
=
1
β
i
v
β
k
+
1
(1.251)
By the operational laws for IFVs, we have
⎛
⎝
1
+
(γ
−
1
)μ
⎞
⎠
1
+
(γ
−
1
)μ
β
k
+
1
k
⊕
1
β
i
i
=
1
+
(γ
−
1
)
1
+
(γ
−
1
)μ
β
k
+
1
i
=
1
(
)μ
β
i
)
−
i
=
1
(
−
μ
β
i
)
i
=
1
(
1
+
(γ
−
1
)μ
β
i
)
+
(γ
−
1
)
i
=
1
(
1
−
μ
β
i
)
1
+
(γ
−
1
1
=
1
−
1
−
μ
β
k
+
1
i
=
1
(
)μ
β
i
)
−
i
=
1
(
1
+
(γ
−
1
1
−
μ
β
i
)
−
i
=
1
(
)
i
=
1
(
1
+
(γ
−
1
)μ
β
i
)
+
(γ
−
1
1
−
μ
β
i
)
)μ
β
i
)
1
)μ
β
k
+
1
−
γ
i
=
1
(
−
μ
β
i
)
1
−
μ
β
k
+
1
γ
i
=
1
(
1
+
(γ
−
1
+
(γ
−
1
1
=
i
=
1
(
1
+
(γ
−
1
)μ
β
i
)
+
(γ
−
1
)
i
=
1
(
1
−
μ
β
i
)
γ
k
+
1
i
=
1
(
1
+
(γ
−
1
)μ
β
i
)
−
γ
k
+
1
1
(
1
−
μ
β
i
)
i
=
1
(
1
+
(γ
−
1
)μ
β
i
)
+
(γ
−
1
)
i
=
1
(
1
−
μ
β
i
)
i
=
(1.252)
=
Search WWH ::
Custom Search