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or there exists a positive integer
s
, such that
1
2
min
{
λ
z
is
,
λ
z
sj
}=
(2.56)
Case 1
If for any
k
,wehave
min
{
λ
z
ik
,
λ
z
kj
}=
0
(2.57)
Then for any
λ
∈[
0
,
1
]
,ityields
max
{
v
z
ik
,
v
z
kj
}
>
1
−
λ
(2.58)
hence
mi
k
{
max
{
v
z
ik
,
v
z
kj
}} ≥
v
z
ij
>
1
−
λ
(2.59)
Considering the arbitrary of
λ
, when
λ
tends to be infinitely small, we get
ma
k
{
min
{
μ
z
ik
,μ
z
kj
}} ≤
1
−
mi
k
{
max
{
v
z
ik
,
v
z
kj
}} =
0
(2.60)
As a result,
ma
k
{
min
{
μ
z
ik
,μ
z
kj
}} ≤
μ
z
ij
(2.61)
Case 2
If there exists a positive integer
k
1
, such that
1
2
min
{
λ
z
ik
1
,
λ
z
k
1
j
}=
(2.62)
and for any
k
=
k
1
,let
min
{
λ
z
ik
,
λ
z
kj
}=
0
(2.63)
Then according to Case 1, we have
k
1
{
{
v
z
ik
,
v
z
kj
}} ≥
min
k
max
v
z
ij
(2.64)
=
max
k
k
1
{
min
{
μ
z
ik
,μ
z
kj
}} ≤
μ
z
ij
(2.65)
=
and when
k
=
k
1
, suppose that
min
{
μ
z
ik
1
,μ
z
k
1
j
}=
μ
z
ik
1
,μ
z
ik
1
>μ
z
ij
(2.66)
Then let
λ
=
(μ
z
ij
+
μ
z
ik
1
)/
2, and thus,
μ
z
ij
≤
λ
≤
μ
z
ik
1
. Accordingly,
z
ik
1
=
λ
z
k
1
j
=
1
,
min
{
λ
z
ik
1
,
λ
z
k
1
j
}=
1
(2.67)
λ
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