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From the known condition, it follows that
ma
k
{
min
{
μ
z
ik
,μ
z
kj
}} ≤
μ
z
ij
<λ
(2.48)
which indicates that the case
λ
≤
ma
k
{
min
{
μ
z
ik
,μ
z
kj
}}
(2.49)
does not exist. Therefore, when
μ
z
ij
<λ
≤
1
−
v
z
ij
,wehave
ma
k
{
min
{
λ
z
ik
,
λ
z
kj
}} ≤
λ
z
ij
(2.50)
Hence,
Z
λ
=
(
λ
z
ij
)
n
×
n
satisfies the transitivity property.
(Sufficiency)
(1) (Reflexivity) Since
λ
z
ii
=
1, then for any
λ
∈[
0
,
1
]
,
λ
≤
μ
z
ii
, and then let
λ
=
.
(2) (Symmetry) Since for any
i
1. Then
z
ii
=
(
1
,
0
)
,
=
λ
z
ki
, if there exists
z
ik
=
k
,
z
ik
z
ki
, i.e.,
λ
μ
z
ik
=
μ
z
ki
or
v
z
ij
=
μ
z
ij
<μ
z
ji
, and let
v
z
ji
, without loss of generality, suppose that
λ
=
(μ
z
ij
+
μ
z
ji
)/
2, then
μ
z
ij
<λ<μ
z
ji
,
z
ik
=
0or1
/
2, and
z
ki
=
1,
z
ik
=
λ
z
ki
,
λ
λ
λ
which contradicts the known condition. Therefore,
Z
=
(
z
ij
)
n
×
n
is symmetry.
(3) (Transitivity) Since for any
i
,
j
,wehave
ma
k
{
min
{
λ
z
ik
,
λ
z
kj
}} ≤
λ
z
ij
(2.51)
and each element in
Z
takes its value from
{
0
,
1
/
2
,
1
}
. Then
λ
(a) When
z
ij
=
1, for any
λ
∈[
0
,
1
]
,wehave
μ
z
ij
≥
λ
, taking
λ
=
1, it can be
λ
obtained that
μ
z
ij
=
1 and
v
z
ij
=
0. Consequently,
ma
k
{
min
{
μ
z
ik
,μ
z
kj
}} ≤
μ
z
ij
,
mi
k
{
max
{
v
z
ik
,
v
z
kj
}} ≥
v
z
ij
(2.52)
(b) When
λ
z
ij
=
1
/
2, for any
λ
∈[
0
,
1
]
,wehave
μ
z
ij
<λ
≤
1
−
v
z
ij
. Also since
1
2
ma
k
{
min
{
λ
z
ik
,
λ
z
kj
}} ≤
(2.53)
then
0or
1
2
ma
k
{
{
λ
z
ik
,
λ
z
kj
}} =
min
(2.54)
Thus, for any
k
, we get
{
λ
z
ik
,
λ
z
kj
}=
min
0
(2.55)
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