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From the known condition, it follows that
ma k {
min
{ μ z ik z kj }} ≤ μ z ij
(2.48)
which indicates that the case
λ
ma k {
min
{ μ z ik z kj }}
(2.49)
does not exist. Therefore, when
μ z ij
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 z ji , and let
v z ji , without loss of generality, suppose that
λ = 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 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
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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