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n
∨
z
(
1
)
ik
z
(
2
)
kj
z
ij
=
1
(
∧
)
=
(
ma
k
{
min
{
μ
z
(
1
)
ik
,μ
z
(
2
)
kj
}
,
mi
k
{
max
{
v
z
(
1
)
ik
,
v
z
(
2
)
kj
}}
)
k
=
=
(
max
{
min
{
μ
z
(
1
)
i
1
,μ
z
(
2
)
1
j
}
,...,
min
{
μ
z
(
1
)
in
,μ
z
(
2
)
nj
}}
,
min
{
max
{
v
z
(
1
)
i
1
,
v
z
(
2
)
1
j
}
,...,
max
{
v
z
(
1
)
in
,
v
z
(
2
)
nj
}}
)
(2.19)
Since
0
≤
max
{
min
{
μ
z
(
1
)
i
1
,μ
z
(
2
)
1
j
}
,...,
min
{
μ
z
(
1
)
in
,μ
z
(
2
)
nj
}} ≤
1
(2.20)
0
≤
min
{
max
{
v
z
(
1
)
i
1
,
v
z
(
2
)
1
j
}
,...,
max
{
v
z
(
1
)
in
,
v
z
(
2
)
nj
}}
)
≤
1
(2.21)
There must exist two positive integers
k
1
and
k
2
such that
max
{
min
{
μ
z
(
1
)
i
1
,μ
z
(
2
)
}
,...,
min
{
μ
z
(
1
)
in
,μ
z
(
2
)
}} =
min
{
μ
z
(
1
)
ik
1
,μ
z
(
2
)
k
1
j
}
(2.22)
1
j
nj
min
{
max
{
v
z
(
1
)
i
1
,
v
z
(
2
)
1
j
}
,...,
max
{
v
z
(
1
)
in
,
v
z
(
2
)
nj
}}
)
=
max
{
v
z
(
1
)
ik
2
,
v
z
(
2
)
k
2
j
}
(2.23)
Accordingly, we have
max
{
min
{
μ
z
(
1
)
i
1
,μ
z
(
2
)
}
,...,
min
{
μ
z
(
1
)
in
,μ
z
(
2
)
}} +
min
{
max
{
v
z
(
1
)
i
1
,
v
z
(
2
)
1
j
}
,...,
1
j
nj
max
{
v
z
(
1
)
in
,
v
z
(
2
)
nj
}}
)
=
min
{
μ
z
(
1
)
ik
1
,μ
z
(
2
)
k
1
j
}+
max
{
v
z
(
1
)
ik
2
,
v
z
(
2
)
k
2
j
}
(2.24)
In the case of
k
1
=
k
2
, we get
{
μ
z
(
1
)
ik
1
,μ
z
(
2
)
k
1
j
}+
{
ik
1
,
k
2
j
}=
{
μ
z
(
1
)
ik
1
,μ
z
(
2
)
k
1
j
}+
{
ik
1
,
k
1
j
}≤
min
max
v
z
(
1
)
v
z
(
2
)
min
max
v
z
(
1
)
v
z
(
2
)
1
(2.25)
Also when
k
1
=
k
2
,ityields
min
{
μ
z
(
1
)
ik
1
,μ
z
(
2
)
k
1
j
}+
max
{
v
z
(
1
)
ik
2
,
v
z
(
2
)
k
2
j
}≤
min
{
μ
z
(
1
)
ik
2
,μ
z
(
2
)
k
2
j
}+
max
{
v
z
(
1
)
ik
2
,
v
z
(
2
)
k
2
j
}≤
1
(2.26)
Hence
max
{
min
{
μ
z
(
1
)
i
1
,μ
z
(
2
)
1
j
}
,...,
min
{
μ
z
(
1
)
in
,μ
z
(
2
)
nj
}}
+
min
{
max
{
v
z
(
1
)
i
1
,
v
z
(
2
)
1
j
}
,...,
max
{
v
z
(
1
)
in
,
v
z
(
2
)
nj
}}
)
≤
1
(2.27)
Consequently, the composition matrix of two intuitionistic fuzzy matrices is also an
intuitionistic fuzzy matrix. This completes the proof.
Definition 2.4
(Zhang et al. 2007) If the intuitionistic fuzzy matrix
Z
=
(
z
ij
)
n
×
n
satisfies the following condition:
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