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f
a
ʵ
f
s
(
y
3
)
s
(
z
3
)
t
(
y
3
)
t
(
z
3
)
M
ʵ
f
s
(
y
2
)
t
(
y
2
)
t
(
z
2
)
f
a
s
(
z
2
)
M
s
(
y
1
)
f
t
(
y
1
)
t
(
z
1
)
f
a
s
(
z
1
)
M
:= mid(row(
t
(
x
)))
M
:= mid(row(
t
(
x
)))
s
=
h
(
f ⓦ t
)
s
:=
h
(
f
a
ⓦ f ⓦ t
), for
a
:=
f ⓦ t
(
z
2
)
Fig. 3
For the first and last equalities note that
f
a
and
f
preserve the rows. For (i)
recall also that
ʵ
is self-inverse.
Let then 1
≤ j ≤ l
be such that
j ∈ I
when both (i) and (ii) and fail for
j
.
Then we obtain
))) and row(
t
(
z
j
))
>
mid(row(
t
(
row(
t
(
y
j
))
>
mid(row(
t
(
x
x
)))
,
or (12)
))) and row(
t
(
z
j
))
<
mid(row(
t
(
row(
t
(
y
j
))
<
mid(row(
t
(
x
x
)))
.
(13)
Assume first that (12) holds and let
i
∈
I
. Then either
(i) row(
t
(
y
i
))
<
mid(row(
t
(
x
)))
<
row(
t
(
y
j
)), or
(ii) row(
t
(
z
i
))
<
mid(row(
t
(
)))
<
row(
t
(
z
j
))
.
x
Since
t
(
y
j
)=
t
(
z
j
),
t
(
y
i
)=
t
(
z
i
), and
f
preserves the rows, in both cases
we conclude that
t
(
z
j
))
t
(
z
i
))
|
row(
f
ⓦ
−
row(
f
ⓦ
|
>
1
.
t
(
z
j
) fixed. By (12) we now have
Therefore
f
a
leaves
f
ⓦ
t
(
z
j
)=
ʵ
t
(
z
j
)=
s
(
z
j
)
.
s
(
y
j
)=
ʵ
ⓦ
f
ⓦ
t
(
y
j
)=
ʵ
ⓦ
f
ⓦ
ⓦ
f
a
ⓦ
f
ⓦ
