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T
1
=
T
1
(
v
,y
1
)
,
T
2
=
T
2
(
v
,y
1
,y
2
)
,
······
T
k
=
T
k
(
v
,y
1
, ··· ,y
k
)
,
where
v
are the indeterminates other than
y
i
. It is obvious that we now only
need to consider triangular systems in the following form
f
1
(
u
,x
1
)=0
,
.
f
s
(
u
,x
1
, ..., x
s
)=0
,
G
1
,
G
2
,
H
.
(2)
Certainly, it can be proven that there exists a correspondence between the
solutions of these triangular sets and
S
's so that we only need to consider the
solutions of these triangular sets in order to deal with
S
's.
Example 1.
Consider an SAS
S
:[
P
,
G
1
,
G
2
,
H
]
in
Q
[
b, x, y, z
]
with
P
=[
p
1
,p
2
,p
3
]
,
G
1
=
∅,
G
2
=[
x, y, z, b,
2
− b
]
,
H
=
∅
,where
p
1
=
x
2
+
y
2
− z
2
,p
2
=(1
− x
)
2
− z
2
+1
,p
3
=(1
− y
)
2
− b
2
z
2
+1
.
The equations
P
can be decomposed into two triangular sets in
Q
(
b
)[
x, y, z
]
T
1
:[
b
4
x
2
−
2
b
2
(
b
2
−
2)
x
+2
b
4
−
8
b
2
+4
, −b
2
y
+
b
2
x
+2
−
2
b
2
,b
4
z
2
+4
b
2
x −
8
b
2
+4]
,
T
2
:[
x
2
−
2
x
+2
,y
+
x −
2
,z
]
,
with the relation
Zero(
T
2
)
where
Zero()
means the set of zeros and
Zero(
T
1
/b
)=Zero(
T
1
)
\
Zero(
b
)
.
Zero(
P
)=Zero(
T
1
/b
)
Second, compute a so-called
border polynomial
from the resulting triangular
systems, say
[
T
i
,
G
1
,
G
2
,
H
]
. We need to introduce some concepts. Suppose
F
and
G
are polynomials in
x
with degrees
m
and
l
, respectively. Thus, they can be
writteninthefollowingforms
F
=
a
0
x
m
+
a
1
x
m−
1
+
···
+
a
m−
1
x
+
a
m
,G
=
b
0
x
l
+
b
1
x
l−
1
+
···
+
b
l−
1
x
+
b
l
.
The following (
m
+
l
)
×
(
m
+
l
) matrix (those entries except
a
i
,b
j
are all zero)
a
0
a
1
···
a
m
a
0
a
1
···
a
m
l
.
.
.
.
.
.
.
.
.
a
0
a
1
···
a
m
,
b
0
b
1
···
b
l
b
0
b
1
···
b
l
m
.
.
.
.
.
.
.
.
.
b
0
b
1
···
b
l
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