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m
with
γ
(
c
), we
cannot uniformly count
Z
+
or
Z
−
for
χ
.Notethat
χ
(
c
) is of real type, i.e., there
are no zeroes in
While the computations used so far are all uniform in
c
∈
R
. For those polynomials we can compute the number of
positive and negative zeroes using Descartes rule of signs. It states that the
positive real zeroes of a polynomial
χ
(
c
) of real type are exactly the number of
sign changes in the list of coecients of
χ
(
c
) ignoring 0. By considering
χ
(
C
\
R
−
c
)
one can also compute the negative zeroes using Descartes rule of signs.
Let
χ
=
i
=0
a
i
/b
i
y
i
,where
a
i
,b
i
∈
Q
[
u
1
,...,u
m
]. For
δ
∈{
<,
=
,>
}
l
let
ϕ
δ
be the formula
a
l
b
l
δ
l
0
.
Using Descartes rule of signs we can now uniformly count the number
Z
+
of
positive and the number
Z
δ
−
a
1
b
1
δ
1
0
∧···∧
m
with
γ
(
c
)
of negative zeroes of
χ
(
c
) for all
c
∈
R
and
ϕ
δ
(
c
).
Finally define
ϕ
by
ϕ
δ
=0
.
Z
+
−
Z
δ
−
δ∈{<,
=
,>}
l
Our formula
ϕ
states that the polynomial
χ
has exactly the same number of
positive as of negative real zeroes. A quantifierfree formula with this property is
called
type formula
for the polynomial
χ
. Recall from our discussion above that
in this situation
G
has no zeroes which satisfy the given side conditions. Thus
our final elimination result is
γ
∧¬
ϕ
.
2.3
Constructing Equations
We enter this case of the Hermitian quantifier elimination if the input formula
does not contain any equation or the dimension of Id
G
(
c
)
is
n
, i.e.,
I
(
G
)=
{
0
}
for
c
with
γ
(
c
). In other words we consider the input formula
x
n
h
=0
.
∃
x
1
···∃
h>
0
∧
f
∈
H
f
∈
F
In this case, we can eliminate one quantifier, say
x
n
, and the other quanti
fiers of the considered block are eliminated by a recursive call of the Hermitian
quantifier elimination.
Let
h
have a representation of the form
d
k
=0
a
h,k
x
n
where each
a
h,k
is a
polynomial in
Q
[
u
1
,...,u
m
,x
1
,...,x
n−
1
]with
a
h,d
h
= 0. Assume for a moment
r
that
H
=
{
h
1
,...,h
r
}
and let
D
=
×
k
=1
{
0
,...,d
h
k
}
.For
δ
∈
D
we denote by
δ
h
the
s
th element of
δ
such that
h
s
=
h
. Define
P
=
{
(
h
i
,h
j
)

1
≤
i<j
≤
r
}
H
2
.
For
h
⊆
let
Γ
d
∈
H
and
d
∈{
0
,...,d
h
}
be the following formula:
d
h
a
h,k
=0
∧
a
h,d
=0
.
k
=
d
+1
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