The idea of global degree shift first appeared in [84, Section 3]. Later it was used as a heuristic to cope with higher than quadratic degrees in the context of quantifier elimination for formulas originating from the area of automated geometric theorem proving [30]. Since this technique is of general interest, and we build on it in the following, we restate the original idea here:
As usual let ϕ be an ∧-∨-combination of Tarski atomic formulas. Let d > 0 be the gcd of all exponents of x in ϕ. We divide all exponents of x in ϕ by d obtaining ϕ0. If d is odd, then we have ∃x(ϕ) ←→ ∃x(ϕ0). If d is even, then we have ∃x(ϕ) ←→ ∃x(x ≥ 0 ∧ ϕ0). For d > 1 this reduces the degree of x in ϕ.
Notice that in order to obtain larger gcds and hence a better degree reduc- tion, we may in advance “adjust” the degree k > 0 of x in an atomic formula of the form cxk % 0, where x does not occur in c, using Proposition 22 as follows:
In equations and negated equations, i.e., % ∈ {=, 6=}, k may be equivalently replaced with any k0 > 0. In ordering inequalities, i.e., % ∈ {<, ≤, ≥, >}, we
may choose any k0 > 0 of the same parity as k. During this adjustment one
should in general use k0 that is as small as possible to keep the degree of x in
ϕ0 low.
In [47, Section 5] we have shown how to realize global degree shift by virtual substitution. This makes an integration of global degree shift into our framework of Chapter 2 even easier. Next we therefore explain and prove the correctness of the realization from [47], because we will build on it later.
Let f % 0 be an atomic formula such that f = ckxk + · · · + c1x + c0, the
coefficients c0, . . . , ck are elements of Z[u], and % ∈ {=, 6=, <, ≤, ≥, >}. As
usual, u = u0, . . . , um−1 are the parameters. Let ˆx be a fresh variable, which
does not occur in f % 0. Let d > 0 be a natural number. We define the virtual substitution [x //√d
ˆ
x] of √d
ˆ
x for x within atomic formula f % 0 as follows:
k X j=0 cjxj% 0 !" x //√dxˆ # = k X j=0 cjxbˆ j dc % 0 ! . (4.1)
The floor function is applied to make the definition complete; we will always ensure that d | j for every j ∈ {0, . . . , k} when applying a degree shift by d to atomic formula Pk
j=0cjxj % 0. Note that the mapping [x // d
√ ˆ
x] naturally
generalizes from atomic to arbitrary quantifier-free formulas.
Now we are ready to prove the semantic correctness of global degree shift realized by virtual substitution as defined in (4.1):
Proposition 68 (Correctness of Global Degree Shift). Let ϕ(u, x) be an ∧-∨-
combination of Tarski atomic formulas. Let d ≥ 1 be a divisor of all exponents of x occurring in ϕ. Let ˆx be a fresh variable that does not occur in ϕ. Then we have:
(i) If d is even, then ∃x(ϕ) ←→ ∃ˆx ˆx ≥ 0 ∧ ϕ[x //√d
ˆ
x].
(ii) If d is odd, then ∃x(ϕ) ←→ ∃ˆx ϕ[x //√d
ˆ
x].
Proof. (i) Assume that d is even. Fix some real values a ∈ Rm for the
parameters u. We show that we have R |= ∃x(ϕ) ←→ ∃ˆx ˆx ≥ 0 ∧ ϕ[x // d √ ˆ x](a).
First assume that there exists b ∈ R such that (a, b) satisfy ϕ. We show that (a, bd) satisfy ˆx ≥ 0 ∧ ϕ[x //√d
ˆ
x]. Since d is even, bd indeed satisfies
ˆ
x ≥ 0. Now it suffices to prove the following: The atomic formula ˆα =
Pk j=0cjxbˆ j dc % 0 at position π in ϕ[x // d √ ˆ
x]—obtained from the atomic
formula α = Pk
j=0cjxj % 0 at position π in ϕ—is satisfied by (a, bd)
whenever (a, b) satisfy the atomic formula α. The fact that ϕ is an ∧-∨- combination of atoms will then imply that (a, bd) satisfy ϕ[x //√dx].ˆ Since we assume that d divides all exponents of x occurring in ϕ, we obtain that cj= 0 for every j ∈ {0, . . . , k} such that d - j. Therefore, we have
k X j=0 cjxj= X 0 ≤ j ≤ k d | j cjxj and k X j=0 cjxbˆ j dc = X 0 ≤ j ≤ k d | j cjxˆ j d. (4.2)
This already ensures that whenever (a, b) satisfy the atomic formula α, then (a, bd) satisfy the atomic formula ˆα.
Now assume that there exists b ∈ R such that (a, b) satisfy the formula ˆ
x ≥ 0 ∧ ϕ[x //√dx]. We show that (a,ˆ √db) satisfy ϕ. Since b satisfies ˆx ≥ 0,
d
√
b ∈ R. Similarly as above, it is sufficient to look at atomic formulas in ϕ and their counterparts in ϕ[x //√d
ˆ
x]. Using equations (4.2) again, we
can show that whenever (a, b) satisfy Pk
j=0cjxbˆ j dc % 0—obtained from Pk j=0cjx j % 0—then (a,√d b) satisfyPk j=0cjx
j% 0. Since we assume that
(a, b) satisfy ˆx ≥ 0 ∧ ϕ[x //√d
ˆ
x], this ensures that (a,√d
b) satisfy ϕ, so the
proof of (i) is finished.
(ii) The proof is similar to the proof of (i), so we omit it.
We illustrate the application of a global degree shift along with adjustments that make it possibly on a simple example.
Example 69. Consider the formula ∃x(ϕ):
∃x(u0x2> 0 ∧ x6+ u1x3+ u0< 0).
As such it does not allow for global degree shift, because the gcd of the degrees of x in ϕ is 1. Observe, however, that ϕ is equivalent to
∃x(u0x6> 0 ∧ x6+ u1x3+ u0< 0).
This formula allows a degree shift with d = 3. Proposition 68 then guarantees that this formula, and therefore also ϕ, is equivalent to
∃ˆx(u0xˆ2> 0 ∧ ˆx2+ u1x + uˆ 0< 0).
Similarly, formula ϕ = u0x14 ≥ 0 ∧ x8+ u1x4+ u0 ≤ 0 allows for global
degree shift with d = 2. Equivalently adjusting the formula beforehand to
u0x12≥ 0 ∧ x8+ u1x4+ u0≤ 0 we see that a global degree shift with d = 4 is
possible, so ∃x(ϕ) is equivalent to ∃ˆx(ˆx ≥ 0 ∧ u0xˆ3≥ 0 ∧ ˆx2+ u1x + uˆ 0≤ 0). 3
To integrate global degree shift into our framework of Chapter 2 we show how to apply definitions and tools developed there to allow for objects like √g
ˆ
x to be
substituted for x into a quantifier-free formula by means of virtual substitution we have just defined and proven correct.
For this, we have to slightly generalize the framework by introducing shadow
quantifiers. Recall that we are considering the elimination of ∃x from the for-
mula ∃x ϕ(u, x), where u are the parameters. As a first step we switch to the equivalent problem ∃ˆx∃x ϕ(u, x), where the shadow variable ˆx does not
occur in {u0, . . . , um−1, x}, so ˆx does not occur in ϕ either. Proceeding in accor-
dance with the framework of Chapter 2 to obtain a quantifier-free equivalent of ∃ˆx∃x(ϕ), the shadow quantifier ∃ˆx imposes a trivial elimination problem that
needs to be “solved” after eliminating ∃x from ∃x(ϕ).
Here notice that strictly following vs-scheme of Chapter 2 one would not simply drop the quantifier ∃ˆx with the argument that the variable ˆx does not
occur in the quantifier-free equivalent of ∃x(ϕ). The scheme would yield a trivial elimination set like E = {±∞}. Moreover, notice that using ∅ as an elimination set E in vs-scheme always yields “false,” which is incorrect.
Observe that in contrast to all elimination sets studied so far we introduce here a variable ˆx which was not present in ϕ before. That variable is bound by
shadow quantifier ∃ˆx. Intuitively, for the elimination of ∃ˆx∃x we switch from
one hard plus one trivial elimination step to two nontrivial elimination steps. Semantically we view ˆx during the elimination of ∃x as a parameter.
The termination of quantifier elimination involving shadow quantifiers fol- lows from the termination of the underlying quantifier elimination method plus the fact that we will always use only finitely many shadow quantifiers for each regular quantifier.
To keep the notation simple, we will in the sequel not formally introduce shadow quantifiers for all quantifiers. Instead, we will always silently assume their presence whenever we perform a degree shift.
Consider now the elimination of ∃x. Assume that we are in the situation of Proposition 68, and d > 1 is the gcd of the degrees of x in ϕ. We use an elimination set that depends on the parity of d: E = √dxˆ , where √dx is aˆ
global degree shift test point such that ˆx is a shadow variable for x and d > 1.
The symbol “√d
ˆ
x” is merely a shorthand notation for the respective parametric
root description: If d is odd, then it stands for parametric root description (f, S) = xd− ˆx, ((−1, 0, 1), 1)
with a guard true. If d is even, then it stands for parametric root description
(f, S) =xd− ˆx,((1, 0, 1), 1), ((1, 0, −1, 0, 1), 2) with a guard ˆx ≥ 0.
Observe here that everything is semantically correct for f = xd− ˆx. For any
value a ∈ Rm+1of the parameters u and ˆx the following holds: f hai is of real
type (−1, 0, 1) when d is odd, and f hai is of real type (1, 0, 1) or (1, 0, −1, 0, 1) if and only if R |= (ˆx ≥ 0)(a). Furthermore, Proposition 68 ensures the semantic
correctness of the virtual substitution of √dx into an atomic formula g % 0ˆ whenever d divides all the degrees of x in g. In such a case it is obvious that (4.1) yields a formula with the following property: Whenever a satisfies a guard of √d
ˆ
x, then a satisfies (g % 0)[x //√d
ˆ
x] if and only if the formula g % 0 holds
at (f, S)hai, which is indeed a well-defined real number. Therefore, using the above-mentioned guards along with (4.1) give us correct realizations of guard and vs-prd-at, respectively, for √d
ˆ
x.