Independence Logic
In this section we define the system Algebraically Closed Field Atomic Absolute Independence Logic (ACFAAIndL) and then prove its soundness and complete- ness. The syntax and deductive apparatus of this system are the same as those of AAIndL.
4.6.1
Semantics
As for the case of vector spaces, the atom⊥(x) has a natural interpretation in this context: the elements inxare algebraically independent elements.
Definition 4.6.1. Let Kbe an algebraically closed field ands: dom(s)→K
with x ⊆ dom(s) ⊆ Var. We say that K satisfies ⊥(x) under s, in symbols
K |=s ⊥(x), if for every x ∈ x we have that s(x) is transcendental over the
subfieldFofKgenerated by{s(z)|z∈x} \ {s(x)}, that is for every non-trivial
polynomial P(X) = Xn+ n−1 X i=0 aiXi
Notice that, because of Lemma 4.5.1, the above condition is equivalent to the classical characterization of algebraic independence, according to which a set of elements is independent if and only if the elements in the set do not satisfy any non-trivial polynomial equation with coefficients in the prime field ofK.
Definition 4.6.2. Let Σ be a set of atoms and let s be such that the set of variables occurring in Σ is included in dom(s). We say thatKsatisfies Σ under
s, in symbolsK|=sΣ, ifKsatisfies every atom in Σ under s.
Definition 4.6.3. Let Σ be a set of atoms. We say that ⊥(x) is a logical consequence of Σ, in symbols Σ |=⊥(x), if for every algebraically closed field
K and s such that the set of variables occurring in Σ∪ {⊥(x)} is included in
dom(s) we have that
if K|=sΣ then K|=s⊥(x).
4.6.2
Soundness and Completeness
Theorem 4.6.4. Let Σ be a set of atoms, then
Σ|=⊥(x) if and only if Σ` ⊥(x).
[The deductive system to which we refer has been defined in Section 2.3.3.]
Proof. (⇐) Obvious.
(⇒) Suppose Σ0⊥(x). Notice that if this is the case thenx6=∅. Indeed if
x=∅then Σ` ⊥(x) because by rule (a2.) ` ⊥(∅).
We can assume thatxis injective. This is without loss of generality because clearlyK|=s⊥(x) if and only ifK|=s⊥(πx), whereπ: Var<ω→Var<ω is the function that eliminates repetitions in finite sequences of variables. Let then
x= (xj0, ..., xjn−1)6=∅be injective.
LetQbe the field of rational numbers andEbe the field of algebraic numbers.
As known,Eis infinite-dimensional overQ. Indeed suppose not, say [E:Q] =n,
letqbe a root of an irreducible polynomial of degreen+1, then [Q(q) :Q] =n+1
but a subspace of a vector space can not be of dimension grater than it. Let then {ai|i∈ω} be an injective enumeration of a basisA of Eover Q
and{wi|i∈ω} an injective enumeration of Var\ {xj0}. LetCbe the field of
complex numbers and letsbe the following assignment:
s(wi) =eai and s(xj0) = 0 ifx={xj0} s(xj0) =e Pn−1 i=1api ifx6={xj 0},
whereeis the Euler number and wpi =xji for everyi∈ {1, ..., n−1}.
We claim that C 6|=s ⊥(x). In accordance to the semantic we then have to show that there is x∈ xsuch that s(x) is algebraic over the subfield of C
generated by {s(z)|z∈x} \ {s(x)}. But xj0 satisfies this condition, indeed
either s(xj0) = 0 or s(xj0) = n−1 Y i=1 s(xji)
and clearly in both casess(xj0) is algebraic over the subfield ofCgenerated by
{s(v)|v∈x} \ {s(xj0)}.
Let now⊥(v)∈Σ, we want to show thatC|=s⊥(v). As before, we assume, without loss of generality, thatv is injective. Notice that ifv =∅, then C|=s
⊥(v). Thus let v = (vh0, ..., vhc−1) 6= ∅. We first state the following deep
theorem.
Theorem 4.6.5(Lindemann–Weierstrass Theorem). If{b0, ..., bd−1}is a set of
algebraic numbers which are linearly independent overQ, then the set
eb0, ..., ebd−1 is algebraically independent over
Q.
Proof. See [3, Theorem 1.4].
Case 1. xj0 ∈/ v.
Letwri =vhi for everyi∈ {0, ..., c−1}, we then have that
ar0, ..., arc−1
is linearly independent overQand so by the theorem
s(vh0) =s(wr0) =e
ar0, ..., s(vh
c−1) =s(wrc−1) =e
arc−1
is algebraically independent overQ.
Case 2. xj0 ∈v. Subcase 1. x\v 6=∅.
Notice that x 6= {xj0} because if not then x\v = {xj0} and so xj0 ∈/ v.
Hence s(xj0) =e Pn−1 i=1 api. Let (v\ {xj0})∩x= n vh0 0, ..., vh0d−1 o ,v\x=nvh00 0, ..., vh00t−1 o ,wr0 i=vh0i for everyi∈ {0, ..., d−1}andwr00 i =vh 00 i for every i∈ {0, ..., t−1} Suppose now that the set
( ar0 0, ..., ar0d−1, n−1 X i=1 api, ar00 0, ..., art00−1 )
is linearly dependent, then there existsf ∈Kd,l∈Kandg∈Ktsuch that
d−1 X i=0 fi(ar0 i) +l( n−1 X i=1 api) + t−1 X i=0 gi(ar00 i) = 0 withf 6= (00, ...,0d−1) orl6= 0 org6= (00, ...,0t−1).
LetV ={i∈ {1, ..., n−1} |xi∈(v\ {xj0})∩x}. In each three of the cases
the linear combination d−1 X i=0 (fi+l)(ar0 i) +l( n−1 X i=1 i /∈V api) + t−1 X i=0 gi(ar00 i) = 0
is non trivial. Thus the set
n
s(xj1), ..., s(xjn−1), ar000, ..., ar00t−1 o
is linearly dependent, which is absurd. Hence again by the theorem, the set
n
ear00, ..., ear0d−1
, ePni=1−1api, ear000, ..., ear00t−1o
is algebraically independent overQ.
Subcase 2. x⊆v.
This case is not possible. Suppose indeed it is, then by rule (c2.) we can
assume that v = x v0 with v0 ⊆ Var\x. Thus by rule (b2.) we have that
Σ` ⊥(x) which is absurd.