Logic: Lecture 10, 3 September 2015
--------------------------------
----------------------------------------------------------------------
FO satisfiability
Theorem: Let X be a set of sentences over L.
The following are equivalent.
1. There is an L-structure M = (S,i) that satisfies X
2. There is an L_H structure M' = (S,i') that satisfies X
3. X U Phi_H U Phi_Q U Phi_Eq is propositonally satisfiable
(3 => 2)
We have a valuation v over all prime formulas in L_H such that
v |= X U Phi_H U Phi_Q U Phi_Eq
Need to define an L_H structure I = (S,i) that satisfies X
First S:
Define an equivalence relation on closed terms
t ~ t' iff v |= t = t'
v |= Phi_Eq guarantees that ~ is an equivalence relation
Elements of S are equivalence classes [t] of closed terms
Define i as follows:
For C : i(c) = [c]
For R : ([c1],...,[cn]) in i(r) iff v |= r(c1,...,cn)
For F :
For closed terms (t1,...,tn), need to identify [t] such that
i(f)([t1],...,[tn]) = [t]
Choose t = f(t1,...,tn), itself a closed term, so
i(f)([t1],...,[tn]) = [f(t1,...,tn)]
The fact that interpretations of F and R are compatible
with ~ follows from Phi_Eq
Claim: For every closed term t, there is a constant c such that
[t] = [c] --- i.e. v(t = c) is true
Proof:
Consider the formula phi(x) = t = x with one free variable.
C_H contains a constant c_phi for the formula Ex phi(x).
Substituting t for x, we get phi(t) to be t = t.
- By the equality axioms v(phi(t)) = v(t = t) = tt.
- By the quantifier axiom v(phi(t) -> Ex phi(x)) = tt.
Hence, v(Ex phi(x)) = tt.
- By the Henkin axiom v(Ex phi(x) -> v(phi(c_phi)) = tt.
Hence v(phi(c_phi)) = v(t = c_phi) = tt
Thus, [t] = [c_phi].
Claim: For all formulas phi in L_H, I |= phi iff v |= phi
Proof:
Construction of S ensures that I |= phi iff v |= phi for
all atomic formulas.
Boolean connectives are easy.
phi = Ex psi
I |= Ex psi => I |= psi(a) for some element a of s.
Every a in S is [t] for some closed term t, and hence (by
the previous claim) [c] for some constant c. Hence
v |= psi(c). From quantifier axiom, v |= Ex psi
v |= Ex psi => v |= psi(c_phi) by Henkin axiom,
so I |= psi(c_phi) so I |= Ex psi.
From the claim, it follows that I |= X
----------------------------------------------------------------------
Lemma: Finite satisfiability
X a set of FO sentences over L. X is satisfiable iff every
finite subset Y of X is satisfiable
(<==)
Every finite subset Y of X is FO satisfiable. To show that X
is FO satisfiable.
Sufficient to show that X U Phi_H U Phi_Q U Phi_Eq is
propositionally satisfiable.
Sufficient to show that every finite subset of
X U Phi_H U Phi_Q U Phi_Eq is propositionally satisfiable
If Y is FO satisfiable, then Y U Phi_H U Phi_Q U Phi_Eq is
propositionally satisfiable
Each finite subset of X U Phi_H U Phi_Q U Phi_Eq is contained
in some Y U Phi_H U Phi_Q U Phi_Eq, hence satisfiable.
So the result follows.
----------------------------------------------------------------------
Theorem: Compactness
X |= phi iff Y |= phi for some finite Y subset of X
X U {~phi} is not satisfiable, so there is finite Y such that
Y U {~phi} is not satisfiable, so Y |= A
----------------------------------------------------------------------
Theorem: Lowenheim-Skolem (downward)
1. If L is finite or countable, then X is satisfiable iff X is
satisfiable in a countable structure.
2. If L is not countable, X is satisfiable in a structure of
cardinality bounded by cardinality of L
Proof:
1. From the size of the witnessing expansion
Countable union of countable sets is countable.
2. Similar argument for larger cardinalities
----------------------------------------------------------------------
Corollary:
No finite/countable language can axiomatize the real numbers
(upto isomorphism). i.e., any countable set of axioms that is
satisfied over the reals will also be satisfied in some
countable structure.
----------------------------------------------------------------------
Finiteness of structures cannot be captured in FO.
Theorem: If X has arbitrarily large finite models (for all n in N
there is a model of at least size n), then X has an infinite
model.
Proof
Let Y = X U {phi_>=n | n >= 2}.
Every finite subset Y_0 of Y is satisfiable by assumption.
By finite satisfiability, Y has a model. This must be
infinite, so X has an infinite model.
----------------------------------------------------------------------
Theorem: Upward Lowenheim-Skolem
If X has an infinite model, then for any set A, X has a model
of cardinality at least the size of A (i.e., there exists an
injection from A to the underlying set of the structure.)
Proof
Add constants c_a for a in A
Y = X U { ~(c_a = c_b) | a =/= b}
Every finite subset Y_0 of Y is satisfiable. Y_0 has a finite
set of inequalities ~(c_a = c_b). Map these constants to
distinct elements in the structure and map the remaining
constants arbitrarily.
By finite satisfiability, Y is satifiable. Each constant c_a,
a in A, is mapped to a distinct element of the underlying
structure for Y. The interpreretation for the constants c_a, a
in A, defines an injection from A to the underlying set.
======================================================================