Logic: Lecture 9, 1 September 2015
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Witnessing expansion
L = (R,F,C)
L_0 = L, C_0 = emptyset
C_1 = {c_phi | Ex phi in Formulas(L_0)}
L_1 is (R,F,C U C_1)
C_2 = {c_phi | Ex phi in Formulas(L_1)}
...
Assume we have L_n = (R,F,C U C_1 U ... U C_n)
C_n+1 = {c_phi | Ex phi in Formulas(L_n) - Formulas(L_n-1)}
L_n+1 is (R,F,C U C_1 U ... U C_n U C_n+1)
C_H = C_1 U ... U C_n ...
L_H = (R,F,C U C_H)
Henkin axiom:
Ex phi(x) => phi(c_phi(x))
Quantifer axiom:
phi(t) => Ex phi(x), t closed
Equality axioms:
reflexitivy, symmetry and transitivity for all terms
functions: equal arguments given equal results
relations: equal tuples have equal membership properties
Phi_H : all instances of Henkin axioms
Phi_Q : all instances of quantifier axioms
Phi_Eq : all instances of equality axioms
Phi_H and Phi_Q are sentences.
Phi_Eq are not sentences but are true under all assignments.
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FO satisfiability
Theorem: Let X be a set of formulas over L.
The following are equivalent.
1. There is an L-interpretation M = (S,i,sigma) that satisfies X
2. There is an L_H interprestructure M' = (S,i',sigma') that satisfies X
3. X U Phi_H U Phi_Q U Phi_Eq is propositonally satisfiable
(2 => 1)
Immediate (restrict i' over C U C_H to i over C)
(1 => 2)
Need to extend M = (S,i,sigma) to M' = (S,i',sigma') to cover L_H
i'(c) = i(c) for c in C
For c = c_phi in C_H,
if M |= Ex phi, map i'(c_phi) to a s.t. I |= phi(a)
otherwise map i'(c_phi) arbitrarily
(need to do this in stages for C_1, C_2, ...)
(2 => 3)
From M' we derive v such that for every prime formula phi,
v(phi) = tt iff M' |= phi
It follows that v |= X U Phi_H U Phi_Q U Phi_Eq
(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 M = (S,i) that satisfies X
First S:
Define and equivalence relation on terms,
t ~ t' iff v |= t = t'
v |= Phi_Eq guarantees that ~ is an equivalence relation
Elements of S are equivalence classes [c] for c in C U C_H
Define i as follows:
For C : i(c) = [c]
For R : ([c1],...,[cn]) in i(r) iff v |= r(c1,...,cn)
For F :
Need to identify c such that i(f)([t1],...,[tn]) = [c]
To show that such a constant c must exist
Define phi(x) to be the formula f(c1,...,cn) = x
If v |= Ey phi(y), then v |= f(c1,...,cn) = c_phi (Phi_H)
If not, the quantifier axiom says
phi(f(c1,...,cn)) -> Ey phi(y)
Since not(v |= Ey phi(y)), we must have
not(v |= phi(f(c1,...,cn)))
But phi(f(c1,...,cn)) is just f(c1,...,cn) = f(c1,...,cn)
which is an instance of the equality axiom t = t!
The fact that interpretations of F and R are compatible
with ~ follows from Phi_Eq
(to be continued ...)
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