Re: Godel's Theorem and Model Theory

Newberry says...

Here is the standard interpretation:
(Ex)(P(x, #("A")) --> A (1)
When you add this to PA you get an immediate contradiction. (1) is
equivalent to assuming the standard model. If (1) is not the case then
it leaves the non-standard models.
What is wrong with this reasoning?

For one thing, your statement (1) doesn't say anything
about the standard model. So stop saying that.

For the second problem: no, it's not true that if you add
(1) to PA you get an immediate contradiction. The proof
predicate P(x,y) is defined for a specific set of axioms,
the axioms of PA. If you add *new* axioms, you have to
make a corresponding change to the proof predicate.

So let's make the dependence on axioms explicit. Instead
of P(x,y), let's introduce a three-place predicate
P(z,x,y) with the interpretation:

P(z,x,y) holds if A is a code for an r.e. set of axioms A,
y is a code for a statement Phi, and x is a code for
a proof of Phi using the axioms in A.

Let #PA be the code for the axioms of PA. Then your
statement (1) becomes:

(1) (Ex)(P(#PA, x, #("A")) --> A

Godel's theorem shows that if PA is consistent, then
PA cannot prove a specific instance of (1), namely
there is a sentence G_PA such that

G_PA <-> not (Ex)P(#PA, x, #G_PA)

is provable in PA. It follows that if PA is consistent,
then PA cannot prove the instance of (1) in which A is
replace by G; PA cannot prove

(Ex) (P(#PA,x,#G_PA)) --> G_PA

Now, you ask what happens if we add all instances of
schema (1) to PA? Well, that means that we are *changing*
the axioms. We are defining a *new* theory, which we can
call PA_2 whose axioms include all axioms of PA plus all
instances of

(Ex) (P(#PA,x,#A)) --> A

Is this new theory consistent? Yes (the proof that it
is consistent can be done in ZFC, for example). Note
that what PA_2 cannot prove is this:

(Ex) (P(#PA_2,x,#A)) --> A

If you want, you can add those as axioms to PA_2 to
get yet another theory PA_3, but then PA_3 will be
unable to prove

(Ex) (P(#PA_3,x,#A)) --> A

There is no way to construct an inconsistent theory
by starting with a theory that is true in the standard
model and adding more statements that are also true
in the standard model.

--
Daryl McCullough
Ithaca, NY

.

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