Re: help with Godel's

herbzet wrote:
David Marcus wrote:
herbzet wrote:

And I suppose in a non-standard model, n might be infinite, so
the proof-predicate would allow proofs of infinite length.

Although ... I find it odd that the syntactic object,
a "proof", would be different in different _models_ of
a theory, because it is the _syntactic_ object that
_determines_ the models.

The syntactic objects are the same. The theorems are the same. We do
math using the standard model of PA. The problem is that a non-standard
model thinks something is the Godel number of a proof that isn't really.

So, why would we say that ~Con(PA) is true in that model?

~Con(PA) is just a statement about numbers. And, in this model there are
these weird numbers that satisfy this statement. Remember: our
interpretation of the statement is based on the standard model.

It is like asking whether there are any real numbers that satisfy x^2 =
-1 and being told that i does.

Or (and perhaps this is a different question) why would we suppose
that ~Con(PA) is a sentence in any sequence of n proof-steps from
the axioms of PA, even if we allow n to be infinite (hyper-finite?)?

There certainly isn't a proof where n is finite. If we have these non-
standard numbers running around, then we'd have to look at what our
"Provable" predicate is saying in this model to understand what the
model thinks is a proof.

In this standard model, Con(PA) is true.

How does this follow from what you said?

Because PA has a model (the standard model), hence is consistent. And,
that's what Con(PA) says.

I suppose you can point to a set constructable in ZFC and say "Hey,
look, a model of PA!" Is this what the assertion of a model for
PA rests on?

Of course. We point to the actual natural numbers.

Do you have more confidence in ZFC then in PA?

Nope. Why is that question relevant? When we study PA, we do so using
ordinary mathematics.

--
David Marcus
.

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