Re: CHAT: mathematics

Dennis Paul Himes wrote:

> > > If G is refutable in, say, Peano Arithmetic, then
> > > PA would be inconsistent, which is a possibility, but one which no
> > > mathematician seriously entertains.
> >
> > No, it would only be omega-inconsistent, which is quite different.
>
>     No, it would be outright inconsistent.  This is one of the things
> Goedel proved.
I don't think so.  Let x be the G-number of a PA proof, and y the G-number of
a (purported) PA theorem.  Let PRF(x,y) be true iff y is the G-number of the
theorem proved by the proof whose G-number is x.

E,A are existential and universal quantification.  Then Ey PRF(x,y) means
(meta-mathematically) that x has a PA proof.  G then is the formula
that asserts "G has no PA proof"; contrariwise, ~G asserts "G has a PA proof".

Assume that G has a proof whose G-number is k.  Then PRF(G,k) is true,
contrary to G's assertion that G has no proof.  In that case, given the
proof of G, we have found a proof of ~G as well, so PA is inconsistent.

But the alternative, that there is a proof of ~G, is not symmetrical with
the preceding case.  If there is a proof of ~G, then there is a proof of
Ey PRF(g, y), but it does not follow that if we attempt to prove
for each k (a natural number) a proof of ~PRF(G,k), that we will necessarily
find a proof of G.  We might not.

The result is very ugly, but not necessarily a direct contradiction,
because "for each k, there is a proof of ~PRF(G,k)" does *not* amount to
a proof of Ay PRF(G,y).  This is omega-inconsistency.

I am leaning here on http://www.ltn.lv/~podnieks/gt5.html, which provides
a lot more detail.

> > GT shows that G is independent of the other axioms of PA,
>
>     Right [...] but this contradicts what you said above.  Specifically it
> contradicts your statement that it "says nothing about the provability of
> "~G".  "G is independent of the other axioms of PA" directly implies "~G is
> not provable in PA".
I should have been more precise:  assuming G has a proof, then PA is inconsistent,
but assuming ~G has a proof, then PA is either inconsistent or omega-inconsistent,
we don't know which.

Since we have no proof of G (and can't), we can assume ~G as an axiom.  Since
we have no proof of ~G either, we can assume G as an axiom.  However, if a proof
for ~G were actually found, then PA *would* be omega-inconsistent.

>     I think you're confusing several results here.  It's certainly true that
> there are models of PA which give us only the usual natural numbers, and
> other models of PA which give us more, but this has nothing to do with
> whether or not you add G to PA.  PA+G is consistent with both sorts of
> models, as is PA+~G.
~G (which says "G has a proof") is consistent with PA, but only at the cost of making
PA+~G omega-inconsistent:  Ey PRF(y) is true, but each of PRF(G,k) is false
for each natural number k.  It must therefore be true of something other
than the natural numbers.  Hence, theories which postulate ~G necessarily
have nonstandard numbers.

If there did turn out to be a proof of G in PA itself, though, then PA
would be omega-inconsistent, and nonstandard numbers would exist in
every PA-compatible number theory.

>     If by "proof of ~G" you mean "proof, in some theory, of that theory's
> ~G", then we know no such proof is possible.
That is what I deny: we do not know that.

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