Is there more than one underivable, true theorem in a formal system under godel?

Question

I've been reading Douglas Hofstadter's excellent book "Godel, Escher, Bach: An eternal golden braid" and I think I understand the proof for Godel's incompleteness theory, but I still have a couple of questions. My first question is this: One can produce a theorem that is true, but is not derivable from the formal system S. In the proof it appears that this theorem is essentially "I am not a theorem of S". My question is thus: Is this the ONLY true statement not in S, or are there more? I'm thinking there should be infinitely more, but I do not know how to express, let alone prove, that feeling mathematically.

The second question, which I'll ask on a separate thread if I need to, is the following. Are there any "non-trivial" true statements not in S? By which I mean, are there any statements that could be used to derive other statements? It seems like there's not a lot you can do with the statement "I am not a member of S".

Answer 1

Once you know one true sentence $A$ that is not provable from $S$, it is easy to produce more of them, such as $$ A \land A, A\lor A, \neg\neg A, \neg(\neg A\lor \neg A), \ldots $$ But these are not very interesting, because $S$ does at least prove that they are all equivalent.

For a more interesting question, let's first answer the second question. Namely, Rosser proved that under similar conditions on $S$ as Gödel's theorem, the sentence "$S$ is consistent" -- that is, "it is impossible to prove a contradiction from $S$" -- is unprovable. And we're already assuming that $S$ is in fact consistent. This is, at least on the face of it, a more interesting statement than just that a particular sentence is not provable. It can be used to prove that certain other theories are consistent, or to prove that $S$ has a model (which is Gödel's completeness theorem).

Now, since "$S$ is consistent" is true it is not, disprovable either, so we get another consistent theory by adding "$S$ is consistent" to $S$ as a new axiom. And then we can do the entire Gödel exercise on the extended theory once again, and get a whole series of true-but-unprovable-in-$S$ setences: $$ A_1 \equiv S\text{ is consistent} \\ A_2 \equiv S\cup\{A_1\}\text{ is consistent} \\ A_3 \equiv S\cup\{A_1,A_2\}\text{ is consistent} \\ \vdots $$ Neither of these sentences is equivalent to any other, though the latter of them do imply each of the earlier ones.

Somewhere I have a seen a description of an infinity of sentences that are all true but unprovable, but where neither of them is implied by any combination of the other sentences ... ah, now I remember: We can take the sentences $A_n \Rightarrow A_{n+1}$ for the above $A_i$s.

Answer 2

We can easily produce infinitely more. If your statement is $T$, we can do $T \wedge 1+1=2$ and so on.

As for statements that are not provable from $PA$, you could see Goodstein's theorem and the Paris-Harrington theorem. Both have a much more "mathematical" than "logical" flavor.