# Natural statements independent from true $\Pi^0_2$ sentences

I am looking for sentences in the language of first order arithmetic ($0,1,+,\cdot,\leq$) which are independent from $\Pi^0_2$ consequences of true arithmetic $\Pi^0_2\text{-}\mathsf{Th}(\mathbb{N})$. I want natural statements, e.g. statements that have been studied in number theory or combinatorics for their own sake. The motivation comes from looking for true statements that are not provable in $\mathsf{I}\Delta_0(L)$ where $L$ contains arbitrary fast growing (computable) functions.

-
I assume the $\Pi^0_2$ in the body of your question is what you intended and the $\Sigma^0_2$ in the title isn't. But just in case you're actually interested in the title question, I think the Paris-Harrington theorem answers that. The point is that true $\Sigma^0_2$ sentences are consequences of true $\Pi^0_2$ ones. – Andreas Blass Nov 18 '11 at 0:29
@Andreas, yes, I fixed the title, thanks. – Kaveh Nov 18 '11 at 1:39
And if comments were editable, the $\Pi^0_2$ at the end of my previous comment would become $\Pi^0_1$. – Andreas Blass Nov 18 '11 at 2:51
Depending on how strict your definition of "natural" is, even Paris-Harrington might not be considered "natural." The condition of having as many elements as the least element was not "studied in combinatorics for its own sake." – Timothy Chow Nov 21 '11 at 23:50

I passed this question on to Harvey Friedman, who provided the following information. Friedman has shown that the following statement is equivalent to the 2-consistency of PA:

For every recursive function $f:{\mathbb N}^k \to {\mathbb N}^k$, there exists $n_1 < \cdots < n_{k+1}$ such that $f(n_1,\ldots,n_k) \le f(n_2, \ldots, n_{k+1})$ coordinatewise.

Friedman also says that there are versions of Paris-Harrington and Kruskal's tree theorem that work. For example, "Every infinite recursive sequence of finite trees has a tree that is inf-preserving-embeddable into a later tree" is equivalent to the 2-consistency of $\Pi^1_2$ bar induction.

-
Thanks a lot for the example. I was guessing that Harvey may have an example. I guess that it is unlikely that I would get an answer which is more natural than this so I accept it. – Kaveh Nov 22 '11 at 0:44

If computability counts, Turing famously showed that if M is a Turing machine equipped with an oracle for the regular halting problem, then M's own halting problem is undecidable by M. And if M2 is a machine with an oracle for M, then M2 can't decide its own halting problem, and so on. If I'm not mistaken, that can be turned into independent statements at every level of the arithmetic hierarchy. Having access to the true $\Pi_2^0$ sentences amounts to having M2. It doesn't help you with M3, etc.

-
Thanks, but your answer is essentially saying that the arithmetic hierarchy doesn't collapse, which I know, that is not what I want. Statements from logic/computability like Soundness, Halting, ... are not what I am looking for. – Kaveh Nov 20 '11 at 21:28