Update. I've improved the argument to use only the consistency of $T$. (2/7/12): I corrected some over-statements previously made about Robinson's Q.
I claim that for every statement $\varphi$, there is a variant way
to express it, $\psi$, which is equivalent to the original
statement $\varphi$, but which is formally independent of any
particular desired consistent theory $T$.
In particular, if $\varphi$ is your favorite natural open question,
whose truth value is unknown, then there is an equivalent
formulation of that question which exhibits formal independence in
the way you had requested. In this sense, every open question is
equivalent to an assertion with the property you have requested. I
take this to reveal certain difficult subtleties with your project.
Theorem. Suppose that $\varphi$ is any sentence and $T$ is any consistent theory containing weak arithmetic. Then there is another sentence $\psi$ such that
- $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent.
- $T$ does not prove $\psi$.
- $T$ does not prove $\neg\psi$.
Proof. Let $R$ be the Rosser sentence for $T$, the self-referential assertion that for any proof of $R$ in $T$, there is a smaller proof of $\neg R$. The Gödel-Rosser theorem establishes that if $T$ is consistent, then $T$ proves neither $R$ nor $\neg R$. Formalizing the first part of this argument shows that $\text{PA}+\text{Con}(T)$ proves that $R$ is not provable in $T$ and hence that $R$ is vacuously true. Formalizing the second part of this argument shows that $\text{Con}(T)$ implies $\text{Con}(T+R)$, and hence by the incompleteness theorem applied to $T+R$, we deduce that $T+R$ does not prove $\text{Con}(T)$. Thus, $T+R$ is a strictly intermediate theory between $T$ and $T+\text{Con}(T)$.
Now, let $\psi$ be the assertion $R\to (\text{Con}(T)\wedge \varphi)$. Since $\text{PA}+\text{Con}(T)$ proves $R$, it is easy to see by elementary logic that $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent.
The statement $\psi$, however, is not provable in $T$, since if it were, then $T+R$ would prove $\text{Con}(T)$, which it does not by our observations above.
Conversely, $\psi$ is not refutable in $T$, since
any such refutation would mean that $T$ proves that the hypothesis
of $\psi$ is true and the conclusion false; in particular, it
would require $T$ to prove the Rosser sentence $R$, which it does not by the Gödel-Rosser theorem. QED
Note that any instance of non-provability from $T$ will require the consistency of $T$, and so one cannot provide a solution to the problem without assuming the theory is consistent.
The observation of the theorem has arisen in some of the philosophical literature you may
have in mind, based on what you said in the question. For example, the claim of the theorem is mentioned in Haim Gaifman's new
paper "On ontology and realism in mathematics," which we read in my course last semester
on the philosophy of set theory; see the discussion on page 24 of Gaifman's paper and specifically footnote 35, where he credits a fixed-point argument to Torkel Franzen, and an independent construction to Harvey Friedman.
My original argument (see edit history) used the sentence $\text{Con}(T)\to(\text{Con}^2(T)\wedge\varphi)$, where $\text{Con}^2(T)$ is the assertion $\text{Con}(T+\text{Con}(T))$, and worked under the assumption that $\text{Con}^2(T)$ is true, relying on the fact that $T+\text{Con}(T)$ is strictly between $T$ and this stronger theory. The current argument uses the essentially similarly idea that $T+R$ is strictly between $T$ and $T+\text{Con}(T)$, thereby reducing the consistency assumption.
The Feferman-Schutte analysis is completely wrong. The problem is that "well-ordered" is not a single predicative concept because predicativists lack strong comprehension axioms. If you can't form the set $\{x \in X: \neg\phi(x)\}$ then you can't go from "every subset of $X$ has a smallest element" to "if $(\forall x < a)\phi(x) \to \phi(a)$ for any $a \in X$,then $(\forall x \in X)\phi(x)$".
This wrecks their whole argument because at each stage they use proof trees of height $\gamma_n$ to prove that $\gamma_{n+1}$ is well-ordered in the weak sense and then claim that this justifies acceptance of proof trees of height $\gamma_{n+1}$, which would require induction up to $\gamma_{n+1}$ for all sentences of second order arithmetic.
The argument is absurd in the other direction as well --- if the predicativist can grasp the jump from $\gamma_n$ to $\gamma_{n+1}$ for each $n$, why can't he grasp the general principle that for all $n$, induction up to $\gamma_n$ implies induction up to $\gamma_{n+1}$, and infer induction up to $\Gamma_0$? No cogent answer to this question was ever given. Believe me, I dug through the literature pretty hard.
I discuss this in more detail than you could possibly want in my paper Predicativity beyond $\Gamma_0$.
Best Answer
Ah, Thomas Forster's 1998 paper
is available on-line at various places including here.
He says it is proved in
Jensen RB, On the consistency of a slight (?) modification of Quine's NF, Synthese 19 1969 pp 25--63, doi:10.1007/978-94-010-1709-1_16
Lake J, Comparing Type theory and Set theory, Zeitschrift fur Matematische Logik 21 1975 pp 355-56. doi:10.1002/malq.19750210144
For a fanatically detailed proof and discussion see