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All of the following concerns Simpson's Subsystems of Second Order Arithmetic (2nd ed.).

In the notes subsequent to lemmas VII.1.6 and VII.1.7 (pp. 245–246), Simpson remarks that both lemmas are provable in $\mathsf{RCA}_0$, although the proofs given require $\mathsf{ACA}_0$, not least because they rely on a formal version of the Kleene normal form theorem for $\Sigma^1_1$ relations: lemma V.1.4 (pp. 169—170).

Is this normal form theorem in fact provable in $\mathsf{RCA}_0$?

(Kleene normal form theorem) Let $\varphi(X)$ be a $\Sigma^1_1$ formula. Then we can find an arithmetical (in fact $\Sigma^0_0$) formula $\theta(\sigma, \tau)$ such that $\mathsf{ACA}_0$ proves

$$\forall{X}(\varphi(X) \leftrightarrow \exists{f}\forall{m} \theta(X[m], f[m])).$$

(Here $f$ ranges over total functions from $\mathbb{N}$ into $\mathbb{N}$. Also

$$X[m] = \langle \xi_0, \xi_1, \dotsc, \xi_{m-1} \rangle$$

where $\xi_i = 1$ if $i \in X$, $0$ if $i \not\in X$. Note that $\varphi(X)$ may contain free variables other than $X$. If this is the case, then $\theta(\sigma, \tau)$ will also contain those free variables.)

Arithmetical comprehension is used in the proof of the normal form theorem to show that $\varphi(X)$ holds iff there exist Skolem functions for $X$. So one way to show that this lemma is provable in $\mathsf{RCA}_0$ would be to show that recursive comprehension suffices to prove this equivalence.

Benedict Eastaugh
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    I doubt that it can be proved in $\mathsf{RCA}_0$, but I don't know a particular counterexample. – Carl Mummert Sep 18 '13 at 17:22
  • Supposing that is the case, I'm interested in how the lemmas I mentioned at the start of the question can be proved while bypassing the normal form theorem. – Benedict Eastaugh Sep 18 '13 at 17:33
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    I don't know the argument, but I would conjecture that, when you are already assuming that the hyperjump exists, the hyperjump has the necessary information inside it to simulate the Skolem functions. The detailed proof would have to show how to extract that information from the hyperjump using just $\mathsf{RCA}_0$. (Note that this is all relative to Simpson's particular definition of the hyperjump; changing the definition to refer to arbitrary $\Sigma^1_1$ formulas would likely solve the problem as well, and should be equivalent over $\mathsf{ACA}_0$.) – Carl Mummert Sep 18 '13 at 18:45
  • Thanks Carl, that's very helpful. The thing I'm actually trying to prove isn't quite that, but it's quite similar, so that gives me a good idea of what my next steps should be. Much appreciated. – Benedict Eastaugh Sep 18 '13 at 22:39
  • @CarlMummert on reflection a simpler strategy would be just to use the hyperjump to prove arithmetical comprehension, and then apply the normal form theorem as usual. – Benedict Eastaugh Sep 30 '13 at 23:15
  • Yes - I think that is basically the method I had in mind. But if you only have the hyperjump of a set $X$, it appears at first glance that you may not be able to prove arithmetical comprehension relative to a set parameter $Y$, e.g. if $Y$ is not arithmetical in the hyperjump of $X$. However, if I remember correctly, it turns out for the result you asked about that you only need Skolem functions relative to the same set parameter $X$ for which you have a hyperjump. – Carl Mummert Sep 30 '13 at 23:28

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Certainly the strong form of the normal form theorem - where we control the complexity of the arithmetic matrix - is not provable in RCA$_0$.

This follows from considering the model $REC$ of recursive sets. Over $REC$, we can convert second-order quantifiers into first-order quantifiers, since it's enough to quantify over indices for total Turing machines. (At a glance, I think that each second-order quantifier turns into three first-order quantifiers.) In particular, every formula is (equivalent to) an analytic formula in the broad, non-bounded-matrix sense. However, an analytic formula whose matrix is $\Sigma^0_n$ is equivalent (in $REC$) to a $\Sigma^0_{2+n}$-formula; the usual arguments for producing a $\Sigma^0_k$ formula not equivalent to any $\Sigma^0_m$ formula for $m<k$ go through here, and so the (strong) normal form theorem can't hold.

Noah Schweber
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