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Theorems II.3.7 and II.3.9 in Simpson's Subsystems of Second-Order Arithmetic appear to be circular. Specifically, theorem II.3.7 seems to make implicit use of theorem II.3.9.

[Theorem II.3.9 is the principle of bounded $\Sigma_1$ comprehension. That is, if $n$ is some fixed number and $Q[n]$ is a $\Sigma_1$ formula, the class $\{i\colon i < n\wedge Q[i]\}$ is a set.]

II.3.7 states that for any $\Sigma_1$ formula $P[i]$, either the set $\{i\colon P[i]\}$ exists and is finite, or else there is a one-to-one function $f$ for which $\forall n\colon P[n] \leftrightarrow \exists m\colon f(m) = n$. The proof proceeds by assuming that the set $\{i\colon P[i]\}$ is not finite and constructing an appropriate function $f$. It is left implicit that if the function $f$ does not exist, then the set $\{i\colon P[i]\}$ must be finite (and in fact this argument is used in the proof of II.3.9).

My problem is that we never actually show that the set $\{i\colon P[i]\}$ exists. The proof of II.3.7 uses the assumption that there is no upper bound on the $i$'s for which $P[i]$ is true to construct the function $f$. So if $f$ does not exist, I can accept that there is a largest $i$ for which $P[i]$ holds. I accept that if $\{i\colon P[i]\}$ were already known to be a set, that it would be finite. But I don't accept that $\{i\colon P[i]\}$ is a set. True, we know that it must be bounded, but we have not yet proven the bounded comprehension principle, so boundedness does not yet imply set existence.

It is even worse if you try to prove II.3.9 directly by writing out all the steps from II.3.7 in-line. If you do this, you start by assuming that the set $\{i: i < n\wedge Q[i]\}$ (for a $\Sigma_1$ formula $Q[i]$) is infinite. But this function can be used to derive a contradiction, from which you conclude that the set $\{i < n\wedge Q[i]\}$ is finite. But there's a missing step: all we've shown is that there is an upper bound on the numbers $i$ for which $i < n \wedge Q[i]$ holds (but we already had a bound--$n$ itself!). We have not yet shown that $\{i < n\wedge Q[i]\}$ is a set.

Am I missing something?

Gro-Tsen figured it out in the comments. I was missing something, but Simpson's proof is missing something, too, specifically the explicit discussion about $j_\hbox{max}$. The bound on $Y$ provides not only a bound on $X$ but also a bound for the existential quantifier. A correct version of II.3.7 follows:

Let $R[j,i]$ be a $\Sigma_0$ formula. Consider the set $Y = \{(j,i)\colon R[j,i]\wedge \neg(\exists k<j\colon R[k,i])\}$. It is either finite or infinite. If it is infinite, then follow Simpson's original proof to construct the one-to-one function $f$. Otherwise $Y$ is finite and there must be bounds $i_\hbox{max}$ and $j_\hbox{max}$ for which $(i,j)\in Y$ implies $i\leq i_\hbox{max}$ and $j\leq j_\hbox{max}$. Discard $i_\hbox{max}$ as irrelevant, and then observe that $\forall i\colon(\exists j\colon R[j,i])\leftrightarrow(\exists j\leq j_\hbox{max}\colon R[j,i])$.

Now by $\Sigma_0$ comprehension the set $X = \{i\colon (\exists j\leq j_\hbox{max}\colon R[j,i])\}$ must exist, and the previous paragraph shows that it can be rewritten as $X=\{i\colon (\exists j\colon R[j,i])\} = \{i\colon P[i]\}$, where $P[i]$ is $\Sigma_1$. QED.

[Somewhat obnoxiously, the bound $j_\hbox{max}$ is not in general computable. The set $X$ is recursive only because it is finite, and every finite set can be recognized by some Turing machine. There's just no way to determine, in general, which Turing machine to use.]

A.C.
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    Which edition of the book do you have? The second edition, which I have, calls II.3.7 a "lemma", and the proof writes $P(n)$ as $\exists j, R(j,n)$ where $R$ is $\Sigma_0$, then considers the set $Y$ of $(j,n)$ such that $R(j,n)\land\neg(\exists i<j)R(i,n)$, which exists by $\Sigma_0$-comprehension, and then discusses whether this $Y$ (not $X:={i:P(i)}$) is infinite. – Gro-Tsen Jun 17 '16 at 19:57
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    I looked at Simpson's errata page for the book to see if he explains the difference between the two editions or if this particular statement is referenced, but neither is true. (Also, I just discovered that I'm responsible for pointing out half the known errors in the second edition!) – Gro-Tsen Jun 17 '16 at 20:06
  • I, too, have the second edition and labeled II.3.7 incorrectly in my post. Nevertheless: I do not challenge the existence of $Y$, and I agree that if $Y$ is infinite then it should be possible to construct the one-to-one function $f$. Thus if $f$ does not exist, then $Y$ must have been finite. The bound on $Y$ can be transformed into a bound on $X$. But we still have yet to prove that $X$ is a set. – A.C. Jun 17 '16 at 21:18
  • But if $Y$ is bounded, say by $q$, then $X={n : (\exists j<q)R(j,q)}$ and $(\exists j<q)R(j,q)$ is $\Sigma_0$, so this exists by $\Sigma_0$-comprehension. Or am I confused? (I know I easily end up saying nonsense in this kind of stuff, so maybe I am.) – Gro-Tsen Jun 17 '16 at 21:58
  • Simpson's proof isn't missing anything, though you always have to be careful to remember definitions precisely when reading SOSOA. Recall that 'finite set' means 'a set whose elements are bounded' (p. 67). So a bound for $Y$ exists if it is finite and, by Lemma II.2.2.1, this bound works for both your $i_{\max}$ and $j_{\max}$. Remembering all this, it's immediate that if $Y$ is finite then so is $X$. "Since the first alternative fails, Y is infinite" summarizes this well enough. – François G. Dorais Jun 18 '16 at 03:36
  • Regarding the final remark, Simpson carefully indicates that the proof is not effective by stating Lemma II.3.7 as a schema. Note that this non-uniformity is also well-known in computability theory, where Lemma II.3.7 corresponds to the fact that the infinite computably enumerable sets are precisely those which are the range of some one-to-one computable function. – François G. Dorais Jun 18 '16 at 03:44
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    It is perhaps moot now that I understand the solution, but I would repeat that the finitude of $X$ was never in question; the problem was proving that $X$ was a set. One could argue, incorrectly, that every bounded comprehension scheme is meaningful in $\mathsf{RCA}0$ with a similar argument: Consider the set ${i\colon i < n\wedge P[i]}$ where $P[i]$ is a fantastically complex formula (e.g. $\Sigma{37}^1$). Now this set is either finite or infinite. If it is infinite, then we have a contradiction because it is bounded above by $n$, so it must be a finite set, and thus a set. – A.C. Jun 18 '16 at 11:41

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