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I know a proof of Hindman's theorem that uses ultrafilters on the natural numbers, and ultimately, the axiom of choice. But the theorem itself is essentially a combinatorial property of the natural numbers, so somehow I expect that there should be a proof which doesn't need choice, especially since it can be written in a totally "finitistic" form. My question is not about Hindman's theorem itself though; I am interested to know if we have a general method of reduction of proof from $\sf ZFC$ to $\sf ZF$, for theorems that can be stated in the language of $\sf PA$.

Mario Carneiro
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Turning my comment into an answer:

Any arithmetic statement provable in $\mathsf{ZFC}$ is provable in $\mathsf{ZF}$. The point is that if $\mathsf{ZFC}$ proves that $\mathbb N\models\phi$, then $\mathsf{ZF}$ proves that $L$ models that $\mathbb N\models\phi$. To see this, note that the $\mathsf{ZFC}$ proof uses only finitely many axioms of $\mathsf{ZFC}$, and it is a theorem of $\mathsf{ZF}$ that $L$ satisfies these axioms.

We conclude by noting that if $L\models \mathbb N\models\phi$, then $\mathbb N\models \phi$.

The argument can be pushed a bit. The Shoenfield absoluteness theorem gives us that the same holds not just for arithmetic statements ($\Sigma^0_n$ for some $n$), but even for $\Sigma^1_2$ statements. (Kanamori's book on large cardinals has a careful discussion of this result.) And this is not the end of the story: Leaving the arithmetic hierarchy, we can still find traces of absoluteness; for instance, see here.

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Andrés E. Caicedo
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  • Where does the last step come from? Is there some sort of "pullback" lemma that allows you to turn a proof that a proof exists into an actual proof? – Mario Carneiro Apr 09 '14 at 09:48
  • Do you mean, the step from $L\models\mathbb N\models\phi$ to $\mathbb N\models\phi$? This is the fact that $\mathbb N^L=\mathbb N$, and $\Delta_1$ statements are absolute between transitive models of $\mathsf{ZF}$. Kunen's set theory book treats this in detail. – Andrés E. Caicedo Apr 09 '14 at 13:27
  • Ah, so I should be reading that as ${\sf ZF}\vdash (L\models\Bbb N\models\phi\to\Bbb N\models\phi)$, right? This is a statement of metamathematics done within ZF. But seems this doesn't give me ${\sf ZF}\vdash\phi$ from ${\sf ZFC}\vdash\phi$, which was my original hope. I've read about weird theories like $\sf PA+\neg Con(PA)$, which necessarily include "natural numbers" which can't really be translated to regular numbers as one would expect. I worry in this context about a "proof of proof existence", which doesn't carry over to an actual proof. – Mario Carneiro Apr 09 '14 at 14:57
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    I do not understand what you are saying. I proved the result you asked for. There is no "proof that there is a proof" here, I directly gave you an algorithm that transforms any proof of an arithmetic statement in $\mathsf{ZFC}$ into a proof of the same statement in $\mathsf{ZF}$. This is even stronger than what you were after. I write "$\mathbb N\models\phi$" rather than $\phi$, because writing $\phi$ alone and claiming $\mathsf{ZFC}\vdash\phi$ makes no sense, as $\phi$ is not even in the language of set theory. – Andrés E. Caicedo Apr 09 '14 at 15:55
  • Ah, okay I see what you mean. I guess when I write ${\sf ZFC}\vdash\phi$ I really mean ${\sf ZFC}\vdash\Bbb N\models\phi$, because I view the PA statements as being translated to equivalent versions in $\omega$, with a slight change in notation. Why do you need a compactness-type argument for $L$? It seems to me that ${\sf ZF}\vdash L\models{\sf ZFC-Rep}$ (this being a finite set of axioms), and I don't see why it wouldn't be the case that ${\sf ZF}\vdash \forall\phi[L\models{\sf Rep}(\phi)]$ is a theorem? – Mario Carneiro Apr 09 '14 at 16:13
  • The issue is that we would need a satisfiability predicate for $L$ in order to formalize that last statement, and we cannot have that working in $\mathsf{ZF}$ alone, by Tarski's theorem on undefinability of truth. (Similarly, we cannot really prove in $\mathsf{ZFC}$ that $V$ satisfies $\mathsf{ZFC}$, though of course we can prove (one by one) that it satisfies every axiom.) – Andrés E. Caicedo Apr 09 '14 at 16:17
  • Can't we construct an inner model for $L$ within $V$? Or is the issue that such an $L$ is not a set? – Mario Carneiro Apr 09 '14 at 16:19
  • Yes, that's the problem. We (provably) have satisfiability predicates for set structures, but $L$ is a proper class. We can build $L$, and for each axiom of $\mathsf{ZFC}$ we can prove that $L$ satisfies that axiom. Of course, if we have a set model of $\mathsf{ZF}$, we can formalize the metamathematical argument and conclude that its $L$ is a set model of $\mathsf{ZFC}$. – Andrés E. Caicedo Apr 09 '14 at 16:22
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    So this isn't really metamathematics at all (i.e. Godel numbers and all that), just regular math inside some definition of $L$. So to actually apply this approach to, say Hindman's theorem, I would just replace "$p$ is an ultrafilter on $\Bbb N$" with "$p$ is an $L$-ultrafilter", where the notion of $L$-ultrafilter is just the same as an ultrafilter with everything relativized to $L$. Then since $L$ is well-ordered, I can find an $L$-ultrafilter with the needed properties, prove what I need to, and return to a statement about $\Bbb N^L$, which is just the same as $\Bbb N$. Is this about right? – Mario Carneiro Apr 09 '14 at 16:27
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    Yes. Precisely. There was a question a while ago about whether choice was needed in the proof of Fermat's Last Theorem. It is not, and the point to highlight is that the translation procedure is explicit, just as you suggest. – Andrés E. Caicedo Apr 09 '14 at 16:30