As discussed in Noah Schweber's answer to What is the proof-theoretic ordinal of true arithmetic?, it is somewhat ambiguous what “the proof-theoretic ordinal of True Arithmetic” might mean. In one sense, it is $\omega_1^\text{CK}$, because e.g. True Arithmetic satisfies $\Sigma^0_1$ induction for every notation for a recursive ordinal. But of course $\Sigma^0_1$ transfinite induction is not the same as full transfinite induction, which can’t even be expressed in the language of first-order arithmetic, so I’d like to ask about a second-order version of this notion.

My question is, which ordinals $\alpha$ can $\text{ACA}_0$ + True Arithmetic prove are well-founded, in the sense that there exists some arithmetically definable well-ordering $R$ with order type $\alpha$ such that $\text{ACA}_0$ + True Arithmetic proves $\operatorname{WO}(R)$? Where $\operatorname{WO}(R)$ is the $\Pi^1_1$ sentence asserting $R$ is well-ordered, i.e. $R$ satisfies transfinite induction for all sets, not merely for e.g. $\Sigma^0_1$ sets.

Is the answer all recursive ordinals? If so, how does that square with the work of Feferman and Schutte, who made infinitary deductive systems for ramified second-order arithmetic containing $\text{ACA}_0$, the omega rule, and various comprehension schemes, and found that the proof-theoretic ordinal of the resultant systems is $\Gamma_0$, the Feferman–Schutte ordinal which is recursive?


1 Answer 1


There is a result of Kreisel (see [1, Theorem 6.7.4,6.7.5]) that for extensions of $\mathsf{ACA}_0$ the $\Pi^1_1$ proof-theoretic ordinal (suprema of the order types of provable recursive well-orderings) doesn't change after additions of true $\Sigma^1_1$-sentences. Also note that for infinitely axiomatized theory $T$ its $\Pi^1_1$ proof-theoretic ordinal is the suprema of proof-theoretic ordinals of its finite subtheories. Thus the $\Pi^1_1$ proof-theoretic ordinal of $\mathsf{ACA}_0+\mathsf{Th}(\mathbb{N};0,1,+,\times)$ is $\varepsilon_0$.

[1] Wolfram Pohlers. Proof theory: The first step into impredicativity. Springer Science, 2008.

  • $\begingroup$ Hmm, does $ACA_0$ + the omega rule have a higher $\Pi^1_1$ proof-theoretic ordinal? $\endgroup$ Dec 23, 2022 at 14:17
  • 2
    $\begingroup$ Sure, the closure of $\mathsf{ACA}_0$ under $\omega$-rule will have $\omega_1^{\mathit{CK}}$ as its $\Pi^1_1$-ordinal. The point is that unlike all first-order truths, when applied to second-order formulas, $\omega$ rule isn't restricted to $\Sigma^1_1$ conclusions. $\endgroup$ Dec 23, 2022 at 14:31
  • $\begingroup$ By the way, there is a citation button in the editor. It's not supported on touch devices or I'd edit the answer myself. $\endgroup$ Dec 23, 2022 at 15:46
  • $\begingroup$ How do you prove that $ACA_0$ + the omega rule can prove that every recursive ordinal is well-founded? $\endgroup$ Dec 23, 2022 at 17:42
  • $\begingroup$ @Keshav Simply by virtue of the fact that $\omega$-logic proves all true $\Pi^1_1$-sentences (for example, you could read about this in Pohlers textbook that I referred to in the answer). $\endgroup$ Dec 23, 2022 at 18:37

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