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In a previous question, I asked whether there can be effectively axiomatizable set theories (at least as strong as, say, ZF) that are complete modulo first-order arithmetic, to which the answer is no; there will be sentences of second-order arithmetic that can't be proved even using the omega-rule. I'm now curious whether there can be set theories that are complete modulo finite-order arithmetic (in the comments of my last question, someone claimed the answer was no without proof, but he has since deleted his comment). That is, for any effectively axiomatizable set theory at least as strong as ZF, will there be a sentence independent of it with no arithmetic consequences?

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    $\begingroup$ Your example is confusing because $CH$ is a finite-order arithmetic sentence. $\endgroup$
    – Rodrigo Freire
    Commented Oct 21, 2020 at 20:19
  • $\begingroup$ Link to OP's previous question: Incompleteness theorems for theories with omega-rule $\endgroup$
    – jeq
    Commented Oct 21, 2020 at 21:10
  • $\begingroup$ I didn't know that CH is a finite-order arithmetic sentence. Where on the arithmetic hierarchy does it lie? $\endgroup$
    – BPP
    Commented Oct 21, 2020 at 22:49
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    $\begingroup$ CH is not in the arithmetic hierarchy, which refers to first-order arithmetic. It's a $\Sigma^2_1$ sentence of third-order arithmetic (unless I've overlooked something). $\endgroup$
    – Andreas Blass
    Commented Oct 22, 2020 at 2:42
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    $\begingroup$ Do I understand correctly that by theory $T\supseteq \mathsf{ZF}$ "complete modulo finite-order arithmetic" you mean that the theory $T^*=(T+\text{true }\Sigma^{<\omega}_{<\omega}\text{-sentences})$ is complete? Note that Godel's second incompleteness theorem is applicable to the case of $T^*$. If $T^*$ is consistent, it couldn't prove its own consistency. The adaptation of the proof is simply reduced to the verification that provability predicate for $T^*$ satisfies Hilbert-Bernays-Lob conditions. By adapted Rosser's theorem we even get that $T^*$ should be either inconsistent or incomplete. $\endgroup$
    – Fedor Pakhomov
    Commented Oct 22, 2020 at 11:15

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