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In the proof that Martin's maximum implies (*), the introduction gives the following theorem:

Assume there is a proper class of Woodin cardinals. Then, the following are equivalent:

  • $(^*)$
  • For any $\Pi_2$ $\mathcal{L}$-sentence $\sigma$, where $\mathcal{L}$ is the language of the structure $(H_{\omega_2}; \in, \mathrm{NS}_{\omega_1}, \mathcal{A})_{\mathcal{A} \in \mathcal{P}(\mathbb{R}) \cap L(\mathbb{R})}$, if $\sigma$ is $\Omega$-consistent, then $\sigma$ is true.

Now, let's define the axiom $(^*)^+$ to be the conjunction of the following two statements:

  • There exists a proper class of Woodin cardinals.
  • For any $\mathcal{L}$-sentence $\sigma$, where $\mathcal{L}$ is the language of the structure $(H_{\omega_2}; \in, \mathrm{NS}_{\omega_1}, \mathcal{A})_{\mathcal{A} \in \mathcal{P}(\mathbb{R}) \cap L(\mathbb{R})}$, if $\sigma$ is $\Omega$-consistent, then $\sigma$ is true.

Namely, the restriction to $\Pi_2$-sentences (statements of the form $\forall x \exists y \varphi(x,y)$) is removed. Now, here's my question.

Is $(^*)^+$ inconsistent? If it is consistent (assuming the consistency of certain large cardinal assumptions), then is it equiconsistent with the regular axiom $(^*)^+$, or strictly stronger?

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  • $\begingroup$ If my memory serves me right, there is a $(*)^+$ axiom, and it is, surprisingly, equivalent to the standard $(*)$ axiom. $\endgroup$
    – Asaf Karagila
    Commented Jul 20, 2022 at 22:04
  • $\begingroup$ I believe it was called $(^*)_+$ axiom, where the + is in the subscript. $\endgroup$
    – Binary198
    Commented Jul 27, 2022 at 10:56

1 Answer 1

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This principle is inconsistent, even if we just look at $(H_{\omega_2};\in)$. This is because - for example - the continuum hypothesis is expressible as a sentence in this structure ("There is an $\omega_1$-sequence of reals such that every real appears in the sequence"), but both $\mathsf{CH}$ and $\neg\mathsf{CH}$ are $\Omega$-consistent.

Indeed, the example of $\mathsf{CH}$ shows that $\Pi_2$ is basically the best we can hope for: already at $\Pi_2\vee\Sigma_2$ we get inconsistency.

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  • $\begingroup$ What about $\Pi_3$? Is that inconsistent as well? $\endgroup$
    – Binary198
    Commented Jul 27, 2022 at 10:55
  • $\begingroup$ @Binary198 Since $\Pi_3$ is broader than $\Pi_2\vee\Sigma_2$, yes. $\endgroup$ Commented Aug 1, 2022 at 23:13

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