What is the error in this disproof of the $\Omega$-conjecture?

Question

I was reading about the $\Omega$-conjecture and have thought of a refutation of it, which seems too simple to not have been noticed since the $\Omega$-conjecture has been around, so i'm skeptical and want to see whether anyone can spot a flaw in it.

I will assume there is a proper class of hyper-huge cardinals. This assumption implies a proper class of Woodin cardinals and by Usuba implies the existence of the bedrock model $W$, and since $V$ is a set-generic extension of $W$ and our assumption (as well as the $\Omega$-conjecture) is invariant throughout the set-generic multiverse, we can work in $W$. Suppose for a contradiction that the $\Omega$-conjecture holds. Since $V = W$ is $\Sigma_2$ definable in any set-forcing $V[G]$, we can uniformly evaluate $\Sigma_2$-truths of $V$ in any set-forcing of $V$ by other recursively given $\Sigma_2$ sentences. Since the $\Sigma_2$-laws of the set-generic multiverse are definable in $H(\delta_0^+)$ where $\delta_0$ is the least Woodin cardinal (by the $\Omega$-conjecture), and they can be used to compute all the $\Sigma_2$ truths of $V$ (including the theory of $H(I_0^{+})$ and beyond), this violates Tarski's undefinability of truth. Thus the $\Omega$-conjecture must fail assuming our large cardinal hypothesis. $\square$

Is this argument really valid? (I'm worried 2016 will end by refuting $V = \text{Ultimate }L$)

Answer 1

Woodin's theorem says that assuming the $\Omega$ Conjecture and the existence of a proper class of Woodin cardinals, the set $\mathcal V_\Omega$ of $\Pi_2$ sentences that hold in every universe of the generic multiverse is lightface definable over $H_{\delta^+}$ where $\delta$ is the least Woodin cardinal. You claim there is a Turing reduction from the $\Pi_2$ theory of the bedrock to $\mathcal V_\Omega$, obtaining in this way a $\Sigma_2$ definition of $\Pi_2$ truth (assuming the Ground Axiom), a contradiction. The proposed reduction sends a formula $\phi$ to the sentence $f(\phi)$ expressing "The bedrock satisfies $\phi$." Tell me if I'm misunderstanding you.

One problem: it isn't clear that $f(\phi)$ is $\Pi_2$. Note that if $W$ is a $\Sigma_2$ or $\Pi_2$ or even $\Delta_2$ inner model and $\phi$ is a $\Pi_2$ sentence, the sentence $W\vDash \phi$ is not obviously $\Pi_2$. We can write $\phi$ as $$\forall x \ (x\notin W \vee \exists y\ (y\in W\wedge \psi(x,y))$$ for some $\Delta_0$ formula $\psi$. This seems to be no simpler than $\Pi_3$ even if $W$ is $\Delta_2$.

For example, we claim that the statement $$\Psi\equiv \text{CH}\text{ fails in the bedrock}$$ is not equivalent over ZFC + the Bedrock Axiom to a $\Pi_2$ formula, even though $\neg\text{CH}$ is $\Delta_2$. (The Bedrock Axiom just asserts that the generic multiverse has a bedrock.) Suppose towards a contradiction that this statement is equivalent to a $\Pi_2$ formula, which we may assume is of the form $\forall \alpha\ (V_\alpha\vDash \psi)$. Fix a model $$M\vDash\text{the Bedrock Axiom}+\neg\text{CH} + \neg\Psi$$ (so the bedrock of the generic multiverse of $M$ satisfies $\text{CH}$). Since $M\vDash \neg \Psi$, there is some ordinal $\alpha$ of $M$ such that $M_\alpha\vDash \neg \psi$. Now pass to a class forcing extension $M[H]$ satisfying the Ground Axiom and such that $M[H]_{\alpha} = M_{\alpha}$ and $M[H]_{\omega+\omega} = M_{\omega+\omega}$. (See Reitz's thesis, Theorem 12.) We have $M[H]\vDash \neg\text{CH}$ (since $M\vDash \neg \text{CH}$ and $M[H]_{\omega+\omega} = M_{\omega+\omega}$). On the other hand since $M[H]_\alpha = M_\alpha\vDash \neg \psi$, $M[H]\vDash \exists \alpha\ (V_\alpha\vDash \neg \psi)$. Thus $M[H]\vDash \neg \Psi,$ so the bedrock of $M[H]$ satisfies $\text{CH}$, which contradicts the fact that $M[H]$ is its own bedrock.