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Is the following a theorem of $\sf ZF+[V=HOD]$?

If $Q$ is a property definable in a parameter free manner, then: $$\operatorname {Con}(\sf ZF + [V=HQD]) \implies V=HQD$$

where $\sf V=HQD$ means:

$$\forall X \exists v_0 \exists v_1: Q(v_0) \land Q(v_1) \land \rho(v_0) > \rho(v_1) \land \exists \varphi:\\ X=\{y \in V_{\rho(v_0)} \mid V_{\rho(v_0)} \models \varphi(y,v_1) \}$$

Where $\rho$ is the known rank function.

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The answer is no. Indeed, one can rarely move from consistency to truth in this way.

For a counterexample, let $Q(v)$ be the property "CH holds and $v$ is an ordinal."

If CH holds, then $Q$ expresses the property of being an ordinal, but if CH fails, then $Q$ never holds. Consider a model of ZF+V=HOD in which CH fails, but Con(ZF) holds. So V=HQD is false in this model, since Q is never true. But from Con(ZF) we can prove Con(ZF+V=HQD), since it can build a model of V=HOD in which CH holds, and that will be a model of V=HQD.

The point is that the consistency statement is true in this countermodel, because it is consistent that CH holds with V=HOD, but the statement itself is not true, because CH fails.

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  • $\begingroup$ what if we add a condition to the antecedent that $Q$ is fulfilled by a proper class of sets, so we have the consistency statement and this proper class assertion at the antecedent. Would that save this approach? $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 27, 2023 at 16:00
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    $\begingroup$ In that case, you can prove V=HOD implies V=HQD, since for every ordinal alpha there will be an alphath rank arising by elements of Q, and so yes. $\endgroup$
    – Joel David Hamkins
    Commented Jun 27, 2023 at 16:05

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