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.



Does V=HOD prove all kinds of consistent universal hereditary definability?
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.


