Is this extension of ZFC known to be outright inconsistent?

Question

Is the following first-order theory known to be outright inconsistent? Adjoin to ZFC a unary function $U$ together with the following axioms.

1. If $\alpha$ is an ordinal, then there exists an (uncountable) strongly inaccessible cardinal $\kappa$ such that $U_\alpha = V_\kappa.$
2. If $\alpha$ and $\beta$ are ordinals satisfying $\alpha<\beta$, then $U_\alpha \in U_\beta$.
3. (Schema; for all sentences $\varphi$ in the first-order language of $\in$): If $\alpha$ is an ordinal, then $\varphi$ holds iff $\varphi$ relativized to $U_\alpha$ holds.

Answer 1

No, this is not inconsistent, at least not relative to the existence of a Mahlo cardinal.

Suppose that $\kappa$ is a Mahlo cardinal, then $V_\kappa$ has a club of ordinals $\alpha$ such that $V_\alpha\prec V_\kappa$. Therefore there is a stationary set of inaccessible cardinals satisfying this.

Simply enumerate these inaccessible cardinals and let $\alpha$ be the enumeration of this stationary set. Then $U_\alpha=V_\mu$ for some inaccessible $\mu$ and $V_\mu\prec V_\kappa$.

Now cut the universe at $V_\kappa$, and this satisfies this extension of $\sf ZFC$ that you suggest. You can probably get away with much less than a Mahlo.

Answer 2

Your theory is equiconsistent with what is known as the Levy scheme, or Ord is Mahlo, which is strictly weaker in consistency strength than the existence of a Mahlo cardinal.

On the one hand, your theory implies Ord is Mahlo, since if $C$ is any proper class definable club (definable with parameters), then it will follow that $C$ is unbounded in the $U_\alpha$ that are above the parameters, and hence $C$ must contain an inaccessible cardinal.

Conversely, assume $M\models\text{ZFC}+\text{Ord is Mahlo}$. Consider any finite subtheory of your theory. This subtheory asks for an unbounded collection of inaccessible cardinals, which have a bounded complexity reflection to $V$. But this is true in $M$, since for any formula $\varphi$, there is a club of ordinals $\alpha$ such that $\varphi$ is absolute between $V_\alpha$ and $V$, and so this club must contain unboundedly many inaccessible cardinals.

Although it may seem that Ord is Mahlo is close in consistency strength to an outright Mahlo cardinal, in fact there is a large hierarchy strictly in between, such as the uplifting and pseudo-uplifting cardinal hierarchy.