What is the strength of claiming that the class of all $V_\kappa$ stages that are $H_\kappa$ when $\kappa$ is regular, is inaccessible?

[EDIT] This posting had been edited to assert that we are speaking about regular mutual stages.

Let $$H_{\kappa}$$ be the set of all sets that are hereditarily strictly smaller in cardinality than cardinality $$\kappa$$. Formally

$$H_\kappa = \{x: |TC(x)|<\kappa\}$$

Where $$TC(x)$$ is the transitive closure of $$x$$ defined as the intersectional set of all transitive supersets of $$x$$.

Now a stage of the cumulative hierarchy $$V_{\kappa}$$ is either identical to $$H_{\kappa}$$ or not. For example $$V_\omega = H_\omega$$, and of course $$V_\emptyset = H_\emptyset$$. Lets call such stages as "mutual" stages.

Define: $$mutual(x) \iff \exists \kappa (x=V_\kappa =H_\kappa)$$

Axiom: The class of all regular mutual stages is proper.

Formally: $$\forall x \exists \kappa (x \in V_\kappa \land mutual(V_\kappa) \land regular(\kappa) )$$

What is the strength of adding this axiom to ZFC?

• @ZuhairAl-Johar If $\kappa$ is infinite and regular, then $V_\kappa=H_\kappa$ iff $\kappa$ is inaccessible; so this principle is just "There is a proper class of inaccessibles." – Noah Schweber Jun 13 at 15:06
• you mean $\kappa$ is strongly inaccessible! (we are of course not including $\omega$). – Zuhair Al-Johar Jun 13 at 15:31