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This situation is a consequence of the existence of $0^\sharp$. To see this, suppose that $0^\sharp$ exists. It follows that every uncountable cardinal is a Silver indiscernible and a limit of Silver …

Under your assumptions, the set $E_\kappa^+$ is empty. Indeed, we don't even need the predicates that you mention. (In any case, since $\kappa$ is definable in $L_{\kappa^+}$, the predicate for $L_\ka …

Let me address the question in the body of the post, rather than the title question. Namely, you asked whether your theory interprets ZFC. Negative answer with axiom as stated. The natural reading of …

Updated answer. I claim that none of the familiar large cardinal notions consistent with $V=L$ are provably $\gamma$-decisive for any $\gamma$. This includes the cases of wordly cardinals, inaccessibl …

No, there are many instances of $L_\alpha\prec L_\beta$ without $L_\alpha\prec L_{\omega_1}$. Here is one easy way to construct one. Consider the smallest $\alpha$ that has an elementary extension $L_ …

Yes, the parameter-free version of $L$ gives rise to the same constructible universe $L$. You will still get all of $L$ this way, but it will come more slowly. The reason is that at stage $\alpha+1$, …

The minimal model, when it exists, also known as the Shepherdson-Cohen model, is the smallest transitive model of ZFC. This model will have the form $L_\alpha$ for some countable ordinal $\alpha$, and …

It seems to me that if you build the constructible universe using $\mathcal{L}_{\omega_1,\omega}$ logic, you will get the inner model $L(\mathbb{R})$. The reason is that every $\mathcal{L}_{\omega_1,\ …

There is no such phenomenom for $n\geq 2$. The reason is that if a model of ZFC has a definable choice function, of any complexity, then it actually has one of complexity $\Delta_2$. This is because t …

The answer is no, you cannot have measurable cardinals consistently with your theory. Your theory includes the axiom "V=L or V=L[c] for an $L$-generic Cohen real $c$". This statement is provable from …

This answer is not correct, because it is a subtler matter to ensure that the levels are closed under relative constructibility while also maintaining $L_{\alpha+1}\cap\mathfrak{L}_\alpha=L_\alpha$. …

Great question! The answer is yes. Start with $V=L$, and fix any ordinal $\alpha$. Now go to the forcing extension $V[G]$ obtained by collapsing $\aleph_\alpha^L$ to $\omega$, so that $(2^\omega)^{V[ …

Every $L^{\Sigma_n}$, for $n\geq 2$ will be the same as HOD, the class of hereditarily ordinal-definable sets. This is a consequence of the Myhill-Scott theorem, which asserts that if you form the con …

This is a fascinating question! I really like your relation. Here is some small progress. (I am hopeful that more definitive answers will appear later). First, you didn't mention it, but for definiten …

$\newcommand\HOD{\text{HOD}}$ There is no such border, because almost all the large cardinal properties, including the very strongest large cardinal axioms, are relatively consistent with $V=\HOD$. F …