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-% Nested recursion is outside the exact direct-inductive fragment.
+\begin{proposition}\label{phase5_nested_guard}
+ For all $A, x$ we have if $x\in\pow{\cumul{A}}$, then
+ $A\in\cumul{A}$.
+\end{proposition}
+\begin{proof}
+ Fix $A, x$.
+ Assume $x\in\pow{\cumul{A}}$.
+\end{proof}
+
+\begin{axiom}\label{phase5_nested_unsafe_support}
+ For all $A$ we have $A=A$.
+\end{axiom}
\begin{inductive}\label{phase5_nested}
Define $\phasefivenested{A}\subseteq\cumul{A}$ inductively as follows.
\begin{enumerate}
\item If $x\in\pow{\phasefivenested{A}}$, then
- $x\in\phasefivenested{A}$.
+ $A\in\phasefivenested{A}$.
\end{enumerate}
\end{inductive}