= Recursive powerset mapping
A map $f:a\to\mathcal P(a)$ is recursive if every map $g:\mathcal P(b)\to b$ admits a unique $h:a\to b$ with $h(y)=g(\{h(x):x\in f(y)\})$. This holds exactly when the predecessor relation $x\in f(y)$ is a <well-founded relation>. For the forward construction, take the union of all consistent partial solutions on downward-closed domains; <well-founded induction> proves compatibility, and a minimal missing point allows extension. Conversely, a nonempty subset with no minimal point has an upward reachability closure $T$ satisfying $y\in T$ exactly when some predecessor is in $T$. With $b=\{0,1\}$ and $g(S)=1$ exactly when $1\in S$, both the zero map and the indicator of $T$ solve the recursion, contradicting uniqueness.
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