= Solution
Suppose a nonempty set $S$ had no minimal member for the <intersection-power-set predecessor relation>. Put $b=\bigcap S$, which is a set by <separation>. For every $y\in S$ there is an $x\in S$ with $\mathcal P(x\cap y)\subseteq y$. Since $b\subseteq x\cap y$, this gives $\mathcal P(b)\subseteq y$. Intersecting over every $y\in S$ yields $\mathcal P(b)\subseteq b$.
Now form the diagonal subset $d=\{u\in b:u\notin u\}$. It belongs to $\mathcal P(b)$, hence to $b$, so its defining condition implies $d\in d\iff d\notin d$, a contradiction. Thus $\boxed{R\text{ is well-founded}.}$ This is a <Cantor diagonal argument>; it uses neither the <Axiom of foundation> nor the <axiom of choice>. In particular, no selection of an infinite descending sequence is needed.
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