 Power-set-predecessor relation

This well-founded relation is well-founded without the axiom of choice or Axiom of foundation. If a nonempty set $A$ lacked a minimal member, put $b=\bigcap A$. For each $y\in A$, some $x\in A$ has $\mathcal{P}(x)\subseteq y$, and $b\subseteq x$ gives $\mathcal{P}(b)\subseteq y$. Hence $\mathcal{P}(b)\subseteq b$. The Cantor diagonal argument applied to $d=\{u\in b:u\notin u\}$ contradicts this inclusion.