Equivalent forms of Foundation

Assume Foundation. Membership on the universe is setlike and has the minimal-element property, so well-founded induction proves (ii). Equivalently the counterexamples inside the transitive set $\mathrm{TC}(\{x\})$ have a minimal member, contradicting progressiveness.

Assume (ii). If a nonempty set $a$ had no member disjoint from $a$, the property $P(x)$ meaning $x\mathrm{\notin }a$ would be progressive: if all $y\in x$ were outside $a$ and $x\in a$, the no-minimal-member assumption would supply $y\in x\cap a$, a contradiction. Membership induction gives $x\mathrm{\notin }a$ for all $x$, impossible since $a$ is nonempty.

Under (ii), prove (iii) by membership induction. If each $y\in x$ lies in a stage, it has a unique least stage index $h(y)$, found by minimizing below any witness. Replacement collects these indices; let $\beta =\sup \{h(y):y\in x\}$. Nesting puts every $y\in x$ in $V_{\beta }$, so $x\in V_{\beta +1}$. For $x=\mathrm{\varnothing }$ the supremum is zero and the same conclusion holds.

Assume (iii) and let $a$ be nonempty. The least stage index $h(x)$ exists for each $x\in a$; it is a successor $\beta +1$, since zero is empty and a limit is a union. If $y\in x$, then $x\in V_{\beta +1}$ implies $y\in V_{\beta }$, hence $h(y)\le \beta <h(x)$. Minimize $h$ on the set $a$ using Replacement. A member $x$ of least height has $x\cap a=\mathrm{\varnothing }$, proving Foundation. These heights were defined from stages alone, without membership rank.