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The membership induction theorem is the following schema, for every formula $\phi (x)$ with any fixed set parameters:Its hypothesis says the property is progressive under membership. In ZF it follows from the Axiom of foundation: if a counterexample $x$ exists, take the nonempty set of counterexamples in the transitive closure $\mathrm{TC}(\{x\})$, which contains $x$. A membership-minimal counterexample supplied by the Axiom of foundation has no counterexample among its elements, contradicting progressivity. The transitive closure ensures that all elements relevant to this argument remain in the set being considered.

Compose the two assumed isomorphisms to obtain a membership-preserving and membership-reflecting bijectionTransitivity of both targets implies, for every $a\in M_1$,For one inclusion use preservation of membership. For the other, if $c\in f(a)$, transitivity puts $c$ in $M_2$; surjectivity gives $c=f(b)$ for $b\in M_1$, and reflection gives $b\in a$.

Apply the membership induction theorem to the assertion that every $a\in M_1$ is fixed. If all members $b$ of $a$ are fixed, the displayed equation gives $f(a)=a$. Thus $f$ is the identity on $M_1$. Its range is $M_2$, so $M_1=M_2$, and the definition of $f$ givesThis proves uniqueness of the Mostowski collapse theorem. Existence of the two isomorphisms already supplies the membership structure needed in this argument.

Assume a nonempty set $a$ has no membership-minimal element, so for every $x\in a$ there is $y\in x\cap a$. Consider the formula $\phi (x)\equiv x\notin a$.

It is progressive: if every element of $x$ is outside $a$ but $x\in a$, the assumed absence of a minimal element supplies $y\in x\cap a$, contradicting that hypothesis. The membership induction theorem therefore implies $x\notin a$ for every set $x$. This contradicts $a\ne \varnothing$.

Thus every nonempty set $a$ contains an element $x$ with $x\cap a=\varnothing$, exactly the Axiom of foundation:Only the induction schema and elementary set reasoning were needed; no prior use of Axiom of foundation occurs in the proof.

The cumulative hierarchy is defined by transfinite recursion:This recursion on ordinals is available without assuming the Axiom of foundation for all sets. The stages are increasing and transitive, by transfinite induction using their definitions.

Work internally in the specified model of ZF without the Axiom of foundation, and assume the membership induction theorem. Apply it toSuppose every member $y$ of $x$ lies in some stage. Assign to each such $y$ the least ordinal ${\alpha }_y$ for which $y\in V_{{\alpha }_y}$. The Axiom schema of replacement gives a set of these ordinals, and let $\beta$ be their supremum, taking $\beta =0$ for the empty collection. Monotonicity gives $x\subseteq V_{\beta }$, and therefore $x\in V_{\beta +1}$.

This proves progressivity of $\phi$, so membership induction gives it for every set of the model. Every stage already consists of sets in that model. ConsequentlyThe equality is internal to the given model; its ordinals and power sets are computed there.