That $\in$-induction fails to be a theorem schema of first order Zermelo + Foundation (see [here](https://mathoverflow.net/questions/314483/is-in-induction-provable-in-first-order-zermelo-set-theory)), then it appears that it is more eligible to replace axiom of Regularity (Foundation) by a Regularity schema:


**Axiom Schema of Regularity:** if $\varphi(x,y\_1,...,y\_n)$ is a formula in which only symbols $``x,y\_1,..,y\_n"$ occur free (and only free), and in which the symbol $``z"$ doesn't occur, and $\varphi(z,y\_1,..,y\_n)$ is the formula obtained from formula $\varphi(x,y\_1,..,y\_n)$ by merely replacing each occurrence of the symbol $``x"$ in it by the symbol $``z"$, then


$\forall y\_1,..,y\_n [\exists x (\varphi(x,y\_1,..,y\_n)) \to \exists x (\varphi(x,y\_1,..,y\_n) \wedge \not \exists z \in x (\varphi(z,y\_1,..,y\_n)))]$


is an axiom.


In short this says: $\exists \varphi(x) \to \neg \forall \varphi(x) \exists \varphi(y) \in x$ (parameters allowed)


It is clear that this is a schematic rendering of axiom of foundation, by re-presenting it in terms of predicates instead of sets.


Now I think that first order Zermelo + Regularity schema would imply $\in$-induction.


Is that correct?



Does Regularity schema imply $\in$-induction when added to first order Zermelo set theory?
**Yes**; the axiom schemas of regularity and $\in$-induction are equivalent, in fact by first-order logic alone.


The argument is clearest presented in the language of classes, in the style of NBG set theory. To make this an argument in $Z$, just replace each mention of a class $C$ with a formula $\varphi(x,y\_1,\ldots,y\_n)$, viewed as a predicate on $x$ with parameters $y\_1,\ldots,y\_n$.


In terms of classes, the instance of $\in$-induction for a given class $C$ says “If $C$ is $\in$-hereditary, then $C$ contains all sets.” Similarly, the instance of regularity for a given class $D$ says: “if $D$ is inhabited, then $D$ has an $\in$-minimal element.”


But “$\in$-hereditary” is dual to “has an $\in$-minimal element”, in the sense that a class $C$ is hereditary precisely if its complement $\bar{C}$ does *not* have an $\in$-minimal element, and “contains all sets” is dual to “is inhabited” in the same sense.


So we have a chain of equivalent statements, using purely first-order logic:


* If $C$ is $\in$-hereditary, then $C$ contains all sets
* (contrapositive) If $C$ does not contain all sets, then $C$ is not $\in$-hereditary.
* (duality as noted above) If $\bar{C}$ is inhabited, then $\bar{C}$ has an $\in$-minimal element.


So this shows: the instance of $\in$-induction for any class $C$ is equivalent to the regularity for its complement $\bar{C}$, and vice versa.


Going back to the first-order axiom-scheme versions, we get that an arbitrary instance of $\in$-induction, for some formula $\varphi(x,y\_1,\ldots,y\_n)$, is equivalent to the instance of regularity for the complementary formula $\lnot \varphi(x,y\_1,\ldots,y\_n)$; and similarly, every instance of regularity for a formula $\varphi$ is equivalent to the instance of $\in$-induction for $\lnot \varphi$.


