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David Roberts
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Here's a more detailed answer:

The above-mentionedabove-mentioned link (recreated below [1]) constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.

[1] Normal derivability and first-order arithmetic. P. Tosi, Normal derivability and first-order arithmetic, Notre Dame J. Formal Logic 21(2): 449-466 (April 1980). DOI doi: 1010.1305/ndjfl/1093883058.1305/ndjfl/1093883058 https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-21/issue-2/Normal-derivability-and-first-order-arithmetic/10.1305/ndjfl/1093883058.full

Here's a more detailed answer:

The above-mentioned link (recreated below [1]) constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.

[1] Normal derivability and first-order arithmetic. P. Tosi Notre Dame J. Formal Logic 21(2): 449-466 (April 1980). DOI: 10.1305/ndjfl/1093883058 https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-21/issue-2/Normal-derivability-and-first-order-arithmetic/10.1305/ndjfl/1093883058.full

Here's a more detailed answer:

The above-mentioned link (recreated below [1]) constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.

[1] P. Tosi, Normal derivability and first-order arithmetic, Notre Dame J. Formal Logic 21(2): 449-466 (April 1980) doi:10.1305/ndjfl/1093883058.

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Here's a more detailed answer:

The above-mentioned link (recreated below [1]) constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.

[1] Normal derivability and first-order arithmetic. P. Tosi Notre Dame J. Formal Logic 21(2): 449-466 (April 1980). DOI: 10.1305/ndjfl/1093883058 https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-21/issue-2/Normal-derivability-and-first-order-arithmetic/10.1305/ndjfl/1093883058.full

Here's a more detailed answer:

The above-mentioned link constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.

Here's a more detailed answer:

The above-mentioned link (recreated below [1]) constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.

[1] Normal derivability and first-order arithmetic. P. Tosi Notre Dame J. Formal Logic 21(2): 449-466 (April 1980). DOI: 10.1305/ndjfl/1093883058 https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-21/issue-2/Normal-derivability-and-first-order-arithmetic/10.1305/ndjfl/1093883058.full

Source Link

Here's a more detailed answer:

The above-mentioned link constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon_0, \in )$. Then, induction up to $\epsilon_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.