Proof of The Principle of Induction for the Natural Numbers Starting at One

Throughout, every element of a class is a set by Class Theory NBG: the Axioms, Standing Conventions and Basic Notation §objects. Membership in a class formed by class abstraction is unfolded by Class Abstraction: the Class of All Sets Satisfying a Predicative Formula §abstraction; class abstractions are unfolded without further mention. Let $\omega$ and $0$ be as in The Class Omega of Natural Numbers with Zero §omega and The Class Omega of Natural Numbers with Zero §zero, and let $A$ be as in the clause induction above.

Induction on $\omega$. Let

$$B=\{n\in\omega:n=0\vee n\in A\},$$

formed by class abstraction with the parameters $\omega$ and $A$; its formula has no quantifier, $0$ being a defined set symbol by The Empty Set, the Unordered Pair and the Singleton §empty, so it is predicative as Class Theory NBG: the Axioms, Standing Conventions and Basic Notation §comprehension requires. Since $0\in\omega$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive and $0=0$, $0\in B$. Let $n\in\omega$ with $n\in B$. Then $S(n)\in\omega$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive. If $n=0$, then $S(n)=S(0)=1$ by The Set of Natural Numbers and the Number One §one, and $1\in A$ by hypothesis; if $n\in A$, then $S(n)\in A$ by hypothesis. In either case $S(n)\in A$, so $S(n)\in B$. By Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §induction, $\omega\subseteq B$.

Equality. Let $n\in\mathbb{N}$. By The Set of Natural Numbers and the Number One §naturals and The Boolean Operations on Classes, Disjointness, and the Universal Class §operations, $n\in\omega$ and $n\notin\{0\}$, so $n\neq0$ by The Empty Set, the Unordered Pair and the Singleton §singleton. Since $\omega\subseteq B$, $n\in B$, so $n=0$ or $n\in A$; hence $n\in A$. Conversely, every element of $A$ lies in $\mathbb{N}$ because $A\subseteq\mathbb{N}$, by Subclasses and Subsets §subclass. Thus for every set $n$, $n\in A$ if and only if $n\in\mathbb{N}$, and $A=\mathbb{N}$ by Class Theory NBG: the Axioms, Standing Conventions and Basic Notation §extensionality.