Proof of Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication

Membership in $\mathbb{N}$. By The Set of Natural Numbers and the Number One §naturals, $\mathbb{N}=\omega\setminus\{0\}$, the difference. Unfolding it by The Boolean Operations on Classes, Disjointness, and the Universal Class §operations and Class Abstraction: the Class of All Sets Satisfying a Predicative Formula §abstraction, and using that the only element of $\{0\}$ is $0$ by The Empty Set, the Unordered Pair and the Singleton §singleton, for every set $x$

$$x\in\mathbb{N}\quad\text{if and only if}\quad x\in\omega\ \text{and}\ x\neq0.\qquad(\ast)$$

In particular $\mathbb{N}\subseteq\omega$ by Subclasses and Subsets §subclass.

Successors. Let $x$ be a set. Suppose $x=S(m)$ for some $m\in\omega$. Then $x\in\omega$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive, and $x\neq0$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §successor-nonzero; so $x\in\mathbb{N}$ by $(\ast)$. Conversely, let $x\in\mathbb{N}$. By $(\ast)$, $x\in\omega$ and $x\neq0$, so by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §cases there is $m\in\omega$ with $x=S(m)$. The particular case is the first implication with $x=S(n)$ and $m=n$.

One. By The Set of Natural Numbers and the Number One §one, $1=S(0)$, and $0\in\omega$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive; so $1\in\mathbb{N}$ by the clause successors above. Let $n\in\mathbb{N}$. By $(\ast)$, $n\in\omega$ and $n\neq0$. By The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §zero-least, $0\le n$; as $0\neq n$, The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §strict gives $0<n$. Since $0,n\in\omega$, The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §successor-below turns $0<n$ into $S(0)\le n$, that is, $1\le n$.

Peano. Let $n\in\mathbb{N}$, and suppose $S(n)=1$. Then $S(n)=S(0)$ by The Set of Natural Numbers and the Number One §one, so $n=0$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §successor-injective, contradicting $n\neq0$, which holds by $(\ast)$. Hence $S(n)\neq1$. For $m,n\in\mathbb{N}$, which are sets, $S(m)=S(n)$ implies $m=n$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §successor-injective.

Predecessor. Let $n\in\mathbb{N}$ with $n\neq1$. By the clause successors above there is $m\in\omega$ with $n=S(m)$. If $m=0$, then $n=S(0)=1$ by The Set of Natural Numbers and the Number One §one, contrary to $n\neq1$; so $m\neq0$, and $m\in\mathbb{N}$ by $(\ast)$.

Plus-one. Let $n\in\omega$. By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §successor, $S(n)=n+S(0)$, and by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §one, $n\cdot S(0)=n$. Since $S(0)=1$ by The Set of Natural Numbers and the Number One §one, $S(n)=n+1$ and $n\cdot1=n$.

Closed. Let $m,n\in\mathbb{N}$. By $(\ast)$, $m,n\in\omega$ and $m\neq0$, $n\neq0$. By Addition on Omega §addition and Multiplication on Omega §multiplication, $m+n\in\omega$ and $m\cdot n\in\omega$. If $m+n=0$, then $m=0$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero-sum, which is false; so $m+n\neq0$. If $m\cdot n=0$, then $m=0$ or $n=0$ by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §no-zero-divisors, which is false; so $m\cdot n\neq0$. By $(\ast)$, $m+n\in\mathbb{N}$ and $m\cdot n\in\mathbb{N}$.