Proof of Order and Arithmetic of ℕ₀ in Flat Form

Throughout, the successor of $n\in\mathbb{N}_{0}$ is $S(n)=n+1$ by 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 (v), and every element of a class is a set (Elements Are Sets, and the Empty Set and the Pair of Two Sets Exist and Are Unique (i)).

For (i): $\le$ is a well-order on $\mathbb{N}_{0}$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (i), hence a total order by Well-Orders on a Set (i), which gives the first four statements by Partial and Total Orders on a Set and the Associated Strict Relation (i) and Partial and Total Orders on a Set and the Associated Strict Relation (ii). Next, $0\le n$ is The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (iii), and $n\le n+m$ is Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (vii) with $j=m$. The description of $<$ is The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (ii), and $m<n+1=S(n)$ if and only if $m\le n$ is The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (iv); taking $m=n$ gives $n<n+1$.

For (ii): $S(n)=n+1$ as noted, and $S(n)\neq0$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity (v). Further $1\in\mathbb{N}\subseteq\mathbb{N}_{0}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (ii), so $n+1\in\mathbb{N}_{0}$ by (iii).

For (iii): $m+n\in\mathbb{N}_{0}$ by Addition on Omega (i), and the four laws are Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (i), Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (iv) and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (iii).

For (iv): the second statement is Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (v), and the first follows from it by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (iv).

For (v): $m\cdot n\in\mathbb{N}_{0}$ by Multiplication on Omega: Products Lie in Omega, Multiplication Is a Binary Operation with Value the Product, and Membership in the Multiplication (i), and $m\cdot1=m$ by 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 (v); the other laws are Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order (vi), Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order (i), Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order (v) and Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order (iv).

For (vi): by The Set of Natural Numbers and the Number One (i), $x\in\mathbb{N}$ means $x\in\mathbb{N}_{0}$ and $x\neq0$; this gives the second statement. If $x\in\mathbb{N}$, then $x=S(j)=j+1$ for some $j\in\mathbb{N}_{0}$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity (vii). Conversely, $j+1\in\mathbb{N}_{0}$ and $j+1\neq0$ by (ii), so $j+1\in\mathbb{N}$. Finally, writing $q=j+1$, we get $q+m=(j+m)+1$ with $j+m\in\mathbb{N}_{0}$ by (iii), so $q+m\in\mathbb{N}$ by the first statement.

For (vii): by Intervals of Natural Numbers (ii) and Intervals of Natural Numbers (i), $k\in[n]$ means $k\in\mathbb{N}_{0}$, $1\le k$ and $k\le n$. So $q\in[q]$, as $1\le q$ by 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 (ii) and $q\le q$ by (i). Let $k\in[n]$. Then $k\le n\le n+1$ by (i), so $k\in[n+1]$. Also $k\neq0$: otherwise $1\le0\le1$, so $1=0$ by (i), whereas $1=0+1\neq0$ by (iii) and (ii); hence $k\in\mathbb{N}$ by The Set of Natural Numbers and the Number One (i). Finally, the identity on $[n]$ is a bijection from $[n]$ onto $[n]$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iii), so $[n]$ is finite by Finite Sets (i), taken with $n$ itself.