Proof of Further Order Facts for ℕ₀ in Flat Form: ≤ as Membership or Equality, Transitivity for Sets, and the Order against +1

Throughout, $\mathbb{N}_{0}$ is another name for $\omega$ (The Class Omega of Natural Numbers with Zero (i)), so facts stated for $\mathbb{N}_{0}$ apply to elements of $\omega$; and for every $k\in\omega$ the successor $S(k)$ of $k$ (The Successor of a Set (i)) equals $k+1$ 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).

A preliminary observation. For all sets $x$ and $y$,

$$x\le y\iff\bigl(x\in\omega\wedge y\in\omega\wedge(x\in y\vee x=y)\bigr).\tag{$*$}$$

Indeed, $x\le y$ means $(x,y)\in{\le}$ by The Order on Omega (i). The ordered pair $(x,y)$ is a set, so by Class Abstraction: the Class of All Sets Satisfying a Predicative Formula (i) it lies in the class ${\le}$ if and only if there are sets $m'$ and $n'$ with $(x,y)=(m',n')$, $m'\in\omega$, $n'\in\omega$ and $m'\in n'$ or $m'=n'$. Now $(x,y)=(m',n')$ holds exactly when $m'=x$ and $n'=y$; so such $m'$ and $n'$ exist if and only if $x\in\omega$, $y\in\omega$ and $x\in y$ or $x=y$.

Proof of (i). Since $m,n\in\omega$, the right-hand side of $(*)$ with $x=m$ and $y=n$ reduces to: $m\in n$ or $m=n$.

Proof of (ii). Let $x\le y$ and $y\le z$. By $(*)$, the first gives $x,y\in\omega$ and the second gives $z\in\omega$. Transitivity of the order on $\omega$, Order and Arithmetic of ℕ₀ in Flat Form (i), now gives $x\le z$.

Proof of (iii). By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (vi) with $k=1$, which lies in $\omega$, we have $m\le n$ if and only if $1+m\le 1+n$. By Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (iv), $1+m=m+1$ and $1+n=n+1$, so $m\le n$ if and only if $m+1\le n+1$.

Proof of (iv). By The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (vi), $m<n$ if and only if $S(m)\le n$, and $S(m)=m+1$.

Proof of (v). 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 (i), $S(n)$ is a natural number, and $S(n)=n+1$. Since $1$ is below every natural number 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), $1\le n+1$.

Proof of (vi). Since $n\in\omega$ and $n\neq0$, $n\in\mathbb{N}=\omega\setminus\{0\}$ by The Set of Natural Numbers and the Number One (i); hence $1\le n$ 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).