Proof of Elements of a Segment Are Natural Numbers, and Mixed Transitivity of the Order on the Natural Numbers

(i). Let $n$ be a set and let $k\in\mathbb{N}_{0}$ with $1\le k\le n$. Suppose $k=0$. Then $1\le0$. Since $1\in\mathbb{N}\subseteq\mathbb{N}_{0}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (ii), also $0\le1$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (iii). By The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (i), $\le$ is a well-order on $\mathbb{N}_{0}$, hence a total order and in particular a partial order, which is antisymmetric. So $1=0$. But $1\in\mathbb{N}$, and $\mathbb{N}$ is the difference $\mathbb{N}_{0}\setminus\{0\}$ by The Set of Natural Numbers and the Number One (i), so $1\notin\{0\}$; as $0$ is the only element of the singleton $\{0\}$, $1\neq0$, a contradiction. Hence $k\neq0$, so $k\notin\{0\}$, and with $k\in\mathbb{N}_{0}$ this gives $k\in\mathbb{N}_{0}\setminus\{0\}=\mathbb{N}$.

For $n\in\mathbb{N}_{0}$, $[n]=\{1,\dots,n\}=\{k\in\mathbb{N}_{0}:1\le k\le n\}$ by Intervals of Natural Numbers (ii) and Intervals of Natural Numbers (i), so every element of $[n]$ is a natural number by what was just shown.

(ii). Let $k,l,m\in\mathbb{N}$. By Arithmetic and Order of the Natural Numbers (vi), for $a,b\in\mathbb{N}$, $a\le b$ holds if and only if $a<b$ or $a=b$, and $<$ is transitive on $\mathbb{N}$.

Suppose $k<l$ and $l\le m$. Then $l<m$ or $l=m$. If $l<m$, then $k<m$ by transitivity of $<$; if $l=m$, then $k<l$ is $k<m$.

Suppose $k\le l$ and $l<m$. Then $k<l$ or $k=l$. If $k<l$, then $k<m$ by transitivity of $<$; if $k=l$, then $l<m$ is $k<m$.