Proof of A Natural Number with Zero Is Less Than Its Sum with a Natural Number

Since $m\in\mathbb{N}$, $m\neq0$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (ii), and $0\le m$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (iii); hence $0<m$ by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (ii). Adding $n$ on the left preserves the strict order, by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (vi), so

$$n=n+0<n+m,$$

using $n+0=n$ from Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order (i).