# Prove If $n, m$ are in the Natural Numbers and $n \ge m$, then $n-m$ is in the natural numbers

I need help proving this theorem using the axioms and theorems shown in the picture: If $$n,m$$ are in the natural numbers and $$n\ge m$$ then $$n-m$$ is in the natural numbers (https://i.stack.imgur.com/I665o.jpg)

• Please don't use pictures. – Dietrich Burde Nov 8 at 16:06
• especially sideways ones – J. W. Tanner Nov 8 at 16:10

1. Since $$n\geq m$$, $$n-m\geq 0$$.
2. Since $$n,m$$ are both integers (being natural numbers), $$n-m$$ is also an integer (because the integers are a closed set).
Combining statements $$1$$ and $$2$$ says that $$n-m$$ is a non-negative integer. Therefore it is a natural number.