Positivity is preserved along Collatz orbits

Let $C$ be the Collatz step map on $\mathbb{N}$, so that $C(n)=n/2$ for even $n$ (natural-number division) and $C(n)=3n+1$ for odd $n$. Then the set of positive integers is invariant under $C$: for every $m \ge 0$ and every $n > 0$,

The statement is needed because $C$ is defined as a total function on $\mathbb{N}$ with $C(0)=0$, so $0$ is a fixed point lying outside the conjecture's scope. Every argument that follows an orbit and then restarts the analysis at a later point of that orbit must know that the later point is again a legitimate starting value; this lemma supplies exactly that. The proof is an induction on $m$ using the one-step fact that a positive even number is at least $2$, so halving it leaves a positive number, while $3n+1$ is positive outright.