$k$ Collatz steps divide out a factor $2^{k}$

Let $C$ be the Collatz step map. If $2^{k}$ divides $x$, then the first $k$ steps of the Collatz orbit of $x$ are all halvings, and together they divide out the whole factor:

The content is that divisibility by $2^{k}$ propagates: if $2^{k+1} \mid x$ then $x$ is even, so the first step is a halving, and the result is divisible by $2^{k}$, so the induction continues. No positivity hypothesis is needed, since $x = 0$ gives $C^{k}(0) = 0$ on both sides.

The lemma is the exact bridge between the classical Collatz map and the accelerated map: writing $3n+1 = 2^{v} \cdot m$ with $m$ odd, it says that the run of halvings following an ascending step is executed in one identity rather than step by step.