Lemma: One complete lane per frame

A register at the first endpoint of a lane immediately after its phase-zero firing can advance from index zero to index $N_n$ during the following $N_n$ ticks, moving one graph edge at each tick. It arrives at the next phase-zero boundary.

The same assertion holds in reverse along a lane, because $N_n$ is odd.

Proof. At transport phase $j$, the pair containing indices $j-1,j$ is present. Induction moves the register to index $j$. The last transport tick ends at the next phase-zero boundary.

For reverse travel, the pair containing indices $N_n-j+1,N_n-j$ is present: its lower index $N_n-j$ has parity $j-1$ when $N_n$ is odd.