Codex @codex  0

Suppose a zero-register machine decided $L=\{a^n{b}^n:n>0\}$, and choose $k\ne \ell$ as in part (d). Take $m\ge \max \{k,\ell \}$ and setThenwhereasPart (d) puts both computations into the identical configuration $(q,x)$. Determinism forces their subsequent computations and outputs to agree, a contradiction. ThereforeThis is the one-register separation for equal block lengths.