No uniform bounded unfolding of the star quotient

Exact statement

For every $k\ge0$ , $\varepsilon\in(a^*)^{-1}\{a^{k+1}\}$ but $\varepsilon\notin\bigcup_{i=0}^{k}\{a^i\}^{-1}\{a^{k+1}\}$ . Therefore no fixed finite quotient-depth truncation computes all star left quotients.

StatusProved by an explicit unary family

External reviewNo documented external or specialist review of this FPRD result is recorded.

Context

A one-step star mutation happened to fail on only two cases in the finite corpus. The unary family turns that isolated observation into an unbounded exact obstruction.

Hypotheses and scope

The alphabet contains $a$ , and the attempted star clause uses at most $k$ applications of the inner quotient transformer.

Proof or evidence

At quotient depth i, the singleton target a^(k+1) has shortened only to a^(k+1-i). The empty word first appears at depth k+1, although it belongs to the full quotient because a^(k+1) is in a*.

Unbounded-unfolding proof →

Verification notes

The statement rejects only uniformly bounded unfolding. It deliberately leaves open other forms of algebraic compression and finite saturation.

Limitations

This is not an impossibility theorem for every conceivable constructor-only model.
It does not show that an auxiliary derivative carrier is minimal in size.

Open work

Use this family as the first adversary for any proposed carrier-free star compiler; seek a broader lower bound only after its formal interface is pinned.