Answer
At most H(X|A), and no more. This is the part of QUESTION-010 that its note of 2026-10-14 called “the one-step, information-theoretic form for one categorical concept”, split off on 2026-10-21 (ADR-049).
THEORY-281 (Active) answers it, from Chowdhury et al. (LIT-963), with the record’s re-derivation of 2026-10-20. The map Z is made from X alone, possibly with independent randomness, so A → X → Z.
- The cap. If I(Z;A) = 0, the chain rule gives I(Z;X) = I(Z;X|A) ≤ H(X|A). Nothing else is assumed.
- When the cap is reached. Exactly when H(X|Z,A) = 0, that is, when X is a function of Z and A.
- Deterministic maps. A deterministic Z = f(X) reaches the cap only if the groups’ distributions are permutations of one another. Under disjoint group supports that condition is also enough.
- Stochastic maps. They reach the cap whenever each group’s distribution is a coarsening of one common law. Under disjoint finite supports such a law always exists, at the cost of a larger output support.
- The price. With disjoint groups, A is a function of X and the cap is H(X) − H(A). Erasing the concept costs at least its entropy.
What it leaves to QUESTION-010
The answer is about one step, an ideal map that knows the distributions, and a concept given by a categorical label. QUESTION-010 asks about a chain, and about the relations a task family needs. The task form, I(Z;B) ≤ H(B|A) for any B, is the record’s step, not the source’s (THEORY-281, “What this does not say”). With overlapping groups, no map that reaches the cap is constructed. THEORY-199, the noise-driven answer of the same shape, is Proposed and stays with the parent.