The objection
The manuscript’s transport is a context map τ with local kernels on outcomes. TERM-008 gives it from A81 §5.2: “K_C : E_o(C) ⇝ E_t(τ(C))”. Its naturality law, from C6 Appendix B, is “restrict_tau(D) K_C = K_D restrict_D, interpreted as equality of kernels”. E_o(C) and E_t(τ(C)) are the outcomes of the judgements elicited in a context. So each K_C maps a source reader’s judgement to a target reader’s judgement.
A translation does something else. It maps a text to a text, T(v | u). TERM-008 says of the two: “the two levels are never formally related, and the manuscript uses both.” No translator, generator or captioner takes readers’ judgements as input. The source and target empirical models are two families of responses, to two texts.
For one pair, the kernels are unconstrained. Take an overlap-consistent source model e, and any overlap-consistent target model e′ on the target cover. Let every kernel ignore its input: K_C(· | s) = e′_τ(C) for every source outcome s. Then:
- K_C# e_C = e′_τ(C). So every term of L_obs(T) = Σ_C w_C d_C[(K_C)#e_C^o, e_τ(C)^t] (CLAIM-005) is d_C(e′_τ(C), e′_τ(C)) = 0, and D_obs = 0.
- Naturality asks that restricting after K_C equal K_D after restricting. The left side is the constant kernel with value e′_τ(C) restricted to τ(D). The right side is the constant kernel with value e′_τ(D). So naturality holds exactly when the restriction of e′_τ(C) to τ(D) is e′_τ(D). That is target overlap consistency, which was assumed.
So Proposition 1’s hypotheses hold (CLAIM-121), and its conclusion returns the target consistency that was put in. Proposition 2 is the same (CLAIM-070). If e′ is noncontextual with global law p′, the constant global kernel K_X ≡ p′ restricts to the constant kernels above, and K_X#p = p′ is a global extension of the target. Every overlap-consistent target is naturally transported from every source at zero observational distortion, whatever the translation did.
If K_C is fitted to make L_obs small, the fit can reach this kernel. That is the case CLAIM-011’s remedy is for: “Require state correspondences to be estimated on one collection of tasks and evaluated on another”.
Separation narrows the freedom and does not remove it. An anchoring condition would exclude the constant kernel. CLAIM-077 quotes C1 Appendix A’s anchor constraint, which “rules out constant transports when the anchors distinguish multiple source states”. It also records that “Neither the anchor nor the separation condition is in the manuscript”, whose “§6 naturality condition constrains the kernels but not their separation”.
Even with separation, a kernel that separates the source outcomes can often still push e_C onto a given target. Take two source outcomes a and b with e_C = (1/2, 1/2), and let e′_τ(C) be uniform on three target outcomes. The rows
- K(· | a) = (2/3, 1/3, 0),
- K(· | b) = (0, 1/3, 2/3)
average to (1/3, 1/3, 1/3). So K# e_C = e′_τ(C), and the D_obs term is zero. The two rows are at total-variation distance 2/3.
No pair of rows does better here, and none is at distance 1. The rows must average to 1/3 in each coordinate, so no row exceeds 2/3 anywhere. Their distance is the sum, over the coordinates where K(· | a) exceeds 1/3, of 2K(i | a) − 2/3. With one such coordinate it is at most 4/3 − 2/3 = 2/3. With two it is at most 2 − 4/3 = 2/3, because their entries sum to at most
- With three it is 0. The separation available is bounded by e′, but it is not zero. Whether separating families like this one also satisfy naturality across a whole cover was not checked.
The constraint appears only across utterances. Suppose one kernel family must serve every source utterance u in a class: e^(t,u) = K# e^(o,u) for every u. Then for each context, the target’s judgement distributions, read as an experiment about u, are a garbling of the source’s (THEORY-156). That is a real condition, and it is a Blackwell condition (on where Blackwell’s order reaches, see CLAIM-258). It makes the transport a property of a translation procedure, not of one rendering, which is also CLAIM-230’s point.
It is still a condition on the outcome-level kernels. Nothing in the record derives those kernels from T(v | u), the channel a translator actually applies, uniformly in u. Until that is done, Propositions 1 and 2 are theorems about a map that translation is never shown to induce.
That bears on two claims. CLAIM-125 says of its bridge question, “Half of that question is answered”, by the simulations of which Propositions 1 and 2 are special cases. The answer is about kernels between empirical models. It reaches translation only once the kernels are derived from the text channel. CLAIM-005’s L_obs, and TERM-021’s D(τ) before it, score one rendering against its source. For one pair it can be made zero by the choice of kernel.
What it does not say
- It does not say CLAIM-121 or CLAIM-070 is false. Both were checked and are Active, and the constant kernel meets their hypotheses honestly. It bears on their use in support, as CLAIM-011 bears on CLAIM-117.
- It does not say that every separating kernel can push e_C onto any target. The three-point example shows the separation is bounded by the target. The main point rests on the constant kernel and on the missing derivation from T(v | u).
- It does not say anyone proposes the constant kernel. It says nothing in the propositions, or in D_obs as the manuscript states it, excludes it.
Note of 2026-10-27: the edge says what the claim uses
The status review of this date (ADR-057, ADR-058) changed rests_on: CLAIM-077 to complements. The proof uses CLAIM-121, CLAIM-070 and arithmetic. CLAIM-077 is cited as the admissibility condition the manuscript lacks, which would exclude the constant kernel.