Affine-inequivalence certificate for the family-hunt pair - #17
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Promotes the 2026-08-12 adversarial review's decision (previously only recorded in an issue #14 comment) to a committed, CI-run certificate: A and B are NOT affinely equivalent, in either direction. Method: strict component-degree separation ((12, 11, 4), asserted) forces triangular slot equations for B = T1 o A o T2; the top form x^3*z of F3 plus unique factorization forces the shape of T2's linear part; slot 1 is solved by constraint propagation with every deduction asserted as one exact coefficient equation and completeness certified by an identically-zero residual on the resulting (al != 0, m4) family; slot 2 eliminates the f's by exact linear algebra and the surviving conditions have lex Groebner basis {1} over QQ after saturating al != 0 -- both directions, so linear conjugation and one-sided affine compositions are excluded as special cases. The invariant tie measured by family_hunt.py is therefore non-trivial against the entire affine group; the README status row now records non-affine tame equivalence as the only remaining triviality risk. Exits nonzero unless both directions are excluded; CI runs it. Co-Authored-By: Claude Fable 5 <noreply@anthropic.com>
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Promotes the adversarial review's affine-equivalence decision into the repo (context: #14 — not closed by this PR):
family_equiv.pydecides B = T1∘A∘T2 exactly over ℚ in both directions and excludes it (slot-1 family certified complete by identically-zero residual; slot-2 Gröbner basis {1} saturated at al ≠ 0). Verdict reproduced from the committed script in 35 s; CI runs it as a gate (nonzero exit unless both directions are excluded). README contents table + family-hunt status table updated: the affine layer is decided, non-affine tame equivalence is the only remaining triviality risk for the invariant tie.🤖 Generated with Claude Code