Cube-freeness passes to subsets
ProvedZ2nFiveEighths.cubeFree_subsetIf and is cube-free, then is cube-free.
Immediate from the definition, and needed by any induction that restricts the set to a coset or a subgroup.
import Mathlib import Definitions.Def_Z2nCubeFreeLayers
namespace Z2nFiveEighths
theorem cubeFree_subset {G : Type} [AddCommGroup G] (A B : Finset G)
(hBA : B ⊆ A) (hA : CubeFree A) : CubeFree B := by sorry
end Z2nFiveEighths
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What the Lean code literally says, in plain math · claude-opus-5
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Let G be an abelian group written additively, of the lowest universe (Type, not Type*; the two auxiliary definitions are universe polymorphic, but the theorem is not). Call a finite subset S of G "rich" if there exist elements x, y, z of G, not required to be distinct and not required to be nonzero, such that all seven of x, y, z, x+y, y+z, z+x, and (x+y)+z lie in S. In other words, S contains every nonempty subset sum of some triple drawn from it. The statement is: for all finite subsets A and B of G with B contained in A (not necessarily properly), if A is not rich then B is not rich. Equivalently, richness is inherited upward by supersets, so its negation passes to subsets. No finiteness, torsion, or decidability assumption is placed on G itself, and no constant, cardinality bound, or density appears anywhere.
QUANTIFIER ORDER G : implicit type in universe 0, outermost. [AddCommGroup G] : instance argument, next. A, B : finite subsets of G, universally quantified, both in scope for the two hypotheses. Inside the definition of richness, applied to a fixed set: x, then y, then z, all existential over G, sharing one scope with the seven membership conditions.
HYPOTHESES [AddCommGroup G]: G is abelian and a group, so subtraction and commutativity are available; rules out monoids, semigroups, and nonabelian groups. Note x+y, y+z, z+x are unordered only because commutativity is assumed. hBA (B subset of A): non-strict containment; B = A is allowed. hA (A not rich): rules out A containing all seven subset sums of any triple. Since x = y = z = 0 is permitted, this forces 0 not in A. It also forces that no a in A has both 2a and 3a in A. Satisfiable, so the theorem is not vacuous: over the integers, A = {1} is not rich.
DEGENERATE CASES B empty: richness fails outright since x in B is unsatisfiable, so the conclusion holds independently of hA. A empty: hA is automatic and B is empty too. B = A: conclusion is literally the hypothesis. Any A containing 0 makes hA unsatisfiable, so all such A are excluded from the statement. The seven witnessed elements need not be distinct, so a rich set can have as few as one element only when that element is 0.
UNREADABLE Nothing. One factual note: the proof body is sorry, so the payload asserts the statement without establishing it.
Confirmed by the mission captain (proposal self-audit).