Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Eventown: distinct A1,,Am[n] with every Ai and every AiAj even satisfy m2n/2

Statement

Let A1,,Am be distinct subsets of [n]. If every Ai is even and every intersection AiAj with ij is even, then

m2n/2.

Facts & Assumptions

Given: distinct subsets A1,,Am[n] with every Ai even and every AiAj even for ij.

[L1]

Over F2, the standard-form values of all the incidence vectors vAi vanish against one another and against themselves (vA,vB is the image of AB in F; over F2 it is 0 or 1 according to the parity of AB).

[L3]

A d-dimensional vector space over F2 has 2d elements (A d-dimensional vector space over a field with q elements has exactly qd elements).

Proof

technique · direct
1.1

Work over F2, and let U be the span of the incidence vectors vA1,,vAm. By [L1], every pairing vAi,vAj is 0.

L1given
2.1

Bilinearity then gives u,u=0 for all u,uU, so UU.

step 1.1
3.1

Writing d=dimU, the inclusion of step 2.1 and [L2] give dnd. Hence 2dn, so dn/2.

L2step 2.1
4.1

The m distinct incidence vectors lie in U, so mU. By [L3], U=2d2n/2, and therefore m2n/2.

L3step 3.1

Remarks

  • The floor enters only because d is an integer and 2dn. The proof is otherwise the same in both parities of n.

Depends on

Used by

Dependency tree · two levels

48 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources