Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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.

GG is HH-free if and only if G\overline G is H\overline H-free

Statement

For finite graphs GG and HH,

G is H-freeG is H-free.G\text{ is }H\text{-free}\quad\Longleftrightarrow\quad \overline G\text{ is }\overline H\text{-free}.

Facts & Assumptions

Given: Finite graphs GG and HH.

[F1]

HH-free means containing no induced copy of HH (HH-free and F\mathcal F-free graphs under the induced-subgraph convention).

[L1]
[F2]

Complementation carries isomorphisms HG[W]H\cong G[W] to isomorphisms HG[W]\overline H\cong\overline G[W] (Graph isomorphisms, automorphisms and graph complements).

Proof

technique · direct
1.1

For every WV(G)W\subseteq V(G), one has G[W]HG[W]\cong H if and only if G[W]=G[W]H\overline G[W]=\overline{G[W]}\cong\overline H.

L1F2
2.1

Thus GG contains an induced HH if and only if G\overline G contains an induced H\overline H. Negating both sides gives the claimed equivalence.

step 1.1F1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 13 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources