Alphabeta Math
LemmaStatement: 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.

Every induced subgraph of an F\mathcal F-free graph is F\mathcal F-free

Statement

If GG is F\mathcal F-free and JJ is an induced subgraph of GG, then JJ is F\mathcal F-free.

Facts & Assumptions

Given: An F\mathcal F-free graph GG and an induced embedding ι:J→G\iota:J\to G.

[F1]

F\mathcal F-free means that no H∈FH\in\mathcal F has an induced embedding into GG (HH-free and F\mathcal F-free graphs under the induced-subgraph convention).

[L1]

Induced embeddings compose (Induced embeddings compose, and the induced-subgraph relation is transitive up to isomorphism).

Proof

technique · contradiction
1.1

Suppose JJ is not F\mathcal F-free. Then some H∈FH\in\mathcal F has an induced embedding φ:H→J\varphi:H\to J.

assume-contraF1
2.1

The composite ι∘φ:H→G\iota\circ\varphi:H\to G is an induced embedding.

step 1.1L1
3.1

This contradicts that GG is F\mathcal F-free. Hence JJ is F\mathcal F-free.

step 2.1F1discharge-contradiction∎

Depends on

Used by

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