LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.

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The complement of $G_{1} [a \to G_{2}]$ is $\overline{G_{1}} [a \to \overline{G_{2}}]$

Statement

Let $G_{1} [a \to G_{2}]$ be a substitution. Then $\overline{G_{1}} [a \to \overline{G_{2}}]$ is also a substitution, and

$\overline{G_{1} [a \to G_{2}]} = \overline{G_{1}} [a \to \overline{G_{2}}] .$

Facts & Assumptions

Given: A substitution $G_{1} [a \to G_{2}]$ , with $U = V (G_{1}) ∖ {a}$ disjoint from $V (G_{2})$ and $V (G_{2}) \neq \emptyset$ .

[F1]

For distinct vertices of $G_{1} [a \to G_{2}]$ : two vertices of $U$ are adjacent exactly when they are adjacent in $G_{1}$ ; two vertices of $V (G_{2})$ are adjacent exactly when they are adjacent in $G_{2}$ ; and $u \in U$ is adjacent to $x \in V (G_{2})$ exactly when $u$ is adjacent to $a$ in $G_{1}$ . The vertex set is $U \cup V (G_{2})$ (Substituting one graph for a vertex of another).

[F2]

The complement of $G = (V, E)$ is $\overline{G} = (V, [V]^{2} ∖ E)$ , so distinct vertices are adjacent in $\overline{G}$ exactly when they are not adjacent in $G$ (Graph isomorphisms, automorphisms and graph complements).

Proof

technique · cases

1.1

The graphs $\overline{G_{1}}$ and $\overline{G_{2}}$ have the same vertex sets as $G_{1}$ and $G_{2}$ , and $a \in V (\overline{G_{1}})$ , so $\overline{G_{1}} [a \to \overline{G_{2}}]$ is a substitution with the same hypotheses and the same vertex set $U \cup V (G_{2})$ as $G_{1} [a \to G_{2}]$ ; hence both sides of the claimed identity are graphs on that set.

F1F2

1.2

First case: distinct $p, q \in U$ . Then ${p, q}$ is an edge of $\overline{G_{1} [a \to G_{2}]}$ exactly when it is not an edge of $G_{1}$ , that is, exactly when it is an edge of $\overline{G_{1}}$ , which is exactly the condition for it to be an edge of $\overline{G_{1}} [a \to \overline{G_{2}}]$ .

assume-case hostF1F2

1.3

Second case: distinct $p, q \in V (G_{2})$ . Then ${p, q}$ is an edge of $\overline{G_{1} [a \to G_{2}]}$ exactly when it is not an edge of $G_{2}$ , that is, exactly when it is an edge of $\overline{G_{2}}$ , which is exactly the condition for it to be an edge of $\overline{G_{1}} [a \to \overline{G_{2}}]$ .

assume-case insertedF1F2

1.4

Third case: $u \in U$ and $x \in V (G_{2})$ . Then ${u, x}$ is an edge of $\overline{G_{1} [a \to G_{2}]}$ exactly when ${u, a} \notin E (G_{1})$ , that is, exactly when ${u, a} \in E (\overline{G_{1}})$ , which is exactly the condition for ${u, x}$ to be an edge of $\overline{G_{1}} [a \to \overline{G_{2}}]$ .

assume-case crossF1F2

2.1

Every pair of distinct vertices of $U \cup V (G_{2})$ falls under exactly one of the three cases, because the union is disjoint, so the three cases are exhaustive.

step 1.2step 1.3step 1.4F1cases-exhaustive

3.1

The two graphs of step 1.1 therefore have the same vertex set and the same edge set, so they are equal.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

T. Huang, Y. Ju and R. Zhou, Erdős–Hajnal beyond the five-vertex path, sec. 1.2 (standard reference, not scraped)

The complement of G1[a→G2]G1​[a→G2​] is G1‾[a→G2‾]G1​​[a→G2​​]