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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Substituting an edge for an endpoint of gives a four-vertex graph with the Erdős–Hajnal property
Example
Let be obtained from the path by substituting the edge for one endpoint. Then is the paw graph, and has the Erdős–Hajnal property.
Facts & Assumptions
Given: The path with vertices , the edge , and the graph formed by substituting for the endpoint of .
Every graph on at most three vertices has the Erdős–Hajnal property (Every graph on at most three vertices has the Erdős–Hajnal property, The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).
Substituting one graph for a vertex of another preserves the Erdős–Hajnal property when both factors have it (Alon–Pach–Solymosi: if and have the Erdős–Hajnal property, so does the graph obtained from by substituting for a vertex, Substituting one graph for a vertex of another).
Verification
By [L1], both and have the Erdős–Hajnal property.
The substitution identifies one endpoint of with an edge, so the result is a triangle with one pendant edge, that is, the paw graph.
Applying [L2] to the two factors from step 1.1 gives the Erdős–Hajnal property for the paw.
Depends on
- Alon–Pach–Solymosi: if $H_1$ and $H_2$ have the Erdős–Hajnal property, so does the graph obtained from $H_1$ by substituting $H_2$ for a vertex
- Every graph on at most three vertices has the Erdős–Hajnal property
- Substituting one graph for a vertex of another
- Empty and complete graphs, complete bipartite graphs, and the convention that $P_n$ and $C_n$ have $n$ vertices
- The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class
Used by
Nothing in the library uses this result yet.
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