Basically, my reasoning is that any two finite graphs with at least three vertices will have at least three vertex-deleted subgraphs, which are also induced subgraphs. Any two graphs which share at least three of these vertex-deleted subgraphs must be isomorphic. The hypomorphism implies the two graphs share three or more of these subgraphs, so they must be isomorphic.
- If $G$ and $H$ are hypomorphic, then both share the same multi-set of vertex-deleted subgraphs, or $D(G) = D(H)$
- A graph will have $|V|$ vertex-deleted subgraphs, therefore $|D(G)| \geq 3$ and $|D(H)| \geq 3$
- If $|D(G) \cap D(H)|=1$, then $G$ and $H$ contain an isomorphic induced subgraph and a single shared missing vertex $v_1$.
- If $|D(G) \cap D(H)|=2$, then they share the same set of edges which connects $v_1$ to the rest of the induced subgraph, except for a possible edge connecting $v_1$ to $v_2$, the missing vertex from the second vertex-deleted subgraph
- If $|D(G) \cap D(H)|\geq3$, then $G \simeq H$ as the third vertex-deleted subgraph would contain both $v_1$ and $v_2$, allowing us to deduce whether both graphs have the edge ($v_1$,$v_2$)
- Finally, because of , and the hypomorphism implies that both graphs have the same multi-set (three or more vertex-deleted subgraphs in common), why wouldn't $G \simeq H$?