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I'm trying to disprove the statement that "If a graph $G$ does not have a $K_{3,3}$ or a $K_5$ as an induced subgraph, then it is planar" by counterexample. I'm thinking that I can just use the Petersen graph as a counterexample and say that the Petersen graph has neither a $K_{3,3}$ nor a $K_5$ as an induced subgraph, but it is non-planar by Kuratowski's theorem.

Is this sufficient to disprove the claim?

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This is sufficient as long as you justify the fact that the Petersen graph has neither $K_{3,3}$ nor $K_5$ as an induced subgraph. It has no vertex of degree $4$, so the $K_5$ half is clear, but the $K_{3,3}$ half is (slightly) more difficult.

Or you could start with some subdivision of $K_{3,3}$ in place of the Petersen graph.

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