# Disprove: “If a graph $G$ does not have a $K_{3,3}$ or a $K_5$ as an induced subgraph, then it is planar”

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?

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.