A counterexample to prism-hamiltonicity of 3-connected planar graphs

The prism over a graph $G$ is the Cartesian product of $G$ with the complete graph $K_2$. A graph $G$ is hamiltonian if there exists a spanning cycle in $G$, and $G$ is prism-hamiltonian if the prism over $G$ is hamiltonian. In [M.~Rosenfeld, D.~Barnette, Hamiltonian circuits in certain prisms, Discrete Math. 5 (1973), 389--394] the authors conjectured that every 3-connected planar graph is prism-hamiltonian. We construct a counterexample to the conjecture

Polyhedra without cubic vertices are prism-hamiltonian

The prism over a graph $G$ is the Cartesian product of $G$ with the complete graph on two vertices. A graph $G$ is prism-hamiltonian if the prism over $G$ is hamiltonian. We prove that every polyhedral graph (i.e. 3-connected planar graph) of minimum degree at least four is prism-hamiltonian