Proof. Since the edges of an inclusion-minimal cut in G correspond to a simple cycle in the planar dual G * of G , there is a planar embedding and connect vertices in V * with vertices in W * such that these edges do not cross. Clearly the resulting join is planar. If |A| = |B|, there is a bijection of cut-edges that can be chosen to preserve the circular orders (in opposite directions). In particular, the |A| = |B| new edges can be inserted to be degree-preserving: to this end, insert an edge between x ∈ V * and y ∈ W * and proceed clockwise along c 1 and c 2 to insert the desired number of (multi)edges for x to y and then the clockwise neighbors of y. Then proceed to the clockwise neighbor of x and continue until all |A| = |B| are inserted. By construction, no two edges between V * and W * cross each other, and thus the resulting embedding of the join is planar.