Does this graph contain $K_5$ or $K_{3,3}$ as subdivision or minor? Does this graph contain subdivision of $K_5$ or $K_{3,3}$?

Does this graph contain  $K_5$ or $K_{3,3}$ as minor?



I'm not sure if I'm correct, but I think the answer is yes for both questions. $K_5$ has 5 vertices of degree 4, this graphs has 6 of them, and $K_{3,3}$ has 6 vertices of degree 3, and this graph also has 5 of them. are these enough to conclude that this graph says $G$ contain both $K_5$ and $K_{3,3}$ as subdivision and minor?
Denote the six inner vertices as $A,B,C,D,E,F$ in anticlockwise order. Denote the outer vertices as $A',B',C',D',E',F'$ accordingly.

Let's try subdivision of $K_5$: To find such, we need five vertices of degree (at least) $4$. Since there are six such vertices, we have littel choice. In fact, by symmetry, all possible choices are the same, so we may pick $A,B,C,D,E$. Then to exhibit a subdivision, we need to have a path from each of these vertices to each other vertex - and these paths must be vertex-disjoint (apart from the end points)!
Since the vertices are only of degree $4$, we must use all edges originating from our chosen vertices. This already completes $6$ of our needed $10$ paths: $AC, AD, AE, BD, BE, CE$. 
Of the paths from $C$ to $B$ and from $C$ to $D$, one must make use of $C'$ and one must ake use of $F$.
Then for the path from $A$ to $B$ we must take  $AA'B'B$ because $F$ is already used.  Similarly we have $EE'D'D$. But then the path from $C$ to $B$ or $D$ that goes via $C$ cannot be comppleted.  We conclude that this graph does not contain a subdivision of $K_5$ as subgraph.
the answer is yes. For more defails see following literature:
http://www.math.ucla.edu/~mwilliams/pdf/petersen.pdf