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, The graph B(12 + , 12 + , x, y, z) is not OL-AP for every x,
, We prove this claim by induction on x + y + z. Let us first suppose that x = y = z = 1 and consider the associated graph B = B(12 +
, Indeed, every possible for S makes B[T ] being either disconnected, or isomorphic to either a tree with maximum degree 4 or a tree having two degree-3 vertices. To complete the base case, observe that B(12 + , 12 + , 1, 2, 1) and B(12 + , 12 + , 1, 1, 2) are not OL-AP for they do not admit an OL-AP-partition for 3: for every coherent choice of S, the subgraph B[T ] is disconnected, or isomorphic to either a tree with maximum degree 4, a tree having two degree-3 vertices, or a non-caterpillar 3-pode different from P (2, 4, 6) Suppose now that this claim holds whenever x + y + z ? k ? 1 for some k ? 5, and consider a balloon B = B(12 + , 12 + , x, y, z) where x + y + z = k. Once again, we consider two main cases: ? z > 1: in this case, B is not OL-AP since it cannot be OL-AP-partitioned for 1. Indeed, observe that removing one vertex from B makes the remaining subgraph being disconnected
, 4} \ {y, z}) In this situation, B cannot be OL-AP for the same reason as above but for an OL-AP-partition of B for ?. Indeed, for every coherent choice of S, the remaining graph B[T ] is not OL-AP either according to the induction hypothesis, or because it is isomorphic to a non-connected graph or a tree with maximum degree at least 4
, The graph B(12 + , 12 + , 12 + , x, y) is not OL-AP for every x
, Once again, we prove this claim by induction on x + y. We first suppose that x = y = 1 and let B = B(12 + , 12 + , 12 + , 1, 1) Similarly as in the proofs of the previous lemmas, B is not OL-AP for it cannot be OL-AP-partitioned for