, Now by Lemma 8, there is at least one big vertex outside of S that has a neighbour in S. Note that if there are at least two of these vertices, or if one of them has at least two neighbours in S

, Therefore S is a bud. In this case, S ends up with a charge of at least ?. We will

, For every big vertex v, v gives 1 ? 2 to each of its sons, does not give anything to its father (if it has one), and gives 1? 2 to its other 2-neighbours

, The common pot gives to every internal 2-vertex

, Every vertex has non-negative charge at the end of the procedure, Lemma, vol.17

. Proof, Note that by what precedes every small 3 +-vertex v has non-negative charge. Every 2-vertex that is not in L receives 1? 2 from each of its neighbours. Every light 2-vertex that is not adjacent to a 2-vertex (i.e. every 2-vertex of L that is not an internal 2-vertex) receives 1 ? 2 from its father and from its other neighbour

+. , Every internal 2-vertex receives 1 ? 2 from its father and from the common pot

, Let c(v) be the initial charge of v, and c (v) be the final charge of v. Suppose by contradiction that c (v) < 0. By Lemma 9, vertex v has at most k ? 1 sons. Moreover, if v has k ? 1 sons, then its last neighbour is either the father of v

, ? 1 sons, then v gives 1 ? 2 to each of its k ? 1 sons, and 2(k ? d ? 1) to the common pot, therefore c (v) = c(v) ? (1 ? 2)(k ? 1) + 2(k ? d ? 1), and thus as c (v) < 0

?. +2(k-?d?1, By Lemma 16, this charge is at least times the number of internal 2-vertices. The common pot gives to each internal 2-vertex, therefore it has non-negative charge at the end of the procedure, the electronic journal of combinatorics, vol.25, issue.1, 2018.

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