. Proof and . First, let us suppose, without loss of generality, that T is an out-BFS-tree in D. The proof when T is an in-BFS-tree is analogous

, Since there is no arc from L i to L j for every j ? i + 2, the strong components of D 1 and D 2 are contained in the levels. Hence, by Lemma 29, ?(D 1 ) = max{ ?(D L i ) | i is odd} and ?(D 2 ) = max{ ?(D L i ) | i is even}. Moreover, Let D 1 and D 2 be the subdigraphs of D induced by the vertices of odd and even levels, respectively

, Let u be a vertex in D and T u an out-BFS-tree with root u. By Lemma 30, By Lemma, vol.29

?. V-(c)-=-{v}, Let T v be an in-BFS-tree in C rooted at v. By Lemma 30, there is a level L v of T v such that ?(D L v ) ? c. Now since mader ? (F ? a) = c, D L v contains a subdivision S of F ? a. With a slight abuse of notation

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