=. R. Disjoint-for-i and T. Thus, Without loss of generality, assume y = z Since ?(G(v 1 )) > k, there exist k + 1 internally disjoint x ? y paths P i , 1 ? i ? k + 1, in G(v 1 ) by Menger's Theorem. Note that at most one of them is a path of length 1. Let P k+1 be such a path if , k be integers such that 0 = k 0 < k 1 < · · · < k = k + 1. Similar to the proofs of Proposition 3.1, we can construct k i ? k i?1 + 1 internally disjoint S-trees T i,ji , 1 ? j i ? k i ? k i?1 + 1, in ( ki j=ki?1+1 P j ) T i for each i, where 1 ? i ? ? 1, and k ? k ?1 internally disjoint S-trees T ,j , 1 ? j i ? k ? k ?1 , in ( k j=k ?1 +1 P j ) T i . By Observation 3.1 and 3.2, T i,ji and T r,jr are internally disjoint for i = r. Thus T i,ji , 1 ? i ? , 1 ? j i ? k i ? k i?1 + 1 are k + internally disjoint S-trees. If all of x, y , z are the same vertex in G(v i ) Since ?(G(v 1 )) ? ?(G(v 1 )) > k, x has k neighbors, say x 1 , x 2 , . . . , x k , in G(v 1 )). Let P i be the path xx i , and let k 0 , k 1 , . . . , k be integers such that 0 = k 0 < k 1 < · · · < k = k. Similar to the proofs of Proposition 3.1, we can construct k i ? k i?1 + 1 internally disjoint S-trees T i,ji , 1 ? j i ? k i ? k i?1 + 1, in ( ki j=ki?1+1 P j ) T i for each i, where 1 ? i ? . By Observation 3.1 and 3.2, T i,ji and T r,jr are internally disjoint for i = r. Thus T i,ji , 1 ? i ? , 1 ? j i ? k i ? k i?1 + 1 are k + internally disjoint S-trees, ? i ? , 1 ? j i ? k i ? k i?1 + 1 are k + internally disjoint S-trees

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