B. If and =. {f, f 0 , p, g}, then we can assume that B is the only bag containing p (otherwise, we delete p from any other bag) The intersection

B. If, 0 = B 00 = {f, f 0 , f 00 , p}, then the intersection of B and any of its neighbors in T is contained in {p}. A desired tree-decomposition can be obtained from Lemma 17

B. If, p}, then we delete f 00 from B 00 and add f 00 to B. We will be back then to case 1. The proof is similar for B = B 00 = {f

B. If, = f 00 , then the intersection of B and any of its neighbors in T is contained in {p, x}. If x is a child of p, then x is also a leaf in G and x play the same role as f 00, We are then in case 1. Therefore, in the following we assume that x is not a child of p

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