“…Therefore, for every dimension j there exists exactly one other dimension k such that e j,k / ∈ F , so all dimensions are split into three pairs {j 1 , k 1 }, {j 2 , k 2 } and {j 3 , k 3 } such that e j 1 ,k 1 , e j 2 ,k 2 , e j 3 ,k 3 / ∈ F . This is satisfied up to isomorphism only by one set of faulty vertices F : the set of all vertices of level 0 or 2 except the vertices e 1,2 , e 3,4 and e 5,6 . By Lemma 5.1 with the assumption (i), it suffices to find a long F 6:L -free cycle in Q 6:L which is presented on Figure 2.…”