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, We finally show that B is universal if, and only if, there is no N -bounded run from q(0) to p(m) in A. Observe first that for every word w ? ? * there is exactly one run of B reading w. If there is an N -bounded run ? from q(0) to p(m) in A, it follows that the run of B reading ? ends in the sink state ?, and thus ? ? L(B), witnessing the fact that B is not universal. If, on the other hand, there is a run of B ending in state ?, it must be reading a word of the form ? where ? is an N -bounded run from q(0) to p(m) in A. Since ? is the sole state which is not accepting and since, as observed before, for all words there is a run