Even though (6) satisfies the property of nonshared quartets, it possibly counts more than the number of nonshared quartets claimed by an internal node in each tree. The problem is that given two internal nodes, they do not nescessarily claim the quartets counted by (6). If we denote the leaves of an nonshared quartet a, b, c and d, the first, second, third and fourth factors in (6) counts the number of choices of a, b, c and d respectively. The first and second factor choose a and b from Fi, while the third and fourth choose c and d from F¯ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacuWGgbGrgaqeaaaa@2DD9@i. In the cases where c and d are chosen from the same subtree Fk, k ≠ i of v, v does not claim the quartet. We must subtract these quartets, which can be counted as: