“…V f ={(0, 0, 2, 0, 1),(11, 8, 2, 1, 1),(0, 0, -1, -1, 0),(8, 11, 2, 1, 1),(0, 0, 2, 0, 1),(10, 9, -1, -1, 3),(7, 12, -1, -1, 2)} V g ={(0, 0, 2, 0, 1),(7, 12, -1, -1, 2),(0, 0, 2, 0, 1),(11, 8, 2, 3, 1), (11,8,2,3,1),(0, 0, -1, -1, 0),(10, 9, -1, -1, 3)} From the above updated SS vectors, we know that these two SS vectors are the same and that one single symmetrymapping set (S 1 = {1 → 3}) and two single-mapping sets (χ 5 = {5 → 6 − 0} and χ 6 = {6 → 1 − 0}) exist. Procedure 2 adds variable mappings 1 → 3 − 0, 3 → 4 − 1, 5 → 6 − 0 and 6 → 1 − 0 to the transformation tree.…”