Thus: π∗(a|gs) = 1 |A∗(gs)| = 1 |A∗(s)| =π ∗(g−1a|s).(46) Ifa /∈ A∗(gs), theng −1a /∈ A∗(s), both sides of the equation equal zero

Since gis a bijection, the cardinality is preserved:|A ∗(gs)|=|gA ∗(s)|=|A ∗(s)| · 2012