Regardless of what type we assign to Ann, she believes that Bob will choose \(l\) since in both matrices, \(\lambda_A(t_1^A)\) and \(\lambda_A(t_2^A)\), the only places where a 1 appears is under the \(l\) column. So, fixing a type for Ann, in all of the situations Ann considers possible it is true that Bob chooses \(l\).