Consider the following premises. forall x_12: P_12(x_12) -> Q_12(x_12). ~Q_12(a_11). forall x_12: Q_12(x_12) -> P_12(x_12). t -> P_12(a_11). Can we infer t from them?
 From forall x_12: P_12(x_12) -> Q_12(x_12), forall x_12: Q_12(x_12) -> P_12(x_12) we can infer forall x_12: P_12(x_12) <-> Q_12(x_12). From forall x_12: P_12(x_12) <-> Q_12(x_12) we can infer P_12(a_11) <-> Q_12(a_11). From P_12(a_11) <-> Q_12(a_11), ~Q_12(a_11) we can infer ~P_12(a_11). Finally, from t -> P_12(a_11), ~P_12(a_11) we can infer ~t, which contradicts t. Therefore, the answer is no.