Consider the following premises. forall y_2: P(y_2) <-> R(y_2). forall y_2: P_1(y_2) -> P(y_2). exists y_2: ~R(y_2). forall y_2: R(y_2) -> P_1(y_2). Can we infer forall y_2: P_1(y_2) from them?
 From forall y_2: P(y_2) <-> R(y_2) we can infer forall y_2: P(y_2) -> R(y_2). From forall y_2: P_1(y_2) -> P(y_2), forall y_2: P(y_2) -> R(y_2) we can infer forall y_2: P_1(y_2) -> R(y_2). From forall y_2: R(y_2) -> P_1(y_2), forall y_2: P_1(y_2) -> R(y_2) we can infer forall y_2: R(y_2) <-> P_1(y_2). Finally, from forall y_2: R(y_2) <-> P_1(y_2), exists y_2: ~R(y_2) we can infer exists y_2: ~P_1(y_2), which contradicts forall y_2: P_1(y_2). Therefore, the answer is no.