Consider the following premises. forall x_18: P_18(x_18) <-> Q_18(x_18). forall x_18: ~Q_18(x_18). forall x_18: P_18(x_18) or Q_12(x_18). forall z: T(z). Can we infer forall z: ~(Q_12(z) and T(z)) from them?
 From forall x_18: P_18(x_18) <-> Q_18(x_18), forall x_18: ~Q_18(x_18) we can infer forall x_18: ~P_18(x_18). From forall x_18: P_18(x_18) or Q_12(x_18), forall x_18: ~P_18(x_18) we can infer forall x_18: Q_12(x_18). From forall x_18: Q_12(x_18) we can infer Q_12(a_10). From Q_12(a_10) we can infer exists z: Q_12(z). Finally, from exists z: Q_12(z), forall z: T(z) we can infer exists z: Q_12(z) and T(z), which contradicts forall z: ~(Q_12(z) and T(z)). Therefore, the answer is no.