Consider the following premises. forall x_6: ~Q_6(x_6). forall x_6: P_6(x_6) -> Q_6(x_6). forall x_6: P_6(x_6) or Q_3(x_6). Can we infer exists y: Q_3(y) from them?
 From forall x_6: P_6(x_6) -> Q_6(x_6), forall x_6: ~Q_6(x_6) we can infer forall x_6: ~P_6(x_6). From forall x_6: P_6(x_6) or Q_3(x_6), forall x_6: ~P_6(x_6) we can infer forall x_6: Q_3(x_6). From forall x_6: Q_3(x_6) we can infer Q_3(b_1). Finally, from Q_3(b_1) we can infer exists y: Q_3(y). Therefore, the answer is yes.