Consider the following premises. forall x_17: P_17(x_17) -> ~Q_17(x_17). forall x_17: P_17(x_17) -> P_11(x_17). exists x_17: P_11(x_17) -> Q_11(x_17). forall x_17: S(x_17) -> P_11(x_17). forall x_17: Q_11(x_17) -> P_17(x_17). forall x_17: Q_17(x_17). Can we infer forall x_17: S(x_17) from them?
 From forall x_17: P_17(x_17) -> ~Q_17(x_17), forall x_17: Q_17(x_17) we can infer forall x_17: ~P_17(x_17). From exists x_17: P_11(x_17) -> Q_11(x_17), forall x_17: Q_11(x_17) -> P_17(x_17) we can infer exists x_17: P_11(x_17) -> P_17(x_17). From exists x_17: P_11(x_17) -> P_17(x_17), forall x_17: P_17(x_17) -> P_11(x_17) we can infer exists x_17: P_11(x_17) <-> P_17(x_17). From exists x_17: P_11(x_17) <-> P_17(x_17), forall x_17: ~P_17(x_17) we can infer exists x_17: ~P_11(x_17). Finally, from forall x_17: S(x_17) -> P_11(x_17), exists x_17: ~P_11(x_17) we can infer exists x_17: ~S(x_17), which contradicts forall x_17: S(x_17). Therefore, the answer is no.