Consider the following premises. forall z: W(z) -> S(z). exists z: W(z) -> Q(z). forall z: ~S(z). forall z: Q(z) -> W(z). Can we infer forall z: Q(z) from them?
 From forall z: W(z) -> S(z), forall z: ~S(z) we can infer forall z: ~W(z). From forall z: Q(z) -> W(z), exists z: W(z) -> Q(z) we can infer exists z: Q(z) <-> W(z). Finally, from exists z: Q(z) <-> W(z), forall z: ~W(z) we can infer exists z: ~Q(z), which contradicts forall z: Q(z). Therefore, the answer is no.