Consider the following premises. forall z_2: S(z_2) -> ~U_1(z_2). forall z_2: U_1(z_2). forall z_2: S(z_2) -> U_3(z_2). forall z_2: U_3(z_2) <-> S(z_2). Can we infer exists z_2: U_3(z_2) from them?
 From forall z_2: U_3(z_2) <-> S(z_2) we can infer forall z_2: U_3(z_2) -> S(z_2). From forall z_2: S(z_2) -> ~U_1(z_2), forall z_2: U_1(z_2) we can infer forall z_2: ~S(z_2). From forall z_2: S(z_2) -> U_3(z_2), forall z_2: U_3(z_2) -> S(z_2) we can infer forall z_2: S(z_2) <-> U_3(z_2). Finally, from forall z_2: S(z_2) <-> U_3(z_2), forall z_2: ~S(z_2) we can infer forall z_2: ~U_3(z_2), which contradicts exists z_2: U_3(z_2). Therefore, the answer is no.