Consider the following premises. forall z: U(z) <-> S(z). r. S(d_3) -> ~r. Can we infer U(d_3) from them?
 From forall z: U(z) <-> S(z) we can infer U(d_3) <-> S(d_3). From S(d_3) -> ~r, r we can infer ~S(d_3). Finally, from U(d_3) <-> S(d_3), ~S(d_3) we can infer ~U(d_3), which contradicts U(d_3). Therefore, the answer is no.