Consider the following premises. R_14(f_1) and q_2. forall x_14: R_14(x_14) -> S_14(x_14). Can we infer ~S_14(f_1) from them?
 From forall x_14: R_14(x_14) -> S_14(x_14) we can infer R_14(f_1) -> S_14(f_1). From R_14(f_1) and q_2 we can infer R_14(f_1). Finally, from R_14(f_1) -> S_14(f_1), R_14(f_1) we can infer S_14(f_1), which contradicts ~S_14(f_1). Therefore, the answer is no.