Consider the following premises. r_1 <-> Q(j). ~q_6. forall x: Q(x) <-> S(x). S(j) -> q_6. Can we infer ~r_1 from them?
 From forall x: Q(x) <-> S(x) we can infer Q(j) <-> S(j). From S(j) -> q_6, ~q_6 we can infer ~S(j). From Q(j) <-> S(j), ~S(j) we can infer ~Q(j). Finally, from r_1 <-> Q(j), ~Q(j) we can infer ~r_1. Therefore, the answer is yes.