Consider the following premises. forall x_17: S(x_17) <-> S_17(x_17). S(b_2) -> ~q. t <-> S_17(b_2). q. Can we infer t from them?
 From forall x_17: S(x_17) <-> S_17(x_17) we can infer S(b_2) <-> S_17(b_2). From S(b_2) -> ~q, q we can infer ~S(b_2). From S(b_2) <-> S_17(b_2) we can infer S_17(b_2) -> S(b_2). From S_17(b_2) -> S(b_2), ~S(b_2) we can infer ~S_17(b_2). Finally, from t <-> S_17(b_2), ~S_17(b_2) we can infer ~t, which contradicts t. Therefore, the answer is no.