Consider the following premises. exists x: ~T(x). forall x: T(x) or ~Q(x). forall x: R_1(x) -> Q(x). forall x: T(x) -> R_1(x). Can we infer forall x: Q(x) from them?
 From forall x: T(x) -> R_1(x), forall x: R_1(x) -> Q(x) we can infer forall x: T(x) -> Q(x). From forall x: T(x) or ~Q(x) we can infer forall x: Q(x) -> T(x). From forall x: Q(x) -> T(x), forall x: T(x) -> Q(x) we can infer forall x: Q(x) <-> T(x). Finally, from forall x: Q(x) <-> T(x), exists x: ~T(x) we can infer exists x: ~Q(x), which contradicts forall x: Q(x). Therefore, the answer is no.