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