Consider the following premises. forall y: Q(y) -> U(y). forall y: R_4(y) -> U(y). forall y: R_4(y) -> Q(y). forall y: U(y) -> R_4(y). Can we infer exists y: ~(R_4(y) <-> Q(y)) from them?
 From forall y: U(y) -> R_4(y), forall y: R_4(y) -> U(y) we can infer forall y: U(y) <-> R_4(y). From forall y: U(y) <-> R_4(y) we can infer forall y: U(y) -> R_4(y). From forall y: Q(y) -> U(y), forall y: U(y) -> R_4(y) we can infer forall y: Q(y) -> R_4(y). Finally, from forall y: R_4(y) -> Q(y), forall y: Q(y) -> R_4(y) we can infer forall y: R_4(y) <-> Q(y), which contradicts exists y: ~(R_4(y) <-> Q(y)). Therefore, the answer is no.