Consider the following premises. ~R_6(e_3). forall x_12: P_12(x_12) <-> Q_12(x_12). forall x_12: Q_12(x_12) -> R_6(x_12). Can we infer P_12(e_3) from them?
 From forall x_12: P_12(x_12) <-> Q_12(x_12) we can infer forall x_12: P_12(x_12) -> Q_12(x_12). From forall x_12: P_12(x_12) -> Q_12(x_12), forall x_12: Q_12(x_12) -> R_6(x_12) we can infer forall x_12: P_12(x_12) -> R_6(x_12). From forall x_12: P_12(x_12) -> R_6(x_12) we can infer P_12(e_3) -> R_6(e_3). Finally, from P_12(e_3) -> R_6(e_3), ~R_6(e_3) we can infer ~P_12(e_3), which contradicts P_12(e_3). Therefore, the answer is no.