Translate the following inference to logic notation: For all x, if x_6 writes a letter, then x_6 is sleeping. For all x, if x_6 were a soldier, then x_6 is writing a letter. There is at least one x for which x_6 is bored if and only if x_6 will sleep. Therefore there is at least one x for which if x_6 were a soldier, then x_6 is bored.
forall x: P(x) -> Q(x). forall x: R(x) -> P(x). exists x: S(x) <-> Q(x). Therefore exists x: R(x) -> S(x).