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