Translate the following inference to logic notation: For all x, x_2 is rich or x_2 is not bored. There is at least one x for which x_2 is bored or x_2 is not rich. Therefore there is at least one x for which if x_2 were rich, then x_2 is bored.
forall x: P(x) or ~Q(x). exists x: Q(x) or ~P(x). Therefore exists x: P(x) -> Q(x).