Translate the following inference to logic notation: For all x, if x_3 plays squash, then x_3 is a poet. There is at least one x for which x_3 is not a poet. For all x, x_3 will play squash. Therefore there is at least one x for which x_3 is bored.
forall x: P(x) -> Q(x). exists x: ~Q(x). forall x: P(x). Therefore exists x: R(x).