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