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