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