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