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