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