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