Translate the following inference to logic notation: For all x, x_5 is not excited. For all x, if x_5 were a cashier, then x_5 is excited. For all x, if x_5 were curious, then x_5 is a cashier. There is at least one x for which if x_5 sleeps, then x_5 is curious. Therefore there is at least one x for which x_5 will not sleep.
forall x: ~P(x). forall x: Q(x) -> P(x). forall x: R(x) -> Q(x). exists x: S(x) -> R(x). Therefore exists x: ~S(x).