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