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