Translate the following inference to logic notation: For all x, x_9 will take a plane if and only if x_9 will sleep. For all x, x_9 will not go running. For all x, x_9 will play squash if and only if x_9 will take a plane. There is at least one x for which if x_9 sleeps, then x_9 is running. Therefore there is at least one x for which x_9 will not play squash.
forall x: P(x) <-> Q(x). forall x: ~R(x). forall x: S(x) <-> P(x). exists x: Q(x) -> R(x). Therefore exists x: ~S(x).