Translate the following inference to logic notation: For all x, if x_5 were an astronaut, then x_5 is playing squash. There is at least one x for which if x_5 plays squash, then x_5 is an astronaut. For all x, if x_5 plays squash, then x_5 is running. For all x, x_5 will not go running. Therefore there is at least one x for which x_5 is not an astronaut.
forall x: P(x) -> Q(x). exists x: Q(x) -> P(x). forall x: Q(x) -> R(x). forall x: ~R(x). Therefore exists x: ~P(x).