Translate the following inference to logic notation: For all x, if x_4 goes running, then x_4 is a mechanic. There is at least one x for which x_4 is not a police officer or x_4 is rich. For all x, if x_4 plays squash, then x_4 is a mechanic. For all x, x_4 is a police officer or x_4 will play squash. For all x, if x_4 were rich, then x_4 is running. Therefore there is at least one x for which x_4 is a mechanic.
forall x: P(x) -> Q(x). exists x: ~R(x) or S(x). forall x: T(x) -> Q(x). forall x: R(x) or T(x). forall x: S(x) -> P(x). Therefore exists x: Q(x).