Translate the following inference to logic notation: For all x, if y plays squash, then y is making tea. For all x, if y works, then y is making tea. There is at least one x for which y will sleep or y will not play squash. For all x, if y makes tea, then y is playing squash. Therefore there is at least one x for which if y works, then y is sleeping.
forall x: P(x) -> Q(x). forall x: R(x) -> Q(x). exists x: S(x) or ~P(x). forall x: Q(x) -> P(x). Therefore exists x: R(x) -> S(x).