Translate the following inference to logic notation: For all x, if y_2 writes a letter, then y_2 is playing squash. For all x, y_2 will not write a letter. There is at least one x for which if y_2 plays squash, then y_2 is sleeping. For all x, if y_2 sleeps, then y_2 is writing a letter. Therefore there is at least one x for which y_2 will not play squash.
forall x: P(x) -> Q(x). forall x: ~P(x). exists x: Q(x) -> R(x). forall x: R(x) -> P(x). Therefore exists x: ~Q(x).