Translate the following inference to logic notation: For all x, if x_8 sleeps, then x_8 is not running. There is at least one x for which x_8 will read a book if and only if x_8 will sleep. For all x, x_8 will go running. Therefore there is at least one x for which x_8 will not read a book.
forall x: P(x) -> ~Q(x). exists x: R(x) <-> P(x). forall x: Q(x). Therefore exists x: ~R(x).