Translate the following inference to logic notation: There is at least one x for which z will sleep if and only if z is a janitor. For all x, if z reads a book, then z is sleeping. For all x, if z were a janitor, then z is working. Therefore there is at least one x for which if z reads a book, then z is working.
exists x: P(x) <-> Q(x). forall x: R(x) -> P(x). forall x: Q(x) -> S(x). Therefore exists x: R(x) -> S(x).