Translate the following inference to logic notation: There is at least one x for which if x_15 sleeps, then x_15 is running. For all x, x_15 is bored if and only if x_15 will go running. For all x, x_15 is a janitor. For all x, if x_15 were a janitor, then x_15 is sleeping. Therefore there is at least one x for which x_15 is bored.
exists x: P(x) -> Q(x). forall x: R(x) <-> Q(x). forall x: S(x). forall x: S(x) -> P(x). Therefore exists x: R(x).