Translate the following inference to logic notation: There is at least one x for which if y sleeps, then y is an office clerk. For all x, y is an office clerk if and only if y is a mechanic. For all x, y is not a mechanic. For all x, if y were an office clerk, then y is sleeping. Therefore there is at least one x for which y will not sleep.
exists x: P(x) -> Q(x). forall x: Q(x) <-> R(x). forall x: ~R(x). forall x: Q(x) -> P(x). Therefore exists x: ~P(x).