Translate the following inference to logic notation: There is at least one x for which y will sleep if and only if y will take a plane. For all x, y will take a plane if and only if y will sleep. Therefore there is at least one x for which if y sleeps, then y is taking a plane.
exists x: P(x) <-> Q(x). forall x: Q(x) <-> P(x). Therefore exists x: P(x) -> Q(x).