Translate the following inference to logic notation: For all x, if x_10 makes tea, then x_10 is sleeping. For all x, x_10 is not excited. There is at least one x for which if x_10 sleeps, then x_10 is making tea. For all x, if x_10 takes a plane, then x_10 is excited. For all x, x_10 will sleep if and only if x_10 will take a plane. Therefore there is at least one x for which x_10 will not make tea.
forall x: P(x) -> Q(x). forall x: ~R(x). exists x: Q(x) -> P(x). forall x: S(x) -> R(x). forall x: Q(x) <-> S(x). Therefore exists x: ~P(x).