Translate the following inference to logic notation: For all x, if x_6 makes tea, then x_6 is rich. For all x, if x_6 were rich, then x_6 is a janitor. For all x, if x_6 were a janitor, then x_6 is rich. For all x, x_6 is not a janitor. For all x, if x_6 were rich, then x_6 is making tea. Therefore for all x, x_6 will not make tea.
forall x: P(x) -> Q(x). forall x: Q(x) -> R(x). forall x: R(x) -> Q(x). forall x: ~R(x). forall x: Q(x) -> P(x). Therefore forall x: ~P(x).