Translate the following inference to logic notation: For all x, if x_10 takes a plane, then x_10 is an office clerk. There is at least one x for which x_10 is not fast. For all x, x_10 will take a plane or x_10 will make tea. For all x, if x_10 were an office clerk, then x_10 is fast. Therefore there is at least one x for which x_10 will make tea.
forall x: P(x) -> Q(x). exists x: ~R(x). forall x: P(x) or S(x). forall x: Q(x) -> R(x). Therefore exists x: S(x).