Translate the following inference to logic notation: There is at least one x for which x_2 is not a lawyer. For all x, if x_2 were bored, then x_2 is a lawyer. For all x, if x_2 were an astronaut, then x_2 is bored. Therefore there is at least one x for which x_2 is not an astronaut.
exists x: ~P(x). forall x: Q(x) -> P(x). forall x: R(x) -> Q(x). Therefore exists x: ~R(x).