Translate the following inference to logic notation: There is at least one x for which x_13 is a mechanic if and only if x_13 is bored. For all x, x_13 is a janitor if and only if x_13 will take a plane. For all x, x_13 is not a mechanic. For all x, x_13 is bored if and only if x_13 is a janitor. Therefore there is at least one x for which x_13 will not take a plane.
exists x: P(x) <-> Q(x). forall x: R(x) <-> S(x). forall x: ~P(x). forall x: Q(x) <-> R(x). Therefore exists x: ~S(x).