Translate the following inference to logic notation: For all x, x_11 is a carpenter or x_11 will go running. There is at least one x for which if x_11 were a janitor, then x_11 is bored. For all x, if x_11 goes running, then x_11 is bored. For all x, x_11 is not a carpenter or x_11 is a janitor. Therefore there is at least one x for which x_11 is bored or x_11 will work.
forall x: P(x) or Q(x). exists x: R(x) -> S(x). forall x: Q(x) -> S(x). forall x: ~P(x) or R(x). Therefore exists x: S(x) or T(x).