Translate the following inference to logic notation: For all x, x_7 is not a doctor. For all x, x_7 is bored if and only if x_7 is happy. For all x, x_7 is a doctor if and only if x_7 is happy. There is at least one x for which x_7 will work if and only if x_7 is bored. Therefore there is at least one x for which x_7 will not work.
forall x: ~P(x). forall x: Q(x) <-> R(x). forall x: P(x) <-> R(x). exists x: S(x) <-> Q(x). Therefore exists x: ~S(x).