Translate the following inference to logic notation: For all x, x_6 is not curious. For all x, if x_6 works, then x_6 is playing tennis. For all x, if x_6 plays tennis, then x_6 is an electrician. There is at least one x for which if x_6 were an electrician, then x_6 is curious. Therefore there is at least one x for which x_6 will not work.
forall x: ~P(x). forall x: Q(x) -> R(x). forall x: R(x) -> S(x). exists x: S(x) -> P(x). Therefore exists x: ~Q(x).