Translate the following inference to logic notation: For all x, if y_3 were an electrician, then y_3 is working. There is at least one x for which y_3 will not work. For all x, if y_3 works, then y_3 is an electrician. For all x, if y_3 were an electrician, then y_3 is working. Therefore there is at least one x for which y_3 is not an electrician.
forall x: P(x) -> Q(x). exists x: ~Q(x). forall x: Q(x) -> P(x). forall x: P(x) -> Q(x). Therefore exists x: ~P(x).