Translate the following inference to logic notation: For all x, x_6 will not go running. There is at least one x for which if x_6 were a cashier, then x_6 is not an electrician. For all x, x_6 is not a politician or x_6 is an electrician. For all x, x_6 is a politician or x_6 will go running. Therefore there is at least one x for which x_6 is not a cashier.
forall x: ~P(x). exists x: Q(x) -> ~R(x). forall x: ~S(x) or R(x). forall x: S(x) or P(x). Therefore exists x: ~Q(x).