Translate the following inference to logic notation: For all x, x_6 is a carpenter. For all x, x_6 will drive a car if and only if x_6 will work. There is at least one x for which x_6 is an office clerk if and only if x_6 will drive a car. For all x, if x_6 works, then x_6 is not a carpenter. Therefore there is at least one x for which x_6 is not an office clerk.
forall x: P(x). forall x: Q(x) <-> R(x). exists x: S(x) <-> Q(x). forall x: R(x) -> ~P(x). Therefore exists x: ~S(x).