Translate the following inference to logic notation: There is at least one x for which if x_10 were an office clerk, then x_10 is sleeping. For all x, if x_10 drinks water, then x_10 is not poor. For all x, x_10 will sleep if and only if x_10 will drink water. For all x, x_10 is poor. Therefore there is at least one x for which x_10 is not an office clerk.
exists x: P(x) -> Q(x). forall x: R(x) -> ~S(x). forall x: Q(x) <-> R(x). forall x: S(x). Therefore exists x: ~P(x).