Translate the following inference to logic notation: For all x, x_15 is a cashier. There is at least one x for which if x_15 reads a book, then x_15 is making tea. For all x, if x_15 reads a book, then x_15 is rich. For all x, if x_15 makes tea, then x_15 is not a cashier. For all x, x_15 will read a book if and only if x_15 is rich. Therefore there is at least one x for which x_15 is not rich.
forall x: P(x). exists x: Q(x) -> R(x). forall x: Q(x) -> S(x). forall x: R(x) -> ~P(x). forall x: Q(x) <-> S(x). Therefore exists x: ~S(x).