Translate the following inference to logic notation: For all x, if x_11 were a poet, then x_11 is not curious. There is at least one x for which x_11 will make tea if and only if x_11 is a cashier. For all x, x_11 is curious and x_11 will play tennis. For all x, x_11 is a poet if and only if x_11 will make tea. Therefore there is at least one x for which x_11 is not a cashier.
forall x: P(x) -> ~Q(x). exists x: R(x) <-> S(x). forall x: Q(x) and T(x). forall x: P(x) <-> R(x). Therefore exists x: ~S(x).