Translate the following inference to logic notation: There is at least one x for which x_10 is poor if and only if x_10 is curious. For all x, if x_10 were poor, then x_10 is making tea. For all x, if x_10 makes tea, then x_10 is fast. Therefore there is at least one x for which if x_10 were curious, then x_10 is fast.
exists x: P(x) <-> Q(x). forall x: P(x) -> R(x). forall x: R(x) -> S(x). Therefore exists x: Q(x) -> S(x).