Translate the following inference to logic notation: For all x, if y goes running, then y is curious. There is at least one x for which if y sleeps, then y is running. For all x, if y were curious, then y is sleeping. Therefore there is at least one x for which y is curious if and only if y will sleep.
forall x: P(x) -> Q(x). exists x: R(x) -> P(x). forall x: Q(x) -> R(x). Therefore exists x: Q(x) <-> R(x).