Translate the following inference to logic notation: There is at least one x for which if x_4 were curious, then x_4 is running. For all x, x_4 is curious if and only if x_4 will go running. Therefore there is at least one x for which x_4 will go running if and only if x_4 is curious.
exists x: P(x) -> Q(x). forall x: P(x) <-> Q(x). Therefore exists x: Q(x) <-> P(x).