Translate the following inference to logic notation: There is at least one x for which if z writes a letter, then z is running. For all x, if z makes tea, then z is writing a letter. For all x, if z writes a letter, then z is making tea. Therefore there is at least one x for which if z makes tea, then z is running.
exists x: P(x) -> Q(x). forall x: R(x) -> P(x). forall x: P(x) -> R(x). Therefore exists x: R(x) -> Q(x).