Translate the following inference to logic notation: There is at least one x for which if x_16 writes a letter, then x_16 is not bored. For all x, if x_16 goes running, then x_16 is bored. For all x, x_16 will work. For all x, x_16 will go running if and only if x_16 is happy. For all x, if x_16 works, then x_16 is happy. Therefore there is at least one x for which x_16 will not write a letter.
exists x: P(x) -> ~Q(x). forall x: R(x) -> Q(x). forall x: S(x). forall x: R(x) <-> T(x). forall x: S(x) -> T(x). Therefore exists x: ~P(x).