Translate the following inference to logic notation: For all x, if x_12 were rich, then x_12 is making tea. For all x, if x_12 makes tea, then x_12 is bored. For all x, if x_12 were bored, then x_12 is not writing a letter. There is at least one x for which x_12 will write a letter. Therefore there is at least one x for which x_12 is not rich.
forall x: P(x) -> Q(x). forall x: Q(x) -> R(x). forall x: R(x) -> ~S(x). exists x: S(x). Therefore exists x: ~P(x).