Translate the following inference to logic notation: For all x, if x_10 were an office clerk, then y is bored. For all x, if y were bored, then x_10 is an office clerk. For all x, if x_10 reads a book, then y is bored. For all x, x_10 is an office clerk or x_10 will read a book. Therefore there is at least one x for which y is bored.
forall x: P(x) -> Q(x). forall x: Q(x) -> P(x). forall x: R(x) -> Q(x). forall x: P(x) or R(x). Therefore exists x: Q(x).