Translate the following inference to logic notation: For all x, if x reads a book, then x is rich. For all x, if x were rich, then x is reading a book. There is at least one x for which if x were rich, then x is reading a book. Therefore there is at least one x for which x will read a book if and only if x is rich.
forall x: P(x) -> Q(x). forall x: Q(x) -> P(x). exists x: Q(x) -> P(x). Therefore exists x: P(x) <-> Q(x).