Translate the following inference to logic notation: For all x, z is bored if and only if z will drink water. For all x, if z drinks water, then z is a sailor. For all x, if z were a sailor, then z is drinking water. For all x, z is a cashier if and only if z is a sailor. There is at least one x for which z is not a cashier. Therefore there is at least one x for which z is not bored.
forall x: P(x) <-> Q(x). forall x: Q(x) -> R(x). forall x: R(x) -> Q(x). forall x: S(x) <-> R(x). exists x: ~S(x). Therefore exists x: ~P(x).