Translate the following inference to logic notation: There is at least one x for which if x_1 were a carpenter, then x_1 is sleeping. For all x, if x_1 sleeps, then x_1 is a carpenter. For all x, x_1 is a carpenter or x_1 will not sleep. Therefore there is at least one x for which x_1 is a carpenter if and only if x_1 will sleep.
exists x: P(x) -> Q(x). forall x: Q(x) -> P(x). forall x: P(x) or ~Q(x). Therefore exists x: P(x) <-> Q(x).