Translate the following inference to logic notation: For all x, if x plays squash, then x is making tea. For all x, x is a carpenter if and only if x will make tea. For all x, if x makes tea, then x is playing squash. There is at least one x for which x is not a carpenter. Therefore there is at least one x for which x will not play squash.
forall x: P(x) -> Q(x). forall x: R(x) <-> Q(x). forall x: Q(x) -> P(x). exists x: ~R(x). Therefore exists x: ~P(x).