Translate the following inference to logic notation: For all x, x_6 will work or x_6 will play squash. There is at least one x for which x_6 will make tea. For all x, if x_6 plays squash, then x_6 is a police officer. For all x, x_6 is a police officer or x_6 will not work. Therefore there is at least one x for which x_6 is a police officer and x_6 will make tea.
forall x: P(x) or Q(x). exists x: R(x). forall x: Q(x) -> S(x). forall x: S(x) or ~P(x). Therefore exists x: S(x) and R(x).