Translate the following inference to logic notation: For all x, x_6 will write a letter. For all x, x_6 is a janitor if and only if x_6 is bored. There is at least one x for which x_6 will make tea if and only if x_6 is bored. For all x, if x_6 were a janitor, then x_6 is not writing a letter. Therefore there is at least one x for which x_6 will not make tea.
forall x: P(x). forall x: Q(x) <-> R(x). exists x: S(x) <-> R(x). forall x: Q(x) -> ~P(x). Therefore exists x: ~S(x).