Translate the following inference to logic notation: For all x, if x were an office clerk, then x is an electrician. For all x, if x were excited, then x is a mechanic. For all x, x is an office clerk or x is not a sailor. For all x, if x were an electrician, then x is excited. There is at least one x for which if x were a mechanic, then x is a sailor. Therefore there is at least one x for which x is an office clerk if and only if x is excited.
forall x: P(x) -> Q(x). forall x: R(x) -> S(x). forall x: P(x) or ~T(x). forall x: Q(x) -> R(x). exists x: S(x) -> T(x). Therefore exists x: P(x) <-> R(x).