 Observe that while \(\forall X(Xx) \rightarrow Px\) is an instance of (\(\forall^2 1\)), the formula \(\forall X(Xx) \rightarrow (Px \wedge Qx)\) is not an instance of the axiom. But whatever one takes the value of a second-order variable to be, you may think that there should be one under which an object falls if and only if it satisfies the condition stated by the formula \((Px \wedge Qx)\). To make sure, we may rely on an axiom of second-order comprehension: