= Solution
Membership of either truth set in a <filter on a set> implies membership of its union, by upward closure. The converse need not hold. For the <cofinite filter>, take $p(x)$ to mean that $x$ is even and $q(x)$ that $x$ is odd. The union of their truth sets is all of $\mathbb N$, while neither truth set is cofinite.
Thus the left side of the printed equivalence is true and its right side false: \b[(ii) can be false]. An <ultrafilter> does satisfy the equivalence, because its dichotomy forces one member of a finite union into the ultrafilter. General filters need not have that dichotomy. This is one aspect of the <Boolean failure of the cofinite-filter quantifier>.
Back to article page
