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The even and odd integers are both absent from the cofinite filter, but their union is present. Thus its filter quantifier need not turn a disjunction into the disjunction of quantified assertions, and failure of a quantified assertion need not imply truth of its quantified negation. An ultrafilter restores both laws by deciding each set against its complement.

The filter quantifier satisfies ${\forall }_{\mathcal{F}}(p\wedge q)$ if and only if both ${\forall }_{\mathcal{F}}p$ and ${\forall }_{\mathcal{F}}q$. Finite-intersection closure gives one direction and upward closure gives the other. Properness ensures contradictory truth sets cannot both be filter members.

If the complement of a truth set belongs to a proper filter on a set, the truth set itself cannot belong: their intersection is empty. Hence the right side always implies the left side.

The reverse implication can fail. In the cofinite filter, the set of even integers is absent, but its odd complement is absent as well. The filter-quantified evenness assertion is false, while the filter-quantified assertion of oddness is also false. Therefore (iii) can be false.

For an ultrafilter the equivalence is true, precisely because it contains exactly one of any set and its complement. The filter quantifier preserves conjunction for arbitrary proper filters, but its full classical Boolean behaviour requires the ultrafilter property.