 Solution

Use the same two-index filter on a set $F=\{\{0,1\}\}$, with a unary predicate $P$ true only in the first one-element factor. In the reduced product, $P(a)$ is false, so $\neg P(a)$ is true. ButLogical negation also breaks the equivalence. For a general proper filter on a set, a set and its complement can both be absent. In an ultrafilter, exactly one belongs, which is the step used to transfer logical negation in the Łoś theorem.