= 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. But
$$
\{i:\mathcal M_i\models\neg P(a_i)\}=\{1\}\notin F.
$$
\b[<Logical 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>.
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