= Solution
Let $S=\{F_i:i\ge1\}$. Every finite <subset> of $S$ is satisfiable: if $F_N$ has the largest index occurring, the successive <semantic entailments> imply $F_N\models F_i$ for every earlier index, so any <Boolean valuation> satisfying $F_N$ satisfies that finite <subset>. The <propositional compactness theorem> therefore also shows that $S$ itself is satisfiable.
Suppose that $G$ had exactly the common models of $S$. Then $S\models G$, and hence $S\cup\{\neg G\}$ is unsatisfiable. By the <propositional compactness theorem>, finitely many $F_i$ together with $\neg G$ are already unsatisfiable. Enlarge the finite list if necessary so it is nonempty, and let $N$ be its greatest index. We obtain
$$
F_N\models G.
$$
On the other hand, the asserted equality of model classes gives $G\models F_{N+1}$. Transitivity of <semantic entailment> would yield
$$
F_N\models F_{N+1},
$$
contradicting the strictness assumption at index $N$. Thus
$$
\boxed{\text{No single propositional formula defines the common models of all }F_i.}
$$
The obstruction is the finite character of <propositional formulas>: a single consequence of an infinite set already follows from finitely many premises, whereas this chain continues to impose strictly stronger requirements forever.
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