 Solution

An exponential increase is unavoidable for an equivalent conjunctive normal form over the original variables. Consider the linear-size Boolean formulaDistributing gives the $2^n$ Boolean clausesTo prove that the growth is necessary, fix one choice $r_i\in \{p_i,q_i\}$. Let ${\alpha }_r$ make every $r_i$ false and its partner true. Each pair then contains a false variable, so $F_n$ is false. Any equivalent conjunctive normal form must contain a Boolean clause $C$ falsified by ${\alpha }_r$ and entailed by $F_n$.

Every individual term $p_i\wedge q_i$ entails $C$, because that term implies $F_n$. After tautological Boolean clauses are discarded, this forces $C$ to contain a positive $p_i$ or positive $q_i$ for every $i$: otherwise set both variables in that pair true and independently falsify all Boolean literals of $C$, contradicting the entailment. Since $C$ is false at ${\alpha }_r$, it must contain the false member $r_i$, never its true partner. Consequently the same $C$ cannot be false at any different ${\alpha }_s$. The $2^n$ different assignments therefore require at least $2^n$ different Boolean clauses.

Writing the answer already takes $\boxed{\Omega (2^n)\text{ time in the worst case}}$. This obstruction applies to first-order logic as well, since the example can be expressed using ground unary atomic formulas. It concerns exact equivalence without auxiliary symbols. A Tseytin transformation can produce a polynomial-size equisatisfiable conjunctive normal form with auxiliary variables, so the statement must not be extended to that different task.