Let I have 3-QCNF formula.Classic recursive algorithm for TQBF takes exponential time. However, following divide & conquer modification allows to do it in quadratic time (not sure about correctness, and that's the question).ctr = 0for i = 0..n    Step 1. Take i'th clause, find E-quantified variable inside it.     If can't, return 'NO'. Otherwise, if it's the clause containing 3     variables, choose that variable v.    Step 2. Find all 3-clauses with variable v. Move them from formula     to formula f_i. Reduce n (length of formula ) by number of moved     clauses.    Step 3. Copy f_i to f'_i.    Step 4. Put v = 1 in f_i. Get 2-QCNF.    Step 5. Put v = 0 in f'_i. Get 2-QCNF.    Step 6. If f_i and f'_i have same clauses, move them to .    Increase n by number of moved clauses.    Step 7. ctr = ctr + 1for i = 0..ctr    Step 1. Solve 2-QCNF '^f_i. //' is changed .    Step 2. Solve 2-QCNF '^f'_i.    Step 3. If both are false, return 'NO'.return 'YES'Notes:Algorithm represents 3-QCNF as $\mathsf{\Phi = TAUT-CNF \land \Phi'(2-QCNF) \land (f_i(2-QCNF) \lor f'_i(2-QCNF))\land\ ..}$Trivially, formula is false if TAUT-CNF contains clauses.So, if TAUT-CNF has no clauses, formula is represented this way:$\mathsf{\Phi' \land (f_i \lor f'_i)\land\ ...\ =((\Phi'\land f_i)\lor(\Phi'\land f'_i))\land\ ...}$The question: have I violated any logical rule in my algorithm?
Correctness of following algorithm for 3-QCNF?
algorithms;satisfiability