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Solve the finite state graph by winning-set fixed-point iteration, treating positions outside both winning regions as draws. A naive recursive minimax search over histories may run forever around a cycle. Store each ownership-sensitive state once, with a player-to-move flag and all legal successors.

Let $T_A$ and $T_B$ be terminal states where the propositional formula is true and false, respectively. Initialize $W_A^0=T_A$. Repeatedly add nonterminal positions as follows:For $W_B$, start at $T_B$ and exchange the players in the existential and universal conditions. Terminal states keep their original labels and are excluded from these predecessor rules, avoiding a vacuous all-successors test at a terminal loss. Each process stabilizes after at most $|V|$ strict growth steps; predecessor counters implement both regions via the winning-position attractor in an infinite game construction in $O(|V|+|E|)$ graph time.

The admission round supplies a decreasing rank. At a state controlled by the winning player, choose a successor of smaller rank; at an opposing state, every successor already has smaller rank. Hence these strategies force a finite terminal win. The two winning regions are disjoint. Outside both, each player can keep the play outside the opponent's winning region, so neither can force a terminal win against best defence. Under the convention in (h), assign those remaining positions value $0$; the other regions have values $+1$ and $-1$.

This is a finite reachability game calculation, with memoryless strategies read from successor choices. Merely calling the first back-edge encountered a draw is unsound: a cycle can have an exit through which one player can force a win.