= Solution
The <axiom of determinacy>, $\mathsf{AD}$, says that every <infinite game of perfect information> in which two players alternately choose <natural numbers> and Player I wins according to a specified payoff $A\subseteq\omega^\omega$ is a <determined infinite game>. A <strategy> specifies a move at every finite position where its player moves; it is winning if every compatible infinite play is won by that player.
Finite games are determined by <backward induction>. For infinite games, <open determinacy> has a direct proof. Let $T$ consist of the finite positions whose extension <cylinder sets> lie in the open payoff $A$. Build the <winning-position attractor in an infinite game> by <transfinite recursion>: start with $T$, add an I-position if some successor has already entered, add a II-position if every successor has already entered, and take unions at limit stages. The process stabilizes because there are only countably many finite positions.
At a position in the attractor but outside $T$, I can choose a successor with strictly smaller entry rank, and every move of II has smaller entry rank. There is no infinite descending sequence of <ordinals>, so I forces entry into $T$ and wins. Outside the attractor, every I-move stays outside it, while II can choose a successor outside it; that strategy avoids $T$ forever. Since $A$ is open, a play belongs to $A$ exactly when some prefix belongs to $T$. Thus II wins from outside the attractor. This proves determination for open payoffs; swapping the players proves it for closed payoffs too. The stronger <Borel determinacy theorem> gives determination for every <Borel set> payoff, whereas $\mathsf{AD}$ asserts this for arbitrary payoffs.
To contradict $\mathsf{AD}$ from <choice>, enumerate both players' strategies as $\langle\sigma_\alpha:\alpha<\mathfrak c\rangle$ and $\langle\tau_\alpha:\alpha<\mathfrak c\rangle$, where $\mathfrak c=2^{\aleph_0}$. There are $\mathfrak c$ strategies because the set of finite positions is countable; there are also $\mathfrak c$ plays following any fixed strategy because the other player's moves are arbitrary. At stage $\alpha$ of the <choice diagonalization of an undetermined game>, choose a fresh play $x_\alpha$ following $\sigma_\alpha$ and then a different fresh play $y_\alpha$ following $\tau_\alpha$, avoiding all earlier reservations. This is possible because fewer than $\mathfrak c$ plays have been reserved at every $\alpha<\mathfrak c$; regularity of $\mathfrak c$ is not required.
Set $A=\{y_\alpha:\alpha<\mathfrak c\}$. Every $\sigma_\alpha$ loses on $x_\alpha\notin A$, and every $\tau_\alpha$ loses on $y_\alpha\in A$. Thus neither player has a <winning strategy>, and \b[AD and AC are inconsistent together.]
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