If the game ends in some finite N steps: suppose player 1 has no never-lose strategy, then for every strategy i that player 1 might choose, player 2 has a strategy 2(i) that makes her win IF PLAYER 1 PLAYS i. A never-lose strategy for player 2 is then to play 2(i) if 1 plays i. (or is this possible? for if i1 and i2 are 2 strategies of 1 with the first n steps the same, then 2(i1) and 2(i2) must also be such that their first n steps are the same, since one cannot see moves that have not yet come! )

(proof for the existence of the set consisting of 2(i)'s would utilize "axiom of choice"?)