The cursed chocolate (Chomp)

Pure logicLevel 5 · Expert · ●●●●●

There is a rectangular chocolate bar of $m\times n$ squares, with $m,n\ge2$. The box in the lower left corner is poisoned.

Two players alternate turns. Each turn, a player chooses a square and eats that square along with all the squares above and to its right.

Whoever is forced to eat the poisoned square loses. With a perfect game, who wins: first or second?

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Hints

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The key is monotonic: having chocolate already removed never harms the player who moves; It only reduces the opponent's options.
Test for strategy theft: Suppose the second player had a winning strategy.
The first player has a winning strategy for all m,nge2.

Solution

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Answer: the first player has a winning strategy for all $m,n\ge2$.
Test for strategy theft:

Suppose the second player had a winning strategy.
The first one makes an “extra” initial move: he eats only the upper right corner.
From there, the first tries to copy the supposed winning strategy of the second.

If at any time that strategy calls for an already occupied square, the first player makes any legal move.
The key is monotonic: having chocolate already removed never harms the player who moves; It only reduces the opponent's options.
Then the former could turn the latter's winning strategy into a winning one of its own. Contradiction.
Then the second cannot have a winning strategy, and in this game without ties that implies that the first wins.

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