As I understand it, the Sprague-Grundy theorem states that, under the operation of combining games in parallel, every impartial game can be reduced to a game of Nim. I see that the definition itself specifies this parallel combination operator explicitly: two impartial games G and G' are assigned the same nimber iff for all games H, G + H and G' + H have the same outcome (here, + denotes the parallel combination operator)---this definition is from [Wikipedia](https://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem#Equivalence).↵
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However, what about other methods of combining games? For instance, we combine two games A and B such that they are still played in parallel, but as soon as a player has no moves left in A, they instantly lose, regardless of the position of B. In general, can we still treat games A and B as their corresponding nimbers? If so, why? If not, what's a counterexample?↵
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EDIT (which I added in a comment below):↵
More precisely, my question is: ↵
Let $n(G)$ be the nimber associated with game G, according to the Sprague-Grundy theorem. Moreover, let $f(G, H)$ be a function of two games that outputs another game. Then, I would like to know: for all possible $f$, does $n(f(G, H)) = f(n(G), n(H))$?
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However, what about other methods of combining games? For instance, we combine two games A and B such that they are still played in parallel, but as soon as a player has no moves left in A, they instantly lose, regardless of the position of B. In general, can we still treat games A and B as their corresponding nimbers? If so, why? If not, what's a counterexample?↵
↵
EDIT (which I added in a comment below):↵
More precisely, my question is: ↵
Let $n(G)$ be the nimber associated with game G, according to the Sprague-Grundy theorem. Moreover, let $f(G, H)$ be a function of two games that outputs another game. Then, I would like to know: for all possible $f$, does $n(f(G, H)) = f(n(G), n(H))$?
