Combining Nim Games I was reading the Wikipedia article on the Sprague-Grundy theorem, and I can't get through their example for combining games.
The example combines two sets of three heaps. The first set has 1, 2 and 2 elements in its heaps, and is described by starting position $\color{blue}S$, and the second set has 1, 1 and 1 elements in its heaps, and is described by starting position $\color{red}{S'}$:
$$ \color{blue}{S = \left\{ 
    \left\{ *1, \left\{ *1 \right\}, *2 \right\},
    \left\{ *2, \left\{ *1, \left\{ *1 \right\}, *2 \right\} \right\},
    \left\{ \left\{ *1 \right\}, \left\{ \left\{ *1 \right\} \right\}, \left\{ *1, \left\{ *1 \right\}, *2 \right\} \right\}
\right\}} $$
$$ \color{red}{S' = \left\{ 
    \left\{ *1 \right\}
\right\}} $$
Then the combined game's starting position is said to be:
$$ \color{blue}S + \color{red}{S'} =
    \left\{
        \left\{
            \color{blue}S,
            \color{red}{\left\{ *1 \right\}}
        \right\}
    \right\}
    \cup
    \left\{
        \left\{
            \color{red}{S'},
            \color{blue}{\left\{ *1, \left\{ *1 \right\}, *2 \right\}}
        \right\},
        \left\{
            \color{red}{S'},
            \color{blue}{\left\{ *2, \left\{ *1, \left\{ *1 \right\}, *2 \right\} \right\}}
        \right\},
        \left\{
            \color{red}{S'},
            \color{blue}{\left\{ \left\{ *1 \right\}, \left\{ \left\{ *1 \right\} \right\}, \left\{ *1, \left\{ *1 \right\}, *2 \right\} \right\}}
        \right\}
    \right\}
$$
I can't make sense of the elements of $\color{blue}{S} + \color{red}{S'}$. For example, if I expressed it as the numbers of objects in every heap, what position/positions does $\left\{ \color{blue}S, \color{red}{\left\{ *1 \right\}} \right\} \in \color{blue}{S} + \color{red}{S'}$ represent? Does this addition of sets represent positions that I could express like this at all? It doesn't seem right to me that, if $\color{blue}{S} + \color{red}{S'}$ is the starting position, I could pick $\left\{ \color{blue}S, \color{red}{\left\{ *1 \right\}} \right\}$ as my next position, then $\color{red}{\left\{ *1 \right\}}$, then $\color{red}{*1}$ and finally end the 6-heap game in four moves, that should be impossible.
 A: Since Mark S. seems like he won't post an answer himself, I'll answer my own question with his correction.
The example is incorrect.
Something that is correctly stated in the article is the definition of the operation $+$ used for combining games. We'll say $S$ represents game one and $S'$ represents game two. This operation is recursively defined as:
$$ S + S' = \left\{ S + s' \mid s' \in S' \right\} \cup \left\{ s + S' \mid s \in S \right\}$$
Each element of $S + s'$ of $\left\{ S + s' \mid s' \in S' \right\}$ represents having picked $s'$ as the next position in game two, and thus combines this new position of game two with an untouched game one. Therefore, $\left\{ S + s' \mid s' \in S' \right\}$ will be the set of positions that can be reached by making a move in game two. The case for $\left\{ S' + s \mid s \in S \right\}$ is analogous.
The recursive expansion reaches its end with $\left\{\right\} + \left\{\right\} = \left\{\right\}$.
Thus, the correct expression for $\color{blue}S + \color{red}{S'}$ is:
$$\color{blue}S + \color{red}{S'} =
    \left\{
            \color{blue}S
            + \color{red}{\left\{ *1 \right\}}
    \right\}
    \cup
    \left\{
            \color{red}{S'}
            + \color{blue}{\left\{ *1, \left\{ *1 \right\}, *2 \right\}},
            \color{red}{S'}
            + \color{blue}{\left\{ *2, \left\{ *1, \left\{ *1 \right\}, *2 \right\} \right\}},
            \color{red}{S'}
            + \color{blue}{\left\{ \left\{ *1 \right\}, \left\{ \left\{ *1 \right\} \right\}, \left\{ *1, \left\{ *1 \right\}, *2 \right\} \right\}}
    \right\} $$
