CGT: value of sum game is sum of values of games I am involved in a little study about combinatorial game theory. The study makes heavy use of the fact that, at least in a simple combinatorial game called domineering, the value of the sum game is equal to the sum of the values of the individual subgames. I am looking for a source which would state this result as a theorem and also prove the theorem.
The theory for the study is from Lessons in play by Albert et al. While the book also makes heavy use the result mentioned above, it does not really provide a direct theorem with a corresponding proof. It also seems that there is no clear theorem with proof in Winning ways by Berlekamp et al. Does somebody know a source with a clearly stated theorem and an accompanying proof?
All the best,
Jarmo
 A: As best as I can tell, the reason you may be having trouble finding this is that, in some sense, there are lots of different theorems, depending on what you want to know about. For example, "positive integer positions add like integers do" may as well be one discrete theorem, a special case of the more general "dyadic rational positions add like rationals do". Then there are tons of other facts relevant to adding games, like "$*+*=0$".
I don't have Lessons in Play in front of me, so I don't know when certain facts/notations are introduced, but maybe I can sketch a proof that at least the standard representatives of the integers add properly. I'll start with the nonnegatives, and then add in the negatives.

A standard nonnegative integer is $\{|\}$ (denoted "$0$") or $\{n|\}$ where $n$ is a standard nonnegative integer. $\{n|\}$ is most often denoted "$n+1$", but that would be assuming some of what we're trying to prove here, so I'll call it $S(n)$ for "successor of $n$" (taking a cue from the Peano axioms). Firstly, $0+n=n$ because $0+G=G$ for any game $G$. The only other thing I need to prove (assuming basic facts like commutativity of game addition, transitivity of =, etc.) is simply that $n+S(m)=S(n+m)$. To prove games are equal, we need to prove that one game plus the negative of the other one is a $\mathcal P$-position. Here a minus sign only means the negative game; I'm not saying anything about negative integers:$$n+S(m)+\left(-S(n+m)\right)=n+\{m|\}+\{|-(n+m)\}$$
Hereditarily (all the way down), Left can only move in the first two components, and Right can only move in the third. Left has exactly $n+m$ moves, and so does Right, so after $2(n+m)$ moves, the first player will be stuck with no move. Therefore, this really is a $\mathcal P$ position.
Standard negative integers are like $\{|x\}$ for $x$ a non-positive integer, but this is exactly the negative game of the absolute value of the integer $x$. So asking a question like "Does the game $5$ plus the game $-6$ equal the game $-1$?" simplifies to "does the game $5$ plus the negative of the game $6$ equal the negative of the game $1$?". Firstly, if we have two negative integers, then $(-n)+(-m)=-(n+m)$ is true because it's true for games in general. And if you add $0$ to $-n$, you get $-n$. So all we have to worry about is adding $n$ to $-m$ ($n$,$m$ positive). 
If $n=m$, then we get $0$ by the definition of negative game. Otherwise, either $n>m$ or $n<m$. If $n>m$, then we want to know if $n+(-m)=(n-m)$ (the first minus sign is game negative, and the second is subtraction of the integers). Well, $n+(-m)+(-(n-m))$ is a game in which Left has n moves, and Right has $m+(n-m)=n$ independent moves, so it's a $\mathcal P$-position. Similarly, if $n<m$, then $n+(-m)+(-(n-m))$ is a game in which Left has $n+(m-n)=m$ moves and Right has $m$ independent moves, so it's also a $\mathcal P$-position.
I think I got all the cases, so standard integer games add like regular integers do. $\square$ 
