Why is $\uparrow+\ast=\{\ast,0|0\}$? On p. 189 of Conway's ONAG, he describes a position in Hackenbush Hotchpotch as $\{\ast,0\:|\:0\}$ and then says that $\{\ast,0\:|\:0\}=\:\uparrow+\:\ast$. (where $\ast\equiv\{0\:|\:0\}$ and $\uparrow\:\equiv\{0\:|\:\ast\}$) Helpfully, he then says "The equation was not supposed to be obvious—it can be verified by computing the simplest form of $\uparrow+\:\ast$."
I'm running into difficulties with that. Here's my step-by-step work so far:
$$
\uparrow+\ast\\
\{0+\ast,\uparrow+\:0\:|\ast+\:\ast,\uparrow+\:0\}\\
\{\ast,\uparrow|\:\{\ast\:|\:\ast\},\uparrow\}
$$
Much earlier, Conway has shown that $\{\ast\:|\:\ast\}=0$, because it's a position in which the first player to move loses, so we have:
$$
\{\ast,\uparrow|\:0,\uparrow\}
$$
Then I get shaky. I can see why we can take $\uparrow$ in $X^L$ as $0$, because L, as the first to move, will always win that exchange (all subsequent moves from either R or L leave L with $0$). I guess we can remove $\uparrow$ from $X^R$ because it's a dumb move for R since L always wins $\uparrow$? That would indeed leave us with $\{\ast,0\:|\:0\}$, but why can't we remove $\ast$ from $X^L$ in the same way we removed $\uparrow$ from $X^R$? It's a dumb move for L since it always leaves an R win.
In other words, if we follow the only logic I see for removing $\uparrow$ from $X^R$, we should also remove $\ast$ from $X^L$, leaving us with:
$$
\{0\:|\:0\}\equiv\ast
$$
which would further imply that:
$$
\uparrow+\:\ast\equiv\ast
$$
which I know (I think) is false. Where's the breakdown in my logic?
 A: You can't remove a position just because it leaves a win for the opponent.  The main idea behind combinatorial game theory (not emphasized in most expositions on the subject, but it is implicit in how the definitions are set up) is that the value of a game $G$ tells us not only how to play $G$, but also how to play $G + X$ for any game $X$. We can only remove a move from $G$ without changing its value if that move would not be a good move in $G + X$ for any game $X$.  
So how do we simplify a game?  In general, there are two ways of simplifying games: Deleting dominated positions, reversing reversible positions.  This is discussed in ONAG, Chapter 10, Theorem 68.
In your example for $\uparrow + \mbox{ }\ast$, everything is correct up to this form:
$$G = \{\ast,\uparrow|\:0,\uparrow\}$$
On the right side of $G$, we can indeed remove $\uparrow$, but the reason isn't because $\uparrow$ is a win for Left.  The reason is because $0 < \,\,\uparrow$, so 0 dominates $\uparrow$ for Right (we say that $\uparrow$ is a dominated option).  In any context, $G + X$, Right should never move to $\uparrow$ if 0 is available as an alternative, because $0 + X$ will always be better than $\uparrow + \,X$ for Right (in the sense that if Right wins the latter then Right also wins the former).
On the left side of $G$, the option $\uparrow$ is a reversible option.  I'll let you look up the details in ONAG (or Winning Ways), but the idea is as follows.  If Left moves to $\uparrow$, Right can follow with a move to $\ast$.  But $\ast < \, G$ (because $G =\,\uparrow +\,\ast$ and $\uparrow\, > 0$), which is saying that if Left moves to $\uparrow$, Right can immediately follow with a move to $\ast$ that leaves Left in a worse position than the original $G$.  This means that there is no reason for Left to move to $\uparrow$ unless he plans to continue following through after Right's response to $\ast$ (as opposed to playing in some other component $X$; remember that we are studying $G+X$ for any game $X$). But after Right's response to $\ast$, Left's only move is to 0.  
The upshot of the previous paragraph is that we can replace Left's option $\uparrow$ with a Left option directly to 0.
In summary, in $G$, we have deleted $\uparrow$ from the right side, and replaced $\uparrow$ with 0 on the left side, giving the simplified form $$\{\ast, 0 | 0 \}$$ as claimed by Conway.
