Red-Blue Hackenbush is a cold game Does the following proof make sense to prove that finite Red(Richard)-Blue(Louise) Hackenbush is a cold game?
If all edges touching the ground are exclusively red or blue, clearly one of the players has a winning strategy always (irrespective of whether they play first or second).
Else, assume Richard has a winning strategy as a first player. Now let Louise make the first move.
If Louise's move removes the edge that Richard wanted to remove as a first player(because of how the edges are connected), it is obvious that Richard has a winning strategy. Similarly, if Louise moves an edge that she couldn't have moved if Richard had moved first, again, Richard has a winning strategy by removing his first edge.
If Louise moves an edge that is not connected to R₁ then Richard assumes that R₁ is not present,  and plays by his strategy.
When Louise, for the first time, removes an edge connected to R₁, Richard removes R₁ and assumes that R₂(the edge he would've removed if he was playing first using his strategy) is not present and follows the strategy.
Continuing this way,

*

*If Louise removes an edge not connected to Rⱼ , Richard plays by his strategy assuming Rⱼ
is not there.

*If Louise removes an edge connected to Rⱼ, Richard removes Rⱼ and here on plays assuming Rⱼ₊₁
(the edge he would've removed if he was playing first using his strategy) is not present.

*If Louise's move removes Rⱼ, Richard can win (starting by removing Rⱼ₊₁)

*If Louise removes an edge that she couldn't have moved if Rⱼ wasn't present, again Richard has a winning strategy (removing Rⱼ next)

In this way, we show that if Richard can win as the first player, then he can win as the second player as well. Hence this Hackenbush game is R type.
A similar argument applies if Louise has a winning strategy as the first player.
 A: Your proof is constructive and clear. Below is an alternative proof.

Given a hackenbush position $G$, and an edge $e$, let $G[e]$ denote the result of moving on edge $e$. We need to show that for any blue edge $b$, $G[b]<G$, or that Richard wins moving first and second on $G[b]-G$. It is obvious that Richard wins moving first on $G[b]-G$ (by moving to $G[b]-G[b]$), so assume Louise goes first. There are two types of options available to Louise.

*

*She can remove another blue edge $b'$ from $G[b]$, resulting in $G[b,b']-G$. I claim that a winning response for red is to remove $b'$ from $-G$, resulting in $G[b,b']-G[b']$.

*

*If $b$ is present in $G[b']$, then this follows by induction, since $G[b,b']$ is a left option of $G[b']$.


*If $b$ is not present in $G[b']$, then $G[b,b']=G[b]$, so Richard has moved to the zero position, and therefore wins.




*Louise's other option is to $G[b]-G[r]$, where $r$ is a red edge.

*

*If removing $b$ kills $r$, then Richard wins with by moving to $G[b]-G[r,b]=G[b]-G[b]=0$. The same goes if removing $r$ kills $b$.


*If removing neither of $b$ or $r$ kills the other, then it follows that $G[b,r]$ is a left option of $G[b]$ and a right option of $G[r]$, so by induction, we have
$$
G[b]<G[b,r]<G[r]
$$
proving Richard wins on $G[b]-G[r]$.
