Clarification in Siegel, combinatorial game theory

Question

On page $65$ of Combinatorial Game Theory by Siegel, under the section of Dominated and Reversible Options, there is this part which I do not understand:

Consider $G^{L_1R_1L}-G$. By assumption $G^{L_1R_1}\leq G$. So we have $G^{L_1R_1L}\triangleleft G^{L_1R_1} \leq G$, where $\triangleleft$ denotes being less than or confused with. This shows that Right must have a winning move on $G^{L_1R_1L}-G$.

I don't get why Right must have a winning move. Sure, if $G^{L_1R_1L}\leq G^{L_1R_1} \leq G$ then by transitivity $G^{L_1R_1L}-G\leq 0$ and thus Right has a winning move. But what if $G^{L_1R_1L}$ is confused with $G^{L_1R_1}$ $?$

Answer 1

The result you want is: If $A \triangleleft B \le C$, then $A \triangleleft C$ (i.e. Right has a winning move on $A-C$).

This follows from: If $X \triangleleft 0$ and $Y \le 0$ then $X+Y \triangleleft 0$. (Take $X = A-B$ and $Y = B-C$).

To prove this last statement, note that $X \triangleleft 0$ means that Right has a winning move on $X$. Hence for some $X^R$, we have $X^R \le 0$. Since $Y \le 0$ also, we have $X^R + Y \le 0$. Thus $X^R + Y$ is a winning move for Right in $X+Y$. Hence $X + Y \triangleleft 0$.