Suppose $a$, $b$, and $c$ are dyadic rationals, where their corresponding numbers (games) are $A$, $B$, and $C$ respectively. Prove that $a+b=c$ if and only if $A+B\sim C$ (or $A+B$ is equivalent to $C$). Moreover, prove that $a \geq b$ if and only if $A \geq B$.
Note that for a dyadic rational $g=\frac{x}{2^p}$ for odd $x$, then its corresponding number would be $G=\{\frac{x-1}{2^p}|\frac{x+1}{2^p}\}$. Two games $G$ and $H$ are equivalent (or $G \sim H$ if $G - H$ is a P-position).
I was trying to do this by induction by assuming that the results are all true for simples dyadic rational games than $a + b$. Then, I said that since $a=\frac{n}{2^i}$ and $b=\frac{m}{2^j}$, if $i>j$, I can easily prove the results by looking at the options for $a+b$. However, I am stuck when $i=j$. Is there any better option to do this? I assumed that we would prove the second statement in a similar way, but I am not sure.