How do I show that is G has no right options, then there exists an n $\in$ $\Bbb Z$, where $G = n$

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I know that if $A+B+C=0 \to a+b+c=0$, and by extrapolating I can get $A+B=C \to a+b=c$, and I know that it will have to be an inductive proof, but I'm not sure how to phrase the proof. I'm assuming that the base case will be the $0+1=1$, but I'm not sure where I would go from there. I know that the whole proof will be true, as (for example) if you add another Left option to a Red-Green-Blue Hackenbush Game, where there are no Right Options, the value of the Game will increase by one. I know that the Inductive Step will be showing that $n + (1) = n+1$, so would my Inductive Hypothesis be:

For all games G, where there are no Right Options, there exists a $k \lt n$, and $k, n \in \Bbb Z$, where if $G = k$, and $H = 1$, then $G + 1 = k+1$.

Any help or hints in the right direction would be greatly appreciated, and sorry if the formatting isn't perfect.