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Consider a game $G$. We have to prove that is $s$ is a Nash Equilibrium of $G$, then it is also a Nash Equilibrium of the game formed by removing strictly dominated strategies of $G$.

I looked at the proof for this in https://homepages.cwi.nl/~apt/stra/ch3.pdf (the proof is described in first two pages of it) They use a restriction $R$ ($R_{i}=$ (possibly empty) set of strategies such that $R_{i} \subseteq S_{i}$) of a game at every point of the proof.

My question is what good does it do to take the $R$ of the game instead of directly taking the set of all possible strategies for all the players $S$?

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It is because the set of strictly dominated strategies changes as we eliminate strategies.

The point of the proof is to show the result for ANY $R$ such that $R \rightarrow_{S}R'$.

Look at Example 7 in the next page of the paper.

In $S$, there is only one strictly dominated strategy, and we remove it and get $R$ (hence $S \rightarrow_S R$). Now look at $R$. There is a strictly dominated strategy that was not one before (in $S$), removing it get $R'$ (hence $R \rightarrow_S R')$.

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