<p>It is because the set of strictly dominated strategies changes as we eliminate strategies.</p>

<p>The point of the proof is to show the result for ANY <span class="math-container">$R$</span> such that <span class="math-container">$R \rightarrow_{S}R'$</span>.</p>

<p>Look at Example 7 in the next page of the paper. </p>

<p>In <span class="math-container">$S$</span>, there is only one strictly dominated strategy, and we remove it and get <span class="math-container">$R$</span> (hence <span class="math-container">$S \rightarrow_S R$</span>).
Now look at <span class="math-container">$R$</span>. There is a strictly dominated strategy that was not one before (in <span class="math-container">$S$</span>), removing it get <span class="math-container">$R'$</span> (hence <span class="math-container">$R \rightarrow_S R')$</span>.</p>

CC BY-SA 4.0
2020-04-04T17:05:42.613Z
2020-04-04T17:05:42.613Z
Walrasian Auctioneer
<p>Consider a game <span class="math-container">$G$</span>. We have to prove that is <span class="math-container">$s$</span> is a Nash Equilibrium of <span class="math-container">$G$</span>, then it is also a Nash Equilibrium of the game formed by removing strictly dominated strategies of <span class="math-container">$G$</span>.</p>

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

<p>My question is what good does it do to take the <span class="math-container">$R$</span> of the game instead of directly taking the set of all possible strategies for all the players <span class="math-container">$S$</span>? </p>

<game-theory><nash-equilibrium>
Why do we need a restriction of a game to prove the given statement?
user185887
# Why do we need a restriction of a game to prove the given statement?

(Asked by: username_0 on 2020-04-04)

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 $S$?

**Tags:** game-theory, nash-equilibrium

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## 1 answers

### Answer by username_1 on 2020-04-04. Score: 1 | Accepted answer

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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