# Showing that strategy is not dominated

I was trying to solve the following problem

Suppose player $$i$$ has a strategy $$s_i$$ that he plays with positive probability in every max-min strategy of his, then show that $$s_i$$ is not weakly dominated by any pure or mixed strategy.

Now I can see that if $$s_i$$ was weakly dominated then $$i$$ wouldn’t put positive probability on it in every max min strategy as he could simpy play the strategy that weakly dominates $$s_i$$ say $$t_i$$ , but I don’t know how to give a formal argument/proof for this. Any help will be highly appreciated.

Suppose $$s_i\in S_i$$ was weakly dominated by some mixed strategy $$\sigma_i\in\Delta S_i$$, i.e. $$u_i(s_i,s_{-i})\le u_i(\sigma_i,s_{-i}),\forall s_{-i}\in S_{-i}$$ with strict inequality holding for some $$s_{-i}\in S_{-i}$$.
By assumption, in the case where the strict inequality holds, player $$i$$ plays a mixed strategy involving $$s_i$$ with positive probability.
But in that case, any mixed strategy involving $$s_i$$ (with positive probability) has strictly less utility than the mixed strategy with $$s_i$$ replaced by $$\sigma_i$$.
So player $$i$$ playing their mixed strategy involving $$s_i$$ contradicts that they are rational.