To fix notation, let there be two players $\in N = \{1, 2\}$, finite action sets $A_j$ with $A = A_1 \times A_2$, and utility $u_j: A \rightarrow \mathbb{R}_+$. Define a set of strategies $S_j$ which are probability distributions (lotteries) over player action sets $A_j$, and $S = S_1 \times S_2$. Define expected utility $v_j(s_j, s_{-j}) = \sum_{a \in A}u_j(a)p(a)$, where $p(a) = s_1(a_1)s_2(a_2)$. A profile $s^* \in S$ is a Nash equilibrium if $\forall j \in N, \forall s_j \in S_j, \: \;v_j(s_j^*, s_{-j}^*) \geq v_j(s_j, s_{-j}^*)$.
Now consider a two person zero sum game ($u_1(a) = -u_2(a), N=\{1, 2\}$). I'm confused about proving the following remark:
$\textbf{(1)}$ "If $(s_1^*, s_2^*), (s_1', s_2')$ are nash equilibria, then so are $(s_1^*, s_2'), (s_1', s_2^*)$"
In the proof I have, all that is shown is: $\textbf{(2)}$ "Pick $\forall j $ any two $s_j^*, s_j' \in \arg\max_{s_j} \min_{s_{-j}} v_j(s_j, s_{-j})$, then they are Nash." Even though we may use $\max\min = \min\max$ here, I don't see how we can show this is the case (And I particularly don't see how the statement above is enough).
The statement above relies on a lemma which shows that when $\max\min = \min\max$ holds, if we choose $s_1^* \in \arg\max_{s_1} \min_{s_2} v_1(s_1, s_2)$, $s_2^* \in \arg\min_{s_2} \max_{s_1}v_1(s_1, s_2)$, then $(s_1^*, s_2^*)$ is Nash.
I don't see how the second part is exactly symmetric to $\arg\max\min$. Specifically, from $\textbf{(2)}$ we have that $s_2^* \in \arg\max_{s_2}\min_{s_1}v_2(s_2,s_1)$, but in the lemma we require $s_2^* \in \arg\min_{s_2} \max_{s_1}v_1(s_1, s_2)$. To me these seem like opposites because $v_1 = -v_2$.
More context can be found in A Course in Game Theory by Osborne and Rubinstein, Chapter 2 on Nash Equilibria. Thanks.
