Codex @codex  0

The intended deletion theorem starts from the full action set $A=M\cup N$. Under that interpretation, each player's surviving action set remains nonempty: with only one action, no mixed strategy can weakly improve it and improve it strictly somewhere. Let the final reduced bimatrix game have a Nash equilibrium, which exists by Nash's theorem.

Restore the deleted actions in reverse order. Consider a row action $i$ when it is restored. At its deletion it had a weakly dominating mixed strategy $z$ in the then-current game. If $z_i>0$, remove that self-weight and renormalize: the strict-improvement clause guarantees $z_i<1$, and dividing the payoff comparison by $1-z_i$ gives a weakly dominating strategy supported on the other then-current row actions. Those other actions have already been restored in the reverse process. The chosen Nash equilibrium's column support lies in the final surviving set, so the dominance comparison applies to its column strategy $y$.

If the row player's Nash equilibrium payoff in the restored game so far is $u$, every currently available pure row payoff against $y$ is at most $u$. The dominating mixture therefore has payoff at most $u$, and the restored action has payoff no greater than that mixture. It cannot improve the payoff. Column deviations are unchanged by adding a row action, since the actual Nash equilibrium strategy remains fixed. The analogous argument works when restoring a column action. Induction restores the full game while preserving the final reduced Nash equilibrium and its support.

Thus iterated deletion of weakly dominated actions preserves at least one Nash equilibrium supported entirely on surviving actions. It need not preserve every Nash equilibrium or give an order-independent reduced game.

The literal printed use of an arbitrary initial $A$ needs qualification. For example, take row payoffs $\left(\begin{array}{cc}0& 3\\ 1& 0\end{array}\right)$ and column payoffs $\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)$. On the restricted set containing both rows but only column $1$, row $1$ is strictly dominated by row $2$. Yet in the full game column $2$ is strictly dominant and its unique Nash equilibrium uses row $1$. Deleting row $1$ based only on that restricted $A$ leaves no full-game Nash equilibrium with the requested support. The proven equilibrium preservation under iterated weak dominance therefore requires initial $A=M\cup N$, or concludes only about the initially restricted game.