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For a finite bimatrix game, let $A,B\in {\mathbb{R}}^{r\times s}$ be the two players' payoff matrices. A pair of mixed strategies $(p,q)\in {\Delta }_r\times {\Delta }_s$ is a Nash equilibrium ifHere ${\Delta }_r$ is the probability simplex on the first player's pure actions, and similarly for ${\Delta }_s$. The two payoff matrices need not be negatives of each other. Nash's theorem asserts that every finite two-person game has an equilibrium in mixed strategies.

We prove it using the Brouwer gain-map proof of bimatrix equilibrium. Put $u=p^{T}Aq$, $v=p^{T}Bq$ and define the positive pure-deviation gainsWrite $G={\sum }_i{g}_i$, $H={\sum }_j{h}_j$ and defineEvery coordinate is nonnegative and each player's coordinates sum to one. The map is continuous, since the payoffs and positive-part function are continuous and both denominators are at least one. Thus $T$ is a continuous self-map of the nonempty compact convex set ${\Delta }_r\times {\Delta }_s$. The Brouwer fixed-point theorem supplies a fixed point $(p,q)$.

At that fixed point the first coordinate equations give $g_i=G{p}_i$. Suppose $G>0$. For every $i$ with $p_i>0$, this makes $g_i>0$, so $(Aq{)}_i-u=g_i=G{p}_i$. But averaging the payoff excess over the support givesa contradiction. Therefore $G=0$, and every $g_i$ is zero. Each pure action has payoff at most $u$, so every mixed deviation, being a convex combination of pure actions, has payoff at most $u$ too. The same argument for $h_j=H{q}_j$ gives $H=0$ and excludes every unilateral deviation by the second player. The fixed point is therefore a Nash equilibrium, proving the stated theorem without a zero-sum assumption.