= Solution
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> if
$$
p^TAq\geq (p')^TAq\quad\text{for all }p'\in\Delta_r,
\qquad p^TBq\geq p^TBq'\quad\text{for all }q'\in\Delta_s.
$$
Here $\Delta_r$ 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 \b[every finite two-person game has an equilibrium in mixed strategies].
We prove it using the <Brouwer gain-map proof of bimatrix equilibrium>. Put $u=p^TAq$, $v=p^TBq$ and define the positive pure-deviation gains
$$
g_i(p,q)=\max\{0,(Aq)_i-u\},\qquad
h_j(p,q)=\max\{0,(p^TB)_j-v\}.
$$
Write $G=\sum_i g_i$, $H=\sum_jh_j$ and define
$$
T(p,q)=\left(\left(\frac{p_i+g_i}{1+G}\right)_{i=1}^r,
\left(\frac{q_j+h_j}{1+H}\right)_{j=1}^s\right).
$$
Every 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 $g_i=Gp_i$. Suppose $G>0$. For every $i$ with $p_i>0$, this makes $g_i>0$, so $(Aq)_i-u=g_i=Gp_i$. But averaging the payoff excess over the support gives
$$
0=\sum_i p_i\bigl((Aq)_i-u\bigr)=G\sum_i p_i^2>0,
$$
a 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 $h_j=Hq_j$ 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.
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