% ===================================================================== \section{Duality and certificates}\label{sec:duality} % ===================================================================== Duality theorems assert the existence of a \emph{certificate}: a dual solution, a separating hyperplane, a vertex cover, a cut. They are, formally, existence theorems, and the existence of a certificate behaves like the existence of an optimal solution: cheap in finite dimensions, expensive when the certificate is an infinite object chosen among infinitely many candidates. \subsection{Finite-dimensional duality}\label{subsec:lp} Finite linear programming is the base case. With real data it is not computable in any uniform sense: solving a bounded feasible system of linear inequalities, and finding a Nash equilibrium of a bimatrix game, are both Weihrauch equivalent to $\pr{AoUC}^\ast_{[0,1]}$, the finite parallelization of all-or-unique choice (for linear inequalities \cite[Theorems 37 and 38]{Pauly2010Nash} with \cite[Proposition 7.60]{BrattkaGherardiPauly2021}; for Nash equilibria \cite[Corollary 40]{Pauly2010Nash} and \cite[Theorem 8.5]{BrattkaGherardiPauly2021}), and no Banach--Mazur computable map produces maximizers of linear programs with computable data \cite{LeeBocheKutyniok2023}. In Bishop's constructive mathematics, Farkas' lemma as an exclusive alternative is equivalent to $\LPO$ \cite{BergerSvindland2022Farkas}. Classical reverse mathematics is kinder. We did not find the following statement in the literature. \begin{theorem}\label{thm:lp-rca} $\RCA$ proves linear programming strong duality: for all $m,n$ and all real $A\in\R^{m\times n}$, $b\in\R^m$, $c\in\R^n$, if $\{x\ge0:Ax\le b\}$ and $\{y\ge0:A^\top y\ge c\}$ are nonempty, then there are $x^\ast$ and $y^\ast$ in them with $c^\top x^\ast=b^\top y^\ast$. Consequently $\RCA$ proves von Neumann's minimax theorem for real matrix games of every size. \end{theorem} \begin{proof}[Proof sketch] \Cref{app:lp} gives the details. Write the primal in equality form with slack variables, $[A\ I]\,(x,s)=b$. First obtain the signs of the $b_i$ (by the comprehension step described below) and multiply the rows with $b_i<0$ by $-1$; Phase~I then adds $m$ artificial variables and minimizes their sum. For each phase let $M$ be the bordered matrix formed by the constraint matrix of that phase, the right-hand side, and the cost row of that phase. For a basis $B$, every entry of the simplex tableau, including the reduced costs and the objective value, is a quotient $\mu/\Delta_B$ of minors of $M$, where $\Delta_B\neq0$ is the basis determinant. Every test that the two-phase simplex method with Bland's rule \cite{Bland1977} performs is therefore the sign of a minor, or of an expression $\mu_1\mu_2-\mu_3\mu_4$ in minors (a ratio test, after multiplying out positive denominators; ties are tests ``$=0$''). Let $(v_i)_{i0$'' is $\Sigma^0_1$ uniformly in $i$. By bounded $\Sigma^0_1$ comprehension, which is provable in $\RCA$ \cite[II.3]{Simpson2009SOSOA}, the finite sets $\{i0\}$ and $\{i