 Solution

With equal substrate abundance and association rates, let $a=k_{-}^S<b=k_{-}^R$ and $c=l_{-}^{S\ast }<d=l_{-}^{R\ast }$. ThenBoth factors increase with their forward rate: for example ${\partial }_m[(a+m)/(b+m)]=(b-a)/(b+m{)}^2>0$. Thus maximum discrimination requires enough waiting at each stage for the less stable incorrect complex to be rejected. In the slow-forward limit,The two rejection stages multiply their discrimination. Fast modification or fast product formation bypasses the corresponding dissociation test and drives that factor towards one.

There is a kinetic proofreading fidelity-throughput tradeoff: taking either forward rate literally to zero gives zero product flux, not a useful finite-output optimum. If dissociation rates are well separated, a practical high-yield discrimination window is $k_{-}^S\ll m\ll k_{-}^R$ and $l_{-}^{S\ast }\ll V_A\ll l_{-}^{R\ast }$. Correct complexes then usually pass both stages while incorrect complexes usually dissociate; its fidelity is not the mathematical infimum obtained with vanishing forward rates. The irreversible modification represents a driven, energy-consuming step, permitting successive nonequilibrium checks rather than merely repeating one equilibrium binding comparison. No finite positive optimum can be selected from minimization of $F$ alone without a speed or yield requirement.