The exhaustive active-set QP solve terminates with the unique minimizer exactly when feasible

The equality reduction is finite: there are finitely many row subsets to check, and exact rank and consistency tests decide each one. Under [A1] it preserves the equality feasible set; if it reports inconsistency, that set is empty and the full QP is infeasible.

For any enumerated subset whose rows are independent, the KKT matrix is nonsingular by [F2], because $B$ is positive definite on every subspace. Thus its equality-constrained candidate is uniquely defined. If that candidate passes the inequality feasibility and multiplier-sign filters, its zero-extended multipliers give a feasible KKT tuple for the original QP, with primal point $\bar{p}$. For every feasible $y$, stationarity and complementarity give $(B\bar{p}+c{)}^T(y-\bar{p})=-{\sum }_i{\lambda }_i{G}_i(y-\bar{p})\ge 0$, since only active inequalities have positive multipliers. Thus $q(y)-q(\bar{p})=(B\bar{p}+c{)}^T(y-\bar{p})+\frac{1}{2}(y-\bar{p}{)}^{T}B(y-\bar{p})\ge 0$, with equality only for $y=\bar{p}$. Hence the accepted candidate is the unique global minimizer.

If the QP is feasible, [F1] supplies an active KKT support whose rows are independent together with the reduced equality rows. [F3] therefore tests that support. The solve in step 1.2 returns its KKT point, and its feasibility and nonnegative multipliers make it pass both filters. Hence the algorithm cannot report QP-infeasible on a feasible subproblem.

There are exactly $2^m$ inequality subsets. Every subset requires only finitely many exact rank, solve, and feasibility operations, so [F3] terminates. If no candidate was found, step 2.1 shows the QP is infeasible; if a candidate was found, step 1.2 shows it is the unique minimizer with the returned full KKT multipliers and support data.