Work with finite graphs, as in the usual matching theorem. Hall's condition is for every . It is necessary because a matching injects into its neighbours.
For sufficiency, induct on . If a nonempty proper subset has , induction matches to . For , Hall's inequality for gives , so induction matches the complement to unused neighbours. Combining the matchings finishes this case. Otherwise every nonempty proper has . Choose an edge , delete its endpoints, and observe that every subset of the remaining still has at least its own number of neighbours. Induction matches the remainder, and adjoining finishes. The one-vertex case and the empty case are immediate.
For , the defect version is
Necessity follows because at most vertices of can be unmatched. For sufficiency add new vertices to , each joined to every vertex of . The displayed inequality makes Hall's condition hold in the enlarged graph. Its full matching uses at most new vertices, giving the required original edges.
To prove the matching-cover equality, let be a maximum matching and its unmatched vertices in . Follow alternating paths from , using nonmatching edges from to and matching edges back. Let be the reachable sets. No reachable is unmatched, since that would give an augmenting path and a larger matching. Matching edges give a bijection . The set covers every edge: an edge from a reachable to an unreachable cannot be a nonmatching edge, and its matching edge, if present, is already reachable. Its size is
Every vertex cover has at least vertices, because the matching edges are disjoint. Therefore maximum matching size equals minimum vertex-cover size, as required.

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