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The following is a collection of combinatorics theorems commonly refereed as "equivalent" due to one being easily derived from another.

Theorem (Hall). A finite bipartite graph $G = (A\sqcup B, E)$ admits a matching saturating $A$ if, and only if, for all $W\subseteq A$ we have $|W|\le|N(W)|$.

Theorem (Kőnig). In a finite bipartite graph, a maximum matching and a minimum vertex cover have the same size.

Theorem (Menger). In a finite graph, the maximum amount of internally disjoint paths between two vertices and the minimum cut separating then have the same size.

Theorem (Dilworth). In a finite partially ordered set, a maximum antichain and a minimum chain decomposition have the same size.

Hall's Theorem Wikipedia page also mentions Birkhoff–von Neumann's Theorem as one of the equivalent theorems in this collection. The Doubly Stochastic Matrix Wikipedia page goes beyond this and claims the Birkhoff–von Neumann's Theorem is a "real valued generalization of Kőnig's Theorem". It states

Theorem (Birkhoff–von Neumann). A real matrix is doubly stochastic iff it is a convex combination of permutation matrices.

Despite the sudden change of field, Birkhoff–von Neumann's Theorem can be elegantly derived from Hall's Theorem.

However, I haven't found a proof of Hall's Theorem (or one of the equivalents) using Birkhoff–von Neumann's Theorem in order to justify the "equivalent" title. I wish to know if one exists.

This question has been asked before here. However, it also asked secondarily about a literature reference treating the combinatorics theorem collection as a whole and the accepted answer adresses only this, not the main question.

Here is what I already know.

  1. Kőnig has proved, before Birkhoff and von Neumann, a non-negative integer matrix with constant sum among its lines and columns is a positive integer combination of permutation matrices. Birkhoff–von Neumann's Theorem is a real valued generalization of this theorem, so maybe the Wikipedia page just linked the wrong "Kőnig's Theorem".
  2. Using the bi-adjacency matrix one can apply Birkhoff–von Neumann's Theorem to prove any $k$-regular bipartite graph can be decomposed into $k$ perfect matchings. From this, it is easy to prove the edge chromatic number of a bipartite graph is its maximum degree. Maybe we can use this to prove Hall's Theorem (or one of the equivalents). Notice this is a min-max type of theorem, such as Kőnig, Menger and Dilworth, since the edge chromatic number is the size of the minimum matching decomposition.
Alma Arjuna
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    For what it's worth, I just edited the Wikipedia article to remove this unjustified claim. – Mike Earnest Oct 22 '25 at 00:42
  • I remember seeing an argument where, assuming Hall's hypotheses, and assigning capacity $1$ to the edges, one runs Ford-Fulkerson to get a flow of $|A|$ (from $A$ to $B$ or the other way around). Then putting the flow in the adjacency matrix (and normalizing) gives a doubly-stochastic. Applying Birkhoff-von Neumann gives a convex combination of permutation matrices and one of these gives the matching in the thesis of Hall's theorem. – beginner Oct 22 '25 at 13:07
  • That brings in yet more stuff that can likely be proven from one another or at least prove the ones above, duality theorems from linear programming. Koning, Menger and Dilworth all have already the structure of a duality theorem. – beginner Oct 22 '25 at 13:26
  • @beginner this sounds quite interesting, but would this be using the min-cut max-flow theorem AND Birkhoff-von Neumann's theorem? Because I believe the min-cut max-flow should imply Hall's Theorem by itself... – Alma Arjuna Oct 22 '25 at 20:30
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    Yes. Unlike proving equivalence of axioms in an underlying theory of which they are independent, this business of showing a theorem follows from another has a subjective component to it. One can prove each one of them without any of the others. It is more whether the proof feels natural or easy to say. Or is the use of one theorem interesting or not. In all of the implications the theorem being used can be eliminated, in principle. – beginner Oct 22 '25 at 20:56
  • @beginner One of the ways this can be formalized is by showing that the two theorems are equivalent in some axiomatic system weaker than that necessary to prove a theorem on its own: this is the business of reverse mathematics. (For Hall's theorem in particular, I'm not sure how well this works, because apparently $\text{RCA}_0$, the weakest option generally studied, is already strong enough to prove it - in the finite case. Maybe we can say something interesting about the infinite version of these theorems.) – Misha Lavrov Oct 23 '25 at 00:27
  • If one considers the biadjacency matrix $M$ of $G$, then Hall's condition implies that the matrix $M$ doesn't have a zero block of size $a\times b$ with $a+b>n$. In matrix scaling business, this implies that there is a way to assign nonnegative weights to edges such that the corresponding weighted biadjacency matrix is doubly stochastic. Then, von Neumann-Birkhoff theorem implies that the graph must contain a perfect matching. I guess this comment is similar to @beginner 's comment in the sense that it does not directly prove that von Neumann-Birkhoff implies Hall's marriage theorem – Levent Oct 23 '25 at 16:15
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    Note: Reichmeider's book does NOT prove that Birkhoff-von Neumann is equivalent to Hall; it only proves that it follows from Hall. The confusion seems to arise from people who inferred the content of the book from its title. – darij grinberg Oct 23 '25 at 18:08
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    In principle there is no confusion, since nearly every theorem can be used to prove nearly every other theorem. – beginner Oct 23 '25 at 18:17
  • @MikeEarnest That's quite on brand for MSE, deleting useful information. Missing from MSE is a very very low bar to prove a negative. Even worse, when there are trivial proofs: $(p\implies q)\implies (p\wedge r \implies q)$. So, the set of proofs that use BvN to prove H is trivially non-empty. The only question is picking one, and polishing it to make it pleasing. – beginner Oct 27 '25 at 11:09
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    @beginner You're being needlessly confrontational, and I don't see the point in mentioning trivial proofs when Wikipedia is specifically discussing nontrivial ones. If there were a citation on Wikipedia with a proof of Hall's theorem productively using BvN, deleting it would be deleting useful information. If someone misinterpreted one of the sources to mean what it doesn't, then deleting the misinterpretation is clearly helpful. – Misha Lavrov Oct 28 '25 at 17:59
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    OT: Can you really use Hall's theorem to prove Dilworth's theorem? – user14111 Oct 29 '25 at 14:13

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