Dilworth's Theorem Implies Konig's Theorem An excercise says "deduce Konig's Theorem on bipartite graphs from Dilworth's theorem on posets". 
Let G be a bipartite graph, G=(A,B). Order the edges left to right. A maximum antichain is a largest independent set in the graph. 
I can see a maximum antichain must have every vertex in G incident with it. But it doesn't follow that every edge is incident with it. That may not be true. 
So how does one proceed? 
 A: Create a poset by setting $a \le b$ for every $a \in A$ and $b \in B$ such that $ab \in E(G)$.  Let any two elements in $A$ or $B$ be incomparable.  In other words, orient each edge so that it goes from $A$ to $B$ and make a partial order based off of this orientation.
Dilworth's Theorem: the minimum cardinality of a collection of chains with union $G$ is the maximum cardinality of an antichain.
Recall that each chain must include $1$ or $2$ members, and we want as many chains with $2$ members as possible.  Thus a minimum collection of chains with union $G$ contains a maximum matching and any leftover vertices.
An antichain here is the complement of a vertex cover.  It contains no edges, so its complement must meet all edges.  Making an antichain maximum makes the vertex cover a minimum.
The rest is just arithmetic.
A: View any n-node bipartite graph G as a bipartite poset. The nodes of one
part are maximal elements, and nodes of the other part are minimal. Every
vertex disjoint path covering of the poset of size n-k uses k chains of size 2,
which is actually a matching. Each antichain of size n-k corresponds to an
independent set in G, and the rest of k nodes forms a vertex cover.
