To finish the proof, you must also define an injection g of G into E, using the condition on the subsets of G and Dilworth's Theorem.
Then for each element a∈G \ f(E), you can redefine f so that a∈ f(E), using the injection g as follows :
Build a set of elements a0, a1, ... of G so that a0 = a and ai = g(f(ai-1)) for i>0.
Let j be the smallest integer so that aj does not belong to G (its existence has to be proved, though).
Redefine f so that f(g(ai)) = ai for each i < j.
The new f is still an injection of E into F such that f(x) ∈ A(x) for each x ∈ E, and besides, a∈ f(E).
Repeat this for each element of G \ f(E) and the problem is solved.
Hope I did not go wrong somewhere. Excuse my poor english, I'm French !
Regards
Alexis
These notes contain a proof. The digraph used in this proof is a little more complicated than the one that you have in mind: each point of the partial order corresponds to two vertices of the digraph.
Let $\langle P,\preceq\rangle$ be the partially ordered set; I’ll write $p\prec q$ to indicate that $p\preceq q$ and $p\ne q$. For each $p\in P$ the digraph $D$ will have two vertices, $p^-$ and $p^+$; in addition, it will have a source vertex $s$ and a sink vertex $t$. For each $p\in P$, $D$ has edges $\langle s,p^-\rangle$ and $\langle p^+,t\rangle$; in addition, for each$p,q\in P$ with $p\prec q$ it has $\langle p^-,q^+\rangle$. Each edge has capacity $1$.
Let $f$ be a maximal flow in $D$, and let $\langle S,T\rangle$ be a minimal cut constructed by the Ford-Fulkerson algorithm. Note that every path from $s$ to $t$ has the form $s\to p^-\to q^+\to t$ for some $p,q\in P$ with $p\prec q$; thus $|f|$ is the number of $p\in P$ such that $f(s,p^-)=1$ there is a $q\in P$ such that $f(p^-,q^+)=1$. Let $C=\{p\in P:f(s,p^-)=0\}$; for future reference note that $|C|=|P|-|f|$.
If $p\in P\setminus C$, then $f(s,p^-)=1$, and there must therefore be a unique ‘successor’ $\sigma(p)\in P$ such that $f(p^-,\sigma(p)^+)=1$, and hence $p\prec\sigma(p)$. Thus, for each $p\in P$ there is a well-defined chain $$p=p_0\prec p_1\prec\ldots\prec p_n\in C$$ in $P$ such that $p_{k+1}=\sigma(p_k)$ for $k=0,\dots,n-1$. ($P$ is finite, so each chain must terminate, and the elements of $C$ are the only elements without successors.) For $p\in P\setminus C$ let $\xi(p)$ be the unique element of $C$ at which the chain from $p$ terminates, and let $\xi(p)=p$ for $p\in C$; then $$\Big\{\{p\in P:\xi(p)=q\}:q\in C\Big\}$$ partitions $P$ into $|C|$ chains. To complete the proof we must show that $P$ also has an antichain of cardinality $|C|$.
The desired antichain is $A=\{p\in P:p^-\in S\text{ and }p^+\in T\}$. Proving that it’s an antichain isn’t too hard. Suppose that $p,q\in A$ with $p\ne q$. Then $p^-\in S$ and $q^+\in T$, so $f(p^-,q^+)\ne0$: either $\langle p^-,q^+\rangle$ isn’t an edge of $D$ at all, or $f(p^-,q^+)=1$, and I leave it to you to show that the latter is impossible: it would unbalance the flow at $p^-$. (The argument here uses the hypothesis that the minimal cut was constructed using the F-F algorithm.) It follows that $p\not\preceq q$, and by symmetry (or a similar argument) $q\not\preceq p$. Thus, $A$ is an antichain.
The last step is to show that the edges contributing capacity towards $c(S,T)$ are preciesly the edges $\langle s,p^-\rangle$ and $\langle p^+,t\rangle$ such that $p\notin A$; this also uses the hypothesis that the minimal cut was constructed using the F-F algorithm, to show that there are no edges of the form $\langle p^-,q^+\rangle$ with $p^-\in S$ and $q^+\in T$.
Once you’ve worked out these details, you’ll have shown that $c(S,T)=|P|-|A|$ and hence that $|A|=|P|-c(S,T)=|P|-|f|=|C|$, as desired.
Best Answer
For this proof, actually two things are shown:
1) Max. Cardinality of an Antichain $\le$ Minimum Number of chains in partition
2) Max. Cardinality of an Antichain $\ge$ Minimum Number of chains in partition
If you have both oft these, equality follows.
One oft them is easy (which one?) and this is explained in the Wikipedia Article. The other one is harder, but for this now it suffices to show the existence oft one Antichain and Partition with the same cardinality, then equality immediately follows, because you already have the other Inequality! (this is the difficult part, have you already read this proof? But worry not, to answer your question we don't have to go into the details, the problems right now are concerned with the proof method)