Mirsky's theorem: the minimum number of antichains covering a finite poset equals its height

For each $x\in P$, let $r(x)$ be the greatest cardinality of a chain whose largest element is $x$. Such a chain exists, since $\{x\}$ is one, and the greatest cardinality exists because the possible values form a nonempty finite subset of $\{1,\dots ,\mathrm{\mid }P\mathrm{\mid }\}$.

Let $C$ be a chain of cardinality $h$, which exists by [F2]. Every antichain contains at most one element of $C$, so any antichain cover of $P$ needs at least $h$ members to cover the $h$ elements of $C$.

If $x<y$, then appending $y$ to a chain of cardinality $r(x)$ ending at $x$ gives a chain ending at $y$, so $r(y)\ge r(x)+1$.

For $i\in \{1,\dots ,h\}$ put $A_i:=\{x\in P:r(x)=i\}$. Each $A_i$ is an antichain, since comparable distinct elements have different $r$-values by step 2.1.

Every $x\in P$ belongs to exactly one $A_i$, and $1\le r(x)\le h$ by the definition of height, so $A_1,\dots ,A_h$ cover $P$.

Step 4.1 gives an antichain cover with $h$ members and step 1.2 rules out every smaller one. Hence the minimum antichain-cover number is $h(P)$.