Prove or disprove: for every P we have PC(P) = PC'(P) Given a partially ordered set (i.e. poset) $P$, let $PC(P)$ be the smallest number of
chains that cover all the elements of $P$ . Let $PC'(P)$ be
the smallest number of pairwise disjoint chains that cover all the elements of $P$ (in other words,
we now want to decompose $P$ into chains, not only cover it). We clearly have $PC(P) ≤ PC'(P)$
Prove or disprove: for every $P$ we have $PC(P) = PC'(P)$
 A: Thank you GroundIns! I was wrong. The answer is that they are indeed equal! I'm guessing that $P$ is a finite poset, just from the wording of the question ("the smallest $\textbf{number}$ of chains").
Suppose $P$ is finite, and suppose we have chains $C_1, \ldots, C_n$ that cover $P$. Let $D_1 = C_1$, and for $k = 2, \ldots, n$, let
$$D_k = C_k \setminus \bigcup_{j=1}^{k-1} C_j.$$
Then $D_k \subseteq C_k$, and since any subset of a chain is a chain, $D_k$ is a chain.
Moreover, $D_1, \ldots, D_n$ covers $P$. To prove this, suppose $x \in P$. Then since $P = \bigcup_{k=1}^n C_k$, there exists some $k$ such that $x \in C_k$. There may be multiple such $k$, and we can assume that this is the least such $k$, that is, if $1 \le j < k$, then $x \notin C_j$. Therefore,
$$x \in C_k \setminus \bigcup_{j=1}^{k-1} C_j = D_k.$$
Thus, $D_1, \ldots, D_n$ is another cover of $P$ by chains.
It is also the case that $D_1, \ldots, D_k$ is pairwise disjoint. Suppose that $x \in D_j \cap D_k$, where $j < k$. Then, since $x \in D_j$, we have $x \in C_j$. But, since $j < k$, $x \in \bigcap_{l = 1}^{k - 1} C_l$, which implies $x \notin D_k$. Thus, we have a contradiction: no such $x$ can exist, and $D_j \cap D_k = \emptyset$.
Since a cover by any number of chains can be matched by a cover of the same number of pairwise disjoint chains, we have $PC'(P) \le PC(P)$, completing the proof.
