We proceed by strong induction on $|P|$.

Base case: If $|P| = 1$, then $w = 1$ and $P$ itself is a single chain. ✓

Inductive step: Assume the theorem holds for all posets with fewer than $n$ elements. Let $|P| = n$ and let $A$ be a maximum antichain in $P$ with $|A| = w$.

Case 1: $A = P$ (all elements are pairwise incomparable). Then $P$ is itself an antichain, and we can partition it into $w$ chains of length 1 (singletons). ✓

Case 2: $A \neq P$ (there exist comparable elements). Define:

• $U = \{x \in P : x > a \text{ for some } a \in A\}$ (elements above $A$)
• $L = \{x \in P : x < a \text{ for some } a \in A\}$ (elements below $A$)

Claim: Every maximal antichain in $U$ has size at most $w$.

Proof of claim: Suppose $B$ is an antichain in $U$ with $|B| > w$. For each $b \in B$, choose some $a_b \in A$ with $a_b < b$. Since $|B| > w = |A|$, by pigeonhole principle, some $a \in A$ satisfies $a < b_1$ and $a < b_2$ for distinct $b_1, b_2 \in B$.

But then $A \cup \{b_1, b_2\} \setminus \{a\}$ contains at least $w + 1$ elements, and we claim it's an antichain:

• Elements of $A \setminus \{a\}$ are pairwise incomparable
• $b_1, b_2 \in U$ are incomparable (both in antichain $B$)
• No element of $A \setminus \{a\}$ is comparable to $b_1$ or $b_2$ (else $b_1, b_2 \not\in U$)

This contradicts maximality of $A$. Thus $B$ has size $\leq w$. ✓

By symmetry, every antichain in $L$ also has size $\leq w$.

Applying induction:

• By induction, $U$ can be partitioned into $\leq w$ chains $C_1^U, \ldots, C_k^U$ where $k \leq w$
• By induction, $L$ can be partitioned into $\leq w$ chains $C_1^L, \ldots, C_m^L$ where $m \leq w$

Constructing the decomposition of $P$: For each $a \in A$, we extend chains:

• Find a chain $C_i^L$ with maximum element $< a$ (if exists in $L$)
• Find a chain $C_j^U$ with minimum element $> a$ (if exists in $U$)
• Combine: $C_i^L \cup \{a\} \cup C_j^U$ forms a single chain through $a$

Since $|A| = w$ and we use at most $w$ chains from $U$ and $w$ from $L$, we can construct a chain decomposition of $P$ using exactly $w$ chains.

The matching of chains from $L$ and $U$ through elements of $A$ uses the fact that $k, m \leq w$ and $|A| = w$, allowing us to merge chains through antichain elements to achieve exactly $w$ total chains.