[proofplan] We order the facets of $\Delta(\overline{P})$ by the lexicographic order of their rooted label words, breaking ties arbitrarily. For a fixed maximal chain $C$, we show that every earlier maximal chain $D$ is separated from $C$ by a codimension-one face obtained by replacing a single descent of $C$ with the unique increasing chain in the corresponding rooted rank-two interval. The CL-labelling condition guarantees that this replacement gives an earlier maximal chain, while the choice of the descent inside the first interval where $C$ and $D$ differ guarantees that the resulting codimension-one face contains $C \cap D$. This verifies the shelling condition for the lexicographic order. [/proofplan] [step:Order maximal chains by rooted label words] Let the rank of $P$ be $n$, so every maximal chain from $\hat{0}$ to $\hat{1}$ has the form $C : \hat{0} = x_0 < x_1 < \cdots < x_{n-1} < x_n = \hat{1}$. The corresponding facet of $\Delta(\overline{P})$ is $F_C := \{x_1,\dots,x_{n-1}\}$. For each maximal chain $C$, define its rooted label word $L(C) = (\lambda_1(C),\dots,\lambda_n(C)) \in \Lambda^n$ as follows: $\lambda_i(C)$ is the label assigned to the rooted cover relation $x_{i-1} \lessdot x_i$, where the root is the initial chain $\hat{0} = x_0 < x_1 < \cdots < x_{i-1}$. Order the maximal chains of $P$ lexicographically by $L(C)$, breaking ties by any fixed total order. Let $C_1, C_2,\dots,C_m$ be the resulting order of maximal chains, and write $F_j := F_{C_j}$. We will prove that $F_1,\dots,F_m$ is a shelling order of $\Delta(\overline{P})$. [/step] [step:Locate a non-increasing segment in the first interval where two chains differ] Fix indices $i < j$, and write $C_j : \hat{0} = x_0 < x_1 < \cdots < x_n = \hat{1}$ and $C_i : \hat{0} = y_0 < y_1 < \cdots < y_n = \hat{1}$. Let $s$ be the smallest index with $x_s \neq y_s$. Since both chains begin at $\hat{0}$, we have $s \geq 1$ and $x_{s-1} = y_{s-1}$. Let $t$ be the smallest index greater than or equal to $s$ such that $x_t$ lies on the chain $C_i$. Such an index exists because $x_n = \hat{1}$ lies on both chains. By minimality of $t$, none of $x_s,\dots,x_{t-1}$ lies on $C_i$. Let $r$ be the common root $r : \hat{0} = x_0 < x_1 < \cdots < x_{s-1}$. Consider the rooted interval $[x_{s-1},x_t]$ with root $r$. The chain segment $x_{s-1} < x_s < \cdots < x_t$ is not increasing. Indeed, if this segment were increasing, then by the CL-labelling property it would be the unique increasing maximal chain in the rooted interval $[x_{s-1},x_t]$, and it would be lexicographically first among all maximal chains of that rooted interval. The segment of $C_i$ from $x_{s-1}$ to $x_t$ is another maximal chain in the same rooted interval. Therefore its rooted label word would be lexicographically greater than or equal to the rooted label word of the segment of $C_j$. If equality held, then the segment of $C_i$ would also be increasing and hence, by uniqueness of the increasing chain, would coincide with the segment of $C_j$, contradicting the choice of $s$. Thus the segment of $C_i$ would be strictly lexicographically greater, forcing $L(C_j) < L(C_i)$, contrary to $i < j$. Hence the segment of $C_j$ is not increasing. [guided] We compare $C_i$ and $C_j$ only after their first disagreement. The index $s$ is the first rank where the two chains differ, so the roots up to rank $s-1$ are identical. The index $t$ is the first later rank where $C_j$ returns to a vertex lying on $C_i$. Thus the two chains give two different maximal chains inside the same rooted interval $[x_{s-1},x_t]$ with the same root $r : \hat{0} = x_0 < x_1 < \cdots < x_{s-1}$. We claim that the segment $x_{s-1} < x_s < \cdots < x_t$ cannot be increasing. Suppose instead that it were increasing. The defining property of a CL-labelling applies to the rooted interval $[x_{s-1},x_t]$ with root $r$: there is a unique increasing maximal chain in this rooted interval, and that chain is lexicographically first among all maximal chains in the rooted interval. Since the segment of $C_j$ is assumed increasing, it must be that unique increasing chain. The segment of $C_i$ from $x_{s-1}$ to $x_t$ is another maximal chain of the same rooted interval. Therefore its rooted label word is lexicographically greater than or equal to the rooted label word of the segment of $C_j$. If the two segment label words were equal, then the segment of $C_i$ would also be increasing, because it would have the same weakly increasing label word as the segment of $C_j$. The uniqueness part of the CL-labelling property would then force the two rooted interval chains to be the same, contradicting the definition of $s$ as the first rank where the chains differ. Hence the segment of $C_i$ must have strictly larger rooted label word than the segment of $C_j$. But the two global chains agree before rank $s$, so this strict comparison inside the rooted interval would imply $L(C_j) < L(C_i)$ in the global lexicographic order. That contradicts $i < j$, since $C_i$ was ordered before $C_j$. Therefore the segment of $C_j$ in $[x_{s-1},x_t]$ is not increasing. [/guided] [/step] [step:Replace one descent by the increasing rooted rank-two chain] We use the standard convention for CL-labellings that an increasing chain means one whose label word is weakly increasing in the total label order. Since the segment $x_{s-1} < x_s < \cdots < x_t$ is not increasing, there exists an index $k$ with $s \leq k \leq t-1$ such that the two consecutive labels on $x_{k-1} < x_k < x_{k+1}$ form a strict descent. Let $r_k$ be the root $r_k : \hat{0} = x_0 < x_1 < \cdots < x_{k-1}$. Apply the CL-labelling property to the rooted rank-two interval $[x_{k-1},x_{k+1}]$ with root $r_k$. There is a unique increasing maximal chain $x_{k-1} < z < x_{k+1}$ in this rooted interval, and this chain is lexicographically first among all maximal chains of the rooted interval. Because the original two-step chain through $x_k$ has a descent, it is not increasing. Hence $z \neq x_k$. Define a new maximal chain $C'$ by replacing $x_k$ with $z$ and leaving all other elements of $C_j$ unchanged: $C' : \hat{0} = x_0 < \cdots < x_{k-1} < z < x_{k+1} < \cdots < x_n = \hat{1}$. The rooted label word of $C'$ agrees with $L(C_j)$ before rank $k$, and at the rooted interval $[x_{k-1},x_{k+1}]$ it is lexicographically smaller because the increasing chain is lexicographically first. Therefore $L(C') < L(C_j)$, so $C'$ appears earlier than $C_j$ in the chosen order. [/step] [step:Show the replacement gives the required codimension-one face] The facets $F_{C'}$ and $F_{C_j}$ differ exactly in the rank-$k$ vertex: $F_{C'}$ contains $z$ in place of $x_k$, while all other vertices of $F_{C_j}$ are unchanged. Therefore $F_{C'} \cap F_{C_j} = F_{C_j} \setminus \{x_k\}$. Also, $x_k \notin F_{C_i}$. Indeed, $s \leq k \leq t-1$, and the definition of $t$ says that none of $x_s,\dots,x_{t-1}$ lies on $C_i$. Hence every vertex common to $F_{C_i}$ and $F_{C_j}$ is still present in $F_{C'} \cap F_{C_j}$. Thus $F_{C_i} \cap F_{C_j} \subseteq F_{C'} \cap F_{C_j} = F_{C_j} \setminus \{x_k\}$. Since $C'$ appears earlier than $C_j$, this proves that for every $i < j$ there is an earlier facet whose intersection with $F_{C_j}$ is a codimension-one face of $F_{C_j}$ containing $F_{C_i} \cap F_{C_j}$. [/step] [step:Conclude that the lexicographic order is a shelling] The order complex $\Delta(\overline{P})$ is pure because $P$ is graded and bounded: every facet is obtained from a maximal chain from $\hat{0}$ to $\hat{1}$ by deleting the two endpoints, so every facet has exactly $n-1$ vertices. The preceding step verifies the standard pairwise shelling criterion: for every $j$ and every $i < j$, there exists $h < j$ and a vertex $v \in F_j$ such that $F_i \cap F_j \subseteq F_h \cap F_j = F_j \setminus \{v\}$. Therefore $F_1,\dots,F_m$ is a shelling order of $\Delta(\overline{P})$. Hence $\Delta(\overline{P})$ is shellable. [/step]