[proofplan] We prove the Cohen-Macaulay link criterion directly. A face of $\Delta(\bar P)$ is a chain in the proper part, and its link decomposes as a join of the order complexes of the open intervals between consecutive elements of the augmented chain obtained by adjoining $\hat{0}$ and $\hat{1}$. Each nonempty interval factor is pure and shellable by the graded hypothesis and the assumption, hence Cohen-Macaulay over $\mathbb Z$ by the standard shellability theorem for pure simplicial complexes; rank-one gaps contribute only the empty $(-1)$-dimensional factor. Since finite joins of Cohen-Macaulay complexes over $\mathbb Z$ are Cohen-Macaulay, every link has reduced homology concentrated in its top dimension, which is exactly the desired condition. [/proofplan]

[step:Reduce the Cohen-Macaulay condition to links of chains in the proper part] Let $K:=\Delta(\bar P)$ denote the order complex of the proper part. A face $F$ of $K$ is either the empty face or a finite chain in $\bar P$. If $F\ne\varnothing$, write it uniquely in increasing order as \begin{align*} F=\{x_1,\dots,x_k\}, \qquad x_1<\cdots <x_k, \end{align*} where $k\ge 1$ and $x_j\in \bar P$ for every $j\in\{1,\dots,k\}$. Define \begin{align*} x_0:=\hat{0}, \qquad x_{k+1}:=\hat{1}. \end{align*} For the empty face, define $k:=0$, $x_0:=\hat{0}$, and $x_1:=\hat{1}$. In both cases, the Cohen-Macaulay criterion required in the statement asks us to prove \begin{align*} \widetilde H_i(\operatorname{link}_{K}F;\mathbb Z)=0 \end{align*} for every integer $i<\dim(\operatorname{link}_{K}F)$. The case in which $\bar P=\varnothing$ is immediate: then $K=\Delta(\varnothing)$ has dimension $-1$, and there is no integer $i<-1$. Hence the required vanishing condition is vacuous. We therefore assume for the rest of the proof that the proper part is nonempty or, more generally, that the displayed link is not being evaluated only in this vacuous rank-one situation. [/step]

[step:Identify each link as a join of interval order complexes]For each $j$ in the relevant index set, define the open interval \begin{align*} I_j:=(x_j,x_{j+1})=\{z\in P:x_j<z<x_{j+1}\}. \end{align*} If $F\ne\varnothing$, then $j$ ranges over $\{0,\dots,k\}$; if $F=\varnothing$, then $j=0$ and $I_0=(\hat{0},\hat{1})=\bar P$. We use the standard link formula for order complexes: for a chain $x_1<\cdots <x_k$ in a poset, the link of that chain in the order complex is naturally isomorphic to the join of the order complexes of the intervals cut out by adjacent elements of the augmented chain. In the present notation, this gives a simplicial isomorphism \begin{align*} \operatorname{link}_{K}F \cong \Delta(I_0)*\Delta(I_1)*\cdots *\Delta(I_k) \end{align*} when $F\ne\varnothing$, and \begin{align*} \operatorname{link}_{K}\varnothing=K=\Delta((\hat{0},\hat{1})) \end{align*} when $F=\varnothing$. For completeness, we spell out the identification in the nonempty case. A face $G$ of $\operatorname{link}_{K}F$ is a chain in $\bar P$ disjoint from $F$ such that $F\cup G$ is again a chain. Therefore every element of $G$ lies in exactly one interval $(x_j,x_{j+1})$: it is either below $x_1$, above $x_k$, or between a unique consecutive pair $x_j<x_{j+1}$. Thus $G$ decomposes uniquely as a disjoint union \begin{align*} G=G_0\cup G_1\cup\cdots\cup G_k, \end{align*} where $G_j$ is a chain in $I_j$. Conversely, any choice of chains $G_j$ in the intervals $I_j$ gives a chain \begin{align*} G_0<x_1<G_1<x_2<\cdots <x_k<G_k \end{align*} after adjoining $F$, so it determines a face of the link. This is precisely the face correspondence defining the join of the complexes $\Delta(I_j)$.[/step]

[guided]We now explain why links in an order complex break into interval pieces. The vertices of $K=\Delta(\bar P)$ are the elements of $\bar P$, and the faces are finite chains. Hence a face \begin{align*} F=\{x_1,\dots,x_k\}, \qquad x_1<\cdots <x_k, \end{align*} cuts the poset into the intervals \begin{align*} (\hat{0},x_1),\ (x_1,x_2),\ \dots,\ (x_k,\hat{1}). \end{align*} To make the notation uniform, define $x_0:=\hat{0}$ and $x_{k+1}:=\hat{1}$, and write \begin{align*} I_j=(x_j,x_{j+1}) \end{align*} for $0\le j\le k$. A face $G$ lies in $\operatorname{link}_K F$ exactly when $G$ is disjoint from $F$ and $F\cup G$ is a face of $K$. Since faces of $K$ are chains, this means that every element of $G$ must be comparable with every element of $F$. Once the chain $x_1<\cdots <x_k$ is fixed, any element of $G$ has a unique position: it lies below $x_1$, above $x_k$, or between exactly one adjacent pair $x_j<x_{j+1}$. Thus $G$ decomposes uniquely as \begin{align*} G=G_0\cup G_1\cup\cdots\cup G_k, \end{align*} where $G_j$ is a chain in $I_j$. Conversely, suppose we choose a face $G_j$ of $\Delta(I_j)$ for each $j$. This means that $G_j$ is a chain inside the interval $I_j$. Because every element of $I_j$ is less than every element of $I_m$ whenever $j<m$, the union \begin{align*} G_0\cup \{x_1\}\cup G_1\cup\cdots\cup \{x_k\}\cup G_k \end{align*} is a chain in $\bar P$. Therefore $G_0\cup\cdots\cup G_k$ is a face of the link of $F$. This two-sided construction is inverse on faces and preserves inclusions of faces, so it is a simplicial isomorphism \begin{align*} \operatorname{link}_{K}F \cong \Delta(I_0)*\Delta(I_1)*\cdots *\Delta(I_k). \end{align*} For $F=\varnothing$, the link of the empty face is the whole complex: \begin{align*} \operatorname{link}_{K}\varnothing=K=\Delta(\bar P)=\Delta((\hat{0},\hat{1})). \end{align*} This is the same interval decomposition with only the single interval $(\hat{0},\hat{1})$.[/guided]

[step:Show that every interval factor is Cohen-Macaulay over $\mathbb Z$] Fix an interval factor $I_j=(x_j,x_{j+1})$. Let \begin{align*} r_j:=\rho(x_{j+1})-\rho(x_j) \end{align*} be its rank gap. If $r_j=1$, then $I_j=\varnothing$, because there is no element of rank strictly between $\rho(x_j)$ and $\rho(x_{j+1})$ in a graded poset. Hence $\Delta(I_j)$ is the empty interval factor of dimension $-1$, and by the convention in the statement it imposes no lower-degree reduced homology condition. Assume now that $r_j\ge 2$. Since $P$ is graded with rank function $\rho$, every maximal chain in the interval $[x_j,x_{j+1}]$ has length $r_j$. It follows that every maximal chain in the open interval $I_j$ has length $r_j-2$, so $\Delta(I_j)$ is pure of dimension $r_j-2$. By the hypothesis of the theorem, $\Delta(I_j)$ is shellable. By the standard theorem that a pure shellable simplicial complex is Cohen-Macaulay over $\mathbb Z$ (citing a result not yet in the wiki: Pure shellable simplicial complexes are Cohen-Macaulay over $\mathbb Z$), we obtain \begin{align*} \widetilde H_i(\operatorname{link}_{\Delta(I_j)}E;\mathbb Z)=0 \end{align*} for every face $E$ of $\Delta(I_j)$ and every integer \begin{align*} i<\dim(\operatorname{link}_{\Delta(I_j)}E). \end{align*} Thus every interval factor $\Delta(I_j)$ is Cohen-Macaulay over $\mathbb Z$, with the stated $(-1)$-dimensional convention for rank-one gaps. [/step]

[step:Pass Cohen-Macaulayness through the join decomposition] The link formula gives \begin{align*} \operatorname{link}_{K}F \cong \Delta(I_0)*\Delta(I_1)*\cdots *\Delta(I_k) \end{align*} for $F\ne\varnothing$, and the same argument with one factor gives the empty-face case. Each factor with rank gap at least $2$ is Cohen-Macaulay over $\mathbb Z$ by the previous step, and each rank-one factor is the $(-1)$-dimensional empty interval factor, which is neutral for the join convention and imposes no lower-degree homology condition. Using the standard theorem that a finite join of Cohen-Macaulay simplicial complexes over $\mathbb Z$ is Cohen-Macaulay over $\mathbb Z$ (citing a result not yet in the wiki: Join of Cohen-Macaulay complexes is Cohen-Macaulay), it follows that $\operatorname{link}_{K}F$ is Cohen-Macaulay over $\mathbb Z$. Therefore \begin{align*} \widetilde H_i(\operatorname{link}_{K}F;\mathbb Z)=0 \end{align*} for every integer \begin{align*} i<\dim(\operatorname{link}_{K}F). \end{align*} Since $F$ was an arbitrary face of $K=\Delta(\bar P)$, the Cohen-Macaulay link criterion holds for every face of $\Delta(\bar P)$. Hence $\Delta(\bar P)$ is Cohen-Macaulay over $\mathbb Z$, as required. [/step]