[proofplan] The rank-zero and rank-one intervals are handled directly from the defining recursion of the Möbius function and the stated convention for falling chains. For rank at least two, the EL-labeling gives a shelling of the order complex of the open interval $(x,y)$, with facets indexed by maximal chains in $[x,y]$. The EL-shelling restriction-face criterion says that precisely the falling maximal chains are the homology facets, so the order complex has reduced Euler characteristic equal to $(-1)^{r-2}f(x,y)$. Finally, Philip Hall's theorem identifies this reduced Euler characteristic with $\mu_P(x,y)$, and $(-1)^{r-2}=(-1)^r$. [/proofplan]

[step:Check the rank-zero and rank-one intervals directly] If $r=0$, then $x=y$. By the defining normalization of the Möbius function, \begin{align*} \mu_P(x,x)=1. \end{align*} By convention, $f(x,x)=1$, so \begin{align*} \mu_P(x,x)=1=(-1)^0 f(x,x). \end{align*} Now suppose $r=1$. Then $x\lessdot y$, and the interval $[x,y]$ has exactly one maximal chain, \begin{align*} x\lessdot y. \end{align*} Its label word has length one, hence is strictly decreasing in the vacuous adjacent-pair sense, so $f(x,y)=1$. The Möbius recursion gives \begin{align*} 0=\sum_{x\le z\le y}\mu_P(x,z)=\mu_P(x,x)+\mu_P(x,y)=1+\mu_P(x,y). \end{align*} Therefore \begin{align*} \mu_P(x,y)=-1=(-1)^1 f(x,y). \end{align*} [/step]

[step:Pass from maximal chains to facets of the open interval order complex]Assume from now on that $r\ge 2$. Let \begin{align*} \Delta=\triangle((x,y)) \end{align*} denote the order complex of the open interval \begin{align*} (x,y)=\{z\in P:x<z<y\}. \end{align*} A facet of $\Delta$ is exactly a maximal chain in $(x,y)$, hence is equivalently a maximal chain \begin{align*} C: x=x_0\lessdot x_1\lessdot \cdots \lessdot x_r=y \end{align*} in $[x,y]$, with the endpoints $x$ and $y$ removed. Since $P$ is graded and the rank difference from $x$ to $y$ is $r$, such a facet contains the $r-1$ vertices \begin{align*} x_1,x_2,\dots,x_{r-1}. \end{align*} Thus every facet of $\Delta$ has dimension $r-2$.[/step]

[guided]We now translate the combinatorics of chains into the topology of a simplicial complex. Define \begin{align*} \Delta=\triangle((x,y)). \end{align*} By definition, the vertices of $\Delta$ are the elements $z\in P$ satisfying $x<z<y$, and a face of $\Delta$ is a finite chain in the open interval $(x,y)$. A facet of $\Delta$ is a maximal face, so it is a maximal chain in the open interval. Because $[x,y]$ is graded of rank difference $r$, every maximal chain from $x$ to $y$ has the form \begin{align*} C: x=x_0\lessdot x_1\lessdot \cdots \lessdot x_r=y. \end{align*} Removing the two endpoints gives the chain \begin{align*} x_1<x_2<\cdots <x_{r-1} \end{align*} inside $(x,y)$, and this chain is a facet of $\Delta$. Conversely, every facet of $\Delta$ is obtained in this way by adjoining $x$ at the bottom and $y$ at the top. Therefore facets of $\Delta$ are indexed by maximal chains in $[x,y]$. The number of vertices in the facet corresponding to $C$ is $r-1$. A simplex with $r-1$ vertices has dimension $r-2$, so $\Delta$ is pure of dimension $r-2$. This dimension is the source of the sign in the final Euler characteristic calculation.[/guided]

[step:Use the EL-shelling criterion to count the homology facets]Since $P$ is finite and graded, the interval $[x,y]$ is finite, bounded by $x$ and $y$, and graded with rank difference $r$. By the theorem statement, the restriction of $\lambda$ to cover relations in $[x,y]$ is the inherited EL-labeling. Therefore the standard EL-shelling theorem applies to this bounded graded interval and induces a shelling of $\Delta$ whose facets are ordered lexicographically by their label words. For a facet $F$ in this shelling, let $R(F)$ denote its shelling restriction face, that is, the unique minimal face of $F$ that is not contained in the union of the facets preceding $F$ in the shelling order. The EL-shelling restriction-face criterion says that, for the facet corresponding to \begin{align*} C: x=x_0\lessdot x_1\lessdot \cdots \lessdot x_r=y \end{align*} the equality $R(F)=F$ holds exactly when every adjacent pair in the label word is a strict descent: \begin{align*} \lambda(x_{i-1},x_i)>\lambda(x_i,x_{i+1}) \text{ for every } i\in\{1,\dots,r-1\}. \end{align*} Equivalently, the full label word is strictly decreasing. By the definition of $f(x,y)$, the number of such facets is $f(x,y)$. Let $d=r-2$. Write the shelling facets as $F_1,\dots,F_t$ in lexicographic shelling order, and write $R(F_k)$ for the shelling restriction face of $F_k$. Let $\widetilde{\chi}(\Delta)$ denote the reduced Euler characteristic of the finite simplicial complex $\Delta$. Since $\Delta$ is pure of dimension $d$, the [Partition of Faces by Restriction Faces]([citetheorem:8128]) applied to this shelling partitions all faces of $\Delta$, including the empty face, into intervals $\{G:R(F_k)\subseteq G\subseteq F_k\}$. If $R(F_k)\subsetneq F_k$, then the interval is a Boolean interval with at least one free vertex, so its alternating contribution to $\widetilde{\chi}(\Delta)$ is zero by cancellation of the Boolean binomial sum. If $R(F_k)=F_k$, then its only face is the facet $F_k$, and its contribution is $(-1)^d$. Therefore \begin{align*} \widetilde{\chi}(\Delta)=(-1)^{r-2}f(x,y). \end{align*}[/step]

[guided]The shelling step is where the word "falling" is converted into topology. We have already verified that $[x,y]$ is finite, bounded, and graded, and the theorem statement gives that the restricted labeling is an EL-labeling of this interval. Hence the standard EL-shelling theorem applies: ordering the facets of the order complex $\Delta=\triangle((x,y))$ lexicographically by their full label words gives a shelling. For a facet $F$ in this shelling, define $R(F)$ to be its shelling restriction face, the unique minimal face of $F$ not contained in the union of the earlier facets. The EL-shelling restriction-face criterion identifies $R(F)$ in terms of descents of the label word. For the maximal chain \begin{align*} C: x=x_0\lessdot x_1\lessdot \cdots \lessdot x_r=y, \end{align*} the internal vertex $x_i$ belongs to $R(F)$ exactly when the adjacent labels form a strict descent: \begin{align*} \lambda(x_{i-1},x_i)>\lambda(x_i,x_{i+1}). \end{align*} Thus $R(F)=F$ exactly when every internal vertex $x_i$ belongs to $R(F)$, which is exactly the condition \begin{align*} \lambda(x_{i-1},x_i)>\lambda(x_i,x_{i+1}) \text{ for every } i\in\{1,\dots,r-1\}. \end{align*} This is the strictly decreasing, or falling, condition in the repaired theorem statement. Therefore the number of facets with $R(F)=F$ is precisely $f(x,y)$. Now let $d=r-2$, so every facet of $\Delta$ has dimension $d$. Write the shelling facets as $F_1,\dots,F_t$, and let $R(F_k)$ be the restriction face of $F_k$. Let $\widetilde{\chi}(\Delta)$ denote the reduced Euler characteristic of $\Delta$. By the [Partition of Faces by Restriction Faces]([citetheorem:8128]), the faces of $\Delta$, including the empty face, are partitioned into the intervals \begin{align*} \{G:R(F_k)\subseteq G\subseteq F_k\}. \end{align*} If $R(F_k)\subsetneq F_k$, then this interval is a Boolean interval with at least one optional vertex. The alternating sum over that Boolean interval cancels to zero. If $R(F_k)=F_k$, then the interval consists only of the facet $F_k$, and since $\dim F_k=d$, its contribution to the reduced Euler characteristic is $(-1)^d$. Summing over all shelling intervals gives \begin{align*} \widetilde{\chi}(\Delta)=(-1)^d f(x,y)=(-1)^{r-2}f(x,y). \end{align*}[/guided]

[step:Convert the reduced Euler characteristic into the Möbius value] Since $P$ is finite, the interval $[x,y]$ is finite, so [Philip Hall Theorem]([citetheorem:8124]) applies to the finite poset interval. Because $x<y$, that theorem gives \begin{align*} \mu_P(x,y)=\widetilde{\chi}(\triangle((x,y))). \end{align*} Using the notation $\Delta=\triangle((x,y))$ from the previous steps and substituting the shelling computation, we obtain \begin{align*} \mu_P(x,y)=\widetilde{\chi}(\Delta)=(-1)^{r-2}f(x,y). \end{align*} Since $(-1)^{r-2}=(-1)^r$, this becomes \begin{align*} \mu_P(x,y)=(-1)^r f(x,y). \end{align*} Together with the rank-zero and rank-one cases, this proves the formula for every interval $[x,y]$. [/step]