[proofplan] We fix a comparable pair $x<y$ and use Philip Hall's theorem to identify the Möbius value $\mu_P(x,y)$ with the reduced Euler characteristic of the order complex $\triangle((x,y))$. The cover case is separated because then the open interval is empty, so the stated convention gives reduced Euler characteristic $-1$. In rank difference at least $2$, the hypothesis gives a homotopy equivalence from $\triangle((x,y))$ to a wedge of $m_{x,y}$ spheres, and homotopy invariance of reduced Euler characteristic reduces the computation to the reduced homology of a wedge of spheres. [/proofplan]

[step:Apply Philip Hall's theorem to the open interval] Fix $x,y\in P$ with $x<y$. Since $P$ is finite, it is locally finite, and the open interval \begin{align*} (x,y)=\{z\in P:x<z<y\} \end{align*} is a finite poset. By [citetheorem:8124], applied to the pair $x<y$ in $P$, the Möbius function satisfies \begin{align*} \mu_P(x,y)=\widetilde{\chi}\bigl(\triangle((x,y))\bigr), \end{align*} where $\widetilde{\chi}$ denotes reduced Euler characteristic. [/step]

[step:Compute the cover-relation case from the empty interval] Assume that $y$ covers $x$. Then there is no element $z\in P$ satisfying $x<z<y$, so \begin{align*} (x,y)=\varnothing. \end{align*} Hence $\triangle((x,y))$ is the order complex of the empty open interval. By the convention stated in the theorem, its reduced Euler characteristic is \begin{align*} \widetilde{\chi}\bigl(\triangle((x,y))\bigr)=-1. \end{align*} Combining this with the formula from Philip Hall's theorem gives \begin{align*} \mu_P(x,y)=-1. \end{align*} [/step]

[step:Compute the reduced Euler characteristic of the wedge of spheres]Assume that $\rho(y)-\rho(x)\ge 2$, and define the integer \begin{align*} d=\rho(y)-\rho(x)-2. \end{align*} Then $d\ge 0$. By hypothesis, $\triangle((x,y))$ is homotopy equivalent to a wedge $W_{x,y}$ of $m_{x,y}$ spheres of dimension $d$. Reduced Euler characteristic is invariant under homotopy equivalence, so \begin{align*} \widetilde{\chi}\bigl(\triangle((x,y))\bigr)=\widetilde{\chi}(W_{x,y}). \end{align*} The reduced homology of $W_{x,y}$ is free of rank $m_{x,y}$ in degree $d$ and vanishes in all other degrees. Therefore \begin{align*} \widetilde{\chi}(W_{x,y})=(-1)^d m_{x,y}. \end{align*} Substituting $d=\rho(y)-\rho(x)-2$ gives \begin{align*} \widetilde{\chi}\bigl(\triangle((x,y))\bigr)=(-1)^{\rho(y)-\rho(x)-2}m_{x,y}. \end{align*}[/step]

[guided]Assume that $\rho(y)-\rho(x)\ge 2$. The rank difference determines the dimension of the spheres appearing in the hypothesis, so we define \begin{align*} d=\rho(y)-\rho(x)-2. \end{align*} Because $\rho(y)-\rho(x)\ge 2$, this integer satisfies $d\ge 0$. The hypothesis gives a homotopy equivalence from the order complex $\triangle((x,y))$ to a wedge $W_{x,y}$ of $m_{x,y}$ copies of the sphere $S^d$. Reduced Euler characteristic is a homotopy invariant, so replacing $\triangle((x,y))$ by the homotopy equivalent space $W_{x,y}$ does not change the value of $\widetilde{\chi}$. Thus \begin{align*} \widetilde{\chi}\bigl(\triangle((x,y))\bigr)=\widetilde{\chi}(W_{x,y}). \end{align*} It remains only to compute the reduced Euler characteristic of $W_{x,y}$. A wedge of $m_{x,y}$ spheres of dimension $d$ has reduced homology concentrated in degree $d$, where the reduced homology group is free of rank $m_{x,y}$. In every other degree the reduced homology group is zero. Since reduced Euler characteristic is the alternating sum of the ranks of reduced homology groups, this gives \begin{align*} \widetilde{\chi}(W_{x,y})=(-1)^d m_{x,y}. \end{align*} If $m_{x,y}=0$, this formula gives $0$, which is exactly the convention that an empty wedge of nonnegative-dimensional spheres contributes reduced Euler characteristic $0$. Finally, substituting the definition of $d$ yields \begin{align*} \widetilde{\chi}\bigl(\triangle((x,y))\bigr)=(-1)^{\rho(y)-\rho(x)-2}m_{x,y}. \end{align*}[/guided]

[step:Substitute the Euler characteristic computation into Philip Hall's formula] For $\rho(y)-\rho(x)\ge 2$, the first step gives \begin{align*} \mu_P(x,y)=\widetilde{\chi}\bigl(\triangle((x,y))\bigr). \end{align*} The previous step computes the right-hand side as \begin{align*} \widetilde{\chi}\bigl(\triangle((x,y))\bigr)=(-1)^{\rho(y)-\rho(x)-2}m_{x,y}. \end{align*} Therefore \begin{align*} \mu_P(x,y)=(-1)^{\rho(y)-\rho(x)-2}m_{x,y}. \end{align*} Together with the cover-relation computation, this proves both asserted formulas for every pair $x<y$ in $P$. [/step]