[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*}
[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]
[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]
