[proofplan]
We use the defining fact that the Möbius element is the two-sided convolution inverse of the zeta element. Expanding the identities $\mu*\zeta=\delta$ and $\zeta*\mu=\delta$ on an interval $[x,y]$ gives the diagonal value and the two recursion formulas. For uniqueness, we use induction on the finite cardinality of $[x,y]$: the first recursion expresses $\mu(x,y)$ in terms of values on strictly smaller intervals.
[/proofplan]
[step:Expand the inverse identities on an arbitrary interval]
Let $\zeta\in I(P;R)$ denote the function with $\zeta(x,y)=1_R$ for all $x\le y$, and let $\delta\in I(P;R)$ denote the function with $\delta(x,x)=1_R$ and $\delta(x,y)=0_R$ for $x<y$. By the hypothesis that $\mu$ is the two-sided convolution inverse of $\zeta$, we have
\begin{align*}
\mu*\zeta=\delta
\end{align*}
and
\begin{align*}
\zeta*\mu=\delta.
\end{align*}
Since $P$ is locally finite, every interval $[x,y]=\{z\in P:x\le z\le y\}$ is finite, so all convolution sums below are finite.
For $x\le y$, the convolution formula gives
\begin{align*}
(\mu*\zeta)(x,y)=\sum_{x\le z\le y}\mu(x,z)\zeta(z,y)=\sum_{x\le z\le y}\mu(x,z)
\end{align*}
because $\zeta(z,y)=1_R$ for each $z$ with $x\le z\le y$. Similarly,
\begin{align*}
(\zeta*\mu)(x,y)=\sum_{x\le z\le y}\zeta(x,z)\mu(z,y)=\sum_{x\le z\le y}\mu(z,y).
\end{align*}
[/step]
[step:Read off the diagonal value and both recursion formulas]
Set $y=x$ in the identity $\mu*\zeta=\delta$. The interval $[x,x]$ has the single element $x$, so
\begin{align*}
\mu(x,x)=\sum_{x\le z\le x}\mu(x,z)=\delta(x,x)=1_R.
\end{align*}
Now let $x,y\in P$ satisfy $x<y$. Since $\delta(x,y)=0_R$, the identity $\mu*\zeta=\delta$ gives
\begin{align*}
\sum_{x\le z\le y}\mu(x,z)=0_R.
\end{align*}
The identity $\zeta*\mu=\delta$ gives, again because $x<y$,
\begin{align*}
\sum_{x\le z\le y}\mu(z,y)=0_R.
\end{align*}
This proves all displayed equations for the Möbius element.
[/step]
[step:Use finite interval induction to prove uniqueness]
Let $\nu\in I(P;R)$ be any element satisfying
\begin{align*}
\nu(x,x)=1_R
\end{align*}
for every $x\in P$, and
\begin{align*}
\sum_{x\le z\le y}\nu(x,z)=0_R
\end{align*}
for every $x<y$. We prove that $\nu=\mu$.
Let
\begin{align*}
C_P=\{(a,b)\in P\times P:a\le b\}
\end{align*}
be the set of comparable ordered pairs in $P$. Define the finite-interval cardinality map $N:C_P\to \mathbb N$ by
\begin{align*}
N(a,b)=|[a,b]|.
\end{align*}
We prove by induction on $N(x,y)$ that $\nu(x,y)=\mu(x,y)$ for every comparable pair $(x,y)\in C_P$.
If $N(x,y)=1$, then $x=y$, and both functions have value $1_R$ on the diagonal:
\begin{align*}
\nu(x,x)=1_R=\mu(x,x).
\end{align*}
Now assume $N(x,y)>1$, so $x<y$, and assume that $\nu(a,b)=\mu(a,b)$ whenever $a\le b$ and $N(a,b)<N(x,y)$. Applying the recursion to $\nu$ gives
\begin{align*}
\nu(x,y)=-\sum_{x\le z<y}\nu(x,z).
\end{align*}
The Möbius element satisfies the same recursion, so
\begin{align*}
\mu(x,y)=-\sum_{x\le z<y}\mu(x,z).
\end{align*}
For every $z$ with $x\le z<y$, the interval $[x,z]$ is a proper subset of $[x,y]$, because $y\in [x,y]$ but $y\notin [x,z]$. Hence
\begin{align*}
N(x,z)<N(x,y).
\end{align*}
The induction hypothesis gives $\nu(x,z)=\mu(x,z)$ for every such $z$, and substituting into the two finite sums yields
\begin{align*}
\nu(x,y)=\mu(x,y).
\end{align*}
Thus $\nu=\mu$ on every comparable pair, so $\nu=\mu$ in $I(P;R)$.
[guided]
The uniqueness argument is an induction over finite intervals. The point of using interval cardinality is that the recurrence for the value at $(x,y)$ only involves values at $(x,z)$ with $z<y$, and each interval $[x,z]$ is strictly smaller than $[x,y]$.
Let $\nu\in I(P;R)$ be another element satisfying the same diagonal condition and the first recursion:
\begin{align*}
\nu(x,x)=1_R
\end{align*}
for every $x\in P$, and
\begin{align*}
\sum_{x\le z\le y}\nu(x,z)=0_R
\end{align*}
whenever $x<y$. Let
\begin{align*}
C_P=\{(a,b)\in P\times P:a\le b\}
\end{align*}
be the set of comparable ordered pairs in $P$. Define the finite-interval cardinality map $N:C_P\to \mathbb N$ by
\begin{align*}
N(a,b)=|[a,b]|.
\end{align*}
This is finite because $P$ is locally finite.
We prove by induction on $N(x,y)$ that $\nu(x,y)=\mu(x,y)$. If $N(x,y)=1$, then the interval has only one element, so $x=y$. The diagonal condition for $\nu$ and the already-proved diagonal condition for $\mu$ give
\begin{align*}
\nu(x,x)=1_R=\mu(x,x).
\end{align*}
Now suppose $N(x,y)>1$, so $x<y$, and assume the equality is known for all comparable pairs whose interval cardinality is smaller than $N(x,y)$. The recursion for $\nu$ says
\begin{align*}
0_R=\sum_{x\le z\le y}\nu(x,z)=\nu(x,y)+\sum_{x\le z<y}\nu(x,z).
\end{align*}
Rearranging in the additive group of the ring $R$ gives
\begin{align*}
\nu(x,y)=-\sum_{x\le z<y}\nu(x,z).
\end{align*}
The same calculation applies to $\mu$, because $\mu$ satisfies the same recursion:
\begin{align*}
\mu(x,y)=-\sum_{x\le z<y}\mu(x,z).
\end{align*}
For every $z$ with $x\le z<y$, the interval $[x,z]$ is strictly smaller than $[x,y]$: it is contained in $[x,y]$, and it does not contain $y$. Therefore
\begin{align*}
N(x,z)<N(x,y).
\end{align*}
The induction hypothesis applies to each pair $(x,z)$ appearing in the finite sum, so $\nu(x,z)=\mu(x,z)$ for all $z$ with $x\le z<y$. Substituting these equalities into the recurrence formulas gives
\begin{align*}
\nu(x,y)=\mu(x,y).
\end{align*}
This completes the induction, and hence $\nu=\mu$ as elements of $I(P;R)$.
[/guided]
[/step]
[step:Conclude that the displayed equations characterize the Möbius element]
The preceding induction shows that any element of $I(P;R)$ satisfying the diagonal equation and the first recursion must equal $\mu$. Therefore the listed equations determine the Möbius element uniquely. Since $\mu$ itself satisfies both recursions by the convolution inverse identities, the theorem follows.
[/step]
