[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 $I(P;R)$.[/step]

[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: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]