[proofplan] We work in the incidence algebra of a finite poset and use the defining identity that the Möbius function is the convolution inverse of the zeta function. Multiplying $\mu_P$ by the zeta function gives a convolution sum over the interval $[x,y]$, and comparing with the delta function gives the diagonal value and the zero-sum relation. Solving the zero-sum relation for the final term gives the displayed recursion. Finally, induction on interval cardinality shows that the recursion only uses the induced order structure of the interval, so $\mu_P(x,y)$ is invariant under interval isomorphism. [/proofplan]

[step:Extract the recursion from the convolution inverse identity]We use the incidence algebra construction for finite posets, as in [citetheorem:8100]. Let $I(P;\mathbb Z)$ denote the incidence algebra of $P$ over $\mathbb Z$: its elements are functions \begin{align*} f:\{(a,b)\in P\times P:a\le b\}\to \mathbb Z. \end{align*} For $f,g\in I(P;\mathbb Z)$, their convolution product $f*g\in I(P;\mathbb Z)$ is defined by \begin{align*} (f*g)(a,b)=\sum_{a\le c\le b}f(a,c)g(c,b). \end{align*} Define the zeta function $\zeta_P\in I(P;\mathbb Z)$ by $\zeta_P(a,b)=1$ for every $a\le b$. Define the delta function $\delta_P\in I(P;\mathbb Z)$ by requiring $\delta_P(a,a)=1$ for every $a\in P$ and $\delta_P(a,b)=0$ for every $a,b\in P$ with $a<b$. By definition, the Möbius function $\mu_P\in I(P;\mathbb Z)$ is the convolution inverse of $\zeta_P$, so \begin{align*} \mu_P*\zeta_P=\delta_P. \end{align*} Evaluating this identity at a comparable pair $x\le y$ gives \begin{align*} \sum_{x\le z\le y}\mu_P(x,z)\zeta_P(z,y)=\delta_P(x,y). \end{align*} Since $\zeta_P(z,y)=1$ for every $z$ with $x\le z\le y$, this becomes \begin{align*} \sum_{x\le z\le y}\mu_P(x,z)=\delta_P(x,y). \end{align*} If $x=y$, the sum has the single term $z=x$, and $\delta_P(x,x)=1$, hence \begin{align*} \mu_P(x,x)=1. \end{align*} If $x<y$, then $\delta_P(x,y)=0$, hence \begin{align*} \sum_{x\le z\le y}\mu_P(x,z)=0. \end{align*} Separating the term $z=y$ from this finite sum gives \begin{align*} \sum_{x\le z<y}\mu_P(x,z)+\mu_P(x,y)=0. \end{align*} Therefore \begin{align*} \mu_P(x,y)=-\sum_{x\le z<y}\mu_P(x,z). \end{align*}[/step]

[guided]The incidence algebra packages the recursion into one multiplication identity. Its elements are functions \begin{align*} f:\{(a,b)\in P\times P:a\le b\}\to \mathbb Z, \end{align*} and the convolution product is \begin{align*} (f*g)(a,b)=\sum_{a\le c\le b}f(a,c)g(c,b). \end{align*} The finiteness of $P$ ensures that every interval $\{c\in P:a\le c\le b\}$ is finite, so the displayed sum is a finite integer sum. Define $\zeta_P\in I(P;\mathbb Z)$ by $\zeta_P(a,b)=1$ for every comparable pair $a\le b$. Define $\delta_P\in I(P;\mathbb Z)$ by requiring $\delta_P(a,a)=1$ for every $a\in P$ and $\delta_P(a,b)=0$ for every $a,b\in P$ with $a<b$. The Möbius function $\mu_P$ is defined as the convolution inverse of $\zeta_P$. Thus \begin{align*} \mu_P*\zeta_P=\delta_P. \end{align*} Now fix $x,y\in P$ with $x\le y$. Evaluating the convolution identity at $(x,y)$ gives \begin{align*} (\mu_P*\zeta_P)(x,y)=\sum_{x\le z\le y}\mu_P(x,z)\zeta_P(z,y). \end{align*} Because $z\le y$ for every index $z$ in this sum, the definition of $\zeta_P$ gives $\zeta_P(z,y)=1$. Therefore the identity $\mu_P*\zeta_P=\delta_P$ becomes \begin{align*} \sum_{x\le z\le y}\mu_P(x,z)=\delta_P(x,y). \end{align*} There are two cases. If $x=y$, then the only element $z$ satisfying $x\le z\le x$ is $z=x$, by antisymmetry of the partial order. Hence the sum reduces to \begin{align*} \mu_P(x,x)=\delta_P(x,x)=1. \end{align*} If $x<y$, then $\delta_P(x,y)=0$, so \begin{align*} \sum_{x\le z\le y}\mu_P(x,z)=0. \end{align*} The term with $z=y$ is exactly $\mu_P(x,y)$. Splitting it off from the finite sum gives \begin{align*} \sum_{x\le z<y}\mu_P(x,z)+\mu_P(x,y)=0. \end{align*} Solving for $\mu_P(x,y)$ yields \begin{align*} \mu_P(x,y)=-\sum_{x\le z<y}\mu_P(x,z). \end{align*} This is the desired interval recursion.[/guided]

[step:Prove interval dependence by induction on interval cardinality] Let $P$ and $Q$ be finite posets, let $x\le y$ in $P$, and let $u\le v$ in $Q$. Suppose \begin{align*} \varphi:[x,y]\to [u,v] \end{align*} is a poset isomorphism. We prove $\mu_P(x,y)=\mu_Q(u,v)$ by induction on $N=|[x,y]|$. If $N=1$, then $x=y$. Since $\varphi$ is an isomorphism, $u=v$, and the diagonal formula gives \begin{align*} \mu_P(x,y)=\mu_P(x,x)=1=\mu_Q(u,u)=\mu_Q(u,v). \end{align*} Assume $N>1$, so $x<y$, and assume the result holds for all isomorphic intervals of cardinality smaller than $N$. A poset isomorphism preserves the unique minimum and maximum of an interval, so $\varphi(x)=u$ and $\varphi(y)=v$. The recursion in $P$ gives \begin{align*} \mu_P(x,y)=-\sum_{x\le z<y}\mu_P(x,z). \end{align*} For each $z\in [x,y)$, the restriction of $\varphi$ defines a poset isomorphism \begin{align*} [x,z]\to [u,\varphi(z)]. \end{align*} The interval $[x,z]$ has smaller cardinality than $[x,y]$ because $y\notin [x,z]$. By the induction hypothesis, \begin{align*} \mu_P(x,z)=\mu_Q(u,\varphi(z)). \end{align*} Since $\varphi$ bijects the set $\{z\in P:x\le z<y\}$ with the set $\{w\in Q:u\le w<v\}$, substitution in the finite sum gives \begin{align*} \mu_P(x,y)=-\sum_{u\le w<v}\mu_Q(u,w). \end{align*} Applying the same recursion in $Q$ gives \begin{align*} -\sum_{u\le w<v}\mu_Q(u,w)=\mu_Q(u,v). \end{align*} Thus $\mu_P(x,y)=\mu_Q(u,v)$, completing the induction. Therefore the value of $\mu_P(x,y)$ depends only on the isomorphism type of the interval $[x,y]$. [/step]