[proofplan] A saturated chain from $x$ to $y$ is a chain whose consecutive elements are cover relations. Since $P$ is graded with rank function $\rho$, each cover relation increases rank by exactly one. Applying this to every adjacent pair in the chain and summing the rank increments gives that the chain length is $\rho(y)-\rho(x)$. A maximal chain in the interval $[x,y]$ is precisely a saturated chain from $x$ to $y$, so the same length formula applies to all such chains. [/proofplan] [step:Compute the rank increase along a saturated chain] Let $x=x_0\lessdot x_1\lessdot \cdots \lessdot x_m=y$ be a saturated chain from $x$ to $y$, where $m\in \mathbb{N}\cup\{0\}$ is its length. By the defining rank property of a graded poset, whenever $u\lessdot v$ in $P$, one has \begin{align*} \rho(v)=\rho(u)+1. \end{align*} Applying this to the cover relation $x_{i-1}\lessdot x_i$ for each $i\in\{1,\dots,m\}$ gives \begin{align*} \rho(x_i)-\rho(x_{i-1})=1. \end{align*} Summing these $m$ equalities and using telescoping yields \begin{align*} \rho(y)-\rho(x)=\rho(x_m)-\rho(x_0)=m. \end{align*} Thus every saturated chain from $x$ to $y$ has length $\rho(y)-\rho(x)$. [guided] Let us spell out why the rank function controls the length. A saturated chain from $x$ to $y$ means a finite sequence $x=x_0\lessdot x_1\lessdot \cdots \lessdot x_m=y$ in which each adjacent relation $x_{i-1}\lessdot x_i$ is a cover relation. The integer $m\in\mathbb{N}\cup\{0\}$ is the number of cover steps in the chain, hence the length of the chain. The hypothesis that $P$ is graded with rank function $\rho:P\to\mathbb{N}\cup\{0\}$ means, in particular, that every cover relation raises rank by exactly one. Therefore, for each $i\in\{1,\dots,m\}$, the cover relation $x_{i-1}\lessdot x_i$ gives \begin{align*} \rho(x_i)=\rho(x_{i-1})+1. \end{align*} Equivalently, \begin{align*} \rho(x_i)-\rho(x_{i-1})=1. \end{align*} Now add these equalities over all indices $i\in\{1,\dots,m\}$. The left-hand side telescopes: \begin{align*} \sum_{i=1}^{m}\bigl(\rho(x_i)-\rho(x_{i-1})\bigr)=\rho(x_m)-\rho(x_0). \end{align*} The right-hand side is the sum of $m$ copies of $1$, so it equals $m$. Since $x_0=x$ and $x_m=y$, we obtain \begin{align*} \rho(y)-\rho(x)=m. \end{align*} This proves that the length of the saturated chain is determined only by the endpoints $x$ and $y$, not by the particular chain chosen. [/guided] [/step] [step:Identify maximal chains in the interval with saturated chains] Let $[x,y]=\{z\in P:x\le z\le y\}$ be the interval from $x$ to $y$, ordered by the restriction of the order on $P$. A chain in $[x,y]$ is maximal in the interval if and only if it begins at $x$, ends at $y$, and no additional element of $[x,y]$ can be inserted between any two consecutive elements. This is exactly the condition that the chain is saturated from $x$ to $y$ in $P$. Therefore every maximal chain in $[x,y]$ is a saturated chain from $x$ to $y$. By the previous step, each such chain has length $\rho(y)-\rho(x)$. Hence all maximal chains between $x$ and $y$ have the same length. [/step]