I have a question about the following proof. Suppose a partially ordered set $(P,\le)$ is graded by its height function $h$. All chains are finite and there is a smallest element $0$. Then the poset should satisfy the JDCP.
The proof goes like this: suppse $a=c_0<c_1<\dots<c_n=b$ a maximal chain from $a$ to $b$, then $c_i$ is covered by $c_{i+1}$ and so $h(c_{i+1})=h(c_i)+1$. This implies $h(b)-h(a)=n$. But why must this true for every chain? Why can't there be a maximal chain $a=d_0<\dots<d_m=b$ where $m<n$?
Thanks for clarification!
If $a=d_0<\ldots<d_m=b$, then certainly $h(d_{k+1})\ge h(d_k)+1$ for $k=0,\ldots,m-1$. Suppose that $h(d_{k+1})\ge h(d_k)+2$ for some $k$. Then $d_k<d_{k+1}$, but $d_{k+1}$ does not cover $d_k$, so there is some $p\in P$ such that $d_k<p<d_{k+1}$, and the chain $a=d_0<\ldots<d_m=b$ is not maximal: it can be extended to $a=d_0<\ldots<d_k<p<d_{k+1}<\ldots<d_m=b$. Thus, $h(d_{k+1})=h(d_k)+1$ for $k=0,\ldots,m-1$, and $h(b)=h(a)+m$, i.e., $m=h(b)-h(a)$. Every maximal chain from $a$ to $b$ has length $h(b)-h(a)$.