Let $(P,\leq)$ be a partial order with $P = \lbrace 0,1,2,3,4 \rbrace$ such that $0 \leq 1 \leq 4$,$0 \leq 2 \leq 4$,$0 \leq 3 \leq 4$. I want to calculate the values of the mobius function $\mu : P \times P \to \mathbb{R}$ of this partial order which is defined as the inverse of the zeta function \begin{equation} \zeta(x,y) = \begin{cases} 1 & x \leq y \\ 0 & \mbox{otherwise} \end{cases}. \end{equation} Now I know that the mobius function must satisfy \begin{align*} \mu(x,x) &= 1, \\ \mu(x,y) &= -\sum_{x < z \leq y}\mu(z,y), \\ \mu(x,y) &= -\sum_{x \leq z < y}\mu(x,z). \end{align*} So I get $\mu(i,i) = 1$ for all $i \in P$ and since $0$ is a minimal element and $4$ a maximal \begin{align} \mu(x,x) &= 1 \mbox{ for all } x \in P, \\ \mu(x,0) &= 0, \mbox{ for all } x \in \lbrace 1,2,3,4 \rbrace, \\ \mu(0,4) &= -(\mu(1,4) + \mu(2,4) + \mu(3,4) + 1), \\ \mu(0,4) &= -(1 + \mu(0,1) + \mu(0,2) + \mu(0,3)). \end{align} Since $i$ and $j$ for $i \neq j$ with $i,j \in \lbrace 1,2,3 \rbrace$ are not comparable we get \begin{equation*} \mu(1,2) = \mu(2,1) = \mu(2,3) = \mu(3,2) = \mu(3,1) = \mu(1,3) = 0. \end{equation*} Now I stuck at calculating $\mu(1,4), \mu(2,4), \mu(3,4), \mu(0,1), \mu(0,2)$ and $\mu(0,3)$.
Edit: I think it works out if I use from the definition of the mobius function \begin{equation} \sum_{x \leq z \leq y}\mu(x,z) = \delta(x,y) \end{equation} and since only $i$ and $4$ are comparable with $i$ in the interval [i,4] and $0$ is the minimal element we get \begin{align*} 0 &= \delta(1,4) = \mu(1,1) + \mu(1,4), \\ 0 &= \delta(2,4) = \mu(2,2) + \mu(2,4), \\ 0 &= \delta(3,4) = \mu(3,3) + \mu(3,4) \end{align*} thus \begin{equation*} \mu(1,4) = \mu(2,4) = \mu(3,4) = -1 \end{equation*} therefore \begin{equation*} \mu(0,4) = -\mu(1,4) - \mu(2,4) - \mu(3,4) - 1 = 2. \end{equation*} With the same strategy since $0$ is the minimal element and $4$ the maximal one has \begin{align*} 0 &= \delta(0,1) = \mu(0,0) + \mu(0,1), \\ 0 &= \delta(0,2) = \mu(0,0) + \mu(0,2), \\ 0 &= \delta(0,3) = \mu(0,0) + \mu(0,3) \end{align*} hence \begin{equation*} \mu(0,1) = \mu(0,2) = \mu(0,3) = -1. \end{equation*}