W(\lambda)^\ge$, where $\star$ denotes the topological join,so $\mu(W(\lambda))=-\mu(W(\lambda)^<)\star\mu( W(\lambda)^\ge)$ by [@Koz08 Theorem 10.23(2)]. Therefore, $$-\mu(W(\lambda))=\mu(\sigma,\lambda)\sum_{S\in EZ^{\lambda,\pi}}(-1)^{|S|}.$$ So we start with $\mu(A^{\sigma,\pi})$, given by Lemma \[lem:muA\], and then subtract $\mu(W(\lambda))$ for each $\lambda\in (\sigma,\pi)$, which gives the desired formula.

\[rem:tests\] Computer tests[^1] indicate that $~95\%$ of intervals $[\sigma,\pi]$, where $|\pi|<9$, satisfy $\mu(\sigma,\pi)=(-1)^{|\pi|-|\sigma|}\operatorname{\text{\rm NE}}(\sigma,\pi)$. Thus, for these intervals the latter term in Equation is zero.

The complexity of counting the number of normal embeddings is polynomial so in the cases where we can show that the latter term of Equation  equals zero we have a polynomial time formula for the Möbius function. This is a dramatic improvement over the original recursive formula that has exponential complexity. However, computing the latter term of Equation also has exponential complexity.

Tests show that using Equation is often much quicker than computing the Möbius function using the recursive formula. When computing the Möbius function of the rank $15$ interval $$[54123, 9\, 7\, 10\, 4\, 8\, 1\, 2\, 6\, 5\, 3\, 19 \,17\, 20\, 14 \,18\, 11\, 12 \,16\, 15\, 13],$$ the formula in Equation took 1.75 minutes and the recursive formula took 13.5 hours. Note that this interval has Möbius function $-3$ but no normal embeddings 