or smaller than that of $B_{i+1}$ (by definition they cannot be equal). Let $j<k$ be such that $B_j$ and $B_k$ are zero colored blocks, but $B_{j+1},\dots,B_{k-1}$ are not. In the factor $B_j\dots B_k$ of $(\sigma,c)$, there is a contribution $$\#\big\{ i \,:\,  j\leq i <k \text{ and } B_i > B_{i+1}  \big\}$$ to $\lceil \operatorname{fdes}(\sigma,c) /r \rceil$. But in the factor $B_j B_{k-1}\dots B_{j+1} B_k$ of $I(\sigma,c)$, there is the same contribution to $\lceil \operatorname{fdes}^*(I(\sigma,c)) /r \rceil$.

Now, let $B_j$ be the first zero colored block of $(\sigma,c)$. If $j>1$, the prefix $B_1\dots B_j$ of $(\sigma,c)$ contributes by $$\#\big\{ i \,:\,  1\leq i <j \text{ and } B_i > B_{i+1}  \big\}$$ to $\lceil \operatorname{fdes}(\sigma,c) /r \rceil$, and the prefix $B_{j-1}\dots B_1 B_j$ of $I(\sigma,c)$ contributes by $$1 + \#\big\{ i \,:\,  j-1 > i \geq 1 \text{ and } B_{i+1} < B_i  \big\}$$ to $\lceil \operatorname{fdes}^*(I(\sigma,c)) /r \rceil$ (the 1 come from the term $c_1$ in the definition of $\operatorname{fdes}^*$ since $B_{j-1}$ is a nonzero colored block). The two numbers are easily seen to be equal.

Similarly, let $B_k$ be the last zero colored block of $(\sigma,c)$. If $k<m$, the suffix $B_k\dots B_m$ of $(\sigma,c)$ contributes by $$1+ \#\big\{ i 