that $r<\ell$. We set $$\bar{P}=\cdots \underbrace{11\cdots 11}_{\ell-1}\underbrace{00\cdots 00}_{r}11\cdots$$ and obtain $\bar{P}\in \mathcal{D}_n^N(s)$ and $\text{comaj}(\bar{P})=\text{comaj}(\tilde{P})+1.$

(0,0) grid (19,7); (0,0) – (3,3) – (6,0) – (10,4)–(13,1)–(17,5)–(19,7); (9,3) – (12,0) – (13,1); at (8,2) [$\ell$]{}; at (12,2) [$r$]{};

To complete the proof in this case, we have to show that $\bar{P}$ is admissible. Again, there is nothing to show if $n<m$; so suppose that $n\geq m$. We denote by $z$ the peak of $\tilde{P}$ after the $\ell$ up–steps and set for simplicity $b:=\text{ht}_{z}(\tilde{P})$. It is clear that $\bar{P}$ has a peak at height $b-1$. If $b\leq m$, there is nothing to show. Otherwise we must show that $\text{ht}_{\tilde z}(\bar{P})>\tilde a$ for all points $\tilde z\in \text{supp}(\bar{P})$ with $\ell_{\tilde z}(\bar{P})\geq \ell_{z}(\tilde{P})-1$, where $\tilde a$ is the unique non–negative integer such that $(\tilde a,b-1)\in A(m,n)$. For a point $\tilde z\in \text{supp}(\bar{P})$ we have that $\tilde z\in \text{supp}(\tilde{P})$ or $(\ell_{\tilde z}(\bar{P})+1,\text{ht}_{\tilde z}(\bar{P})+1)\in \text{supp}(\tilde{P})$. In either case, $$\text{ht}_{\tilde z}(\bar{P})> a-1\geq \tilde a,$$ where first inequality follows by our assumption $\tilde{P}\in \mathcal{D}_{m,n}(s)$ and the second inequality follows from .

*Case 2:* In this case we can suppose that each peak in $\tilde{P}$ returns to the $x$–axis. Since $\tilde{P}\neq P$, we have that $\tilde{P}$ contains at least two down–steps. 