Grothendieck polynomials.

\[lem.dominance\] Let $\deg G_{h,\lambda}(\x)=|\lambda|+N$, and let $\{\lambda^{(0)}, \ldots, \lambda^{(N)}\}$ be the sequence of dominating partitions for $G_{h,\lambda}(\x)$.

1.  For $k=0,\ldots, N$, the partition $\lambda^{(k)}$ dominates all other partitions $\mu\in A(h,\lambda)$ such that $\mu \vdash |\lambda|+k$.

2.  The partition $\lambda^{(N)}$ is the unique partition $\mu\in A(h,\lambda)$ such that $\mu \vdash |\lambda|+N$.

3.  $A(h,\lambda) = \{\mu \mid \lambda \subseteq \mu \subseteq \lambda^{(N)} \}. $

Let $\mu\in A(h,\lambda)$ such that $\mu \vdash |\lambda|+k$. Suppose for contradiction that $\lambda^{(k)}$ does not dominate $\mu$, so that there exists a minimum $s>1$ such that $\mu_1+\cdots+\mu_{s-1} \leq \lambda_1^{(k)} + \cdots +\lambda_{s-1}^{(k)}$ but $\mu_1+\cdots+\mu_s > \lambda_1^{(k)} + \cdots +\lambda_s^{(k)}$. This implies $\mu_s > \lambda_s^{(k)}$.

The partition $\lambda^{(k)}$ was obtained by adding a box to $\lambda^{(k-1)}$ in the $r_k$-th row. If $s< r_k$, then $$\lambda_s^{(k-1)}-\lambda_s \leq \lambda_s^{(k)}-\lambda_s < \mu_s - \lambda_s \leq h(s-1),$$ so a box would have been added to $\lambda^{(k-1)}$ in the $s$-th row to obtain $\lambda^{(k)}$, contradicting the construction of $\lambda^{(k)}$. Thus $s\geq r_k$. But then by the construction of $\lambda^{(k)}$, for all $j\geq r_k$, $$(\lambda_1^{(k)}+\cdots+\lambda_{j}^{(k)}) - \left(\lambda_1+\cdots+\lambda_{j} \right) = k 
\geq (\mu_1+\cdots+\mu_{j}) - \left(\lambda_1+\cdots+\lambda_{j} \right),$$ which contradicts the existence of $s$, so part (a) holds.

Parts (b) and (c) follow from the maximality 