the $(r-1)$-th and $r$-th rows of $\lambda^{(N)}$ have the same length. In other words, $\lambda_{r-1}+a_{r-1} = \lambda_r + a_r$. But by the induction hypothesis, $$\begin{aligned}
(t\lambda)_{r-1}^{(N')} 
    &= t\lambda_{r-1} +a_{r-1}'
    = t\lambda_{r-1} +ta_{r-1}
    = t\lambda_{r} +ta_{r}
    < t\lambda_r + a_r'
    = (t\lambda)_r^{(N)},\end{aligned}$$ which contradicts the fact that $(t\lambda)^{(N)}$ is a partition. Therefore, $ta_r = a_r'$ for $r=1,\ldots,m$. Since $t\lambda^{(N)} \vdash t|\lambda|+N'$, it follows from Lemma \[lem.dominance\](b) that $t\lambda^{(N)}=(t\lambda)^{(N')}$. The result follows.

Let $\lambda$ be a partition with at most $m$ parts and let $\x = (x_1,\ldots, x_m)$. Then the Newton polytope ${\mathrm{Newt}}(G_{h,\lambda}(\x))$ has the integer decomposition property.

Let $\mathcal{P} = {\mathrm{Newt}}(G_{h,\lambda}(\x))$ and $\mathcal{Q} = {\mathrm{Newt}}(G_{th,t\lambda}(\x))$. By Proposition \[prop.dilate\], $t\mathcal{P} = \mathcal{Q}$.

Let $\nu = t\lambda$ and let $\p$ be a lattice point in the $t$-th dilate $t\mathcal{P} = \mathcal{Q}$. The polynomial $G_{th,t\lambda}(\x)$ has SNP, and by Proposition \[proposition\_vertices\_of\_sgp\], the point $\p$ is a lattice point in ${\mathrm{Newt}}(s_{\nu^{(k)}})$ for some $k$, so $\p$ is the content vector of a semistandard Young tableau $T$ of shape $\nu^{(k)} \in A(th,t\lambda)$. The tableau $T$ decomposes into $t$ semistandard Young tableaux $T_1,\ldots, T_t$, where the tableau $T_i$ of shape $\theta(i)$ is obtained by taking the $j$-th columns of $T$ for $j \equiv i \mod t$. We 