is equivalent to $f=1$.

Let us prove the different cases.

1.  Assume that there exists a partition ${\mathbf g}=(g_1, \dots,  g_{m+1})$ such that $\sum_{i=1}^{m+1} g_i=S$, ${\mathbf g}\prec'{\mathbf c}$ and ${\mathbf g}\prec'{\mathbf d}$.

    Let $h=\min\{i \; : c_i<g_i\}$ and $h'=\min\{i \; : d_i<g_i\}$. As $g_{\ell+1}\leq d_\ell<c_\ell$, by Remark \[aux\], item \[aux2\], we have $\ell < h$. In the same way, as $g_{f}\leq c_f<d_{f-1}$, $f-1<h'$.

    Therefore, $$\begin{array}{ll}
    g_i\leq\min\{c_i, d_i\}, & 1\leq i \leq f-1,\\
    g_f\leq c_f, \\
    g_{i+1}\leq d_{i}, &f\leq i \leq m,
    \end{array}$$ from where we obtain (\[eqGcdbis\]).

    Conversely, let us assume that (\[eqGcdbis\]) holds.

    -   If $S<\sum_{i=1}^{f-1}\min\{c_i, d_i\}$ then $f>1$. Let $$k=\min\{i\in\{1, \dots f-1\}\; : \; S<\sum_{i=1}^{k}\min\{c_i, d_i\}\},$$ i.e., $\sum_{i=1}^{k-1}\min\{c_i, d_i\}\leq S<\sum_{i=1}^{k}\min\{c_i, d_i\}$ and define $$\begin{array}{ll}
        g_i=\min\{c_i, d_i\}, & 1\leq i \leq k-1,\\
        g_k=S-\sum_{i=1}^{k-1}\min\{c_i, d_i\}, \\
        g_{i}=0, &k+1\leq i \leq m+1.
        \end{array}$$ Then $\sum_{i=1}^{m+1} g_i=S$ and $g_k<\min\{c_k, d_k\}$. Therefore, $g_1\geq \dots \geq g_{k-1}>g_k\geq 0=g_{k+1}= \dots =g_{m+1}$. Thus, ${\mathbf g}=(g_1, \dots, g_{m+1})$ is a partition. As $g_i \leq \min\{c_i, d_i\}$, $1\leq i \leq m$, by Remark \[aux\], item \[aux1\], we have ${\mathbf g}\prec'{\mathbf c}$ and ${\mathbf g}\prec'{\mathbf d}$.

    -   If $S\geq\sum_{i=1}^{f-1}\min\{c_i, d_i\}$, let $\overline S=S- \sum_{i=1}^{f-1}\min\{c_i, d_i\}\geq 0$. Then $\overline S\leq c_f+\sum_{i=f}^{m}d_i.$ We define $
        \overline d_i=d_{f-1+i}$, $1\leq i \leq m-f+1,$ 