i \leq f-1,\\
        g_i\leq d_i\leq c_i, & f\leq i \leq h'-1,\\
        g_i\leq d_{i-1}\leq c_i, & f+1\leq i \leq \ell.\\
        \end{array}$$ Thus, $h>\max\{h'-1, \ell\}$ and, as a consequence, $c_i=d_i=g_{i+1}$ for $h\leq i \leq m$. Therefore ${\mathbf g}\prec' {\mathbf c}$.

2.  Assume that $f>1$ (hence $f'>1$) and there exists a partition ${\mathbf e}=(e_1, \dots,  e_{m-1})$ such that $\sum_{i=1}^{m-1} e_i=E$, ${\mathbf c}\prec' {\mathbf e}$ and ${\mathbf d}\prec' {\mathbf e}$. Then $e_i\geq c_{i+1},$ $1\leq i \leq m-1.$ Moreover, $e_{\ell-1}\geq c_\ell>d_\ell$, by Remark \[aux\], item \[aux2\], $e_{i}\geq d_i,$ $1\leq i \leq \ell-1$. Hence, $e_{f-1}\geq d_{f-1}>c_f$, and as before it means that $e_{i}\geq c_{i},$ $1\leq i \leq f-1.$ Thus, $$\begin{array}{ll}
    e_i\geq \max\{c_i,d_i\} & 1\leq i \leq f-1,\\
    e_i\geq  c_{i+1}, & f\leq i \leq m-1,\\
    \end{array}$$ and we obtain (\[eff’m1bis\]).

    Conversely, let us assume that $f>1$ and (\[eff’m1bis\]) holds. Let us define $$\begin{array}{ll}
    e_i=\max\{c_i, d_i\}, & 2\leq i\leq f-1,\\
    e_i=c_{i+1}, & f\leq i\leq m-1,\\
    e_1=E-\sum_{i=2}^{m-1}e_i.
    \end{array}$$ Then, $E=\sum_{i=1}^{m-1}e_i$, $e_2\geq \dots \geq e_{f-1}\geq c_{f-1} \geq c_{f+1} =e_f\geq \dots \geq e_{m-1}$ and, from (\[eff’m1bis\]) we derive $ e_1\geq \max\{c_1, d_1\}\geq e_2$. Therefore $(e_1, \dots,  e_{m-1})$ is a partition. Let $h=\min\{i \; : e_i<c_i\}$ and $h'=\min\{i \; : e_i<d_i\}$. It is clear that $h\geq f$ and $h'\geq f$, hence $e_i=c_{i+1},$ 