$h\leq i\leq m-1$, which means that ${\mathbf c}\prec' {\mathbf e}$. Moreover, for $f \leq i \leq \ell-1$, $e_i=c_{i+1}\geq d_i$, thus $h'\geq \ell$ and $e_i=c_{i+1}=d_{i+1}$, $h'\leq i\leq m-1.$ Therefore ${\mathbf d}\prec' {\mathbf e}$.

3.  Let us assume that $f=1$ 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 ${\mathbf d}\prec' {\mathbf e}$. From Lemma \[propbat2\] we obtain (\[eff’=1bis\]).

    Conversely, let us assume that $f=1$ and (\[eff’=1bis\]) holds. By Lemma \[propbat2\], there exists a partition ${\mathbf e}=(e_1, \dots,  e_{m-1})$ such that $\sum_{i=1}^{m-1} e_i=E$ and ${\mathbf c}\prec' {\mathbf e}$. Therefore, $e_i\geq c_{i+1}, \ 1\leq i\leq m-1$.

    Let $h=\min\{i \; : e_i<c_i\}$ and $h'=\min\{i \; : e_i<d_i\}$. As $f=1$, we have $ c_{i}\geq d_{i-1}\geq d_i$, $2\leq i \leq  \ell$. Therefore, $e_i\geq c_{i+1}\geq d_i$, $1\leq i<\ell$ and $h'\geq \ell$. Since $e_{h'}<d_{h'}=c_{h'}$, $h\leq h'$ and $e_{i}=c_{i+1}=d_{i+1}$, $h'\leq i \leq m-1$. Hence, ${\mathbf d}\prec' {\mathbf e}$.

$\Box$

\[remarkL54\] Observe that condition (\[eff’=1\]) implies condition (\[eff’m1\]).

\[lemmacard\] Let ${\mathbf a}=(a_1, \dots,  a_{m})$, ${\mathbf e}=(e_1, \dots,  e_{m-1})$ be partitions of nonnegative integers such that ${\mathbf a}\prec' {\mathbf e}$ and $
 \sum_{i=1}^{m-1}e_i \leq \sum_{i=1}^{m}a_i.
$ Let $\theta=\#\{i \; : \; e_i>0\}$ and $\overline \theta=\#\{i \; : \; a_i>0\}$. Then $\overline \theta 