fix a general curve $[C]\in \cM_g$ and a globally generated rank $2$ vector bundle $E$ on $C$ with $\mathrm{Cliff}(C)+2\leq \mu(E)\leq g-1$ and $L:=\mbox{det}(E)\in \mbox{Pic}^d(C)$. Let us assume that inequality (\[mercat1\]) does not hold, that is, $$\label{contradiction}
d<2\bigl(h^0(C, E)-2+\mathrm{Cliff}(C)\bigr).$$ Then, as pointed out, $E$ admits no sub-pencils and $h^0(C, L)\geq 2h^0(C, E)-3$. Since $C$ satisfies the Brill-Noether theorem, one writes $\rho\bigl(g, 2h^0(C, E)-4, d\bigr)\geq 0$. Coupled with assumption (\[contradiction\]), this forces $h^0(C, E)\leq 5$, and then, $h^0(C, E)=5, g=15$ and $d\leq 19$. There is no harm in assuming $d=19$, because if $\mathcal{BN}_C(19, 5)=\emptyset$, the same statement holds for lower degree by carrying out generic elementary transformations.

Therefore $E\in \mathcal{BN}_C(19, 5)$ and from Proposition \[restrquadrics\], one finds that $$\mbox{dim } \mbox{Ker } \nu_2(L)\geq 5=\mbox{dim } H^0\bigl(\PP^9, \I_{G(2, 5)/\PP^9}(2)\bigr).$$ Using again that $C$ is Brill-Noether general, we observe that $h^0(C, L)=7$,  $h^0(C, L^{\otimes 2})=\chi(C, L^{\otimes 2})=24$ and $A:=K_C\otimes L^{\vee}\in W^1_9(C)$ is a pencil of minimal degree. We infer that $\nu_2(L)$ is not surjective, and there exists a vector bundle $F\in \mathcal{SU}^s_C(2, K_C)$ in an extension $$0\rightarrow A\rightarrow F\rightarrow L\rightarrow 0,$$ such that $h^0(C, F)=h^0(C, A)+h^0(C, L)=2+7=9$. The proof that $F$ is stable is standard, cf. [@L] Prop. V.4. Applying [@T3], one can assume 