Lemma \[l:max\]. Let $s,v \in V$. It is easy to see that $f(I)=f_{sv}(I)=\lambda^q_{G+I}(s,v)$ is non-decreasing, and that the second statement in the lemma holds. Thus it remains to prove that $f(I)$ is submodular. For that, we will use the following known characterization of submodularity, c.f. [@Sch]:\
[*A set-function $f$ on $F$ is submodular if, and only if*]{} $$f(I_0 \cup \{e\})+f(I_0 \cup \{e'\}) \geq f(I_0)+f(I_0 \cup \{e,e'\})  \ \ \ \forall I_0 \subset F, e,e' \in F \setminus I_0 $$ Let us fix $I_0 \subseteq F$. Revising our notation to $G \gets G+I_0$, $F \gets F \setminus I_0$, and denoting $h(I)=f(I_0 \cup I)-f(I_0)$, we get that $f$ is submodular if, and only if $$h(\{e\})+h(\{e'\}) \geq h(\{e,e'\})  \ \ \ \forall e,e' \in F  \ .$$ In our setting, $h(I)=\lambda^q_{G+I}(s,v)-\lambda^q_G(s,v)$ is the increase in the $(s,v)$-$q$-connectivity as a result of adding $I$ to $G$. Thus $0 \leq h(I) \leq |I|$ for any $I \subseteq F$, so $0 \leq h(\{e,e'\}) \leq 2$. If $h(\{e,e'\})=0$, then we are done; if $h(\{e,e'\})=1$, then we need to show that $h(\{e\})=1$ or $h(\{e'\})=1$; and if $h(\{e,e'\})=2$, then we need to show that $h(\{e\})=1$ and $h(\{e'\})=1$. We prove the following general statement, that implies the above.

\[l:J\] Let 