Given an undirected graph $G=(V,E)$, and a permutation $\pi$ of the vertices, denote by $\Delta_\pi$ the $\max_{1\leq i\leq n-1} \{\deg_{\{v_{i+1},...,v_n\}}(v_i)\}$. That is, we look only on the edges of $v_i$ that reach a vertex in $\{ v_{i+1},...,v_n\}$ Now for all the permutations, we define: $\mathbb{A}(G) = \min_{\pi} \{ \Delta_\pi \}$.
Another thing is that for a given sub-set of vertices $U\subseteq V$, I'll denote by $\gamma (U)$ the $\min_{u\in U} \{ \deg_{G^{'}=(U,E\uparrow_U)} (u) \}$. Meaning, the degree of $u$ in the sub-graph $G^{'}=(U,E\uparrow_U)$. Now I'll define: $\mathbb{B}(G)=\max_,_{U\subseteq V} {\gamma(U)}$.
I'd like to show that $\mathbb{B} (G) \leq \mathbb{A} (G)$. Honestly, I'm lacking any intuition as to why this even holds?
Take any subset $U \subseteq V$ and any permutation $\pi$. Define $v_k$ as the first vertex of $U$ in $\pi$, that is,
$$ k = \min \big\{i \in \mathbb{N} \mid v_i \in U\big\}. $$
Observe that $$\Delta_\pi \geq \deg_{\{v_{k+1},\ldots\}}(v_k) \geq \deg_{G'}(v_k) \geq \min_{u\in U} \big\{ \deg_{G'} (u) \big\} = \gamma(U).$$
As that works for all $U$'s and all $\pi$'s, we have $\mathbb{B}(G) \leq \mathbb{A}(G)$.
I hope this helps $\ddot\smile$