\bigoplus_{D \cap A_i \neq \varnothing} D^\prime_i $ and $ D^\prime_i \cap A_j = \varnothing $ for $ j > i $.

We may prove it by induction on the number of indices $ i $ such that $ D \cap A_i \neq \varnothing $. If such number is $ 1 $, the result is trivial by taking $ D^\prime_i = D $, since $ D = \langle D \cap A_i \rangle $.

Assume that it is larger than $ 1 $ and $ i $ is the smallest index such that $ D \cap A_i \neq \varnothing $. Define $ \widetilde{D} = \sum_{j = i+1}^l \langle D \cap A_j \rangle \neq \{ \mathbf{0} \} $, and let $ D^\prime_i \neq \{ \mathbf{0} \} $ by one of its complementaries in $ D $. It follows that $ D^\prime_i \subseteq \langle D \cap A_i \rangle $ and $ D^\prime_i \cap A_j = \varnothing $, for $ j > i $.

Now, by induction hypothesis, $ \widetilde{D} $ has a decomposition as in the theorem, which together with $ D^\prime_i $ gives the desired decomposition of $ D $.

\[Proof of Theorem \[grw bounds\]\] We first prove (\[lower bound\]). Take an $ 