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Thanks for the question! You are correct that there is a bug here. Indeed, the sentence "choose $\mathbf{b}_i$ s.t. $\ldots$" makes no sense: the LHS is in $H$, but the RHS may not be. Fortunately, there is a simple fix which guarantees $\mathbf{y}'_i \in H$. (This must have been what we intended in the first place, based on how the rest of the proof ...