Use the prepared blockade from Claim 7.1.1. A vertex mixed on at least a $1/Q$ fraction of the blocks cannot meet a complete pair among them, because anticonnectivity would supply four vertices that, together with it, induce a house. Their union gives the sparser outcome. Otherwise, double-counting mixed incidences finds a block with few mixed outside vertices. The original sparsity bound and the size of the prepared blockade leave a large set anticomplete to that block, giving the peel outcome.