Yes, see @mauvehaus's top-level comment [1] for a proof sketch. The idea is to either treat the folded shade as an unfolded shade and repeat the process, or to

Yes, see @mauvehaus's top-level comment [1] for a proof sketch. The idea is to either treat the folded shade as an unfolded shade and repeat the process, or to make one loop sufficiently large that it can be folded again independent of the other loops. Either way, imagine taping the original loops together, since they do not participate in future folds. [1] https://news.ycombinator.com/item?id=24407572 https://news.ycombinator.com/item?id=24407572