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Aug 13, 2023 at 6:01 comment added Pietro Majer Yes sorry, I mean no "inclusion up to rigid movement". Here the 1x1 square is included in any of the 1x2 rectangles, and each of them is included in the others.
Aug 13, 2023 at 4:09 comment added Gerry Myerson @Pietro, take a $3\times3$ square, cut it into a $1\times1$ square in the middle, and four $1\times2$ rectangles surrounding the small square. That's five rectangles, no one of which is included in some other. But perhaps I misunderstand your criteria.
Aug 12, 2023 at 10:18 comment added Pietro Majer it seems quite hard even to cut a square into a finite "inclusion co-chain" of rectangles, that is in such a way that none of them is included in some other. I guess not less than 7 rectangles are needed.
Aug 12, 2023 at 9:43 history asked Nandakumar R CC BY-SA 4.0