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3 points by mattjones 6127 days ago | link | parent What if x.y!z expanded to ((x y) z)? Then if x were '(1 2 (3 4) 5), x.2!1 would give you 4. If (x 1 3) did work like cut, then you could do x.1.3!1!1, which would also give 4 in this case.