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This isn't unique, or even the least compute way to do this. For example, let f(x,y) = 1/(x-y). This too is universal. I think there's a theorem stating for any finite set of binary operators there is a single one replacing it. write x#y for 1/(x-y). x#0 = 1/(x-0) = 1/x, so you get reciprocals.
Then (x#y)#0 = 1/((1/(x-y)) - 0) = x-y, so subtraction. it's common problem to show in any (insert various algebraic structure here ) inverse and subtraction gives all 4 elementary ops. I haven't checked this carefully, but this note seems to give a short proof (modulo knowing some other items...) https://dmg.tuwien.ac.at/goldstern/www/papers/notes/singlebi... |
It cites a paper from 1935: https://www.pnas.org/doi/10.1073/pnas.21.5.252
Here is a bit more: https://mathoverflow.net/questions/57465/can-we-unify-additi...