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.
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...