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Fascinating! A couple of questions:

Are there any non-singleton sets with the fixpoint property?

> a ↦ f (e a a)

Don't you mean

   a ↦ e a a
? Because later you expand (g a0) to (e a0 a0).



No, g is given by a ↦ f (e a a). The expansion you refer to works because e a0 = g.




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