3 ms·
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 l
by ScottBurson 10y ago
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).
- mafribe 10y agoNo, g is given by a ↦ f (e a a). The expansion you refer to works because e a0 = g.