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> (λ x . (x x)) (λ x . (x x)) This shows an infinite loop in lambda calculus evaluation. It can be arrived at independently (and I'd be surprised if it hadn't
by anatoly 12y ago
> (λ x . (x x)) (λ x . (x x))
This shows an infinite loop in lambda calculus evaluation. It can be arrived at independently (and I'd be surprised if it hadn't been) just by trying to build the simplest expression that doesn't reduce. Why do you need the Y combinator for this?
- tel 12y agoIt had been. Everyone was just talking about the Y-combinator so I derived from there, but Omega was first. Once you have any fixed-point, though, deriving the others is trivial. The explosive self-referential nature of your system has been uncovered.