12 Apr
2019
12 Apr
'19
7:32 p.m.
I don't know the historical answer, but I think it's because the true fixity can't be expressed in Haskell. No, the historical answer is that with lazy evaluation the shortcutting happens in the expected order. We did think about that.
I don't understand how laziness enters the picture: (False && ⊥) && ⊥ ≡ False False && (⊥ && ⊥) ≡ False in both cases we get the same result. Stefan