Open neongreen opened 4 years ago
RemoveAccessTo l ((l := t) ': lts) = RemoveAccessTo l lts RemoveAccessTo q ((l := t) ': lts) = (l := t) ': RemoveAccessTo l lts RemoveAccessTo q '[] = '[]
Shouldn't it be RemoveAccessTo q lts in the second equation?
RemoveAccessTo q lts
Ah, nice catch... Yes!
Given that it doesn't work, I wonder whether it was necessary at all..?
Yeah afaik it is only used in type instances so maybe we can nix it.
Shouldn't it be
RemoveAccessTo q lts
in the second equation?