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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
The text was updated successfully, but these errors were encountered:
Ah, nice catch... Yes!
Sorry, something went wrong.
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.
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Shouldn't it be
RemoveAccessTo q lts
in the second equation?The text was updated successfully, but these errors were encountered: