I have an extremely-newbie question about monads and how to interpret the monadic laws; I asked that same question yesterday on IRC and the answers were interesting but non-conclusive (to me anyway). I'm trying to learn monads by reading "All About Monads", version 1.0.2. I though of defining a monad like that: data Counted a = Ct Int a instance Monad (Counted) where return x = Ct 1 x Ct n x >>= f = let (Ct n' x') = f x in Ct (n+n') x' -- fail as default The intent is that Counted objects "count" the number of times an operation is applied to them. (For the purpose of the question, that's irrelevant anyway: the problem would be the same if I assigned a random number to the Int on each calling of >>=.) According to "All About Monads", the first monadic law says: (return x) >>= f == f x In my case, let's suppose I define: reverseCounted :: [a] -> Counted [a] reverseCounted x = return (reverse x) so I do: c1 = return "xxx" >>= reverseCounted -- c1 == Ct 2 "xxx" c2 = reverseCounted "xxx" -- c2 == Ct 1 "xxx" Now comes the question: In what sense should I interpret the "==" in the monadic law? Obviously, c1 and c2 are not structurally equal. If I can accept that two Counted things are "equal" even if they are not identical, is enough for me to define: instance Eq a => Eq (Counted a) where Ct _ x == Ct _ y = x == y to satisfy the first law? Yesterday I was said that it'd work if c1 and c2 are mutually substitutable. They are, for the purposes of equality, though evidently not for every purpose. A simple: count (Ct n _) = n allows me to distinguish between them. Juanma