Basically there's nothing to be done in haskell for this quiz.

As the List Monad already provides non-deterministic evaluation with "guard" as a description of constraints, you really just have to write the problem in the List Monad and be done with it.

Of course one could write all kinds of wrapping hackery, but I think from the standpoint of usability and conciseness the built-in behaviour of haskell is already pretty optimal.

```constr = do a <- [0..4]
b <- [0..4]
c <- [0..4]
guard (a < b)
guard (a + b == c)
return ("a:",a,"b:",b,"c:",c)

{-
> constr
[("a:",0,"b:",1,"c:",1)
,("a:",0,"b:",2,"c:",2)
,("a:",0,"b:",3,"c:",3)
,("a:",0,"b:",4,"c:",4)
,("a:",1,"b:",2,"c:",3)
,("a:",1,"b:",3,"c:",4)]
-}
```

Obviously, solving a problem with built-in functionality always feels a little like cheating, because it's "pure chance" that the solution to the problem is already built-in.

So, for the sake of argument and because I was bored here's the same solution under the premise that haskell's List Monad didn't already exist.

Unfortunately I "had" to reimplement a fair bit of the Prelude, but that's what happens if you start creating unrealistic scenarios :). Of course, the "do" notation is also kind of a built-in, but representing it with bind (>>=) only made is less nice to look at.

```data List a = a ::: (List a) | Empty deriving (Show)

foldrL :: (a->b->b) -> b -> List a -> b
foldrL _ start Empty       = start
foldrL f start (x ::: xs) = f x (foldrL f start xs)

appendL :: List a -> List a -> List a
appendL xs ys = foldrL (:::) ys xs

concatL :: List (List a) -> List a
concatL = foldrL appendL Empty

instance Functor List where
fmap f = foldrL (\a b -> f a ::: b) Empty

return x = x ::: Empty
l >>= f  = concatL . fmap f \$ l

mzero = Empty
mplus = appendL

range :: (Integral a) => a -> a -> List a
range from to | from > to  = Empty
| from == to = to ::: Empty
| otherwise  = from ::: range (from+1) to

constr' = do a <- range 0 4
b <- range 0 4
c <- range 0 4
guard (a < b)
guard (a + b == c)
return (a,b,c)
```