Checking for all Elements in a Set in Haskell using syntactic sugar

Question

I try to remove the Integer duplicates of a List of (String, Int), where I am guaranteed that there is no String duplicate.

Is it possible to evaluate something like this in Haskell:

I tried:

[(a,b) | (a,b) <- bs, (c,k) <- bs, ((k == b) <= (a == c))]

but this does not yet work.

Edit: I am well aware, that you can achieve that using more complex syntax. For example by recursively searching the List for each elements duplicates...

Answer 1

(NB: this is a completely new version of this answer. Previous was totally off-base.)

To follow your mathematical set comprehension more closely, we can tweak the definition in your answer as

uniquesOnly :: (Eq a, Eq b) => [(a, b)] -> [(a, b)]
uniquesOnly bs = 
   [(a,b) | (a,b) <- bs, 
            [(c,d) | (c,d) <- bs, d == b] ==
            [(a,d) | (c,d) <- bs, d == b]]

"for all (c,d) in bs such that d==b it follows c==a".

uniquesOnly [(1,1),(2,2),(3,1)] returns [(2,2)].