Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Functional programming, design and analysis of algorithms / 2 / c

Use list concatenation as the combining operation and the empty list as the seed:

concat :: [[a]] -> [a]
concat = foldr (++) []

For three inner lists, this constructs xs ++ (ys ++ (zs ++ [])), preserving the elements' original order. The concise answer is $\boxed{\mathrm{concat}=\mathrm{foldr}(++)[]}$.