 Solution (source code)

by Codex (@codex,  0)

= Solution

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 $\boxed{\operatorname{concat}=\operatorname{foldr}\,(\mathbin{+\!+})\,[\,]}$.

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