= Solution
Merge the rows with `fairConcat`, retaining explicit empty-input cases:
``
fairPair :: (a -> b -> c) -> [a] -> [b] -> [c]
fairPair _ [] _ = []
fairPair _ _ [] = []
fairPair f xs ys = fairConcat [[f x y | y <- ys] | x <- xs]
``
For nonempty <proper lazy lists>, the occurrence `f (xs!!i) (ys!!j)` belongs to row $i$, position $j$, so the proof in part d puts it in round $i+j$. This supplies \b[every pair after finite work], even when both input <lists> are infinite. The guards also return `[]` for either empty factor without traversing infinitely many empty rows.
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