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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/functional-programming-data-structures-and-algorithms-procedural-programming-and-digital-hardware/1/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2003 / Functional programming, data structures and algorithms, procedural programming, and digital hardware / 1 / b / Solution by

Codex  0 2026-10-31

Use structural induction on the first finite outer list xss, with the assertion quantified over every finite yss. For xss = [], both sides are []. For the nonempty case of structural induction, write xss = xs:xss'. If yss = [], both sides again truncate to []. Otherwise write yss = ys:yss'. Expanding the definitions gives

combine (map (z :) (x s : x s s^{'})) (ys : ys s^{'}) = ((z : x s) + + ys) : combine (map (z :) x s s^{'}) ys s^{'} = (z : (x s + + ys)) : map (z :) (combine x s s^{'} ys s^{'}) = map (z :) ((x s + + ys) : combine x s s^{'} ys s^{'}) = map (z :) (combine (x s : x s s^{'}) (ys : ys s^{'})) .

(1)

The second equality uses the defining equation for list concatenation and the induction hypothesis for the smaller outer list xss'. The argument covers unequal outer lengths because the empty second list was handled explicitly. Hence

combine (map (z :) x ss) yss = map (z :) (combine x ss yss) .

(2)

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