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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/functional-programming-design-and-analysis-of-algorithms/1/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Functional programming, design and analysis of algorithms / 1 / b / Solution by

Codex  0 2026-10-31

Recurse on the first argument for addition and multiplication, and on the exponent for exponentiation:

add :: Natural -> Natural -> Natural
add Zero     n = n
add (Succ m) n = Succ (add m n)

mul :: Natural -> Natural -> Natural
mul Zero     n = Zero
mul (Succ m) n = add n (mul m n)

power :: Natural -> Natural -> Natural
power m Zero     = Succ Zero
power m (Succ n) = mul m (power m n)

The defining arithmetic identities are

0 + n = n, (m + 1) + n = 1 + (m + n), 0 n = 0, (m + 1) n = n + mn, m^{0} = 1, m^{n + 1} = m m^{n} .

(1)

Each recursive call consumes one constructor. The base of exponentiation makes power Zero Zero equal to one, the empty-product convention. Applying convert and using structural induction verifies all three operations.

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