Prove by induction consider an inductive definition of a version of Ackermann’s function. A(m, n)= 2n, if m = 0 0, if m ≥ 1, n = 0 2, if m ≥ 1, n = 1 A(m − 1, A(m, n − 1)), if m ≥ 1, n ≥ 2   1. Find A(1, 1). 2. Find A(1, 3). 3. Show that A(1, n) = 2n whenever n ≥ 1. 4. Find A(3, 4).

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Q: How do I prove this?

A: Consider

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