In mcs.pdf, it has Problem 7.25. (I only solve somewhat important problems referred to in the chapter contents because I have learnt one Discrete Mathematics book before and read mcs to ensure no importants things are omitted.)
One version of the the Ackermann function $A:\mathbb{N}^2 \to \mathbb{N}$ is defined recursively by the following rules: $$ \begin{align*} A(m, n)&::=2n &&\text{if } m = 0 \text{ or } n \le 1, &(\text{A-base})\\ A(m, n)&::=A(m-1,A(m,n-1)) &&\text{otherwise} &(AA) \end{align*} $$ Prove that if $B:\mathbb{N}^2 \to \mathbb{N}$ is a partial function that satisfies this same definition, then $B$ is total and $B=A$.
This problem is a bit weird since we can directly prove $A$ is total based on the definition and get the unique value of $A(m,n)$ as what wikipedia says by decreasing one of $m,n$ at each step until $m=0$ or $n=1$ where we can use the base case). Then why is the (partial) function $B$ needed?
(This question is edited after hinted by JulioDiEgidio's comment. As this comment says, I directly edit the question instead of using "Edited:" delimeter)