[Hercuros]

The specification proves a property about an algorithm/function, namely the equivalence between a more complicated memoizing implementation and a simpler direct recursive implementation.

It is also true that no numerical reasoning is happening: the memoized version of any recursive relation will return the same result as the original function, assuming the function is not stateful and will return the same outputs given the same inputs.

However, it is not true to say that it does this by exhaustion, since there are infinitely many possible outputs and therefore it cannot be exhaustively checked by direct computation. The “n” for which we are “taking the proof” is symbolic, and hence symbolic justification and abstract invariants are used to provide the proof. It is the symbolic/abstract steps that are verified by the type checker, which involves only finite reasoning.

Of course, the symbolic steps somewhat mirror the concrete computations that would happen if you build the table, especially for a simpler proof like this. But it also shouldn’t be surprising that a program correctness proof would look at the steps that a program takes and reason about them in an abstract way to see that they are correct in all cases.