> For an arbitrary n, compute the table full of values and their proofs, and just pull out the nth proof
if you thought harder about it you'd realize what you're suggesting is impossible
For example, in a theorem prover, you can write an inductive proof that x^2 + x is even for all x. And you can write this via a computation that demonstrates that it's true for zero, and if it's true for x, then it's true for x + 1. However, you don't need to run this computation in order to prove that it's true for large x. That would be computationally intractable, but that's okay. You just have to typecheck to get a proof.
my friend you should either read the article more closely or think harder. he's not proving that the recurrence relation is correct (that would be meaningless - a recurrence relation is just given), he's proving that DP/memoization computes the same values as the recurrence relation.
the obvious indicator is that no property of any numbers is checked here - just that one function agrees with another:
theorem maxDollars_spec_correct : ∀ n, maxDollars n = maxDollars_spec n
Hmmm.
> no property of any numbers is checked here - just that one function agrees with another:
theorem maxDollars_spec_correct : ∀ n, maxDollars n = maxDollars_spec n
> https://en.wikipedia.org/wiki/Richardson%27s_theorem
Given that both `maxDollars n` and `maxDollars_spec n` are defined to be natural numbers, I'm not sure why Richardson's theorem is supposed to be relevant.
But even if it was, the structure of the proof is to produce the fact `maxDollars_spec n = maxDollars n` algebraically from a definition, and then apply the fact that equality is symmetric to conclude that `maxDollars n = maxDollars_spec n`. And once again I'm not sure how you could possibly fail to conclude that two quantities are equal after being given the fact that they're equal.
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.