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> However the theory itself allows that all code could be mathematically proven I don’t think this is true, something as “trivial” as halting already can’t be

• bluGill 3y ago
  But we never need to prove the general case, only specific cases. The general case includes an infinite number of inputs, but in reality there is only a finite number of states a physical computer can have (the number of states far exceeds the number of atoms in the universe, but it is still finite)
  • alpaca128 3y ago
    I don't think a theoretical computer scientist would agree.
    • bluGill 3y ago
      So long as the number of states is finite theory says we can prove by exhaustion. Either a state terminates, or it becomes a different state. Then you do a tree seach until you have reached a termination or a state already in the tree. You need more memory than the number of states to do this, but theroy allows for infine memory in the proof.
      • bluGill 3y ago
        Remember the halting problem is proving within a system, we can often step outside the system to prove it. Which sometimes allowes for more properties to be proved'
        • drdeca 3y ago
          The halting problem, as I generally think of it, is to receive a program/specification-of-a-machine and output whether it halts. In this formulation, there is no system in which things need proving. Instead, the analogous thing is that, for any program that returns one of “halts”, “doesn’t halt”, or “unknown” and only returns the first two in cases where those are correct, there are other machines which also do that, but where they answer “unknown” for a smaller set of inputs. (E.g. you can just augment the program by hard-coding in the correct value for some input that it previously answered “unknown” for.) Of course, what you are saying about “stronger systems can prove the answers to the questions of the whether the machines from larger sets, halt” is also true, I just, wouldn’t describe this as “the halting problem”, even though they are I guess basically equivalent.
      • deterministic 3y ago
        Computer scientists routinely prove theories with an infinite number of states. It is actually usually the easiest way to prove things because you don’t have to handle edge cases. A typical theory says: For all x:A, P x In other words, given any x (from an infinite set A) P x is true. Where P is a proposition. You typically prove it using straightforward induction/recursion.
  • kaba0 3y ago
    General case in what meaning? You do have to prove (the interesting properties of) a given algorithm to each possible input, e.g. no one would want to use a sort algorithm that loops forever on a specific list of integers. And the number of possible states increase so fast that you might as well think of it as infinite.
    • bluGill 3y ago
      So long as the list is finite and no being modified outside of sort that can be done for some sort algorithms. Bogo sort wouldn't. If < works as expected (a<b means b<a) most algorithms will sort algorithms terminate, and a few don't even need that (though < working is needed to prove the result is ordered ). Gödell showed there are algorithms that we cannot prove, but he did that with self referencing algorithms which we rarely use.
      • kaba0 3y ago
        Of course the most commonly used CS algorithm is proved correct in various properties, that wasn’t my point. Still, you can’t even in theory prove every interesting property about any problem/algorithm.
        • bluGill 3y ago
          Sure, but i'm contending those algorithms we cannot prove are mostly not used in programming in the real world.