Why we cannot prove that a computable function is total?

Question

We know that we cannot find an algorithm that would prove that a computable function "f" is total if it IS total. How come?

When a function is total, it must have a proof (derived from soundness and correctness of logic). And proof is a finite sequence of symbols, so why does simple enumeration through all the sequences (which ends in finite time) not provide us with a proof?

Answer 1

You misunderstand soundness and completeness of logic. You think that it says:

A statement is true if, and only if, it is provable.

But it really says:

A statement is true in every model if, and only if, it is provable.

It can happen that there is a model in which a statement is true but not provable. What if you live in such a model? Then it could happen that in the model in which you live there is a total function $f$, but there is no proof that $f$ is total.

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Here is a precise mathematical reason why your thinking does not work. I am going to show that, whatever formal system you are using (so long as it is a reasonable one), there is a total function which your formal system does not prove to be total.

You do proofs in some formal system $T$ (for instance, $T$ could be first-order logic and Zermelo-Fraenkel set theory) whose axioms are computably enumerable (or else you're a mystic guru). Let us also assume that $T$ contains arithmetic, and that $T$ is consistent (or else you're mad).

Define the following function f

def f(n):
    if n encodes a proof of 0 = 1 in formal system T:
        while True: pass
    else:
        return 0

Observe that f is total if, and only if, $T$ is consistent:

• if $T$ is consistent then it does not prove $0 = 1$, hence f never enters the infinite while loop.
• if f is total, then it never enters the infinite while loop, therefore no $n$ encodes a proof of $0 = 1$ in $T$, therefore there is no proof of $0 = 1$ in $T$, which means that $T$ is consistent.

Because $T$ is consistent, f is total.

But $T$ does not prove that f is total: if it did, then it would prove that, for every $n$, $n$ does not encode a proof of $0 = 1$, but this would imply that $T$ proves its own consistency, which it cannot by Gödel's incompleteness theorem.

Supplemental: Ideas about having multiple theories $T_0, T_1, T_2, \ldots$ are quashed by (where unpair is a bijection $\mathbb{N} \to \mathbb{N} \times \mathbb{N}$):

def g(n):
    (i, j) = unpair(n)
    if i encodes a proof of 0 = 1 in formal system T(j):
        while True: pass
    else:
        return 0

Now $g$ is total if and only if all $T_i$'s are consistent, but no $T_j$ can prove that they are all consistent (if it could, it would prove its own consistency as well).

Answer 2

so why does simple enumeration through all the sequences (which ends in finite time) not provide us with a proof?

That idea could maybe give you a semi-decider, but certainly not a decider -- you don't know a size bound on the proof! That is, you can never say "no, it's not total" after finite time.

Note, though, that the idea has to be fundamentally flawed; the set of total functions is not enumerable.

Answer 3

"It must have a proof" - no, not necessarily. For example, write a program that for each even integer n ≥ 4 finds the smallest prime p such that n-p is also a prime, and doesn't halt if such a p does not exist. This is very, very likely to be a total function. But currently there is no prove known. And it is quite possible that no prove exists.

Someone remarked that this was a weak example and not convincing. OK, the fact is that with this easy to understand problems the worlds greatest mathematicians don't agree at all that a total function would have a proof of totality. So "it must have a proof" is a claim that you are more clever than the worlds greatest mathematicians.

And someone will be able to find a total function where the existence of a proof that it is total leads to a contradiction.