Updating on the observations "ϕ(n) for all 0≤n≤N", the probability of ∀nϕ(n) goes to 1 as N→∞

Logical induction doesn't have this property. No alternatives can either.

To make sense of this answer, I recommend reading

https://www.lesswrong.com/posts/3FoMuCLqZggTxoC3S/logical-pinpointing

and

https://www.lesswrong.com/posts/i7oNcHR3ZSnEAM29X/standard-and-nonstandard-numbers

In short, logical induction can't be sure if its operating in a nonstandard context or not. So suppose that ϕ is true for all standard numbers, but false for some nonstandard number. Logical Induction has no way to know whether it is operating in the standard or non-standard numbers, so assigns a probability strictly between 0 and 1 to it.

Suppose you had an alternate algorithm that didn't do that. A Turing machine can be formally specified in PA, so we can talk about the output of the algorithm. The concept of limits can be discussed in PA using ∀ϵ∃δ of foundational calculus. (The ϵ and δ actually need to be Natural numbers, but you can encode rational numbers in them with prime factors)

So for every formula ϕ in PA there exists another formula ϕ′ that says that your algorithm assigns probability 1 to ϕ in the limit. (And ϕ′ is computable from ϕ)

Let K be a PA formula with one free variable. K takes in a number a. If a does not encode a PA formula with one free variable, then K is false. If a encodes A, then K=¬(A(a))′. Now consider k an integer encoding K. And consider K(k) .

This is a straightforward modification of godels incompleteness theorem. If you have any procedure that can perfectly distinguish between true and false statements within the structure of PA, then you can make a "this statement is false" and get a contradiction.

Technical Note. I am also assuming that your proposed alternative to logical induction respects negation and implication in the limit. (Ie if limn→∞P(ψ)=1⟺limn→∞P(¬ψ)=0 and ( limn→∞P(ψ)=1 ∧ limn→∞P(ψ⟹ϕ)=1)⟹limn→∞P(ϕ)=1 .)

Some expressions in PA have multiple qualifiers ∀a∃b∀c... Suppose we know that P(ϕ(a))=1 for every particular a. Then by finite combinations of probabilities, P(∀a<N:ϕ(a))=1 and so P(∀a:ϕ(a))=1. As ∃a=¬∀a¬, then it follows that any attempt at logical induction with this property must fail.

EDIT:

I have realized that I was confusing two separate concepts. You can't have any function from formulas to booleans that has the property ∀x:f(′ϕ(x)′)⟹f(′∀x:ϕ(x)′). and the trivial preservation of logical operations. (definable within a formal first order theory with the same symbols)

However, if you have a process that returns True, False or Maybe, and is never wrong, then you can extend it to another process like this. Take f(ϕ) the first function. If f(ϕ) is True or False, define g(ϕ)=f(ϕ) otherwise, if ϕ=∀x:ψ(x) and ∀x:f(ψ(x)) then g(ϕ)=True. Otherwise Maybe.

You can't always combine infinite sequences of your own beliefs, but you can combine infinite sequences of beliefs from some weaker system, and each time you do that you move further up the arithmetic hierarchy.

https://en.wikipedia.org/wiki/Arithmetical_hierarchy

As totally computable functions are in Δ0, the limit of a sequence of boolean functions is in Σ2. So as taking a forall over a function makes it a Π the highest you can get while still having

Updating on the observations "ϕ(n) for all 0≤n≤N", the probability of ∀nϕ(n) goes to 1 as N→∞

Is Π1. This means that you can have your asymtotic property compared to any Π1 function.

Limits in the rationals add a ∀ϵ>0 to the front, puting them in Π3. Meaning that your asymtotic property holds for all Π3 sets.

What this means is that if you have a computable function px(n)∈[0,1] which tends to 1 as n→∞ if and only if ϕ(x) then you can construct q(n)=∑nx=12−xpx(n) which tends to 1 if and only if ∀x:ϕ(x).

The downside of this approach is that it assigns probability almost 1 to false statements.

The boolean function case is to take a function that searches for an explicit counterexample, and return a probability that tends to one if no counter example has been found yet.

Sorry if these are stupid questions...

The question specified "all my variables are implicitly natural numbers". Why can't there be traders that specialize on questions specifically about standard numbers and ignore others? (I assume that the natural numbers are standard numbers, correct?) Also, what's the connection between nonstandard numbers and your Godel-like proof?

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