This notion, by the argument you give before the final theorem, implies belief (with probability 1) in all true sentences -- therefore it must disbelieve true sentences. This is quite intriguing, but likely means it's not quite what we want.
If uniformly coherent predictors exist, their self-referential knowledge would obey Paul Christiano's reflection principle (which follows directly from believing true sentences).
It does not imply belief in all true sentences. If is true for all , the probability of will go to 1 as goes to infinity, but the probability of may stay bounded away from 1.
EDIT: This post is out of date, the new, better definition is here.
This post is part of the Asymptotic Logical Uncertainty series. Here, I give a concrete proposal for a definition of Uniform Coherence, as mentioned here.
This is only a proposal for a definition. It may be that this definition is bad, and we would rather replace it with something else
Let M be a Turing machine which on input N runs for some amount of time R(N) then outputs a probability, representing the probability assigned to ϕN.
In the following definition, we fix a function T(N) (e.g. 2N). We say that a sequence {sn} is quickly computable if it is an increasing sequence and there exists a Turing machine which determines whether or not an input N is of the form sn in time T(N).
We say that M is Uniformly Coherent if
limn→∞M(┌(¬¬)n⊥┐)=0
If {sn} is quickly computable and PA⊢ϕsn→ϕsn+1 for all n, then limn→∞M(sn) exists.
If {qn}, {rn}, and {sn} are quickly computable and PA⊢(ϕqi∨ϕrn∨ϕrn)∧¬(ϕqn∧ϕrn)∧¬(ϕqn∧ϕsn)∧¬(ϕrn∧ϕsn) for all n, then limn→∞M(qn)+M(rn)+M(sn)=1
Open Question 1: Does there exist a uniformly coherent logical predictor M?
Open Question 2: Does there exist a uniformly coherent logical predictor M which also passes the Generalized Benford Test? (Here, we mean the generalization of the Benford Test to all irreducible patterns)
Theroem: If M is uniformly coherent, then if we define P(ϕ)=limn→∞M(┌(¬¬)nϕ┐), then P(ϕ) is defined for all ϕ and is a computably approximable coherent probability assignment. (see here for definitions.)
Proof: Computable approximability is clear. For coherence, it suffices to show that P(ϕ) is well defined, P(ϕ)=1 for provable ϕ, P(ϕ)=0 for disprovable ϕ, and P(ϕ∧ψ)+P(ϕ∧¬ψ)=P(ϕ).
The fact that P(ϕ) is well defined comes from applying 2 to the sequence sn=┌(¬¬)nϕ┐.
The fact that P(ϕ)=1 for provable ϕ, comes from applying 3 to qn=┌(¬¬)n⊥┐, rn=┌(¬¬)n⊥┐, and sn=┌(¬¬)nϕ┐. The fact that P(ϕ)=0 for disprovable ϕ, comes from applying 3 to qn=┌(¬¬)n⊥┐, rn=┌(¬¬)n⊤┐, and sn=┌(¬¬)nϕ┐, since we already know that P(⊤)=1.
The fact that P(ϕ∧ψ)+P(ϕ∧¬ψ)=P(ϕ) comes from applying 3 to qn=┌(¬¬)nϕ∧ψ┐, rn=┌(¬¬)nϕ∧¬ψ┐, and sn=┌(¬¬)n¬ϕ┐, then applying 3 again to qn=┌(¬¬)n⊥┐, rn=┌(¬¬)nϕ┐, and sn=┌(¬¬)n¬ϕ┐, to get that P(ϕ∧ψ)+P(ϕ∧¬ψ)+P(ϕ)=1=P(ϕ)+P(ϕ). □
Uniform Coherence is however much stronger than coherence. To see an example, consider an infinite sequence of sentences ϕsn="PA is consistent on proofs of length up to n." Uniform coherence can be shown to imply that limn→∞M(sn)=1. (Each of the sentences is provable, so you can apply 3 to this sequence and a pair of ┌(¬¬)n⊥┐ sequences.)
Theorem: If we take a quickly computable sequence of mutually exclusive sentences, {ϕsn}, then if M is uniformly coherent, we have limn→∞M(sn)=0.
Proof: Let ϕrn be the disjunction of all the ϕsi for i<n. Let ϕqn=¬ϕrn+1. By 2, limn→∞M(qn) converges to some p. Applying 3 to {qn}, {rn+1} and {┌(¬¬)n⊥┐} shows that limn→∞M(rn) converges to 1−p. Therefore, applying 3 to {qn}, {rn}, and {sn} shows that limn→∞M(sn) converges to 0.(Note that I assumed here that T was reasonable enough that qn and rn ane also quickly computable)□