All of SamEisenstat's Comments + Replies

Nice, I like this proof also. Maybe there's a clearer way to say thing, but your "unrolling one step" corresponds to my going from $u$ to $v$. We somehow need to "look two possible worlds deep".

Here's a simple Kripke frame proof of Payor's lemma.

Let $⟨W,R,⊩⟩$ be a Kripke frame over our language, i.e. $W$ is a set of possible worlds, $R$ is an accessibility relation, and $⊩$ judges that a sentence holds in a world. Now, suppose for contradiction that $W⊩x\leftrightarrow \square (\square x\to x)$ but that $W⊮x$, i.e. $x$ does not hold in some world $u\in W$.

A bit of De Morganing tells us that the hypothesis on $x$ is equivalent to $\neg x\leftrightarrow ◊(\square x\wedge \neg x)$, so $u⊩◊(\square x\wedge \neg x)$. So, there is some world $v$ with $uRv$ ... (read more)

Theorem. Weak HCH (and similar proposals) contain EXP.

Proof sketch: I give a strategy that H can follow to determine whether some machine that runs in $O\left(2^{c{n}^k}\right)$ time accepts. Basically, we need to answer questions of the form "Does cell $x$ have value $y$ at time $t$." and "Was the head in position $x$ at time $t$?", where $x$ and $t$ are bounded by some function in $O\left(2^{c{n}^k}\right)$. Using place-system representations of $x$ and $t$, these questions have length in $O\left(n^k\right)$, so they can be asked. Further, each question is a simple function of a constant number of other... (read more)

As you say, this isn't a proof, but it wouldn't be too surprising if this were consistent. There is some $k\in N$ such that ${\square }_n\varphi \to \varphi$ has a proof of length $n^k$ by a result of Pavel Pudlák (On the length of proofs of finitistic consistency statements in first order theories). Here I'm making the dependence on $n$ explicit, but not the dependence on $\varphi$. I haven't looked at it closely, but the proof strategy in Theorems 5.4 and 5.5 suggests that $k$ will not depend on $\varphi$, as long as we only ask for the weaker property that ${\square }_n\varphi \to \varphi$ will only be provable in length $n^k$ for sentenc

I misunderstood your proposal, but you don't need to do this work to get what you want. You can just take each sentence ${\square }_n\varphi \to \varphi$ as an axiom, but declare that this axiom takes $n$ symbols to invoke. This could be done by changing the notion of length of a proof, or by taking axioms ${\psi }_{\varphi ,n}\to ({\square }_n\varphi \to \varphi )$ and ${\psi }_{\varphi ,n}$ with ${\psi }_{\varphi ,n}$ very long.

I'm having trouble thinking about what it would mean for a circuit to contain daemons such that we could hope for a proof. It would be nice if we could find a simple such definition, but it seems hard to make this intuition precise.

For example, we might say that a circuit contains daemons if it displays more optimization that necessary to solve a problem. Minimal circuits could have daemons under this definition though. Suppose that some function $f$ describes the behaviour of some powerful agent, a function $\tilde{f}$ is like $f$ with noise added, and our proble... (read more),,,,,

Two minor comments. First, the bitstrings that you use do not all correspond to worlds, since, for example, $Con(PA+Con(PA\left)\right)$ implies $Con\left(PA\right)$, as $PA$ is a subtheory of $PA+Con\left(PA\right)$. This can be fixed by instead using a tree of sentences that all diagonalize against themselves. Tsvi and I used a construction in this spirit in A limit-computable, self-reflective distribution, for example.

Second, I believe that weakening #2 in this post also cannot be satisfied by any constant distribution. To sketch my reasoning, a trader can try to buy a sequence of sentences ${\varphi }_1$

I at first didn't understand your argument for claim (2), so I wrote an alternate proof that's a bit more obvious/careful. I now see why it works, but I'll give my version below for anyone interested. In any case, what you really mean is the probability of deciding a sentence outside of $\Phi$ by having it announced by nature; there may be a high probability of sentences being decided indirectly via sentences in $\Phi$.

Instead of choosing $\Phi$ as you describe, pick $\Phi$ so that the probability $\mu \left(\Phi \right)$ of sampling something in $\Phi$ is greater than $1-\mu \left(\psi \right)\cdot \epsilon /2$. Then, the probabili

A few thoughts:

I agree that the LI criterion is "pointwise" in the way that you describe, but I think that this is both pretty good and as much as could actually be asked. A single efficiently computable trader can do a lot. It can enforce coherence on a polynomially growing set of sentences, search for proofs using many different proof strategies, enforce a polynomially growing set of statistical patterns, enforce reflection properties on a polynomially large set of sentences, etc. So, eventually the market will not be exploitable on all these things simu

Universal Prediction of Selected Bits solves the related question of what happens if the odd bits are adversarial but the even bits copy the preceding odd bits. Basically, the universal semimeasure learns to do the right thing, but the exact sense in which the result is positive is subtle and has to do with the difference between measures and semimeasures. The methods may also be relevant to the questions here, though I don't see a proof for either question yet.

Yeah, I like tail dependence.

There's this question of whether for logical uncertainty we should think of it more as trying to "un-update" from a more logically informed perspective rather than trying to use some logical prior that exists at the beginning of time. Maybe you've heard such ideas from Scott? I'm not sure if that's the right perspective, but it's what I'm alluding to when I say you're introducing artificial logical uncertainty.

Yeah, the 5 and 10 problem in the post actually can be addressed using provability ideas, in a way that fits in pretty natually with logical induction. The motivation here is to work with decision problems where you can't prove statements $A=a\to U=u$ for agent $A$, utility function $U$, action $a$, and utility value $u$, at least not with the amount of computing power provided, but you want to use inductive generalizations instead. That isn't necessary in this example, so it's more of an illustration.

To say a bit more, if you make logical inductors propositionally con

It's hard to analyze the dynamics of logical inductors too precisely, so we often have to do things that feel like worst-case analysis, like considering an adversarial trader with sufficient wealth. I think that problems that show up from this sort of analysis can be expected to correspond to real problems in superintelligent agents, but that is a difficult question. The malignancy of the universal prior is part of the reason.

As to your proposed solution, I don't see how it would work. Scott is proposing that the trader makes conditional contracts, which

In counterfactual mugging with a logical coin, AsDT always uses a logical inductor’s best-estimate of the utility it would get right now, so it sees the coin as already determined, and sees no benefit from giving Omega money in the cases where Omega asks for money.

The way I would think about what's going on is that if the coin is already known at the time that the expectations are evaluated, then the problem isn't convergent in the sense of AsDT. The agent that pays up whenever asked has a constant action, but it doesn't receive a constant expected util

This isn't too related to your main point, but every ordered field can be embedded into a field of Hahn series, which might be simpler to work with than surreals.

That page discusses the basics of Hahn series, but not the embedding theorem. (Ehrlich, 1995) treats things in detail, but is long and introduces a lot of definitions. The embedding theorem is stated on page 23 (24 in the pdf).

Nicely done. I should have spent more time thinking about liar sentences; you really can do a lot with them.

Follow-up question - is the set of limits of logical inductors convex? (Your proof also makes me curious as to whether the set of "expanded limits" $\left\{{liminf}_{n\to \infty }{P}_n\right({\varphi }_n){\}}_{\overline{\varphi }\text{efficiently computable}}$ is convex, though that question is a bit less nice than the other).

I'm copying over some (lightly edited) conversation about this post from Facebook.

Sam Eisenstat: In the original counterexample to the Trolljecture, I guess you're proposing that $Con\left(PA\right)$ be logically prior to $A\left(\right)=1$, so we can go back and replace $A\left(\right)=1$ with $A\left(\right)=2$, but still have $Con\left(PA\right)$ to use in order to derive $A\left(\right)=2\square \to U\left(\right)=10$? Here I'm just using "$\square \to$" to mean "counterfactually results in" without meaning to imply that this should be defined modally.

I agree that this is intuitively sensible, but it is difficult to give a general definition that agrees with

[EDIT: This all deals with measures, not semimeasures; see below.]

For $P$ to dominate $Q$ in the old sense means that its Bayes score can only be some constant worse than $Q$, no worse, regardless of environment. My definition here implies that, but I think it’s a stricter requirement.

Your definition is equivalent to the standard definition. Li and Vitányi say that $P$ dominates $Q$ iff there is some $\alpha \in R$ such that for any binary string $x$, we have $\alpha P\left(x\right)\ge Q\left(x\right)$. Li and Vitányi's "probability distributions" take a binary string as input while the probability distribu

Haskell doesn't actually let you do this as far as I can tell, but the natural computational way to implement a function $Bot->A$ is with a case expression with no cases. This is sensible because $Bot$ has no constructors, so it should be possible to pattern match it with such an expression. Another way of thinking of this is that we can use pattern matching on sum types and $Bot$ is just the $0$-ary sum type.

Condition 4 in your theorem coincides with Lewis' account of counterfactuals. Pearl cites Lewis, but he also criticizes him on the ground that the ordering on worlds is too arbitrary. In the language of this post, he is saying that condition 2 arises naturally from the structure of the problem and that condition 4 is derives from the deeper structure corresponding to condition 2.

I also noticed that the function $f$ and the partial order $\succ$ can be read as "time of first divergence from the real world" and "first diverges before", respectively. This makes the t

We can also construct an example where $PA⊢\varphi \to \psi$ with a short proof and $PA$ also proves $\varphi \to \neg \psi$, but any such proof is much longer. We only need to put a bound on the proof length in A's proof search. Then, the argument that $A\left(\right)=1\wedge U\left(\right)\ne 0$ proves its own consistency still works, and is rather short: $O\left(\log n\right)$ as the proof length bound $n$ increases. However, there cannot be a proof of $A\left(\right)=1\to U\left(\right)=10$ within $A$'s proof length bound, since if it found one it would immediately take action 1. In this case $PA$ can still prove that $A\left(\right)=1\to U\left(\right)=10$ simply by running the agent, but thi

The argument that I had in mind was that if $PA⊢A\left(\right)\ne 1$, then $PA⊢\square ┌A\left(\right)\ne 1┐$, so $PA⊢A\left(\right)=1$ since PA knows how the chicken rule works. This gives us $PA⊢\perp$, so PA can prove that if $PA⊢A\left(\right)\ne 1$, then PA is inconsistent. I'll include this argument in my post, since you're right that this was too big a jump.

Edit: We also need to use this argument to show that the modal UDT agent gets to the part where it iterates over utilities, rather than taking an action at the chicken rule step. I didn't mention this explicitly, since I felt like I had seen it before often enough, but

I (just yesterday) found a counterexample to this. The universe is a 5-and-10 variant that uses the unprovability of consistency:

def U():
  if A() == 2:
    if PA is consistent:
      return 10
    else:
      return 0
  else:
    return 5

The agent can be taken to be modal UDT, using PA as its theory. (The example will still work for other theories extending PA, we just need the universe's theory to include the agents. Also, to simplify some later arguments we suppose that the agent uses the chicken rule, and that it checks action 1 first, then action 2