Alex Mennen

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Yeah, sorry that was unclear; there's no need for any form of hypercomputation to get an enumeration of the axioms of U. But you need a halting oracle to distinguish between the axioms and non-axioms. If you don't care about distinguishing axioms from non-axioms, but you do want to get an assignment of truthvalues to the atomic formulas Q(i,j) that's consistent with the axioms of U, then that is applying a consistent guessing oracle to U.

I see that when I commented yesterday, I was confused about how you had defined U. You're right that you don't need a consistent guessing oracle to get from U to a completion of U, since the axioms are all atomic propositions, and you can just set the remaining atomic propositions however you want. However, this introduces the problem that getting the axioms of U requires a halting oracle, not just a consistent guessing oracle, since to tell whether something is an axiom, you need to know whether there actually is a proof of a given thing in T.

I think what you proved essentially boils down to the fact that a consistent guessing oracle can be used to compute a completion of any consistent recursively axiomatizable theory. (In fact, it turns out that a consistent guessing oracle can be used to compute a model (in the sense of functions and relations on a set) of any consistent recursively axiomatizable theory; this follows from what you showed and the fact that an oracle for a complete theory can be used to compute a model of that theory.)

I disagree with

Philosophically, what I take from this is that, even if statements in a first-order theory such as Peano arithmetic appear to refer to high levels of the Arithmetic hierarchy, as far as proof theory is concerned, they may as well be referring to a fixed low level of hypercomputation, namely a consistent guessing oracle.

The translation from T to U is computable. The consistent guessing oracle only came in to find a completion of U, but it could also find a completion of T (in fact, a completion of U can be computably translated to a completion of T), so the consistent guessing oracle doesn't really have anything to do with the relationship between T and U.

a consistent guessing oracle rather than a halting oracle (which I theorize to be more powerful than a consistent guessing oracle).

This is correct. Or at least, the claim I'm interpreting this as is that there exist consistent guessing oracles that are strictly weaker than a halting oracle, and that claim is correct. Specifically, it follows from the low basis theorem that there are consistent guessing oracles that are low, meaning that access to a halting oracle makes it possible to tell whether any Turing machine with access to the consistent guessing oracle halts. In contrast, access to a halting oracle does not make it possible to tell whether any Turing machine with access to a halting oracle halts.

I don't understand what relevance the first paragraph is supposed to have to the rest of the post.

Oh, derp. You're right.

I think the way I would rule out my counterexample is by strengthening A3 to if  and  then there is ...

Answer by Alex Mennen102

Q2: No. Counterexample: Suppose there's one outcome  such that all lotteries are equally good, except for the lottery than puts probability 1 on , which is worse than the others.

It would kind of use assumption 3 inside step 1, but inside the syntax, rather than in the metalanguage. That is, step 1 involves checking that the number encoding "this proof" does in fact encode a proof of C. This can't be done if you never end up proving C.

One thing that might help make clear what's going on is that you can follow the same proof strategy, but replace "this proof" with "the usual proof of Lob's theorem", and get another valid proof of Lob's theorem, that goes like this: Suppose you can prove that []C->C, and let n be the number encoding a proof of C via the usual proof of Lob's theorem. Now we can prove C a different way like so:

  1. n encodes a proof of C.
  2. Therefore []C.
  3. By assumption, []C->C.
  4. Therefore C.

Step 1 can't be correctly made precise if it isn't true that n encodes a proof of C.

It sounds to me like, in the claim "deep learning is uninterpretable", the key word in "deep learning" that makes this claim true is "learning", and you're substituting the similar-sounding but less true claim "deep neural networks are uninterpretable" as something to argue against. You're right that deep neural networks can be interpretable if you hand-pick the semantic meanings of each neuron in advance and carefully design the weights of the network such that these intended semantic meanings are correct, but that's not what deep learning is. The other things you're comparing it to that are often called more interpretable than deep learning are in fact more interpretable than deep learning, not (as you rightly point out) because the underlying structures they work with is inherently more interpretable, but because they aren't machine learning of any kind.

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