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Comment by Flying Pen and Paper (flying-pen-and-paper) on Circular Reasoning · 2024-08-09T19:45:14.473Z · LW · GW

Avtur Chin writes:

'From A follows A' can be in two meta-stable situations; If A is false and if A is true. But typically circular logic is used to prove that A is true and ignores the other situation.

which I agree with. This can be contrasted with “From A does not follow not A”, which I believe entails A (as a false statement implies everything?).

When trying to prove a logical statement B from A, we generally have a sense of how much B resembles A, which we could interpret as a form of distance. Both A and not A resemble A very much.

I’ll formalise this mathematically in the following sense. The set of logical statements form a metric space where the metric is (lack of) resemblance. The metric is normalised so that A and not A lie in the unit ball centred at A, denoted by U. When reading a proof from premise A, we move from one point to the next, describing a walk.

Therefore, we might think of circular reasoning as an excursion (that is to say, a walk that returns to where it started) of logical statements. A proof that A implies not A is then an almost-excursion; it returns to U.

If there is reason to believe that (1) excursions and almost-excursions are comparably likely in some sense not defined (2) lots of excursions are recorded but not any almost-excursions, then this seems to lend evidence to the proposition that A does not imply not A i.e. A is true.

The issue is obviously that it’s not clear that (1) is true. However, intuitively, it does seem at least more likely to be true if conditioned on the walk becoming very distant from A before its return to U.

Comment by Flying Pen and Paper (flying-pen-and-paper) on When is a mind me? · 2024-07-09T23:43:36.911Z · LW · GW

Yes, it would imply the observer is external, but then it also would not change anything about how the brain functions. (Or vice versa, but I prefer this one.) I am unconvinced of the truth of what you say in the last sentence of your second paragraph.

Either way, whether or not it might seem implausible, my question is why it is, or is not, implausible. Why exactly, based on what we currently know, is this extremely unlikely?

Comment by Flying Pen and Paper (flying-pen-and-paper) on When is a mind me? · 2024-07-09T04:49:52.863Z · LW · GW

I don’t see why split-screen mode is crazy talk at all. Is it just because it would imply faster-than-light communication? With our understanding of physics incomplete, I remain agnostic on the existence of FTL, so I wouldn’t rule this out. But even more than that, I’d propose that if there is one observer, there does not even need to be FTL communication in the first place, because it is just that the observer is in more than one place at once, similarly to how a wormhole does not necessitate true FTL. What are the other objections?

The belief system which seems most coherent to me is that we are thinking organisms, where the thought is mediated by our brains, and our internal experience is a way for the brain to refer to itself – it provides a handhold for the useful concept of “I” to latch onto. This has the added benefit that I find this idea rather cute. In this frame, you die only if your brain dies (or a weaker claim: you don’t die if you don’t undergo significant brain trauma).

On a last meta-philosophical point, which is not necessarily direct relevant to your post, it increasingly strikes me as unwise to use reasoning from external perceptions (e.g. the results of neuroscientific experiments) to attack at how we internally perceive the experience of consciousness. If neuroscience proved that the self is dying every second, then I would say with no reservation that there is either an error in the experiment, or that the self is not what it was thought to be. I genuinely believe that the answer to such philosophical questions is (perhaps even must be) “if it is felt to be true, it is true”.

Comment by Flying Pen and Paper (flying-pen-and-paper) on Hyperreals in a Nutshell · 2023-11-07T17:59:46.957Z · LW · GW

Observe that  is a set of natural numbers. If  then  cannot be finite, and it seems pretty obvious that almost all the elements in  are the same (they only disagree at a finite number of places after all). 

The bracketed remark doesn't appear to be true. Why can we not have  or ? Indeed, by the definition of an ultrafilter, we must have one of them in . Also, in the post, you use  for two different purposes, which makes the post slightly less clear.