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Andreas_Giger comments on How to Convince Me That 2 + 2 = 3 - Less Wrong

53 Post author: Eliezer_Yudkowsky 27 September 2007 11:00PM

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Comment author: Andreas_Giger 14 September 2011 06:20:58AM *  0 points [-]

A lot of people seem to have trouble imagining what it means to consider the hypothesis that SS0+SS0 = SSS0 is true in all models of arithmetic

Maybe I'm misinterpreting you, but could you explain how any non-symmetric equation can possibly be true in all models of arithmetic?

Comment author: Eliezer_Yudkowsky 14 September 2011 06:46:09AM 2 points [-]

SS0 isn't a free variable like "x", it is, in any given model of arithmetic, the unique object related by the successor relation to the unique object related by the successor relation to the unique object which is not related by the successor relation to any object, which is how mathematicians say "Two".

Comment author: ec429 14 September 2011 06:59:07AM 2 points [-]

Although as a mathmo myself I should point out that, to save time, we usually pronounce it "Two". :)

Comment author: Andreas_Giger 14 September 2011 07:10:50AM *  1 point [-]

I am quite familiar with TNT. However either you are talking about models of arithmetic based on peano axioms, in which case e.g. SS0 + SS0 = SSS0 simply cannot be true, for it contradicts these axioms and if both the peano axioms and said equation were true, you wouldn't have a model of arithmetics; or (what I'm assuming) you are actually talking about non-peano arithmetics, in which case there is no compelling reason why any equation of this kind should generally be true anyway.

On another note, it seems that bayesianism is heavily based on peano arithmetic, so refuting peano arithmetic by means of bayesianism seems like refuting bayesianism rather than refuting peano arithmetic, at least to me.

Comment author: Tyrrell_McAllister 14 September 2011 08:03:35PM *  7 points [-]

Maybe I'm misinterpreting you, but could you explain how any non-symmetric equation can possibly be true in all models of arithmetic?

The purpose of the article is only to describe some subjective experiences that would cause you to conclude that SS0+SS0 = SSS0 is true in all models of arithmetic. But Eliezer can only describe certain properties that those subjective experiences would have. He can't make you have the experiences themselves.

So, for example, he could say that one such experience would conform to the following description: "You count up all the S's on one side of the equation, and you count up all the S's on the other side of the equation, and you find yourself getting the same answer again and again. You show the equation to other people, and they get the same answer again and again. You build a computer from scratch to count the S's on both sides, and it says that there are the same number again and again."

Such a description gives some features of an experience. The description provides a test that you could apply to any given experience and answer the question "Does this experience satisfy this description or not?" But the description is not like one in a novel, which, ideally, would induce you to have the experience, at least in your imagination. That is a separate and additional task beyond what this post set out to accomplish.

Comment author: Andreas_Giger 14 September 2011 09:55:33PM *  1 point [-]

Yes, I am aware of that. However, I don't think two pebbles on the table plus another two pebbles on the table resulting in three pebbles on the table could cause anyone sane to conclude that SS0 + SS0 = SSS0 is true in all models of arithmetic. In order to be convinced of that, you would have to assign "PA doesn't apply to pebbles" a lower prior probability than "PA is wrong".

The statement "PA applies to pebbles" (or anything else for that matter) doesn't follow of the peano axioms in any way and is therefore not part of peano arithmetic. So what if peano arithmetic doesn't apply to pebbles, there are other arithmetics that don't either, and that doesn't make them any wrong. You're using them everyday in situations where they do apply.

A mathematical theory doesn't consist of beliefs that are based on evidence; it's an axiomatic system. There is no way any real-life situation could convince me that PA is false. Saying "SS0 + SS0 = SSS0 is true in all models of arithmetic" sounds like "0 = S0" or "garble asdf qwerty sputz" to me. It just doesn't make any sense.

Mathematics has nothing to do with experience, only to what extent mathematics applies to reality does.

Comment author: Tyrrell_McAllister 14 September 2011 10:08:03PM 1 point [-]

Mathematics has nothing to do with experience.

That you have certain mathematical beliefs has a lot to do with the experiences that you have had. This applies in particular to your beliefs about what the theorems of PA are.

Comment author: Andreas_Giger 14 September 2011 10:11:55PM *  1 point [-]

Sorry, I edited the statement in question right before you posted that because I anticipated a similar reaction. However, you're still wrong. It has only to do with my beliefs to what extent peano arithmetic applies to reality, which is something completely different.

Edit: Ok, you're probably not wrong, but it rather seems we are talking about different things when we say "mathematical beliefs". Whether peano arithmetic applies to reality is not a mathematical belief for me.

Comment author: Andreas_Giger 14 September 2011 10:17:14PM 0 points [-]

And another thing: It might be possible that if peano arithmetic didn't apply to reality I wouldn't have any beliefs about peano arithmetic because I might not even think of it. However there is no way I could establish the peano axioms and then believe that SS0 + SS0 = SSS0 is true within peano arithmetic. It's just not possible.

Comment author: Tyrrell_McAllister 14 September 2011 10:18:39PM *  1 point [-]

Consider the experiences that you have had while reading and thinking about proofs within PA. (The experience of devising and confirming a proof is just a particular kind of experience, after all.) Are you saying that the contents of those experiences have had nothing to do with the beliefs that you have formed about what the theorems of PA are?

Suppose that those experiences had been systematically different in a certain way. Say that you consistently made a certain kind of mistake while confirming PA proofs, so that certain proofs seemed to be valid to you that don't seem valid to you in reality. Would you not have arrived at different beliefs about what the theorems of PA are?

That is the sense in which your beliefs about what the theorems of PA are depend on your experiences.

Comment author: Andreas_Giger 14 September 2011 10:23:47PM 0 points [-]

I'm not sure I 100% understand what you're saying, but the question "which beliefs will I end up with if logical reasoning itself is flawed" is of little interest to me.

Comment author: Tyrrell_McAllister 14 September 2011 10:26:58PM 1 point [-]

Is the question "Which beliefs will I end up with if my faculty of logical reasoning is flawed" also of little interest to you?

Comment author: Andreas_Giger 14 September 2011 10:32:59PM *  0 points [-]

Yes, because if I assume that my faculty of logical reasoning is flawed, no deductions of logical reasoning I do can be considered certain, in which case everything falls: Mathematics, physics, bayesianism, you name it. It is therefore (haha! but what if my faculty of logical reasoning is flawed?) very irrational to assume this.

Comment author: Tyrrell_McAllister 14 September 2011 10:39:06PM *  1 point [-]

But you know that your faculty of logical reasoning is flawed to some extent. Humans are not perfect logicians. We manage to find use in making long chains of logical deductions even though we know that they contain mistakes with some nonzero probability.