Viliam_Bur comments on Logical Pinpointing - Less Wrong

62 Post author: Eliezer_Yudkowsky 02 November 2012 03:33PM

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Comment author: Incorrect 03 November 2012 09:34:29AM *  2 points [-]

"Because if you had another separated chain, you could have a property P that was true all along the 0-chain, but false along the separated chain. And then P would be true of 0, true of the successor of any number of which it was true, and not true of all numbers."

But the axiom schema of induction does not completely exclude nonstandard numbers. Sure if I prove some property P for P(0) and for all n, P(n) => P(n+1) then for all n, P(n); then I have excluded the possibility of some nonstandard number "n" for which not P(n) but there are some properties which cannot be proved true or false in Peano Arithmetic and therefore whose truth hood can be altered by the presence of nonstandard numbers.

Can you give me a property P which is true along the zero-chain but necessarily false along a separated chain that is infinitely long in both directions? I do not believe this is possible but I may be mistaken.

Comment author: Viliam_Bur 03 November 2012 05:49:09PM *  1 point [-]

Not sure if I understand the point of your argument.

Are you saying that in reality every property P has actually three outcomes: true, false, undecidable? And that those always decidable, like e.g. "P(n) <-> (n = 2)" cannot be true for all natural numbers, while those which can be true for all natural numbers, but mostly false otherwise, are always undecidable for... some other values?

Can you give me a property P which is true along the zero-chain but necessarily false along a separated chain that is infinitely long in both directions? I do not believe this is possible but I may be mistaken.

I don't know.

Let's suppose that for any specific value V in the separated chain it is possible to make such property PV. For example "PV(x) <-> (x <> V)". And let's suppose that it is not possible to make one such property for all values in all separated chains, except by saying something like "P(x) <-> there is no such PV which would be true for all numbers in the first chain and false for x".

What would that prove? Would it contradict the article? How specifically?

Comment author: Incorrect 03 November 2012 06:29:50PM 1 point [-]

Are you saying that in reality every property P has actually three outcomes: true, false, undecidable?

By Godel's incompleteness theorem yes, unless your theory of arithmetic has a non-recursively enumerable set of axioms or is inconsistent.

And that those always decidable, like e.g. "P(n) <-> (n = 2)" cannot be true for all natural numbers, while those which can be true for all natural numbers, but mostly false otherwise, are always undecidable for... some other values?

I'm having trouble understanding this sentence but I think I know what you are asking about.

There are some properties P(x) which are true for every x in the 0 chain, however, Peano Arithmetic does not include all these P(x) as theorems. If PA doesn't include P(x) as a theorem, then it is independent of PA whether there exist nonstandard elements for which P(x) is false.

Let's suppose that for any specific value V in the separated chain it is possible to make such property PV. What would that prove? Would it contradict the article? How specifically?

I think this is what I am saying I believe to be impossible. You can't just say "V is in the separated chain". V is a constant symbol. The model can assign constants to whatever object in the domain of discourse it wants to unless you add axioms forbidding it.

Honestly I am becoming confused. I'm going to take a break and think about all this for a bit.

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