Eugine_Nier 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: Eugine_Nier 03 November 2012 11:03:28PM 1 point [-]

But the axiom schema of induction does not completely exclude

Eliezer isn't using an axiom schema, he's using an axiom of second order logic.

Comment author: Incorrect 03 November 2012 11:57:49PM *  1 point [-]

I don't see what the difference is... They look very similar to me.

At some point you have to translate it into a (possibly infinite) set of first-order axioms or you wont be able to perform first-order resolution anyway.

Comment author: Eugine_Nier 04 November 2012 03:18:08AM 0 points [-]

What's wrong with second order resolution?

Comment author: Incorrect 04 November 2012 04:56:55AM *  0 points [-]

There's no complete deductive system for second-order logic.