Eliezer_Yudkowsky comments on Logical Pinpointing - Less Wrong

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

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Comment author: Eliezer_Yudkowsky 02 November 2012 08:10:02AM 2 points [-]

Let the property P be, "is a standard number".

Then P would be true of 0, true along the successor-chain of 0, and false at a separated infinite successor-chain.

Thus P would be a property that was true of 0 and true of the successor of every object of which it was true, yet not true of all numbers.

This would contradict the second-order axiom, so this must not be a model of second-order PA.

Comment author: Decius 02 November 2012 02:07:20PM -1 points [-]

And why aren't -1, every predecessors of -1, and every successor of -1 a standard number? Or are we simply using second-order logic to declare in a roundabout way that there is exactly one series of numbers?

Let us postulate a predecessor function: Iff a number x is a successor of another number, then the predecessor of x is the number which has x as a successor. The predecessor function retains every property that is defined to be retained by the successor function. The alternate chain C' has o: o has EVERY property of 0 (including null properties) except that it is the successor to -a, and the successor to o is a. Those two properties are not true of S(x) given that they are true of x. every successor of o in the chain meets the definition of number; I can't find a property that is not true of -a and the predecessors of a but that is true for every natural number.

Comment author: nshepperd 03 November 2012 02:51:45AM *  0 points [-]

P(k) := R(k, 0)

R(k, n) := (k = n) ∨ R(k, Sn)

P(0) and P(k) => P(Sk) can be easily proved, so

P(k) for all k

Or something like that.

Comment author: Decius 03 November 2012 06:47:00PM 0 points [-]

In that case, P(o) is true, and P(k)->P(Pk) is equally provable.

Comment author: nshepperd 04 November 2012 02:12:27AM *  0 points [-]

No...?

The above basically says that P(k) is "is within the successor chain of 0". Note that the base case is k = 0, not k = o. Anyway, the point is, since such a property is possible (including only the objects that are some n-successor of 0), the axiom of induction implies that numbers that follow 0 are the only numbers.

ETA: Reading your parent post again, the problem is it's impossible to have an o that has every property 0 does. As a demonstration, Z(k) := (k = 0) is a valid property. It's true only of 0. R(k, 0) is similarly a property that is only true of 0, or SS..0.