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.
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.
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.
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.
In that case, P(o) is true, and P(k)->P(Pk) is equally provable.
No...?
The above basically says that
P(k)
is “is within the successor chain of 0”. Note that the base case isk = 0
, notk = 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 property0
does. As a demonstration,Z(k) := (k = 0)
is a valid property. It’s true only of0
.R(k, 0)
is similarly a property that is only true of0
, orSS..0
.