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.
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 kOr 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
othat has every property0does. 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.