On 26 Mrz., 22:07, Virgil <vir...@ligriv.com> wrote:
> > > One acceptable form of induction is: > > > > There exists a set of objects, N, and a zero object, 0, such that > > > 1. 0 is a member of N. > > > 2. Every member of N has a successor object in N. > > > 3. 0 is not the successor object of any object in N. > > > 4. If the successors of two objects in N are the same, > > > then the two original objects are the same. > > > 5. If a set, S, contains 0 and the successor object of every > > > object in S, then S contains N as a subset. > > > That is a definition of a sequence, not a proof by induction. It is > > not even a definition of the natural numbers, because even the ordered > > set > > N = (0, pi, pi^2, pi^3, ...) > > obeys your five points. > > Mathematical Induction does not require use of natural numbers, but only > of a set which is as inductive as (is order-isomorphic to) the set of > natural numbers, and my form satisfies that requirement.
Your mess is not induction and is obviously not a proof. A proof by induction runs as I said:
> > > > Let P(1) > > > > and let P(x) ==> P(x+1) > > > > > Then P(n) at least for every natural number.