The case of two horses is special here because the sets 1..n and 2..n+1 don’t overlap if n+1 = 2, and not because of some fundamental property of every induction hypothesis, but that—along with some arbitrary large n, and maybe the next case if I’m using any parity tricks—is one of the first cases I’d check when verifying a proof by induction.
The case of P(n) → P(n+1) (i.e., the second part of the induction argument) that fails is n=1. (In other words n+1 = 2).
The second part of the induction argument must begin (i.e., include n >= n0) at the value n0 that you have proven in the first part to be true from 1 to n0. In this case n0 = 1, so you must begin the induction at n = 1.
You’re right, of course. I was trying to describe the flaw in the set-overlap assumption without actually going through an inductive step, on the assumption that that would be clearer, but in retrospect my phrasing muddled that.
The case of two horses is special here because the sets 1..n and 2..n+1 don’t overlap if n+1 = 2, and not because of some fundamental property of every induction hypothesis, but that—along with some arbitrary large n, and maybe the next case if I’m using any parity tricks—is one of the first cases I’d check when verifying a proof by induction.
The case of P(n) → P(n+1) (i.e., the second part of the induction argument) that fails is n=1. (In other words n+1 = 2).
The second part of the induction argument must begin (i.e., include n >= n0) at the value n0 that you have proven in the first part to be true from 1 to n0. In this case n0 = 1, so you must begin the induction at n = 1.
I have edited my comment to avoid this confusion.
You’re right, of course. I was trying to describe the flaw in the set-overlap assumption without actually going through an inductive step, on the assumption that that would be clearer, but in retrospect my phrasing muddled that.
I’ll see if I can fix that.