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Nornagest comments on How to Not Lose an Argument - Less Wrong
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And the point that I am trying to get you to understand, is that you do not need special rule to always check P(2) when making a proof by induction, in this case where the induction fails at P(1) -> P(2), carefully trying to prove the induction step will cause you to realize this. More generally you cannot rigorously prove that for all integers n > 0, P(n) -> P(n+1) if it is not true, and in particular if P(1) does not imply P(2).
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.