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 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.