I’m trying to prove backward induction, which I’ll state as follows:
Consider the set $mathsf{A}$, where $nin{mathsf{A}}$, and $m+1in{mathsf{A}}$ $implies$$min{mathsf{A}}$. Then $mathsf{A}={0,…,n}$.
I’m attempting to proceed with the well-ordering principle, since I’m a little confused when trying to prove it using ordinary induction. I’ll also be tremendously happy if you could show me a clear prove using ordinary induction.
Proof: Suppose $mathsf{A}$ doesn’t contain all the elements in ${0,…,n}$. In other words, there is at least one element which fails to be in $mathsf{A}$, but satisfies the backward induction hypothesis. Consider the set $mathsf{B}$ which comprises of those elements which ‘fail to be in $mathsf{A}$’. By the well-ordering principle, $mathsf{B}$ must contain a smallest element, let’s call it $k$. But by our hypothesis of backward induction, $k-1$ must also be in $mathsf{B}$, and we reach a contradiction, since $k$ was our supposed smallest element.
I’m feeling quite uneasy about this, and I need guidance…