hi guys I have a question.
SECOND PROOF. We will repeat the argument, but without the trees. Consider an arbitrary wff α. It is the last member of some construction sequence ε1,...,εn . By ordinary strong numerical induction on the number i, we see that each εi ∈ S, i ≤ n. That is, we suppose, as our inductive hypothesis, that ε j ∈ S for all j < i. We then verify that εi ∈ S, by considering the various cases. So by strong induction on i, it follows that εi ∈ S for each i ≤ n. In particular, the last member α belongs to S.
this is the example from the book mathematical introduction to logic by herbert enderton.
Usually, the case of using induction or strong induction is to prove the property P for all natural numbers. But this example uses induction for finite cases from 1 to n. How is this possible?
To do induction for a finite number of $n$ cases make your Property statement be: P(k) is "If $k\le n$ than $Q(k)$"
Then when you do induction proof you will have proven that for all natural numbers $m$, that if $m \le n$ then $Q(m)$. The thing is that as $m\le n$ for only $0,....m$ we don't have to worry about the rest.
And if you are worried about how "If $k \le n$" will effect your induction step, well, don't worry. This is what happens:
Induction step: Assume $P(k)$ that is assume if $k \le n$ then $Q(k)$. We need to prove that if $k+1 \le n$ then $Q(k)$. If $k+1 > n$ then we are done and have nothing to prove. And if $k+1\le n$ we .... do our induction step just like normal.
That condition doesn't hurt us in any way.