In normal induction we proved that if base case is true then we assume some number n to be true then we prove n+1 is true.
As written, this is inaccurate, though that may be simply a result of poor phrasing.
In standard induction, we do two things:
- First we prove the "base case". That the result holds for $n=1$.
- Then we prove that for every positive integer $n$, if the result holds for $n$, then it also holds for $n+1$.
(This is different from what you wrote; what you wrote is that one proves that if the result holds for $1$, then if it holds for some $n$, then it holds for $n+1$.)
In strong induction, we only need to prove one thing:
- For every positive integer $n$, if the result holds for all positive integers $k\lt n$, then the result holds for $n$.
This is enough to establish the result holds for all positive integer: if the result does not hold for all positive integers, then it fails to hold for some. Take the smallest integer $n$ for which the result does not hold. Then it holds for all strictly smaller integers; but by the implication above, this would imply that it holds for $n$ as well, a contradiction. So it holds for all positive integers.
(I used what is called the "Well Ordering Principle for Positive Integers": every nonempty collection of positive integers has a smallest element; this is in fact equivalent to induction).
Caveat: In strong induction, it is often the case that the general argument proving the implication does not hold for all $n$, but only for all "sufficiently large" $n$. In that case, we need to establish the implication for some $n$ "by hand". This is often, incorrectly, called a "base" of the induction. In fact, it is a "special case" of the proof of the single inductive step.
Here, the proposition being proven is:
Either $n\lt 12$, or else an $n$ cent stamp can be made using only $3$- and $7$-cent stamps.
The strong inductive step is:
Assume that the result is true for all $k$ strictly smaller than $n$. Then it holds for $n-3$; we can make an $n-3$-cent stamp using $3$- and $7$-cent stamps. Then we can make an $n$-cent stamp by making an $n-3$-cent stamp, and adding a $3$-cent stamp. So we can make an $n$-cent stamp. QED
The problem is that this argument works if $n$ is "sufficiently large", but it does not work if $n\lt 12$ (because that is not what we need to prove for $n\lt 12$) and it does not work if $n=12$, $n=13$, or $n=14$, because then $n-3\lt 12$, so our inductive hypothesis does not guarantee that we can make an $n-3$-cent stamp (the proposition "works" for any $n\lt 12$ by default). So the argument above is not complete. We still need to make sure everything works for $n\lt 12$, $n=12$, $n=13$, and $n=14$. By "everything works", we mean "if the proposition is true for all $k$ strictly smaller than $n$, then it holds for $n$.
If $n\lt 12$, then this is true simply because the proposition is true for $n$, so the consequent is true.
If $n=12$, this is true because we can verify that we can make a $12$-cent stamp (four $3$-cent stamps). So the implication is true because the consequent is true.
If $n=13$, the implication is true because the consequent is true: we can make a $13$-cent stamp (a $7$-cent stamp and two $3$-cent stamps).
If $n=14$, the implication is true because the consequent is true: we can make a $14-$cent stamp (two $7$-cent stamps).
And if $n\geq 15$, the argument we had before already worked.
So now we have established the strong inductive step for every positive integer $n$, and so by strong induction we have established the desired proposition for all positive integers.
(The $3+n-2$ came from applying the inductive argument to $n+1$).
Personally, I prefer to do proofs by strong induction by first doing the "general case", and then doing the "special cases", as the latter are only revealed after we examine the general proof and see if it works for all $n$ or not. This also helps draw the distinction between proofs by strong induction and proofs by regular induction, specifically that the latter need a base and an inductive step, while the former only needs an inductive step (but may require special cases).
Added. See also this previous question
You seem to be worrying about the explicit algebraic formulation, but not understanding how/why induction works. I'll try to explain without using mathematical symbols, in order to clarify.
Induction works as follows. You prove two things.
- Prove it for the base case
- Prove that if it is true for some case, then it must be true for the next case.
Why does this work? Let's say, for concreteness, the base case is 0. Then, by (1), it (whatever we are trying to prove) is true for 0. By (2), the fact that it is true for zero implies that it is true for 1. Again by (2), the fact that it is true for 1 implies that it is true for 2. Again by (2), the fact that it is true for 2 implies that it is true for 3. The process continues.
Now the question is, how do we formalize the statement of (2)? The answer is to replace "some case" and "the next case" by numbers. If we assign "some case" the number $n$, then "the next case" would be the number $n+1$. This is how you seem to be thinking about it.
If, on the other hand, we assign "some case" to be the number $n-1$, then "the next case" would be $n$. If we think about it this way, maybe a better way to write (2) would have been:
- Prove that it is true for this case whenever it is true for the previous case.
In plain english, we can understand that the two versions of (2) are the same, but people get caught up in the symbols when they try to translate it into math.
Best Answer
There are two equivalent common formulations of the second step: one is to assume that the claim is true for $n$ and trying to prove it's true for $n+1$. The other is to assume it's true for $n-1$ and try to prove it for $n$. Either way, we are not assuming what we're trying to prove, we're assuming the case before what we're trying to prove.
Perhaps a more intuitive way to explain the induction technique is that we're demonstrating that there cannot be any counterexamples. We do this by showing that no case can be the lowest counterexample; neither the base case, nor any case after that. The axiom of induction that others have referenced says that this absence of a lowest counterexample is enough to conclude that the statement is true in all cases.