Thm: for all natural numbers $n≥12$, $n = 3a +7b$, for some natural numbers $a$, and $b$. (Natural numbers include 0 here).
My question about the following proof has to do with why we need to show 4 different cases, instead just the last one. I expand this question further at the end.
Proof: By strong induction
For all natural numbers $n$, such that $n≥12$, let $P(n)$ be the statement "$n = 3a +7b$."
Let $n$ be an arbitrary natural number such that $n≥12$. Suppose for every $12 ≤ k < n, P(k)$ is true.
We consider 4 cases.
Case#1: $n = 12$ $$n = 3(4) + 0(7)$$
Case#2: $n = 13$ $$n = 3(2) + 7(1)$$
Case#3: $n = 14$ $$n = 3(0) + 7(2)$$
Case#4: $n≥15$ Then $(n−3)≥12$ and $(n−3)<n$. It follows from the induction hypothesis that $P(n-3)$ is true, and so we can choose some $a$ and $b$ such that $3a + 7b = (n-3)$. Thus, $n = 3a +7b +3 = 3(a+1) +7b$. Since $a+1$ is a natural number, it follows that $P(n)$ is true, and the implication follows. Since $n$ was arbitrary, it is true for all such, and the theorem follows.
My question: Why do we need cases $1, 2,$ and $3$? Why can't I just assume that $P(k)$ is true for all $12 ≤ k < n, P(k)$, and then start with case 4 that says $n≥15$. Since I have assumed $P(k)$ for all values less than $n$, I should be able to draw the same conclusions, as none seem dependent on the previous 3 cases.
If we need a "base case" to start the induction process, then why can't we just use case 1 and case 4, cutting out 2 and 3? Any help understanding this would be greatly appreciated.
Edit: changed $c$ to $b$, also added MathJax delimiters