Proving the division theorem with strong induction The exercise goes like this:

Prove the division theorem using strong induction. That is, prove that for $a \in \mathbb{N}$, $b \in \mathbb{Z}^+$ there always exists $q, r \in \mathbb{N}$ such that $a = qb + r$ and $r < b$. In particular, give a proof that does not use $P(n−1)$ to prove $P(n)$ when $b > 1$.

I have done a few proofs with strong induction before, but never with a predicate with multiple variables, so I'm unsure how to approach this.
One idea I had, was to use the following as my predicate:
$$P(a,b):= \exists r,q\in\mathbb{N}(a=b\cdot q+r)$$
and then use $\forall b \in \mathbb{Z}^+ .\forall i < a(P(i, b))$ as my first inductive hypothesis, and $\forall a \in \mathbb{N} .\forall i < b(P(a, i))$ as my second, proving them separately. But I'm not sure this is right, as I can't seem to prove it this way.
Am I even on the right track here? Any help would be much appreciated!

You can prove it by strong induction on $a$.
For $a=0$, it is trivial.
Now, consider an arbitrary $a\in\mathbb N$ and assume that each $a'<a$ can be written as $qb+r$, with $r<b$. Now, if $a<b$, you can write $a$ as $0\times b+a$. Otherwise, consider $a-b$. By the induction hypothesis, it can be written as $bq+r$, with $r<b$. But then $a=(q+1)b+r$.
