i was wandering if is there an extension of the induction principle whether number the integer variables are more than 1?
Example... if i need to prove that $p(n)$ is true then i would start by prove $p(0)$ and assuming for each $k <= n$ is true i'd try to prove $p(n+1)$ is also true...
Is there some generalization whether the statement is of the form $p(n,m)$? and so on and so forth...
As the comments indicate, it is often possible to do induction on each variable separately. Where that isn't useful, you have to define some "successor" relation between $(m, n)$ tuples that allow the induction to go through. One example is the following. Define:
$\begin{align} C(m, n) = \frac{(2 m)! (2 n)!}{n! m! (m + n)!} \end{align}$
Prove that $C(m, n)$ is always an integer.
To do this, we first prove that $C(m, 0)$ is an integer for $m \ge 0$, then prove a relation between $C(m, n), C(m + 1, n), C(m, n + 1)$ that implies that if the first two are integers, so is the third. But:
$\begin{align} C(m, 0) &= \frac{(2 m)! 0!}{0! m! m!} = \binom{2 m}{m} \end{align}$
which certainly is an integer.
Now do an induction over $n$ to prove $C(m, n)$ is always an integer.
Base: If $n = 0$, we are done by the above.
Induction: We assume $C(m, n)$ is an integer for all $m \ge 0$, and consider:
$\begin{align} C(m + 1, n) + C(m, n + 1) &= \frac{(2 m + 2)! (2 n)!}{(m + 1)! n! (m + n + 1)!} + \frac{(2 m)! (2 n + 2)!}{m! (n + 1)! (m + n + 1)!} \\ &= \frac{(2 m)! (2 n)!}{m! n! (m + n + 1)!} \cdot \left( \frac{(2 m + 1) (2 m + 2)}{m + 1} + \frac{(2 n + 1) (2 n + 2)}{n + 1} \right) \\ &= \frac{(2 m)! (2 n)!}{m! n! (m + n + 1)!} \cdot 4 \cdot (m + n + 1) \\ &= 4 C(m, n) \end{align}$
As claimed, if $C(m, n)$ and $C(m + 1, n)$ are integers, so is $C(m, n + 1)$. Together with the base, this implies $C(m, n)$ is always an integer.
As a final remark, I haven't been able to find an example of a direct "successor" relation between pairs. This is the closest I've seen.
