I've been looking at examples of problems using double induction and have found one that has stumped me.
Here is the problem: Let $n,m\in \mathbb{N}$. Let $P(n,m)$ denote the statement $n>m \implies n-m \in \mathbb{N}$
I've shown the base case when both $n$ and $m$ are equal to $1$, i.e., $P(1,1)$
My goal is to show $P(n,m) \implies P(n+1,m)$ and that $P(n,m) \implies P(n,m+1)$, but I'm having a bit of trouble.
The first implication shouldn’t be a problem. For the second you need to show that if $P(n,m)$ is true, and $n>m+1$, then $n-(m+1)\in\Bbb N$. From the induction hypothesis you know that $n-m\in\Bbb N$, and there are two possibilities: either $n-m=1$, or $n-m>1$. In the first case $P(n,m+1)$ is vacuously true, and in the second case $n-m-1\ge 1$.