Proof by induction on two variables (forward and backward at the same time?) Function $f(t,n,m)$ is defined for $0\le t \le T$, $n\in\{0,\ldots\}$ and $m\in\{2,\ldots\}$.
We would like to show some properties are true for any $t$ and all $n$ and $m$.
I thought perhaps I can show this by induction. So I started by the base case of $f(t, 0, 2)$ and it turned out we can show such properties hold for $f(t,n,2)$, $n\in\{0,\ldots\}$.
But to generalize our results for $f(t,n,m)$, we need to have some properties to be true for $f(t,n + 1,m - 1)$.
So I'm confused as to how the induction assumption should be here.
 A: Any reason why you want to descend on the $m$, given that you need to prove the result for $m$ being $2$ and up?
Anyway, all you need to do is to have some scheme that will make 'all the dominoes fall over'.  For 2 variables, there are many such schemes, but the simplest is probably to start with base case $f(t,0,2)$, and then show the following two things:

*

*For any $n\geq 0$ and $m \geq 2$: if $f(t,n,m)$, then $f(t,n+1,m)$


*For any $n\geq 0$ and $m \geq 2$: if $f(t,n,m)$, then $f(t,n,m+1)$
Notice that in effect something like $f(t,1,3)$ is now shown in two different ways: because we have $f(t,0,2)$, we have by 1 that $f(t,1,2)$, and thus by 2 that $(t,1,3)$.  But you can also say that we have $f(t,0,2)$, and thus by 2 you have $f(t,0,3)$, and thus by 1 you have $f(t,1,3)$.
In other words, this method seems like it's overkill.  Is that bad? Shouldn't you find some scheme that gets to every point through one and only one path?  No, there is no need for that at all: you just want to make all the dominoes fall over, and this scheme works just fine .... unless of course the intricacies of what you need to prove naturally suggest a different scheme.
