Aaaah! Math problem!

[moches]

m and n are both natural numbers that satisfy these three conditions:

1. [1, n] = n+1
2. [m+1, 1] = [m, 2]
3. [m+1, n+1] = [m, [m+1, n]]

What does [2, n] equal?

I got the answer correct just by adding the two, but I don't understand the problem at all, and if this problem comes up on my final it'll likely be twisted so that you have to unravel it. Can anybody shine some light on how this is supposed to work?

[dag12]

Start with equation 3.
If we want to evaluate [2,n], then m=1 and we can let the n in the equation be replaced with n-1, assuming that n>1, so that n-1 is also a natural number. This yields
[2,n]=[1+1,(n-1)+1]=[1,[2,n-1]].
Using equation 1 now,
[1,[2,n-1]]=[2,n-1]+1

This gives us a recursive relation that lowers n by 1. Since n is a natural number, we can repeat this a total of (n-1) times until [2,n-1]=[2,1]. If we repeat this n-1 times, the equation becomes
[2,n-1]+1=[2,1]+(n-1)

Now we can apply equation 2, giving us
[2,1]+(n-1)=[1,2]+(n-1)

Application of equation 1 finally gives
[1,2]+(n-1)=3+(n-1)=n+2.

Now we have to handle the exception when n=1, because if n=1, we cannot use equation 3.
But this is a trivial case, for
[2,1]=[1,2]=3=n+2.

So for all natural numbers n, [2, n]=n+2.

[moches]

Thank you I get it now!