[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.