Checking a list containing another list, and checking if the elements in the list are equal problem

So it is a bit hard to explain but I am reading through a list containing another list. So lets say I have this

FirstList[0].SecondList[0] = 0;
FirstList[0].SecondList[1] = 0;

FirstList[1].SecondList[0] = 1;
FirstList[1].SecondList[1] = -1;

So basically I want to be able to compare:

    FirstList[0].SecondList[0] = 0;
    FirstList[0].SecondList[1] = 0;

By themselves and then after that compare:

    FirstList[1].SecondList[0] = 1;
    FirstList[1].SecondList[1] = -1;

by themselves. I have been going through it all night and have come up with nothing. All I want is a push so I can find the answer. Thank you in advance for any reply. Cheers!

The items that you want to compare are lists of intergers, regardless of where in your FirtList, they are located.
I would first generate a simple CompareIntLists function, that takes two lists as a parameter.
e.g. (untested/uncompiled)

bool CompareIntLists (List<int> a, List<int> b)// warning: assumes the lists are in the same order!
{
   if(a.Count!=b.Count) return false;
   for(int i=0;i<a.Count;i++)  if(a_!=b*) return false*_

return true;
}
Now, you can simply pass whichever lists you want compared to this function… e.g.
bool compareResult0001=CompareIntLists(FirstList[0].SecondList[0],FirstList[0].SecondList[1]);
bool compareResult1011=CompareIntLists(FirstList[1].SecondList[0],FirstList[1].SecondList[1]);
Though I suspect, you may wish to implement a nested-loop, rather than hard-coding which element indices to compare.
e.g.
for(int i=0;i<FirsList.Count;i++)
{
for(int j=0;j<FirstList*.SecondList.Count;j++)*
{
for(int k=0;k<FirstList*.SecondList.Count;k++)//note: both these inner-loops use “i” as the index for FirstList. To compae against OTHER firstList elements, you will need to nest this stuff inside yet ANOTHER loop.*
{
if(k!=j) //dont compare against itself
if(!CompareIntLists(FirstList_.SecondList[j],FirstList*.SecondList[k]))
return false
}
}
}*

return true;_