Linear transformation proof Ok, so I have this question and I honestly have no idea how to prove it. I know that the function is linear and I know it works for every possible combination, I just don't know how to prove it in an arbitrary manner. Here is the question:

Suppose that instead of using those S-boxes, DES would just substitute a 6-bit string with its first 4 bits, that is T(b1 b2 b3 b4 b5 b6 ) = b1 b2 b3 b4. We have called this transformation T. Show that T is a linear transformation (Note: you need to prove that T(x xor y) = T(x) xor T(y), for all 6-bit strings x and y, not just for a particular pair x and y; so write down a proof that works for arbitrary x and y).

I have no idea where to start since this class isn't about writing proofs.
Well, we start with a small definition of XOR for bit strings larger than zero bits. $x$ and $y$ are bit strings with the lenght $n$ and $x_{k}$ denoting the $k^{th}$ bit (from $0$ to $lenght - 1$). $||$ represents concatenation and $\oplus$ means XOR. So we get:
  $$  x \oplus y = ((x_{0} \oplus y_{0}) || (x_{1} \oplus y_{1}) || ... || (x_{n-1} \oplus y_{n-1})) $$
Our function $T$ just returns the first 4 bits from a 6 bit string (or an arbitrarily large bit string longer or equal to 4 bits, it doesn't really matter), so we can write it as:
  $$ T(b) = (b_{0} || b_{1} || b_{2} || b_{3}) $$

Now we want to show that $T(x \oplus y) = T(x) \oplus T(y)$

How could we do that? Maybe we could insert the different definitions of our functions and see where it goes from there..?

(That's maybe not the most optimal way to answer this question, but (I think...) it works.)