The following seems to me to be relevant to this question, but to me is an interesting exercise, especially since I have not formally worked with complexity before, but I want to learn more:
Suppose that A is NP-Complete, but B is in P. I claim that A\B is NP-Complete and B\A is NP-Complete as well. To see this, assume first that A\B is in P, and let X and Y be polynomial-time algorithms for B and for A\B, respectively. "Concatenating" X and Y as follows yields an algorithm Z for A:
Given L, test L using X; if X outputs "yes", test using Y; if Y yields "yes", output "no" and stop; if X yields "no", output "no" and stop; output "yes" otherwise and stop.
This algorithm Z runs in polynomial time, because if the (polynomial time) complexity exponent of X is k and the (polynomial time) complexity exponent of Y is n, then this algorithm clearly has (polynomial time) complexity exponent m=max(k,n). This would provide proof that P=NP, so A\B is NP-Complete.
Now suppose that B\A is in P. This time, let Y' be a polynomial-time algorithm for B\A and let X be as above. We construct an algorithm Z' for A, as follows:
Given L, test L using X; if X outputs "yes", test using Y'; if Y yields "no", output "yes" and stop; if X yields "no", test using Y'; if Y yields "no", output "yes" and stop; output "no" otherwise and stop.
This yields a polynomial-time algorithm for A, and so again, this would entail that P=NP, so B\A also is NP-Complete.
+++++++++End of Example++++++++++++
While I don't see anything wrong with the above at the moment, perhaps I have a mistake or complexity miscalculation? ...because for a while, as I was writing the second algorithm, I began to think it was odd and perhaps impossible that I can be right about B\A also being NP-Complete...
As I said, I'm somewhat new to this area, so feedback would be appreciated.