Yes. For X and Y being of the same type, defining the < operator should be enough to derive all other operators: Define X < Y, and then derive the rest: X > Y

Yes. For X and Y being of the same type, defining the < operator should be enough to derive all other operators: Define X < Y, and then derive the rest: X > Y : X < Y X == Y: !(X < Y) && !(Y > X) X != Y: !(X == Y) X >= Y: !(X < Y) X <= Y: !(Y < X) That's why it's the only operator that needs to be defined for std::map/set (which are rb-trees) in C++ Now, if X and Y aren't the same type, the only thing you can expect to get out of the system is a cronenberg.