>Edit: I just saw your comment about O. I think O might just work! So instead of remembering big-O, Theta, and Omega, I can just use > O(f) <= O(g) for f

>Edit: I just saw your comment about O. I think O might just work! So instead of remembering big-O, Theta, and Omega, I can just use > O(f) <= O(g) for f is O(g) > O(f) = O(g) for f is Theta(g) > O(f) >= O(g) for f is Omega(g) >Is that true? Yeah, I'm pretty sure that will work (haven't checked it very rigorously though). The definition of O(f) <= O(g) would simply be O(f) \subset O(g), which immediately satisfies almost all criteria (only tricky bit is proving that either O(f) is a subset of O(g) or the other way around). For little o I think you should be able to use O(f) < O(g), using the ordering defined above.