The greater than theorem on some sets and their complements:

Let N be the set of natural numbers. Let A and B be non-empty proper subsets of N such that A is not a subset of B nor is B a subset of A.
Let f(x) be a one-to-one function with domain N and range N where f(x+1)>f(x) for all x. Let R be the set of natural numbers in the range [f(x)+1, f(x+1)]. Let k(x)=f(x+1)-f(x), which is the total number of elements in the range R. Let a(x) be the number of elements in A which are also in R; similarly, let b(x) be the number of elements in B which are also in R.

Let Ac and Bc be the complements of A and B in N respectively, and define ac(x) and bc(x) similar to a(x) and b(x) for Ac and Bc respectively.

Then, a(x)>=b(x) for all x iff bc(x)>=ac(x) for all x.

Proof:

k(x) is the number of elements in R (a finite number since it is a range with definite endpoints in N). 0<=a(x)<=k(x) for all x since a(x) is the number of elements in common between A and R. 0<=b(x)<=k(x) for all x since b(x) is the number of elements in common between B and R.
By the definition of complement set, if an element t is in A, then it is not in Ac, so ac(x)=k(x)-a(x). Similarly, bc(x)=k(x)-b(x). Solving for k(x),
k(x)=a(x)+ac(x)=b(x)+bc(x).
This means that a(x)-b(x)=bc(x)-ac(x). Therefore, a(x)>=b(x) for all x iff bc(x)>=ac(x) for all x.

This theorem is obscure to say the least, but it may prove useful very soon.