Q:

(a,b) ≤ (a',b') iff b ≤ a' or (b' = b and a ≤ a') where a≤b and a'≤b'. Has any one seen this order?

I have been mucking round with orders, and this is the order that I found I needed. I would like to know if it is defined somewhere else, it is kind of difficult to search for such things.
*EDIT*
The only application I have used it for at the moment is a totally ordered domain (e.g. '$a,b \in R $'), but I think it would extend to a partially ordered lattice. 
Consider two intervals (a,b) (a',b'), if one is clearly better than the other, i.e. b≤a' then a≤b≤a'≤b' it is better.
But if the interval has the same least upper bound where b=b', then only the greatest lower bound matters, so a ≤ a'.
The order that this creates is why I defined it. for instance, given 3 reals {1,2,3} we use a function that defines an interval on each where f(1) = [0,2] f(2) = [1,3] and f(3) = [2,4]. Using this order we can then state that f(1) || f(2) and f(2) || f(3) but f(1) < f(3).
The goal of this order is that it should create a lattice as well.
Thanks for your help.
EDIT
I missed a really important part, sorry. This is based on intervals so a ≤ b and a' ≤ b' must be true. 
(a,b) ≤ (a',b') iff b ≤ a' or (b' = b and a < a') where a ≤ b AND a' ≤ b'

A:

If the ambient order is a lattice, they your order is indeed a lattice order. To see this, suppose that we have intervals (a,b) and (a',b'), in your sense that $a\le b$ and $a'\le b'$. If $b\lt b'$, then the least upper bound, written $(a,b)\vee(a',b')$, is $(a'\vee b,b')$. If $b=b'$, then $(a,b)\vee(a',b')=(a\vee a',b)$. Similarly, the case $b'\le b$ is symmetric, and so the remaining case is when $b\perp b'$, in which case $(a,b)\vee(a',b')=(b\vee b',b\vee b')$. A symmetric argument shows $(a,b)\wedge(a',b')$ exists, so it is a lattice.
This argument shows that if the ambient order is an upper semi-lattice, then so is your order, and the smae for lower semi-lattice, since we only needed $\vee$ in the underlying order to define $\vee$ in your order, and the same for $\wedge$. 
Also, if the underlying order is well-founded, then your order is also well-founded, since 
any descending sequence must descend infinitely in one of the coordinates. If the rank of the ambient order is $\omega$, then the new rank is also $\omega$, since every pair would have only have finitely many predecessors in your order. But for larger ordinal ranks, it seems to go up, and one could calculate some bounds if this was a case you were interested in. 

